Microelectronics Circuit Analysis and Design - Solution Manual [4 ed.]

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Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 1 1.1

ni = BT 3 / 2 e (a) Silicon

− Eg / 2 kT

⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦ = 2.067 × 1019 exp [ −25.58] ni = 1.61× 108 cm −3

(i)

ni = ( 5.23 × 1015 ) ( 250 )

(ii)

ni = ( 5.23 × 1015 ) ( 350 )

(b)

GaAs

(i)

ni = ( 2.10 × 1014 ) ( 250 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦ = 3.425 × 1019 exp [ −18.27 ] ni = 3.97 ×1011 cm −3 3/ 2

3/ 2

⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦

= ( 8.301× 1017 ) exp [ −32.56] ni = 6.02 × 103 cm −3

(ii)

ni = ( 2.10 × 1014 ) ( 350 )

3/ 2

⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦

= (1.375 × 1018 ) exp [ −23.26] ni = 1.09 × 108 cm −3

______________________________________________________________________________________ 1.2

a.

⎛ − Eg ⎞ ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠ ⎛ ⎞ −1.1 1012 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎟ −6 ⎝ 2(86 × 10 )(T ) ⎠

⎛ 6.40 × 103 ⎞ 1.91× 10−4 = T 3 / 2 exp ⎜ − ⎟ T ⎝ ⎠ By trial and error, T ≈ 368 K b. ni = 109 cm −3 ⎛ ⎞ −1.1 ⎟ 109 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎜ 2 ( 86 × 10−6 ) (T ) ⎟ ⎝ ⎠ ⎛ 6.40 × 103 ⎞ 1.91× 10−7 = T 3 / 2 exp ⎜ − ⎟ T ⎝ ⎠ By trial and error, T ≈ 268° K ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.3

Silicon (a)

ni = ( 5.23 × 1015 ) (100 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) (100 ) ⎥⎦

= ( 5.23 × 1018 ) exp [ −63.95] ni = 8.79 ×10−10 cm −3

(b)

ni = ( 5.23 × 1015 ) ( 300 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦

= ( 2.718 × 1019 ) exp [ −21.32] ni = 1.5 × 1010 cm −3

(c)

ni = ( 5.23 × 1015 ) ( 500 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 500 ) ⎥⎦

= ( 5.847 × 1019 ) exp [ −12.79] ni = 1.63 × 1014 cm −3

Germanium. (a)

ni = (1.66 × 1015 ) (100 )

3/ 2

⎡ ⎤ −0.66 ⎥ = (1.66 × 1018 ) exp [ −38.37 ] exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) (100 ) ⎥⎦

ni = 35.9 cm −3 (b)

ni = (1.66 × 1015 ) ( 300 )

3/ 2

⎡ ⎤ −0.66 ⎥ = ( 8.626 × 1018 ) exp [ −12.79] exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦

ni = 2.40 × 1013 cm −3 (c)

ni = (1.66 × 1015 ) ( 500 )

3/ 2

⎡ ⎤ −0.66 ⎥ = (1.856 × 1019 ) exp [ −7.674] exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 500 ) ⎥⎦

ni = 8.62 ×1015 cm −3 ______________________________________________________________________________________ 1.4

(a) n-type; no = 10

15

(

)

(

)

n2 2.4 × 1013 cm ; po = i = no 1015 −3

2

2

= 5.76 × 1011 cm −3

ni2 1.5 × 1010 = = 2.25 × 10 5 cm −3 no 1015 ______________________________________________________________________________________

(b) n-type; no = 1015 cm −3 ; po =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.5

(a) p-type; p o = 1016 cm −3 ; no =

(

ni2 1.8 × 10 6 = po 1016

)

(

2

= 3.24 × 10 − 4 cm −3

)

2

ni2 2.4 × 1013 = = 5.76 × 1010 cm −3 po 1016 ______________________________________________________________________________________

(b) p-type; p o = 1016 cm −3 ; no =

1.6

(a) (b)

n-type no = N d = 5 × 1016 cm −3 10 ni2 (1.5 × 10 ) po = = = 4.5 × 103 cm −3 no 5 × 1016 2

(c)

no = N d = 5 × 1016 cm −3

From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3

( 3.97 × 10 ) =

11 2

= 3.15 × 106 cm −3 5 × 1016 ______________________________________________________________________________________ po

1.7

(a) p-type; p o = 5 × 1016 cm −3 ; no =

(

)

(

)

ni2 1.5 × 1010 = po 5 × 1016

2

= 4.5 × 10 3 cm −3

2

ni2 1.8 × 10 6 = = 6.48 × 10 −5 cm −3 po 5 × 1016 ______________________________________________________________________________________

(b) p-type; p o = 5 × 1016 cm −3 ; no =

1.8 (a) Add boron atoms (b) N a = po = 2 × 1017 cm −3

(

)

2

ni2 1.5 × 1010 = = 1.125 × 10 3 cm −3 po 2 × 1017 ______________________________________________________________________________________

(c) no =

1.9

(a)

no = 5 × 1015 cm −3 10 n 2 (1.5 × 10 ) po = i = ⇒ po = 4.5 × 104 cm −3 no 5 × 1015 2

(b)

n o > p o ⇒ n-type

(c) no ≅ N d = 5 × 1015 cm −3 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.10 a.

b.

Add Donors N d = 7 × 1015 cm −3 Want po = 106 cm −3 = ni2 / N d

So ni2 = (106 )( 7 × 1015 ) = 7 × 10 21 ⎛ − Eg ⎞ = B 2T 3 exp ⎜ ⎟ ⎝ kT ⎠

⎛ ⎞ 2 −1.1 ⎟ 7 × 1021 = ( 5.23 × 1015 ) T 3 exp ⎜ 6 − ⎜ ( 86 × 10 ) (T ) ⎟ ⎝ ⎠ By trial and error, T ≈ 324° K ______________________________________________________________________________________

1.11 (a) I = Aσ Ε = 10 −5 (1.5)(10) ⇒ I = 0.15 mA

( )

(

)

Iρ 1.2 × 10 −3 (0.4) = 2.4 V/cm = A ρ 2 × 10 − 4 ______________________________________________________________________________________

(b) I =

AΕ

⇒Ε=

(

)

1.12

J 120 −1 = = 6.67 (Ω − cm) Ε 18 σ (6.67) σ ≅ eμ n N d ⇒ N d = = = 3.33 × 1016 cm −3 eμ n 1.6 × 10 −19 (1250) ______________________________________________________________________________________ J =σΕ ⇒σ =

(

)

1.13

1 1 1 ⇒ Nd = = = 7.69 × 1015 cm −3 eμ n N d eμ n ρ 1.6 × 10 −19 (1250)(0.65) Ε (b) J = ⇒ Ε = ρ J = (0.65)(160 ) = 104 V/cm ρ ______________________________________________________________________________________ (a) ρ ≅

(

)

1.14

σ 1.5 = = 9.375 × 1015 cm −3 eμ n 1.6 × 10 −19 (1000) σ 0.8 = = 1.25 × 1016 cm −3 (b) N a = −19 eμ p 1.6 × 10 (400) ______________________________________________________________________________________ (a) σ ≅ eμ n N d ⇒ N d =

(

(

)

)

1.15 (a) For n-type, σ ≅ eμ n N d = (1.6 × 10 −19 ) ( 8500 ) N d

For 1015 ≤ N d ≤ 1019 cm −3 ⇒ 1.36 ≤ σ ≤ 1.36 × 104 ( Ω − cm )

−1

(b) J = σ E = σ ( 0.1) ⇒ 0.136 ≤ J ≤ 1.36 ×103 A / cm2 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.16 Dn = (0.026)(1250) = 32.5 cm 2 /s; D p = (0.026 )(450 ) = 11.7 cm 2 /s J n = eDn

⎛ 1016 − 1012 dn = 1.6 × 10 −19 (32.5)⎜⎜ dx ⎝ 0 − 0.001

(

J p = −eD p

)

⎞ ⎟⎟ = −52 A/cm 2 ⎠

⎛ 1012 − 1016 dp = − 1.6 × 10 −19 (11.7 )⎜⎜ dx ⎝ 0 − 0.001

(

)

⎞ ⎟⎟ = −18.72 A/cm 2 ⎠

Total diffusion current density J = −52 − 18.72 = −70.7 A/cm 2 ______________________________________________________________________________________ 1.17 J p = −eD p

dp dx

⎛ −1 ⎞ ⎛ −x ⎞ = −eD p (10 15 ) ⎜ ⎟ exp ⎜ ⎟ ⎜ Lp ⎟ ⎜ Lp ⎟ ⎝ ⎠ ⎝ ⎠

(1.6 ×10 ) (15) (10 ) exp ⎛ − x ⎞ −19

Jp =

15

⎜⎜ ⎟⎟ ⎝ Lp ⎠

10 × 10 −4

J p = 2.4 e

− x / Lp

J p = 2.4 A/cm2

(a)

x=0

(b)

x = 10 μ m

J p = 2.4 e−1 = 0.883 A/cm 2

x = 30 μ m J p = 2.4 e−3 = 0.119 A/cm 2 ______________________________________________________________________________________ (c)

1.18 a.

N a = 1017 cm −3 ⇒ po = 1017 cm −3 n 2 (1.8 × 10 no = i = 1017 po

b.

)

6 2

⇒ no = 3.24 × 10−5 cm −3

n = no + δ n = 3.24 × 10−5 + 1015 ⇒ n = 1015 cm −3 p = po + δ p = 1017 + 1015 ⇒ p = 1.01× 1017 cm −3

______________________________________________________________________________________ ⎛N N 1.19 Vbi = VT ln⎜⎜ a 2 d ⎝ ni

(a) (i)

⎞ ⎟ ⎟ ⎠

(

)(

⎡ 5 × 1015 5 × 1015 Vbi = (0.026 ) ln ⎢ 2 ⎢⎣ 1.5 × 1010

(

(

)

)⎤⎥ = 0.661 V ⎥⎦

)( ) ( ) ⎡ (10 )(10 ) ⎤ = (0.026 ) ln ⎢ ⎥ = 0.937 V ⎣⎢ (1.5 × 10 ) ⎦⎥

(ii)

⎡ 5 × 1017 1015 ⎤ Vbi = (0.026 ) ln ⎢ ⎥ = 0.739 V 10 2 ⎥⎦ ⎣⎢ 1.5 × 10

(iii)

Vbi

18

18

10 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(

)(

⎡ 5 × 1015 5 × 1015 Vbi = (0.026 ) ln ⎢ 2 ⎢⎣ 1.8 × 10 6

(b) (i)

(

(

)

)⎤⎥ = 1.13 V ⎥⎦

)( )⎤⎥ = 1.21 V ( ) ⎥⎦ ⎡ (10 )(10 ) ⎤ = (0.026 ) ln ⎢ ⎥ = 1.41 V ⎢⎣ (1.8 × 10 ) ⎥⎦

(ii)

⎡ 5 × 1017 1015 Vbi = (0.026 ) ln ⎢ 2 ⎢⎣ 1.8 × 10 6

(iii)

Vbi

18

18

6 2

______________________________________________________________________________________ 1.20 ⎛N N Vbi = VT ln⎜⎜ a 2 d ⎝ ni

or

⎞ ⎟ ⎟ ⎠

(n ) exp⎛⎜ V

(1.5 × 10) exp⎛ 0.712 ⎞ = 1.76 × 1016 cm −3 bi ⎞ ⎜ ⎟ ⎜ V ⎟⎟ = 1016 Nd ⎝ 0.026 ⎠ ⎝ T ⎠ ______________________________________________________________________________________ Na =

1.21

2 i

2

⎡ N a (1016 ) ⎤ ⎛N N ⎞ ⎥ Vbi = VT ln ⎜ a 2 d ⎟ = ( 0.026 ) ln ⎢ 10 2 ⎢⎣ (1.5 × 10 ) ⎥⎦ ⎝ ni ⎠

For N a = 1015 cm −3 , Vbi = 0.637 V For N a = 1018 cm −3 , Vbi = 0.817 V

______________________________________________________________________________________ 1.22

⎛ T ⎞ kT = (0.026) ⎜ ⎟ ⎝ 300 ⎠ 200 250 300 350 400 450 500

kT 0.01733 0.02167 0.026 0.03033 0.03467 0.0390 0.04333

(T)3/2 2828.4 3952.8 5196.2 6547.9 8000.0 9545.9 11,180.3

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ ⎞ −1.4 ⎟ ni = ( 2.1× 1014 )(T 3 / 2 ) exp ⎜ ⎜ 2 ( 86 × 10−6 ) (T ) ⎟ ⎝ ⎠ ⎛N N ⎞ Vbi = VT ln ⎜ a 2 d ⎟ ⎝ ni ⎠ T ni 200 1.256 250 6.02 × 103 300 1.80 × 106 350 1.09 × 108 400 2.44 × 109 450 2.80 × 1010 500 2.00 × 1011

Vbi 1.405 1.389 1.370 1.349 1.327 1.302 1.277

______________________________________________________________________________________ 1.23

⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠

−1/ 2

⎡ (1.5 × 10 16 )( 4 × 10 15 ) ⎤ ⎥ = 0.684 V Vbi = ( 0.026 ) ln ⎢ ⎢⎣ (1.5 ×10 10 ) 2 ⎥⎦ 1 ⎞ ⎛ C j = ( 0.4 ) ⎜ 1 + ⎟ 0.684 ⎝ ⎠

−1/ 2

(a)

3 ⎞ ⎛ C j = ( 0.4 ) ⎜ 1 + ⎟ ⎝ 0.684 ⎠

−1/ 2

(b)

= 0.255 pF

= 0.172 pF −1/ 2

5 ⎞ ⎛ = 0.139 pF C j = ( 0.4 ) ⎜ 1 + ⎟ ⎝ 0.684 ⎠ ______________________________________________________________________________________

(c)

1.24

(a)

⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠

−1 / 2

5 ⎞ ⎛ For VR = 5 V, C j = (0.02) ⎜ 1 + ⎟ ⎝ 0. 8 ⎠

−1 / 2

⎛ 1. 5 ⎞ For VR = 1.5 V, C j = (0.02) ⎜1 + ⎟ ⎝ 0. 8 ⎠

= 0.00743 pF −1 / 2

= 0.0118 pF

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 0.00743 + 0.0118 = 0.00962 pF 2 vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ

C j (avg ) =

where τ = RC = RC j (avg ) = (47 × 103 )(0.00962 × 10−12 ) or

τ = 4.52 ×10−10 s Then vC ( t ) = 1.5 = 0 + ( 5 − 0 ) e−ti / τ

5 + r /τ ⎛ 5 ⎞ = e 1 ⇒ t1 = τ ln ⎜ ⎟ 1.5 ⎝ 1.5 ⎠ −10 t1 = 5.44 × 10 s (b)

For VR = 0 V, Cj = Cjo = 0.02 pF −1/ 2 ⎛ 3.5 ⎞ For VR = 3.5 V, C j = ( 0.02 ) ⎜ 1 + = 0.00863 pF ⎟ ⎝ 0.8 ⎠ 0.02 + 0.00863 C j (avg ) = = 0.0143 pF 2 τ = RC j ( avg ) = 6.72 ×10−10 s vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ

(

3.5 = 5 + (0 − 5)e − t2 /τ = 5 1 − e − t2 /τ

so that t2 = 8.09 × 10

−10

)

s

______________________________________________________________________________________ 1.25

⎛ V C j = C jo ⎜⎜1 + R ⎝ Vbi

⎞ ⎟⎟ ⎠

−1 / 2

(

)( ) )

⎡ 5 × 1015 1017 ⎤ ; Vbi = (0.026 ) ln ⎢ ⎥ = 0.739 V 10 2 ⎥⎦ ⎣⎢ 1.5 × 10

(

For V R = 1 V, Cj =

0.60 1 1+ 0.739

= 0.391 pF

For VR = 3 V, Cj =

0.60 3 1+ 0.739

= 0.267 pF

For V R = 5 V, 0.60

Cj =

(a)

fo =

(b) f o =

1 2π LC

(

5 1+ 0.739

=

(

= 0.215 pF

2π 1.5 × 10

2π 1.5 × 10

1 −3

)(0.391× 10 ) −12

1 −3

)(0.267 × 10 ) −12

⇒ f o = 6.57 MHz

⇒ f o = 7.95 MHz

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

fo =

1

⇒ f o = 8.86 MHz 2π 1.5 × 10 0.215 × 10 −12 ______________________________________________________________________________________ (c)

(

−3

)(

)

1.26

a.

⎡ ⎛V ⎞ ⎤ ⎛V ⎞ I = I S ⎢ exp ⎜ D ⎟ − 1⎥ − 0.90 = exp ⎜ D ⎟ − 1 ⎝ VT ⎠ ⎦ ⎝ VT ⎠ ⎣ ⎛V ⎞ exp ⎜ D ⎟ = 1 − 0.90 = 0.10 ⎝ VT ⎠

VD = VT ln ( 0.10 ) ⇒ VD = −0.0599 V

b. IF IR

⎡ ⎛ VF ⎢ exp ⎜ I ⎝ VT = S ⋅⎣ IS ⎡ ⎛ VR ⎢exp ⎜ ⎝ VT ⎣ =

⎞ ⎤ ⎛ 0.2 ⎞ ⎟ − 1⎥ exp ⎜ ⎟ −1 0.026 ⎠ ⎠ ⎦ ⎝ = ⎞ ⎤ exp ⎛ −0.2 ⎞ − 1 ⎜ ⎟ ⎟ − 1⎥ ⎝ 0.026 ⎠ ⎠ ⎦

2190 −1

IF = 2190 IR

______________________________________________________________________________________

⎡ ⎛V 1.27 I D = I S ⎢exp⎜⎜ D ⎣⎢ ⎝ VT (a) (i) (ii) (iii) (iv)

⎞ ⎤ ⎟⎟ − 1⎥ ⎠ ⎦⎥

(

)

(

)

(

)

(

)

(

)

⎛ 0.3 ⎞ I D = 10 −11 exp⎜ ⎟ ⇒ 1.03 μ A ⎝ 0.026 ⎠ ⎛ 0.5 ⎞ I D = 10 −11 exp⎜ ⎟ ⇒ 2.25 mA ⎝ 0.026 ⎠ ⎛ 0.7 ⎞ I D = 10 −11 exp⎜ ⎟ ⇒ 4.93 A ⎝ 0.026 ⎠ ⎡ ⎛ − 0.02 ⎞ ⎤ −12 A I D = 10 −11 ⎢exp⎜ ⎟ − 1⎥ = −5.37 × 10 ⎣ ⎝ 0.026 ⎠ ⎦

(v)

⎡ ⎛ − 0.20 ⎞ ⎤ −11 I D = 10 −11 ⎢exp⎜ ⎟ − 1⎥ ≅ −10 A 0 . 026 ⎠ ⎦ ⎣ ⎝

(vi)

I D = − 10 −11 A

(b) (i) (ii) (iii)

(

)

(

)

(

)

(

)

⎛ 0.3 ⎞ I D = 10 −13 exp⎜ ⎟ ⇒ 0.0103 μ A ⎝ 0.026 ⎠ ⎛ 0.5 ⎞ I D = 10 −13 exp⎜ ⎟ ⇒ 22.5 μ A ⎝ 0.026 ⎠ ⎛ 0.7 ⎞ I D = 10 −13 exp⎜ ⎟ ⇒ 49.3 mA ⎝ 0.026 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(

)

(iv)

⎡ ⎛ − 0.02 ⎞ ⎤ −14 I D = 10 −13 ⎢exp⎜ A ⎟ − 1⎥ = −5.37 × 10 0 . 026 ⎝ ⎠ ⎣ ⎦

(v)

I D ≅ −10 −13 A

I D ≅ −10 −13 A (vi) ______________________________________________________________________________________ ⎛I 1.28 V D = VT ln⎜⎜ D ⎝ IS

⎞ ⎟ ⎟ ⎠

⎛ 10 × 10 −6 (a) (i) V D = (0.026 ) ln⎜⎜ −11 ⎝ 10

⎞ ⎟⎟ = 0.359 V ⎠

⎛ 100 × 10 −6 V D = (0.026 ) ln⎜⎜ −11 ⎝ 10

⎞ ⎟⎟ = 0.419 V ⎠

⎛ 10 −3 ⎞ V D = (0.026) ln⎜⎜ −11 ⎟⎟ = 0.479 V ⎝ 10 ⎠ ⎡ ⎛ V ⎞ ⎤ (ii) − 5 × 10 −12 = 10 −11 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = −0.018 V ⎣ ⎝ 0.026 ⎠ ⎦ ⎛ 10 × 10 −6 (b) (i) V D = (0.026 ) ln⎜⎜ −13 ⎝ 10

⎞ ⎟⎟ = 0.479 V ⎠

⎛ 100 × 10 −6 V D = (0.026 ) ln⎜⎜ −13 ⎝ 10 ⎛ 10 −3 V D = (0.026) ln⎜⎜ −13 ⎝ 10

⎞ ⎟⎟ = 0.539 V ⎠

⎞ ⎟⎟ = 0.599 V ⎠

⎡ ⎛ V ⎞ ⎤ (ii) − 10 −14 = 10 −13 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = −0.00274 V ⎣ ⎝ 0.026 ⎠ ⎦ ______________________________________________________________________________________

1.29

(a)

(b)

VD

⎛ 0.7 ⎞ 10−3 = I S exp ⎜ ⎟ ⎝ 0.026 ⎠ I S = 2.03 × 10 −15 A I D ( A ) ( n = 1)

I D ( A )( n = 2 )

0.1 9.50 ×10 1.39 ×10 −14 0.2 4.45 ×10 −12 9.50 ×10 −14 0.3 2.08 ×10 −10 6.50 ×10 −13 − 9 0.4 9.75 ×10 4.45 ×10 −12 0.5 4.56 ×10 −7 3.04 ×10 −11 − 5 0.6 2.14 ×10 2.08 ×10 −10 0.7 10 −3 1.42 ×10 −9 ______________________________________________________________________________________ −14

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.30 (a) I S = 10 −12 A

VD(v) 0.10 0.20 0.30 0.40 0.50 0.60 0.70 (b)

ID(A) 4.68 ×10−11 2.19 ×10−9 1.03 ×10−7 4.80 ×10−6 2.25 ×10−4 1.05 ×10−2 4.93 ×10−1

log10ID −10.3 −8.66 −6.99 −5.32 −3.65 −1.98 −0.307

I S = 10 −14 A

VD(v) ID(A) log10ID −13 −12.3 0.10 4.68 ×10 − 11 −10.66 0.20 2.19 ×10 − 9 −8.99 0.30 1.03 ×10 −8 −7.32 0.40 4.80 ×10 − 6 −5.65 0.50 2.25 ×10 −4 −3.98 0.60 1.05 ×10 − 3 −2.31 0.70 4.93 ×10 ______________________________________________________________________________________ 1.31 a. ⎛ V − VD1 ⎞ ID2 = 10 = exp ⎜ D 2 ⎟ I D1 ⎝ VT ⎠ ΔVD = VT ln (10) ⇒ ΔVD = 59.9 mV ≈ 60 mV

b.

ΔVD = VT ln (100 ) ⇒ ΔVD = 119.7 mV ≈ 120 mV

______________________________________________________________________________________ 1.32

⎛ 2 ⎞ (a) (i) V D = (0.026) ln⎜ ⎟ = 0.539 V −9 ⎝ 2 × 10 ⎠ ⎛ 20 ⎞ (ii) V D = (0.026) ln⎜ ⎟ = 0.599 V −9 ⎝ 2 × 10 ⎠ ⎛ 0.4 ⎞ (b) (i) I D = 2 × 10 −9 exp⎜ ⎟ ⇒ 9.60 mA ⎝ 0.026 ⎠ ⎛ 0.65 ⎞ (ii) I D = 2 × 10 −9 exp⎜ ⎟ ⇒ 144 A ⎝ 0.026 ⎠ ______________________________________________________________________________________

(

)

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.33 ⎛I ⎞ ⎛ 2 × 10−3 ⎞ = 0.6347 V VD = Vt ln ⎜ D ⎟ = (0.026) ln ⎜ −14 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠ ⎛ 2 × 10−3 ⎞ = 0.5150 V VD = (0.026) ln ⎜ −12 ⎟ ⎝ 5 × 10 ⎠ 0.5150 ≤ VD ≤ 0.6347 V

______________________________________________________________________________________ 1.34

⎛ 0.30 ⎞ −8 (a) 1.5 × 10 −3 = I S exp⎜ ⎟ ⇒ I S = 1.46 × 10 A ⎝ 0.026 ⎠ ⎛ 0.35 ⎞ (b) (i) I D = 1.462 × 10 −8 exp⎜ ⎟ ⇒ I D = 10.3 mA ⎝ 0.026 ⎠ ⎛ 0.25 ⎞ (ii) I D = 1.462 × 10 −8 exp⎜ ⎟ ⇒ I D = 0.219 mA ⎝ 0.026 ⎠ ______________________________________________________________________________________

(

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(

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1.35

(

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(

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(

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(

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⎛ 0.8 ⎞ (a) I D = 10 − 22 exp⎜ ⎟ ⇒ 2.31 nA ⎝ 0.026 ⎠ ⎛ 1.0 ⎞ I D = 10 − 22 exp⎜ ⎟ ⇒ 5.05 μ A ⎝ 0.026 ⎠ ⎛ 1.2 ⎞ I D = 10 − 22 exp⎜ ⎟ ⇒ 11.1 mA ⎝ 0.026 ⎠ ⎡ ⎛ − 0.02 ⎞ ⎤ − 23 I D = 10 − 22 ⎢exp⎜ ⎟ − 1⎥ = −5.37 × 10 A 0 . 026 ⎠ ⎦ ⎣ ⎝

For V D = −0.20 V, I D = −10 −22 A For V D = −2 V, I D = −10 −22 A (b) ⎛ 0.8 ⎞ I D = 5 × 10 −24 exp⎜ ⎟ ⇒ 115 pA ⎝ 0.026 ⎠ ⎛ 1.0 ⎞ I D = 5 × 10 − 24 exp⎜ ⎟ ⇒ 0.253 μ A ⎝ 0.026 ⎠ ⎛ 1.2 ⎞ I D = 5 × 10 − 24 exp⎜ ⎟ ⇒ 0.554 mA ⎝ 0.026 ⎠

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(

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(

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(

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⎡ ⎛ − 0.02 ⎞ ⎤ − 24 A I D = 5 × 10 − 24 ⎢exp⎜ ⎟ − 1⎥ = −2.68 × 10 0 . 026 ⎠ ⎦ ⎣ ⎝

For V D = −0.20 V, I D = −5 × 10 −24 A For V D = −2 V, I D = −5 × 10 −24 A ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

1.36

IS doubles for every 5C increase in temperature. I S = 10 −12 A at T = 300K For I S = 0.5 × 10 −12 A ⇒ T = 295 K For I S = 50 × 10 −12 A, (2) n = 50 ⇒ n = 5.64 Where n equals number of 5C increases. Then ΔT = ( 5.64 )( 5 ) = 28.2 K So 295 ≤ T ≤ 328.2 K ______________________________________________________________________________________ 1.37 I S (T ) = 2ΔT / 5 , ΔT = 155° C I S (−55) I S (100) = 2155 / 5 = 2.147 × 109 I S (−55) VT @100°C ⇒ 373°K ⇒ VT = 0.03220 VT @ − 55°C ⇒ 216°K ⇒ VT = 0.01865

I D (100) = (2.147 × 109 ) × I D (−55)

⎛ 0.6 ⎞ exp ⎜ ⎟ ⎝ 0.0322 ⎠ ⎛ 0.6 ⎞ exp ⎜ ⎟ ⎝ 0.01865 ⎠

( 2.147 ×10 )(1.237 ×10 ) ( 9.374 ×10 ) 9

=

8

13

I D (100) = 2.83 × 103 I D (−55) ______________________________________________________________________________________ 1.38 (a) V PS = I D R + V D

( )

2.8 = I D 10 6 + VD ;

(

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⎛ V ⎞ I D = 5 × 10 −11 exp⎜ D ⎟ ⎝ 0.026 ⎠

By trial and error, V D = 0.282 V, I D = 2.52 μ A (b) I D ≅ −5 × 10 −11 A, VD = −2.8 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

1.39 ⎛ I ⎞ 10 = I D ( 2 × 10 4 ) + VD and VD = ( 0.026 ) ln ⎜ D−12 ⎟ ⎝ 10 ⎠ Trial and error. VD(v) ID(A) VD(v) −4 0.50 0.5194 4.75 ×10 − 4 0.517 0.5194 4.7415 ×10 −4 0.5194 0.5194 4.740 ×10

VD = 0.5194 V I D = 0.4740 mA ______________________________________________________________________________________ 1.40 I s = 5 × 10 −13 A

⎛ R2 VTH = ⎜ ⎝ R1 + R2

⎞ ⎛ 30 ⎞ ⎟ (1.2) = ⎜ ⎟ (1.2) = 0.45 V 80 ⎠ ⎝ ⎠

⎛I ⎞ 0.45 = I D RTH + VD , VD = VT ln ⎜ D ⎟ ⎝ IS ⎠ By trial and error: I D = 2.56 μ A, VD = 0.402 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

1.41 (a) I D1 = I D 2 = 1 mA

(i)

⎛ 10 −3 V D1 = V D 2 = (0.026 ) ln⎜⎜ −13 ⎝ 10

⎞ ⎟⎟ = 0.599 V ⎠

(ii)

⎛ 10 −3 V D1 = (0.026 ) ln⎜⎜ −14 ⎝ 5 × 10

⎞ ⎟⎟ = 0.617 V ⎠

⎛ 10 −3 V D 2 = (0.026 ) ln⎜⎜ −13 ⎝ 5 × 10

⎞ ⎟⎟ = 0.557 V ⎠

(b) V D1 = V D 2

V D1 = V D 2

(ii)

Ii = 0.5 mA 2 ⎛ 0.5 × 10 −3 = (0.026 ) ln⎜⎜ −13 ⎝ 10

I D1 = I D 2 =

(i)

⎞ ⎟⎟ = 0.581 V ⎠

I I D1 5 × 10 −14 = S1 = = 0.10 I D 2 I S 2 5 × 10 −13 So I D1 = 0.10 I D 2

I D1 + I D 2 = 1.1I D 2 = 1 mA

So I D 2 = 0.909 mA, I D1 = 0.0909 mA Now ⎛ 0.0909 × 10 −3 ⎞ ⎟⎟ = 0.554 V V D1 = (0.026 ) ln⎜⎜ −14 ⎝ 5 × 10 ⎠ ⎛ 0.909 × 10 −3 ⎞ ⎟⎟ = 0.554 V V D 2 = (0.026 ) ln⎜⎜ −13 ⎝ 5 × 10 ⎠ ______________________________________________________________________________________

1.42

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⎛ 0.635 ⎞ (a) I D 3 = 6 × 10 −14 exp⎜ ⎟ ⇒ 2.426 mA ⎝ 0.026 ⎠ 0.635 = 0.635 mA IR = 1 I D1 = I D 2 = 2.426 + 0.635 = 3.061 mA ⎛ 3.061 × 10 −3 V D1 = V D 2 = (0.026 ) ln⎜⎜ −14 ⎝ 6 × 10 V I = 2(0.641) + 0.635 = 1.917 V

⎞ ⎟⎟ = 0.641 V ⎠

(b) I D 3 = 2.426 mA 0.635 = 1.27 mA IR = 0.5 I D1 = I D 2 = 2.426 + 1.27 = 3.696 mA ⎛ 3.696 × 10 −3 V D1 = V D 2 = (0.026 ) ln⎜⎜ −14 ⎝ 6 × 10 V I = 2(0.6459 ) + 0.635 = 1.927 V

⎞ ⎟⎟ = 0.6459 V ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 1.43 (a) Assume diode is conducting. Then, VD = Vγ = 0.7 V 0. 7 ⇒ 23.3 μ A 30 1.2 − 0.7 ⇒ 50 μ A I R1 = 10 Then I D = I R1 − I R 2 = 50 − 23.3

So that I R 2 =

Or I D = 26.7 μ A (b) Let R1 = 50 k Ω Diode is cutoff. 30 ⋅ (1.2) = 0.45 V 30 + 50 Since VD < Vγ , I D = 0 VD =

______________________________________________________________________________________ 1.44

At node VA: 5 − VA V = ID + A (1) 2 2 At node V B = V A − Vγ (2)

5 − (VA − Vr )

+ ID =

(VA − Vr )

2 2 5 − (VA − Vr ) ⎡ 5 − VA VA ⎤ VA − Vr +⎢ − ⎥= So 3 2⎦ 2 ⎣ 2 Multiply by 6: 10 − 2 (VA − Vr ) + 15 − 6VA = 3 (VA − Vr )

25 + 2Vr + 3Vr = 11VA

(a)

Vr = 0.6 V

11VA = 25 + 5 ( 0.6 ) = 28 ⇒ VA = 2.545 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5 − VA VA − = 2.5 − VA ⇒ I D Neg. ⇒ I D = 0 2 2 Both (a), (b) I D = 0

From (1) I D =

2 ⋅ 5 = 2 V ⇒ VD = 0.50 V 5 ______________________________________________________________________________________

VA = 2.5, VB =

1.45 (a) VO = I i (1) ; I D = 0 ; for 0 ≤ I i ≤ 0.7 mA VO = 0.7 V; I D = (I i − 0.7 ) mA; for I i ≥ 0.7 mA

(b) VO = I i (1) ; I D = 0 ; for 0 ≤ I i ≤ 1.7 mA VO = 1.7 V; I D = (I i − 1.7 ) mA; for I i ≥ 1.7 mA (c) VO = 0.7 V; I D1 = I i ; I D 2 = 0 ; for 0 ≤ I i ≤ 2 mA ______________________________________________________________________________________ 1.46 Minimum diode current for VPS (min) I D (min) = 2 mA, VD = 0.7 V I2 =

0.7 5 − 0.7 4.3 , I1 = = R2 R1 R1

We have I1 = I 2 + I D 4.3 0.7 = +2 R1 R2 Maximum diode current for VPS (max) P = I DVD 10 = I D ( 0.7 ) ⇒ I D = 14.3 mA

so (1)

I1 = I 2 + I D

or (2)

9.3 0.7 = + 14.3 R1 R2

Using Eq. (1),

9.3 4.3 = − 2 + 14.3 ⇒ R1 R1

R1 = 0.41 kΩ

Then R2 = 82.5Ω 82.5Ω ______________________________________________________________________________________ 1.47

5 − 0.7 = 0.215 mA, VO = 0.7 V 20 5 − 0.6 = 0.220 mA, VO = 0.6 V (ii) I = 20 5 − 0.7 − (− 5) = 0.2325 mA, VO = (0.2325)(20 ) − 5 = −0.35 V (b) (i) I = 40 5 − 0.6 − (− 5) I= = 0.235 mA, VO = (0.235)(20 ) − 5 = −0.30 V (iii) 40 (a) (i) I =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

2 − 0.7 − (− 8) = 0.372 mA, VO = 2 − (0.372 )(5) = 0.14 V 25 2 − 0.6 − (− 8) = 0.376 mA, VO = 2 − (0.376 )(5) = 0.12 V (ii) I = 25 (d) (i) I = 0 , VO = −5 V (ii) I = 0 , VO = −5 V ______________________________________________________________________________________ (c) (i) I =

1.48

(a) I =

5 − VO ⎛ V ⎞ , I = 5 × 10 −14 exp⎜ D ⎟ 20 ⎝ 0.026 ⎠ By trial and error, V D = VO = 0.5775 V, I = 0.221 mA

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10 − V D , VO = 5 − I (20 ) − V D 40 I = 0.2355 mA, V D = 0.579 V, VO = −0.289

(b) I =

10 − V D , VO = 2 − I (5) 25 I = 0.3763 mA, V D = 0.5913 V, VO = 0.1185

(c) I =

(d) I = −5 × 10 −14 A, VO ≅ −5 V ______________________________________________________________________________________ 1.49 (a)

Diode forward biased VD = 0.7 V 5 = (0.4)(4.7) + 0.7 + V ⇒ V = 2.42 V (b) P = I ⋅ VD = (0.4)(0.7) ⇒ P = 0.28 mω ______________________________________________________________________________________ 1.50

(a)

0.65 = 0.65 mA = I D1 1 = 2(0.65) = 1.30 mA

I R 2 = I D1 = ID2

ID2 =

(b)

VI − 2Vr − V0 5 − 3(0.65) = = 1.30 ⇒ R1 = 2.35 K R1 R1

0.65 = 0.65 mA 1 8 − 3(0.65) ID2 = ⇒ I D 2 = 3.025 mA 2 I D1 = I D 2 − I R 2 = 3.025 − 0.65 I D1 = 2.375 mA IR2 =

______________________________________________________________________________________ 1.51

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V (0.026) τd = T = = 0.026 kΩ = 26Ω a. 1 I DQ

id = 0.05 I DQ = 50 μ A peak-to-peak vd = idτ d = (26)(50) μ A ⇒ vd = 1.30 mV peak-to-peak b.

(0.026) = 260Ω 0. 1 = 5 μ A peak-to-peak

For I DQ = 0.1 mA ⇒ τ d = id = 0.05 I DQ

vd = idτ d = (260)(5) μ V ⇒ vd = 1.30 mV peak-to-peak ______________________________________________________________________________________ 1.52

(a) rd =

VT 0.026 = =1 kΩ I DQ 0.026

0.026 ⇒ 100 Ω 0.26 0.026 ⇒ 10 Ω (c) rd = 2.6 ______________________________________________________________________________________ (b) rd =

1.53

a.

diode resistance rd = VT /I ⎛ ⎜ VT /I ⎛ rd ⎞ vd = ⎜ ⎟ vS = ⎜ V ⎝ rd + RS ⎠ ⎜⎜ T + RS ⎝ I ⎛ VT ⎞ vd = ⎜ ⎟ vs = vo ⎝ VT + IRS ⎠

b.

⎞ ⎟ ⎟ vS ⎟⎟ ⎠

RS = 260Ω

⎞ v0 ⎛ VT v 0.026 =⎜ ⇒ 0 = 0.0909 ⎟= vS ⎝ VT + IRS ⎠ 0.026 + (1)(0.26) vS v v 0.026 I = 0.1 mA, 0 = ⇒ 0 = 0.50 vs 0.026 + ( 0.1)( 0.26 ) vS I = 1 mA,

I = 0.01 mA.

v0 v 0.026 = ⇒ 0 = 0.909 vS 0.026 + (0.01)(0.26) vS

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

1.54 pn junction diode ⎛ 0.72 × 10 −3 ⎞ ⎟ = 0.548 V V D = (0.026 ) ln⎜⎜ −13 ⎟ ⎝ 5 × 10 ⎠ Schottky diode ⎛ 0.72 × 10 −3 ⎞ ⎟⎟ = 0.249 V V D = (0.026 ) ln⎜⎜ −8 ⎝ 5 × 10 ⎠ ______________________________________________________________________________________

1.55

⎛V ⎞ Schottky: I ≅ I S exp ⎜ a ⎟ ⎝ VT ⎠

⎛ I ⎞ ⎛ 0.5 × 10−3 ⎞ Va = VT ln ⎜ ⎟ = (0.026) ln ⎜ −7 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠ = 0.1796 V Then Va of pn junction = 0.1796 + 0.30

= 0.4796 I 0.5 × 10−3 = ⎛V ⎞ ⎛ 0.4796 ⎞ exp ⎜ a ⎟ exp ⎜ ⎟ ⎝ 0.026 ⎠ ⎝ VT ⎠ I S = 4.87 × 10 −12 A

IS =

______________________________________________________________________________________ 1.56

(a)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I1 + I 2 = 0.5 × 10 −3

⎛V 5 × 10 −8 exp ⎜ D ⎝ VT

⎞ ⎛ VD ⎞ −12 −3 ⎟ + 10 exp ⎜ ⎟ = 0.5 × 10 V ⎠ ⎝ T ⎠

⎛V 5.0001× 10 −8 exp ⎜ D ⎝ VT

⎞ −3 ⎟ = 0.5 × 10 ⎠

⎛ 0.5 × 10−3 ⎞ ⇒ VD = 0.2395 VD = (0.026) ln ⎜ −8 ⎟ ⎝ 5.0001 × 10 ⎠ Schottky diode, I 2 = 0.49999 mA

pn junction, I1 = 0.00001 mA (b)

⎛V ⎞ ⎛V ⎞ I = 10 −12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜ D 2 ⎟ ⎝ VT ⎠ ⎝ VT ⎠ VD1 + VD 2 = 0.9

⎛V ⎞ ⎛ 0.9 − VD1 ⎞ 10−12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜ ⎟ ⎝ VT ⎠ ⎝ VT ⎠ ⎛ 0.9 ⎞ ⎛ −VD1 ⎞ = 5 ×10−8 exp ⎜ ⎟ ⎟ exp ⎜ V ⎝ T ⎠ ⎝ VT ⎠ ⎛ 2V ⎞ ⎛ 5 × 10−8 ⎞ ⎛ 0. 9 ⎞ exp ⎜ D1 ⎟ = ⎜ exp ⎜ ⎟ −12 ⎟ V 10 ⎝ 0.026 ⎠ ⎠ ⎝ T ⎠ ⎝ ⎛ 5 × 10−8 ⎞ + 0.9 = 1.1813 2VD1 = VT ln ⎜ −12 ⎟ ⎝ 10 ⎠ VD1 = 0.5907 pn junction

VD 2 = 0.3093 Schottky diode ⎛ 0.5907 ⎞ I = 10−12 exp ⎜ ⎟ ⇒ I = 7.35 mA ⎝ 0.026 ⎠ ______________________________________________________________________________________ 1.57

VZ = VZ 0 = 5.6 V at I Z = 0.1 mA rZ = 10Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I Z rZ = ( 0.1)(10 ) = 1 mV VZ0 = 5.599

a.

RL → ∞ ⇒ IZ =

10 − 5.599 4.401 = = 8.63 mA R + rZ 0.50 + 0.01

VZ = VZ 0 + I Z rZ = 5.599 + ( 0.00863)(10 ) VZ = V0 = 5.685 V b.

11 − 5.599 = 10.59 mA 0.51 VZ = V0 = 5.599 + ( 0.01059 )(10 ) = 5.7049 V VPS = 11 V ⇒ I Z =

9 − 5.599 = 6.669 mA 0.51 VZ = V0 = 5.599 + ( 0.006669 )(10 ) = 5.66569 V

VPS = 9 V ⇒ I Z =

ΔV0 = 5.7049 − 5.66569 ⇒ ΔV0 = 0.0392 V c.

I = IZ + IL V V − V0 V − VZ 0 , IZ = 0 I L = 0 , I = PS RL R rZ 10 − V0 V0 − 5.599 V0 = + 0.50 0.010 2 10 5.599 1 1⎤ ⎡ 1 + = V0 ⎢ + + ⎥ 0.50 0.010 0 . 50 0 . 010 2⎦ ⎣ 20.0 + 559.9 = V0 (102.5) V0 = 5.658 V

______________________________________________________________________________________ 1.58

10 − 6.8 = 6.4 mA 0.5 P = I Z VZ = (6.4)(6.8) = 43.5 mW (b) I Z = (0.1)(6.4) = 0.64 mA I L = 6.4 − 0.64 = 5.76 mA (a) I Z =

VZ V 6.8 ⇒ RL = Z = = 1.18 k Ω RL I Z 5.76 ______________________________________________________________________________________ IL =

1.59

I Z rZ = ( 0.1)( 20 ) = 2 mV VZ 0 = 6.8 − 0.002 = 6.798 V

a.

RL = ∞ IZ =

10 − 6.798 ⇒ I Z = 6.158 mA 0.5 + 0.02

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V0 = VZ = VZ 0 + I Z rZ = 6.798 + ( 0.006158)( 20 )

V0 = 6.921 V

b.

I = IZ + IL 10 − V0 V0 − 6.798 V0 = + 0.50 0.020 1 10 6.798 1 1⎤ ⎡ 1 + = V0 ⎢ + + ⎥ 0.30 0.020 ⎣ 0.50 0.020 1⎦ 359.9 = V0 (53) V0 = 6.791 V ΔV0 = 6.791 − 6.921

ΔV0 = −0.13 V ______________________________________________________________________________________ 1.60 For VD = 0, I SC = 0.1 A

⎛ 0.2 ⎞ + 1⎟ For ID = 0 VD = VT ln ⎜ −14 ⎝ 5 × 10 ⎠ VD = VDC = 0.754 V ______________________________________________________________________________________ 1.61 V D = 0, I D = 0.2 A V D = 0.60 V, I D = 0.1995 A V D = 0.65 V, I D = 0.1964 A V D = 0.70 V, I D = 0.1754 A V D = 0.72 V, I D = 0.1468 A V D = 0.74 V, I D = 0.0853 A V D = 0.7545 V, I D = 0 ______________________________________________________________________________________ 1.62 ⎡ ⎛ V ⎞ ⎤ (a) 0.16 = 0.20 − 5 × 10 −14 ⎢exp⎜ D ⎟ − 1⎥ ⇒ V D = 0.7126 V ⎣ ⎝ 0.026 ⎠ ⎦ (b) P = (0.16 )(0.7126 ) = 0.114 W ______________________________________________________________________________________

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Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 2 2.1 ⎛ 1000 ⎞ (a) For υ I > 0.6 V, υ O = ⎜ ⎟(υ I − 0.6 ) ⎝ 1020 ⎠ For υ I < 0.6 V, υ O = 0

⎛ 1000 ⎞ (b) (ii) υ O = 0 = ⎜ ⎟[10 sin (ω t )1 − 0.6] ⎝ 1020 ⎠ 0. 6 Then sin (ω t )1 = = 0.06 ⇒ (ω t )1 = 3.44° ⇒ 0.01911π rad 10 Also (ω t )2 = 180 − 3.44 = 176.56° ⇒ 0.9809π rad Now

υ O (avg ) = =

T

1 1 υ O (t )dt = T 0 2π

∫

1 2π

2π

∫ [10 sin x − 0.6]dx 0

0.9809π ⎤ ⎡ − 0.6 x ⎢− 10 cos x ⎥ 0.01911π 0.01911π ⎦ ⎣ 0.9809π

1 [(− 10 )(− 0.9982 − 0.9982 ) − 0.6(0.9809π − 0.01911π )] 2π υ O (avg ) = 2.89 V

=

⎤ ⎛ 1000 ⎞ ⎡ ⎛π ⎞ (iii) υ O ( peak ) = ⎜ ⎟ ⎢10 sin ⎜ ⎟ − 0.6⎥ = 9.2157 V; i d (max ) = 9.2157 mA 1020 2 ⎝ ⎠⎣ ⎝ ⎠ ⎦ (iv) PIV = 10 V ______________________________________________________________________________________

2.2

v0 = vI − vD ⎛i ⎞ v vD = VT ln ⎜ D ⎟ and iD = 0 I R ⎝ S⎠ ⎛ v ⎞ v0 = vI − VT ln ⎜ 0 ⎟ ⎝ IS R ⎠

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.3 ⎛ 1⎞ (a) υ S = 120 2 ⎜ ⎟ = 16.97 V (peak) ⎝ 10 ⎠ υ O ( peak ) = 16.27 V

16.27 = 8.14 mA 2 (c) υ O = 16.97 sin ω t − 0.7

(b) i D ( peak ) =

sin (ω t )1 =

0. 7 = 0.04125 ⇒ (ω t )1 = 2.364° 16.97 (ω t )2 = 180 − 2.364 = 177.64°

⎛ 177.64 − 2.364 ⎞ %=⎜ ⎟ × 100% = 48.7% 360 ⎝ ⎠

(d)

υ O (avg ) =

1 2π

0.9869π

∫ [16.97 sin x − 0.7]dx

0.01313π

0.9869π 0.9869π ⎤ 1 ⎡ = − 0. 7 x (− 16.97 ) cos x ⎢ ⎥ 2π ⎣ 0.01313π 0.01313π ⎦ 1 [(− 16.97 )(− 0.99915 − 0.99915) − 0.7(0.9738π )] = 2π υ O (avg ) = 5.06 V

υ O (avg )

5.06 = 2.53 mA 2 2 ______________________________________________________________________________________

(e) i D (avg ) =

=

2.4 (a) υ R (t ) = 15 sin ω t − 0.7 − 9 = 15 sin ω t − 9.7

(ω t )1 = sin −1 ⎛⎜ 9.7 ⎞⎟ = 40.29° ⇒ 0.2238π

(ω t )2

rad ⎝ 15 ⎠ = 180 − 40.29 = 139.71° ⇒ 0.7762π rad

υ R (avg ) =

1 2π

0.7762π

∫ [15 sin x − 9.7]dx

0.2238π

0.7762π 0.7762π ⎤ 1 ⎡ 1 (− 15) cos x − 9.7 x ⎢ ⎥ = 2π [(− 15)(− 0.7628 − 0.7628 ) − 9.7(0.5523π )] 2π ⎣ 0.2238π 0.2238π ⎦ υ R (avg ) = 0.9628 V

=

i D (avg ) = 0.8 =

0.9628 ⇒ R = 1.20 Ω R

(b) ⎛ 139.71 − 40.29 ⎞ %=⎜ ⎟ × 100% = 27.6% 360 ⎝ ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.5

(a) i ( peak ) =

υ R ( peak )

⇒R= R (b) υ R (t ) = 15 sin ω t − 9.7

15 − 9.7 = 4.417 Ω 1. 2

(ω t )1 = 0.2238π ; (ω t )2 = 0.7762π

υ R (avg ) =

1

π

0.7762π

∫ [15 sin x − 9.7]dx

0.2238π

Or from Problem 2.4, υ R (avg ) = 2(0.9628) = 1.9256 V υ (avg ) 1.9256 = = 0.436 A i D (avg ) = R R 4.417 (c) ⎛ 139.71 − 40.29 ⎞ %=⎜ ⎟ × 100% = 27.6% 360 ⎝ ⎠ ______________________________________________________________________________________

2.6 (a) υ S ( peak ) = 12 + 0.7 = 12.7 V

N 1 120 2 = = 13.4 N2 12.7 12 = 60 Ω 0 .2 VM 12 = ⇒ 6667 μ F C= 2 fRV r 2(60 )(60 )(0.25)

(b) R =

(c) PIV = 2υ S (max ) − Vγ = 2(12.7 ) − 0.7 = 24.7 V ______________________________________________________________________________________ 2.7

v0 = vS − 2Vγ ⇒ vS ( max ) = v0 ( max ) + 2Vγ a.

b.

v ( max ) = 25 V ⇒ vS ( max ) = 25 + 2 ( 0.7 ) = 26.4 V For 0 N1 160 N = ⇒ 1 = 6.06 N 2 26.4 N2 v ( max ) = 100 V ⇒ vS ( max ) = 101.4 V For 0 N1 N 160 = ⇒ 1 = 1.58 N 2 101.4 N2

PIV = 2vS ( max ) − Vγ = 2 ( 26.4 ) − 0.7 From part (a) PIV = 2 (101.4 ) − 0.7 or PIV = 52.1 V or, from part (b) or PIV = 202.1 V ______________________________________________________________________________________ 2.8

(a)

vs (max) = 12 + 2(0.7) = 13.4 V 13.4 vs ( rms ) = ⇒ vs (rms) = 9.48 V 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) Vr =

VM VM ⇒C = 2 f RC 2 f Vr R

C=

12 ⇒ C = 2222 μ F 2 ( 60 )( 0.3)(150 )

(c) ⎡ 2VM ⎤ ⎢1 + π ⎥ Vr ⎥⎦ ⎢⎣ 2 (12 ) ⎤ 12 ⎡ ⎢1 + π ⎥ = 150 ⎢ 0.3 ⎥ ⎣ ⎦ id , peak = 2.33 A id , peak =

VM R

______________________________________________________________________________________ 2.9

(a)

vS ( max ) = 12 + 0.7 = 12.7 V vS ( rms ) = Vr =

(b)

vS ( max ) 2

⇒ vS ( rms ) = 8.98 V

VM V 12 ⇒C = M = fRC fRVr ( 60 )(150 )( 0.3)

or

C = 4444 μ F

⎛ VM ⎞ 12 ⎛ 12 ⎞ ⎜ 1 + 4π ⎟ ⎜⎜ 1 + 4π ⎟⎟ = iD , max = 4.58 A 2Vr ⎠ 150 ⎜⎝ 2 ( 0.3) ⎟⎠ ⎝ or (c) For the half-wave rectifier ______________________________________________________________________________________ iD , max =

VM R

2.10 (a) υ O ( peak ) = 10 − 0.7 = 9.3 V VM 9.3 = ⇒ 620 μ F fRV r (60 )(500 )(0.5) (c) PIV = 10 + 9.3 = 19.3 V ______________________________________________________________________________________

(b) C =

2.11 (a) 10.3 ≤ υ O ≤ 12.3 V

(b) Vr =

VM 12.3 = = 0.586 V fRC (60 )(1000 ) 350 × 10 − 6

(

)

10.3 = 0.490 V (60)(1000) 350 × 10 −6 So 0.490 ≤ V r ≤ 0.586 V Vr =

(

)

VM 12.3 = ⇒ 513 μ F fRV r (60)(1000 )(0.4) ______________________________________________________________________________________

(c) C =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.12

( )

(a) υ S ( peak ) = 8.5 2 = 12.02 V VO

max

= 12.02 − 0.7 = 11.32 V

VM 11.32 = = 0.03773 F 2 f RV r 2(60 )(10 )(0.25) (c) PIV = 2υ S ( peak ) − Vγ = 2(12.02 ) − 0.7 = 23.34 V

(b) C =

______________________________________________________________________________________ 2.13 (a)

vs ( peak ) = 15 + 2 ( 0.7 ) = 16.4 V vs ( rms ) = C=

16.4 2

= 11.6 V

VM 15 = = 2857 μ F 2 f RVr 2 ( 60 )(125 )( 0.35 )

(b) ______________________________________________________________________________________ 2.14

______________________________________________________________________________________ 2.15 (a) υ S = 12.8 V

N 1 120 2 = = 13.3 N2 12.8 12 = 24 Ω 0 .5 V r = 3% ⇒ V r = (0.03)(12 ) = 0.36 V

(b) R =

C=

VM 12 = = 0.0116 F 2 fRV r 2(60 )(24 )(0.36 )

⎛ ⎜1 + π 2V M ⎜ Vr ⎝ i D ( peak ) = 13.3 A

(c) i D ( peak ) =

VM R

⎞ 12 ⎛ ⎞ ⎟= ⎜1 + π 2(12 ) ⎟ ⎜ ⎟ 24 0.36 ⎟⎠ ⎝ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (d) i D (avg ) =

1

π

2V r V M ⋅ VM R

⎛ π ⎜1 + ⎜ 2 ⎝

2V M Vr

⎞ 1 ⎟= ⎟ π ⎠

2(0.36 ) ⎛ 12 ⎞⎛⎜ π ⎜ ⎟ 1+ 12 ⎝ 24 ⎠⎜⎝ 2

2(12 ) ⎞⎟ 0.36 ⎟⎠

i D (avg ) = 0.539 A (e) PIV = 12.8 + 12 = 24.8 V ______________________________________________________________________________________

2.16 (a) υ S = 9 + 2(0.8) = 10.6 V

N 1 120 2 = = 16 N2 10.6 9 = 90 Ω 0. 1 VM 9 = ⇒ 4167 μ F C= 2(60)(90)(0.2) 2 fRV r

(b) R =

(c) i D ( peak ) =

VM R

(d) i D (avg ) =

1

π

⎛ ⎜1 + π 2V M ⎜ Vr ⎝ 2V r V M ⋅ VM R

⎞ 9 ⎛ ⎞ ⎟= ⎜1 + π 2(9 ) ⎟ = 3.08 A ⎟ 90 ⎜ 0.2 ⎟⎠ ⎝ ⎠

⎛ π ⎜1 + ⎜ 2 ⎝

2V M Vr

⎞ 1 ⎟= ⎟ π ⎠

i D (avg ) = 0.1067 A (e) PIV = υ S (max ) − Vγ = 10.6 − 0.8 = 9.8 V

2(0.2 ) ⎛ 9 ⎞⎛⎜ π ⎜ ⎟ 1+ 9 ⎝ 90 ⎠⎜⎝ 2

2(9 ) ⎞⎟ 0.2 ⎟⎠

______________________________________________________________________________________ 2.17

For vi > 0 Vγ = 0

Voltage across RL + R1 = vi

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ RL ⎞ 1 ⇒ v0 = ⎜ ⎟ vi = vi R + R 2 ⎝ L 1 ⎠ Voltage Divider

______________________________________________________________________________________ 2.18

For

vi > 0, (Vγ = 0 )

a. ⎛ R2 || RL ⎞ v0 = ⎜ ⎟ vi ⎝ R2 || RL + R1 ⎠ R2 || RL = 2.2 || 6.8 = 1.66 kΩ ⎛ 1.66 ⎞ v0 = ⎜ ⎟ vi = 0.43 vi ⎝ 1.66 + 2.2 ⎠

v0 ( rms ) =

v0 ( max )

⇒ v0 ( rms ) = 3.04 V 2 b. ______________________________________________________________________________________

2.19 3 .9 = 0.975 mA 4 20 − 3.9 II = = 1.342 mA 12 I Z = I I − I L = 1.342 − 0.975 = 0.367 mA

(a) I L =

PZ = I Z V Z = (0.367 )(3.9) = 1.43 mW

3.9 = 0.39 mA 10 I Z = 1.342 − 0.39 = 0.952 mA

(b) I L =

PZ = (0.952 )(3.9 ) = 3.71 mW ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.20 (a) 40 − 12 = 0.233 A 120 P = ( 0.233)(12 ) = 2.8 W

IZ =

(b) So

IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A 12 0.21 = ⇒ RL = 57.1Ω RL

P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W (c) ______________________________________________________________________________________ 2.21 (a) PZ = I Z V Z 4 = I z (15.4 ) ⇒ I Z (max ) = 259.74 mA So 15 ≤ I z ≤ 259.74 mA 60 − 15.4 = 297.33 mA 0.15 So I L (max ) = 297.33 − 15 = 282.33 mA

(b) I I =

I L (min ) = 297.33 − 259.74 = 37.59 mA

15.4 = 54.55 Ω 0.28233 15.4 R L (max ) = = 410 Ω 0.03759 So 54.55 ≤ R L ≤ 410 Ω ______________________________________________________________________________________

Then R L (min ) =

2.22 a.

20 − 10 ⇒ I I = 45.0 mA 222 10 IL = ⇒ I L = 26.3 mA 380 I Z = I I − I L ⇒ I Z = 18.7 mA II =

b. PZ ( max ) = 400 mW ⇒ I Z ( max ) = ⇒ I L ( min ) = I I − I Z ( max ) = 45 − 40 ⇒ I L ( min ) = 5 mA =

400 = 40 mA 10

10 RL

⇒ RL = 2 kΩ (c)

I = 57.1 mA I L = 26.3 mA For Ri = 175Ω I

I Z = 30.8 mA

I Z ( max ) = 40 mA ⇒ I L ( min ) = 57.1 − 40 = 17.1 mA RL =

10 ⇒ RL = 585Ω 17.1

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.23 a. From Eq. (2.30) 500 [ 20 − 10] − 50 [15 − 10] I Z ( max ) = 15 − ( 0.9 )(10 ) − ( 0.1)( 20 ) 5000 − 250 4 I Z ( max ) = 1.1875 A I Z ( min ) = 0.11875 A =

From Eq. (2.28(b)) b.

Ri =

20 − 10 ⇒ Ri = 8.08Ω 1187.5 + 50

PZ = (1.1875 )(10 ) ⇒ PZ = 11.9 W PL = I L ( max )V0 = ( 0.5 )(10 ) ⇒ PL = 5 W

______________________________________________________________________________________ 2.24 (a) I L = 0 10 − 5.6 ⇒ 83.0 mA 50 + 3 V Z = 5.6 + (0.083)(3) = 5.85 V = V L IZ =

PZ = I Z V Z = (0.083)(5.85) = 0.486 W

(b)

10 − V L V L − 5.6 V L = + 50 3 200 0.20 + 1.867 = V L (0.02 + 0.3333 + 0.005)

So V L = 5.769 V 5.769 = 28.84 mA 0.2 10 − 5.769 II = = 84.62 mA 0.050 And I Z = I I − I L = 55.8 mA

Then I L =

PZ = (0.0558)(5.769 ) = 0.322 W (c) I L = 0 12 − 5.6 ⇒ 120.8 mA 50 + 3 V Z = V L = 5.6 + (0.1208)(3) = 5.962 V IZ =

PZ = (0.1208)(5.962 ) = 0.72 W

(d)

12 − V L V L − 5.6 V L = + 50 3 200 0.24 + 1.867 = V L (0.02 + 0.333 + 0.005)

So V L = 5.88 V 5.88 12 − 5.88 = 29.4 mA; I I = = 122.4 mA 0.20 0.05 I Z = 122.4 − 29.4 = 93 mA

Then I L =

PZ = (0.093)(5.88) = 0.547 W ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.25

(a) Set I Z = 10 mA; I L =

V L 7.5 = = 7.5 mA 1 RL

I I = 10 + 7.5 = 17.5 mA I I = 17.5 =

12 − 7.5 ⇒ Ri = 257 Ω Ri

(b) 7.5 = V ZO + (0.01)(12 ) ⇒ V ZO = 7.38 V For V I = (1.1)(12 ) = 13.2 V 13.2 − V L V L − 7.38 V L = + 257 12 1000 0.05136 + 0.615 = V L (0.00389 + 0.0833 + 0.001) ⇒ V L = 7.556 V For V I = (0.9 )(12 ) = 10.8 V 10.8 − V L V L − 7.38 V L = + 257 12 1000 0.04202 + 0.615 = V L (0.08819 ) ⇒ V L = 7.450 V ⎛ 7.556 − 7.450 ⎞ Then, Source Reg = ⎜ ⎟ × 100% = 4.42% ⎝ 13.2 − 10.8 ⎠ (c) For R L = 1 k Ω , V L = 7.50 V 12 − 7.38 ⇒ 17.17 mA 257 + 12 V L = 7.38 + (0.01717 )(12 ) = 7.586 V

For R L = ∞ , I Z =

⎛ 7.586 − 7.50 ⎞ Then , Load Reg = ⎜ ⎟ × 100% = 1.15% 7.50 ⎠ ⎝ ______________________________________________________________________________________

2.26 % Reg = =

So

VL ( nom )

× 100%

VL ( nom ) + I Z ( max ) rz − (VL ( nom ) + I Z ( min ) rz ) VL ( nom )

⎡ I Z ( max ) − I Z ( min ) ⎤⎦ ( 3 ) =⎣ = 0.05 6 I Z ( max ) − I Z ( min ) = 0.1 A

Now

I L ( max ) =

Ri = Now 280 =

or

VL ( max ) − VL ( min )

6 6 = 0.012 A, I L ( min ) = = 0.006 A 500 1000

VPS ( min ) − VZ

I Z ( min ) + I L ( max )

15 − 6 ⇒ I Z ( min ) = 0.020 A I Z ( min ) + 0.012

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Then

I Z ( max ) = 0.1 + 0.02 = 0.12 A VPS ( max ) − 6

Ri = and

VPS ( max ) − VZ

I Z ( max ) + I L ( min )

⇒ VPS ( max ) = 41.3 V 0.12 + 0.006 or ______________________________________________________________________________________ 280 =

2.27 Using Figure 2.19 VPS = 20 ± 25% ⇒ 15 ≤ VPS ≤ 25 V a.

For

VPS ( min ) :

I I = I Z ( min ) + I L ( max ) = 5 + 20 = 25 mA Ri =

b. For

VPS ( min ) − VZ II

VPS ( max )

=

15 − 10 ⇒ Ri = 200Ω 25

⇒ I I ( max ) =

25 − 10 ⇒ I I ( max ) = 75 mA Ri

For I L ( min ) = 0 ⇒ I Z ( max ) = 75 mA

VZ 0 = VZ − I Z rZ = 10 − ( 0.025 )( 5 ) = 9.875 V V0 ( max ) = 9.875 + ( 0.075 )( 5 ) = 10.25 V0 ( min ) = 9.875 + ( 0.005 )( 5 ) = 9.90

ΔV0 = 0.35 V % Reg =

ΔV0 × 100% ⇒ % Reg = 3.5% V0 ( nom )

c. ______________________________________________________________________________________ 2.28 From Equation (2.28(a)) VPS ( min ) − VZ 24 − 16 Ri = = I Z ( min ) + I L ( max ) 40 + 400 Also Vr =

or

Ri = 18.2Ω

VM VM ⇒C = 2 fRC 2 fRVr

R ≅ Ri + rz = 18.2 + 2 = 20.2Ω

Then C=

24 ⇒ C = 9901 μ F 2 ( 60 )(1)( 20.2 )

______________________________________________________________________________________ 2.29

VZ = VZ 0 + I Z rZ VZ ( nom ) = 8 V 8 = VZ 0 + ( 0.1)( 0.5 ) ⇒ VZ 0 = 7.95 V Ii =

VS ( max ) − VZ ( nom ) Ri

=

12 − 8 = 1.333 A 3

For I L = 0.2 A ⇒ I Z = 1.133 A For I L = 1 A ⇒ I Z = 0.333 A

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VL ( max ) = VZ 0 + I Z ( max ) rZ = 7.95 + (1.133)( 0.5 ) = 8.5165 VL ( min ) = VZ 0 + I Z ( min ) rZ = 7.95 + ( 0.333)( 0.5 ) = 8.1165 ΔVL = 0.4 V % Reg = Vr =

ΔVL 0.4 = ⇒ % Reg = 5.0% V0 ( nom ) 8 VM VM ⇒C = 2 fRC 2 fRVr

R = Ri + rz = 3 + 0.5 = 3.5Ω C=

12 ⇒ C = 0.0357 F 2 ( 60 )( 3.5 )( 0.8 )

Then ______________________________________________________________________________________ 2.30 For −6.3 ≤ υ I ≤ 3 V, υ O = υ I υ −3 For υ I > 3 V, I = I and υ O = υ I − I (0.5) 1.5 ⎛υ − 3⎞ υ O = υ I − (0.5)⎜⎜ I ⎟⎟ = 0.667υ I + 1.0 ⎝ 1. 5 ⎠

υ I + 6.3

and υ O = υ I − I (0.5) 2.5 ⎛ υ + 6.3 ⎞ υ O = υ I − (0.5)⎜⎜ I ⎟⎟ = 0.8υ I − 1.26 ⎝ 2. 5 ⎠ ______________________________________________________________________________________ For υ I < −6.3 V, I =

2.31

For −10 ≤ vI ≤ 0, both diodes are conducting ⇒ vO = 0 i = 0, vO = 0 For 0 ≤ vI ≤ 3, Zener not in breakdown, so 1 v −3 For vI > 3 i1 = I mA 20 1 ⎛ v −3⎞ vo = ⎜ I ⎟ (10 ) = vI − 1.5 2 ⎝ 20 ⎠ At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA (a)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

For

For vI < 0, both diodes forward biased 0 − vI −i1 = . 10 At vI = −10 V , i1 = −1 mA vI > 3, i1 =

vI − 3 . 20 At vI = 10 V , i1 = 0.35 mA

______________________________________________________________________________________ 2.32 (a)

For

1 V1 = × 15 = 5 V ⇒ for vI ≤ 5.7, v0 = vI 3 vI > 5.7 V

vI − (V1 + 0.7 ) 15 − V1 V1 + = , v0 = V1 + 0.7 1 2 1 vI − v0 15 − ( v0 − 0.7 ) v0 − 0.7 + = 1 2 1 vI 15.7 0.7 ⎛1 1 1⎞ + + = v0 ⎜ + + ⎟ = v0 ( 2.5 ) 1 2 1 ⎝1 2 1⎠ 1 vI + 8.55 = v0 ( 2.5 ) ⇒ v0 = vI + 3.42 2.5 vI = 5.7 ⇒ v0 = 5.7 vI = 15 ⇒ v0 = 9.42

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

iD = 0 for 0 ≤ vI ≤ 5.7

Then for vI > 5.7 V

v −v iD = I O = 1

⎛v ⎞ vI − ⎜ I + 3.42 ⎟ 0.6vI − 3.42 2.5 ⎝ ⎠ iD = 1 1 or

For vI = 15, iD = 5.58 mA

______________________________________________________________________________________ 2.33 (a) (i) V B = 1.8 V For υ I ≥ 1.1 V, υ O = υ I For υ I ≤ 1.1 V, υ O = 1.1 V (ii) V B = −1.8 V For υ I ≥ −2.5 V, υ O = υ I For υ I ≤ −2.5 V, υ O = −2.5 V (b) (i) V B = 1.8 V For υ I ≥ 2.5 V, υ O = 2.5 V For υ I ≤ 2.5 V, υ O = υ I (ii) V B = −1.8 V For υ I ≥ −1.1 V, υ O = −1.1 V For υ I ≤ −1.1 V, υ O = υ I ______________________________________________________________________________________ 2.34

For

vI = 30 V, i =

30 − 10.7 = 0.175 A 100 + 10

v0 = i(10) + 10.7 = 12.5 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b.

______________________________________________________________________________________ 2.35 (a) (i) V B = 5 V For υ I ≥ 5.7 V, υ O = υ I − 5.7 For υ I ≤ 5.7 V, υ O = 0 (ii) V B = −5 V For υ I ≥ −4.3 V, υ O = υ I + 4.3 For υ I ≤ −4.3 V, υ O = 0 (b) (i) V B = 5 V For υ I ≥ 4.3 V, υ O = 0 For υ I ≤ 4.3 V, υ O = υ I − 4.3 (ii) V B = −5 V For υ I ≥ −5.7 V, υ O = 0 For υ I ≤ −5.7 V, υ O = υ I + 5.7 ______________________________________________________________________________________ 2.36 a. Vγ = 0

Vγ = 0.6

b. Vγ = 0

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Vγ = 0.6

______________________________________________________________________________________ 2.37

______________________________________________________________________________________ 2.38 One possible example is shown.

L will tend to block the transient signals Dz will limit the voltage to +14 V and −0.7 V. Power ratings depends on number of pulses per second and duration of pulse. ______________________________________________________________________________________ 2.39 (a) Square wave between +40 V and 0. (b) Square wave between +35 V and −5 V. (c) Square wave between +5 V and −35 V. ______________________________________________________________________________________ 2.40

a.

For

Vγ = 0 ⇒ Vx = 2.7 V

V = 0.7 V ⇒ Vx = 2.0 V b. For γ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.41 Circuit similar to Figure 2.31(a) with V B = −10 V. ______________________________________________________________________________________ 2.42 In steady-state, υ O = (10 sin ω t + 5) V ______________________________________________________________________________________ 2.43 (i) V B = 5 V, In steady-state, υ O = (10 sin ω t − 5) V (ii) V B = −5 V, In steady-state, υ O = (10 sin ω t − 15) V ______________________________________________________________________________________ 2.44 a. 10 − 0.6 ⇒ I D1 = 0.94 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) ⇒ V0 = 8.93 V

I D1 =

ID2 = 0

b. 5 − 0.6 ⇒ I D1 = 0.44 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) ⇒ V0 = 4.18 V

I D1 =

c. d.

ID2 = 0

Same as (a) 10 =

(I ) 2

( 0.5 ) + 0.6 + I ( 9.5 ) ⇒ I = 0.964 mA

V0 = I ( 9.5 ) ⇒ V0 = 9.16 V I ⇒ I D1 = I D 2 = 0.482 mA 2 ______________________________________________________________________________________ I D1 = I D 2 =

2.45 a.

I = I D1 = I D 2 = 0 V0 = 10 b.

10 = I ( 9.5 ) + 0.6 + I ( 0.5) ⇒ I = I D 2 = 0.94 mA

I D1 = 0

V0 = 10 − I ( 9.5 ) ⇒ V0 = 1.07 V c.

10 = I ( 9.5 ) + 0.6 + I ( 0.5) + 5 ⇒ I = I D 2 = 0.44 mA I D1 = 0 V0 = 10 − I ( 9.5 ) ⇒ V0 = 5.82 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ d. 10 = I ( 9.5 ) + 0.6 +

I ( 0.5) ⇒ I = 0.964 mA 2

I ⇒ I D1 = I D 2 = 0.482 mA 2 V0 = 10 − I ( 9.5 ) ⇒ V0 = 0.842 V

I D1 = I D 2 =

______________________________________________________________________________________ 2.46 a. V1 = V2 = 0 ⇒ D1 , D2 , D3 , on V0 = 4.4 V 10 − 4.4 ⇒ I = 0.589 mA 9.5 4.4 − 0.6 = ID2 = ⇒ I D1 = I D 2 = 7.6 mA 0.5 = I D1 + I D 2 − I = 2 ( 7.6 ) − 0.589 ⇒ I D 3 = 14.6 mA

I= I D1 I D3

b. V1 = V2 = 5 V D1 and D2 on, D3 off I 10 = I ( 9.5 ) + 0.6 + ( 0.5 ) + 5 ⇒ I = 0.451 mA 2 I I D1 = I D 2 = ⇒ I D1 = I D 2 = 0.226 mA 2 I D3 = 0 V0 = 10 − I ( 9.5 ) = 10 − ( 0.451)( 9.5 ) ⇒ V0 = 5.72 V

c.

V = 4.4 V V1 = 5 V, V2 = 0 D1 off, D2, D3 on 0 10 − 4.4 I= I = 0.589 mA ⇒ 9.5 4.4 − 0.6 ID2 = I D 2 = 7.6 mA ⇒ 0.5 I D1 = 0 I D 3 = I D 2 − I = 7.6 − 0.589 ⇒ I D 3 = 7.01 mA

d.

V1 = 5 V, V2 = 2 V D1 off, D2, D3 on 10 − 4.4 ⇒ 9.5 4.4 − 0.6 − 2 = ⇒ 0.5

I= ID2

V0 = 4.4 V

I = 0.589 mA I D 2 = 3.6 mA I D1 = 0

I D 3 = I D 2 − I = 3.6 − 0.589 ⇒ I D 3 = 3.01 mA

______________________________________________________________________________________ 2.47

(a) V1 = 4.4 V, I D1 = 0.2 =

10 − 0.6 − 4.4 ⇒ R1 = 25 k Ω R1

I R 2 = 0.2 + 0.3 = 0.5 mA V 2 = −0.6 V, I R 2 = 0.5 =

4.4 − (− 0.6 ) ⇒ R 2 = 10 k Ω R2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I R 3 = 0.5 + 0.5 = 1.0 mA

− 0.6 − (− 5) = 4.4 k Ω 1. 0 (b) Assume all diodes conducting 10 − 0.6 − 4.4 V1 = 4.4 V, I D1 = = 0.5 mA 10 4.4 − (− 0.6 ) V 2 = −0.6 V, I R 2 = = 1.25 mA 4 Then I D 2 = 1.25 − 0.5 = 0.75 mA R3 =

I R3 =

− 0.6 − (− 5) = 2 mA 2.2 = 2 − 1.25 = 0.75 mA

Then I D 3 (c) Diode D 2 cutoff ⇒ I D 2 = 0 V 2 = −0.6 V, I D1 =

10 − 0.6 − (− 0.6 ) 10 = = 1.11 mA R1 + R 2 9

V1 = 10 − 0.6 − (1.11)(3) = 6.07 V I R3 =

− 0.6 − (− 5) = 1.76 mA 2. 5 = 1.76 − 1.11 = 0.65 mA

Then I D 3 (d) Diode D3 cutoff ⇒ I D 3 = 0

10 − 0.6 − 4.4 = 0.833 mA 6 4.4 − (− 5) 9.4 = = = 1.044 mA R 2 + R3 9

V1 = 4.4 V, I D1 = I R2

V 2 = (1.044 )(6 ) − 5 = 1.27 V

Then I D 2 = 1.044 − 0.833 = 0.211 mA ______________________________________________________________________________________ 2.48 (a) I D1 = I D 2 = 2.5 mA

0.7 − 0.7 − (− 2) ⇒ R = 0.8 k Ω R = 0.2 I D 2 , I D1 + I D 2 = 5

I D 2 = 2.5 =

(b) I D1

0.2 I D 2 + I D 2 = 5 ⇒ I D 2 = 4.167 mA 2 ⇒ R = 0.48 k Ω R = 5I D 2 , I D1 + I D 2 = 5

I D 2 = 4.167 =

(c) I D1

5 I D 2 + I D 2 = 5 ⇒ I D 2 = 0.833 mA 2 ⇒ R = 2.4 k Ω R ______________________________________________________________________________________ I D 2 = 0.833 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.49 (a) D1 and D 2 on 5 − 0.7 − V A V A − 0.7 − (− 5) 5+ = 2 1. 1 ⎛1 1 ⎞ 5 + 2.15 − 3.909 = V A ⎜ + ⎟ ⇒ V A = 2.30 V ⎝ 2 1 .1 ⎠ 5 − 0.7 − 2.3 = 1.0 mA 2 2.3 − 0.7 − (− 5) I D2 = = 6.0 mA 1.1 (b) D1 cutoff, I D1 = 0

Then I D1 =

I D 2 = 5 mA, V A = 0.7 + (5)(2.5) − 5 = 8.2 V 5 − 0.7 − 0 = 2.15 mA 2 = 5 + 2.15 = 7.15 mA

(c) V A = 0 , I D1 = Then I D 2

0 − 0.7 − (− 5) ⇒ R 2 = 0.60 k Ω R2 ______________________________________________________________________________________ I D 2 = 7.15 =

2.50 (a) (i) υ I = 5 V, D1 and D 2 on 5 − (υ O + 0.6 ) 5 − υ O υ O υ O − 0.6 + = + 5 5 0. 5 0. 5 0.88 + 1.0 + 1.2 = υ O (0.20 + 0.20 + 2.0 + 2.0) ⇒ υ O = 0.7 V (ii) υ I = −5 V ⎛ 0.5 ⎞ ⎟υ I = −0.455 V ⎝ 0.5 + 5 ⎠

υO = ⎜

(b) (i) υ I = 5 V, υ O = 4.4 V (ii) υ I = −5 V, υ O = −0.6 V ______________________________________________________________________________________ 2.51

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For vI > 0. when D1 and D4 turn off 10 − 0.7 I= = 0.465 mA 20 v0 = I (10 kΩ ) = 4.65 V

v0 = vI for − 4.65 ≤ vI ≤ 4.65 ______________________________________________________________________________________ 2.52 (a) All diodes on 15 − V A V A − 0.7 V A − 0.7 − (− 5) V A − 0.7 − (− 10 ) = + + 6.15 2 14 24 2.439 + 0.35 − 0.307 − 0.3875 = V A (0.1626 + 0.50 + 0.0714 + 0.0417 ) ⇒ V A = 2.70 V 2.70 − 0.7 = 1.0 mA 2 2.70 − 0.7 − (− 5) I D2 = = 0.50 mA 14 2.70 − 0.7 − (− 10 ) I D3 = = 0.50 mA 24 (b) D1 cutoff, I D1 = 0

Then I D1 =

15 − V A V A − 0.7 − (− 5) V A − 0.7 − (− 10 ) = + 6.15 3.3 5. 2 2.439 − 1.303 − 1.788 = V A (0.1626 + 0.303 + 0.1923) ⇒ V A = −0.991 V

− 0.991 − 0.7 − (− 5) = 1.0 mA 3. 3 − 0.991 − 0.7 − (− 10 ) I D3 = = 1.60 mA 5. 2 (c) D1 and D 2 cutoff, I D1 = I D 2 = 0

Then I D 2 =

I D3 =

15 − 0.7 − (− 10 ) 24.3 = = 3.25 mA R1 + R 4 6.15 + 1.32

V A = 15 − (3.25)(6.15) = −5 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.53 a.

R1 = 5 kΩ, R2 = 10 kΩ ⇒ V0 = 0 D1 and D2 on

10 − 0.7 0 − ( −10 ) − = 1.86 − 1.0 5 10 = 0.86 mA

I D1 = I D1

b.

R1 = 10 kΩ, R2 = 5 kΩ, D1 off, D2 on I D1 = 0 I=

10 − 0.7 − ( −10 )

= 1.287 15 V0 = IR2 − 10 ⇒ V0 = −3.57 V

______________________________________________________________________________________ 2.54 If both diodes on (a) VA = −0.7 V, VO = −1.4 V I R1 = IR2 I R1 + I D1

10 − ( −0.7 )

= 1.07 mA 10 −1.4 − ( −15 ) = = 2.72 mA 5 = I R 2 ⇒ I D1 = 2.72 − 1.07

I D1 = 1.65 mA (b) D1 off, D2 on 10 − 0.7 − ( −15 ) = 1.62 mA I R1 = I R 2 = 5 + 10 VO = I R 2 R2 − 15 = (1.62 )(10 ) − 15 ⇒ VO = 1.2 V VA = 1.2 + 0.7 = 1.9 V ⇒ D1 off , I D1 = 0 ______________________________________________________________________________________

2.55 (a)

D1 on, D2 off 10 − 0.7 I D1 = = 0.93 mA 10 VO = −15 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ D1 on, D2 off 10 − 0.7 I D1 = = 1.86 mA 5 VO = −15 V ______________________________________________________________________________________ (b)

2.56

15 − (V0 + 0.7 )

V + 0.7 V0 = 0 + 10 20 20 15 0.7 0.7 1 ⎞ ⎛ 4.0 ⎞ ⎛ 1 1 − − = V0 ⎜ + + ⎟ = V0 ⎜ ⎟ 10 10 20 ⎝ 20 ⎠ ⎝ 10 20 20 ⎠ V0 = 6.975 V V0 ⇒ I D = 0.349 mA 20 ______________________________________________________________________________________ ID =

2.57 (a) Diode is cutoff, I D = 0 , V D = 0 V A = VB = 3 V (b) Diode is conducting, V D = 0.7 V 5 − V B V B V B − 0.7 V B − 0.7 − 2 = + + 10 10 10 10 0.50 + 0.07 + 0.27 = V B (0.10 + 0.10 + 0.10 + 0.10 ) ⇒ V B = 2.1 V and V A = 1.4 V 5 − VB VB = + ID 10 10 5 − 2.1 2.1 So I D = − = 0.08 mA 10 10 (c) Diode is cutoff, I D = 0 1 (5) = 2.5 V, V B = 1 (4) = 2.0 V 2 2 V D = 2 − 2.5 = −0.5 V VA =

(d) Diode is conducting, V D = 0.7 V 8 − V B V B V B − 0. 7 V B − 0 . 7 − 2 = + + 10 10 10 10 0.80 + 0.07 + 0.27 = V B (0.40 ) ⇒ V B = 2.85 V and V A = 2.15 V 8 −VB VB 1 Then I D = − = [8 − 2(2.85)] = 0.23 mA 10 10 10 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.58 vI = 0, D off, D on 1 2

I vo

10 − 2.5 0.5 mA 15 = 10 − ( 0.5 )( 5 ) ⇒ vo = 7.5 V for 0 ≤ vI ≤ 7.5 V

=

For vI > 7.5 V , Both D1 and D2 on vI − vo vo − 2.5 vo − 10 = + vI = vo ( 5.5 ) − 33.75 15 10 5 or When vo = 10 V, D2 turns off vI = (10 )( 5.5 ) − 33.75 = 21.25 V For vI > 21.25 V, vo = 10 V ______________________________________________________________________________________ 2.59 (a) For υ I = 0.5 V, I D1 = I D 2 = I D 3 = 0 , υ O = 0.5 V (b) For υ I = 1.5 V, D1 on; I D 2 = I D 3 = 0 1.5 − 0.7 = 0.0667 mA 4+8 υ O = 0.7 + (0.0667 )(8) = 1.23 V I D1 =

(c) For υ I = 3 V, D1 and D 2 conducting, I D 3 = 0 3 − υ O υ O − 0. 7 υ O − 1. 7 = + 4 8 6 0.75 + 0.0875 + 0.2833 = υ O (0.25 + 0.125 + 0.1667 ) ⇒ υ O = 2.069 V 2.069 − 0.7 = 0.171 mA 8 2.069 − 1.7 I D2 = = 0.0615 mA 6 (d) For υ I = 5 V, all diodes conducting

Then I D1 =

5 − υ O υ O − 0.7 υ O − 1.7 υ O − 2.7 = + + 4 8 6 4 1.25 + 0.0875 + 0.2833 + 0.675 = υ O (0.25 + 0.125 + 0.1667 + 0.25)

So υ O = 2.90 V 2.90 − 0.7 = 0.275 mA 8 2.90 − 1.7 I D2 = = 0.20 mA 6 2.90 − 2.7 I D3 = = 0.05 mA 4 ______________________________________________________________________________________

Then I D1 =

2.60 (a) I D 2 = 0 for υ I < 4.5 V I D 2 = 100 mA for υ I > 4.5 V (b) I D 2 = 0 for υ I < 9 V I D 2 = 100 mA for υ I > 9 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.61

a. b. c.

V01 = V02 = 0 V01 = 4.4 V, V02 = 3.8 V V01 = 4.4 V, V02 = 3.8 V

Logic “1” level degrades as it goes through additional logic gates. ______________________________________________________________________________________ 2.62

a. b. c.

V01 = V02 = 5 V V01 = 0.6 V, V02 = 1.2 V V01 = 0.6 V, V02 = 1.2 V

Logic “0” signal degrades as it goes through additional logic gates. ______________________________________________________________________________________ 2.63

(V1 AND V2 ) OR (V3 AND V4 )

______________________________________________________________________________________ 2.64 10 − 1.5 − 0.2 = 12 mA = 0.012 R + 10 8.3 R + 10 = = 691.7Ω 0.012 R = 681.7Ω ______________________________________________________________________________________ I=

2.65 10 − 1.7 − VI =8 0.75 VI = 10 − 1.7 − 8 ( 0.75 ) ⇒ VI = 2.3 V I=

______________________________________________________________________________________ 2.66 5 − 1.7 = 0.220 k Ω 15 r f = 20 Ω ⇒ R = 200 Ω

R + rf =

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.67

VR = 1 V, I = 0.8 mA VPS = 1 + ( 0.8 )( 2 ) VPS = 2.6 V

______________________________________________________________________________________ 2.68

I Ph = η eΦA

0.6 × 10−3 = (1) (1.6 × 10−19 )(1017 ) A A = 3.75 × 10−2 cm 2 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 3 3.1 Kn =

k n′ W 120 ⎛ 10 ⎞ 2 ⋅ = ⎜ ⎟ ⇒ 0.75 mA/V 2 L 2 ⎝ 0. 8 ⎠

(a) (i) I D = 0

[ ] = (0.75)[2(2 − 0.4)(0.1) − (0.1) ] = 0.2325 mA = (0.75)[2(3 − 0.4)(0.1) − (0.1) ] = 0.3825 mA

(ii) I D = (0.75) 2(1 − 0.4)(0.1) − (0.1) ⇒ 82.5 μ A 2

(iii) I D

2

2

(iv) I D (b) (i) I D = 0

(ii) I D = (0.75)(1 − 0.4) = 0.27 mA 2

(iii) I D = (0.75)(2 − 0.4) = 1.92 mA 2

(iv) I D = (0.75)(3 − 0.4) = 5.07 mA ______________________________________________________________________________________ 2

3.2

[ 0.5 = K [2(0.6)V 1.0 = K [2(1.0)V

2 I D = K n 2(VGS − VTN )V DS − V DS n

DS

−V

2 DS

] ]

]

n DS − V Take ratio 2 1.2V DS − V DS 2 2 ⇒ V DS − 0.5V DS = 1.2V DS − V DS 0.5 = 2 2V DS − V DS 2 DS

or 1 − 0.5V DS = 1.2 − V DS which yields V DS = 0.4 V

[

]

Then 0.5 = K n (1.2)(0.4) − (0.4) ⇒ K n = 1.56 mA/V 2 ______________________________________________________________________________________ 2

3.3 (a) Enhancement-mode (b) From Graph VT = 1.5 V Now 2 0.03 = K n ( 2 − 1.5 ) = 0.25 K n ⇒ K n = 0.12

0.15 = K n ( 3 − 1.5 ) = 2.25 K n

K n = 0.0666

0.39 = K n ( 4 − 1.5 ) = 6.25 K n

K n = 0.0624

0.77 = K n ( 5 − 1.5 ) = 12.25 K n

K n = 0.0629

2

2

2

From last three, K n (Avg) = 0.0640 mA/V iD (sat) = 0.0640(3.5 − 1.5) 2 ⇒ iD (sat) = 0.256 mA for VGS = 3.5 V 2 (c) iD (sat) = 0.0640(4.5 − 1.5) ⇒ iD (sat) = 0.576 mA for VGS = 4.5 V 2

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.4

a.

i.

VGS = 0 VDS ( sat ) = VGS − VTN = 0 − ( −2.5 ) = 2.5 V VDS = 0.5 V ⇒ Biased in nonsaturation 2 I D = (1.1) ⎡ 2 ( 0 − ( −2.5) )( 0.5 ) − ( 0.5 ) ⎤ ⇒ I D = 2.48 mA ⎣ ⎦

VDS = 2.5 V ⇒ Biased in saturation I D = (1.1) ( 0 − ( −2.5 ) ) ⇒ I D = 6.88 mA 2

ii.

iii. b.

⇒ I D = 6.88 mA VDS = 5 V Same as (ii) VGS = 2 V VDS ( sat ) = 2 − ( −2.5) = 4.5 V VDS = 0.5 V ⇒ Nonsaturation

i.

ii.

I D = (1.1) ⎡⎣ 2(2 − (−2.5))(0.5) − (0.5) 2 ⎤⎦ ⇒ I D = 4.68 mA

VDS = 2.5 V ⇒ Nonsaturation I D = (1.1) ⎡⎣ 2(2 − ( −2.5))(2.5) − (2.5) 2 ⎤⎦ ⇒ I D = 17.9 mA

VDS = 5 V ⇒ Saturation

I D = (1.1) ( 2 − ( −2.5) ) ⇒ I D = 22.3 mA iii. ______________________________________________________________________________________ 2

3.5 (a) V DS (sat ) = VGS − VTN = 2.2 − 0.4 = 1.8 V 2.2 = V DS > V DS (sat ) = 1.8 ⇒ Saturation (b) V DS (sat ) = VGS − VTN = 1 − 0.4 = 0.6 V V DS = −0.6 − (− 1) = 0.4 V < V DS (sat ) = 0.6 V ⇒ Nonsaturation (c) VGS = 1 − 1 = 0 ⇒ Cutoff ______________________________________________________________________________________ 3.6 (a) V SG = 2.2 − 2.2 = 0 ⇒ Cutoff (b) V SG = 2 V, V SD = 2 − (− 1) = 3 V V SD (sat ) = V SG + VTP = 2 + (− 0.4) = 1.6 V So V SD = 3 > V SD (sat ) = 1.6 ⇒ Saturation (c) V SG = 2 V, V SD = 2 − 1 = 1 V V SD (sat ) = V SG + VTP = 2 + (− 0.4) = 1.6 V So V SD = 1 < V SD (sat ) = 1.6 ⇒ Nonsaturation ______________________________________________________________________________________ 3.7 ID =

k n′ ⎛ W ⎞ 2 ⎜ ⎟(VGS − VTN ) 2 ⎝L⎠

0.12 ⎛ W ⎞ ⎛W ⎞ 2 ⎜ ⎟[0 − (− 1.2 )] ⇒ ⎜ ⎟ = 5.79 2 ⎝L⎠ ⎝L⎠ ______________________________________________________________________________________ 0. 5 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.8 kn′ = μ n Cox =

μ n ∈ox

=

tox

( 600 )( 3.9 ) (8.85 ×10−14 ) tox

(a)

500 A

k n′ = 41.4 μ A/V

(b)

250

kn′ = 82.8 μ A/V 2

(c)

100

k n′ = 207 μ A/V 2

(d)

50

k n′ = 414 μ A/V 2

=

2.071× 10 −10 tox

2

k n′ = 828 μ A/V 2 (e) 25 ______________________________________________________________________________________

3.9

(a) K n =

(

)

(

)

Wμ n ∈ox 20 × 10 −4 (650 )(3.9 ) 8.85 × 10 −14 = = 1.40 mA/V 2 2 L t ox 2 0.8 × 10 − 4 200 × 10 −8

(

)(

)

(b) I D = K n (VGS − VTN ) = (1.40)(2 − 0.4) Or I D = 3.58 mA (c) V DS (sat ) = VGS − VTN = 2 − 0.4 = 1.6 V ______________________________________________________________________________________ 2

2

3.10

(a) I D =

k n′ ⎛ W ⎞ 2 ⎜ ⎟(VGS − VTN ) 2 ⎝L⎠

⎛ 0.12 ⎞⎛ W ⎞ ⎛W ⎞ 2 0.6 = ⎜ ⎟⎜ ⎟(1.4 − 0.8) ⇒ ⎜ ⎟ = 27.8 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ Or W = (27.8)(0.8) = 22.2 μ m

[

]

⎛ 0.12 ⎞ 2 (b) I D = ⎜ ⎟(27.8) 2(1.4 − 0.8)(0.4 ) − (0.4 ) = 0.534 mA 2 ⎝ ⎠ (c) V DS (sat ) = VGS − VTN = 1.4 − 0.8 = 0.6 V ______________________________________________________________________________________

3.11 k n′ = μ n C ox =

μ n ∈ox t ox

=

(600 )(3.9)(8.85 ×10 −14 ) 200 × 10 −8

k n′ = 0.1035 mA/V 2 ID =

k n′ ⎛ W ⎞ 2 ⎜ ⎟(VGS − VTN ) 2 ⎝L⎠

⎛W ⎞ ⎛ 0.1035 ⎞⎛ W ⎞ 2 1. 2 = ⎜ ⎟⎜ ⎟(3 − 0.6 ) ⇒ ⎜ ⎟ = 4.026 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ Then W = (4.026)(0.8) = 3.22 μ m ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.12

I D = WC ox (VGS − VTN )υ sat C ox =

(

)

)(

)

∈ox (3.9 ) 8.85 × 10 −14 = = 1.726 × 10 − 7 F/cm −8 t ox 200 × 10

(

(

)

I D = 3.22 ×10 −4 1.726 ×10 −7 (3 − 0.6) 2 ×10 7 I D = 2.67 mA ______________________________________________________________________________________ 3.13 ID =

k ′p ⎛ W ⎞ 2 ⎜ ⎟(V SG + VTP ) 2 ⎝L⎠

⎛ 0.05 ⎞⎛ W ⎞ 2 0.225 = ⎜ ⎟⎜ ⎟(2 + VTP ) 2 L ⎝ ⎠⎝ ⎠ ⎛ 0.05 ⎞⎛ W ⎞ 2 0.65 = ⎜ ⎟⎜ ⎟(3 + VTP ) 2 L ⎝ ⎠⎝ ⎠

Then

3 + VTP 0.65 = = 1.70 ⇒ VTP = −0.571 V 0.225 2 + VTP

⎛ 0.05 ⎞⎛ W ⎞ ⎛W ⎞ 2 And 0.225 = ⎜ ⎟⎜ ⎟(2 − 0.571) ⇒ ⎜ ⎟ = 4.41 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ ______________________________________________________________________________________

3.14 VS = 5 V, VG = 0 ⇒ VSG = 5 V VTP = −0.5 V ⇒ VSD ( sat ) = VSG + VTP = 5 − 0.5 = 4.5 V

VD = 0 ⇒ VSD = 5 V ⇒ Biased in saturation I D = 2 ( 5 − 0.5) ⇒ I D = 40.5 mA 2

a.

b.

c.

VD = 2 V ⇒ VSD = 3 V ⇒ Nonsaturation 2 I D = 2 ⎡ 2 ( 5 − 0.5 )( 3) − ( 3) ⎤ ⇒ I D = 36 mA ⎣ ⎦

VD = 4 V ⇒ VSD = 1 V ⇒ Nonsaturation 2 I D = 2 ⎡ 2 ( 5 − 0.5 )(1) − (1) ⎤ ⇒ I D = 16 mA ⎣ ⎦

VD = 5 V ⇒ VSD = 0 ⇒ I D = 0 d. ______________________________________________________________________________________ 3.15 (a) (b)

Enhancement-mode From Graph VTP = + 0.5 V 2 0.45 = k p ( 2 − 0.5 ) = 2.25 K p ⇒ K p =

0.20

1.25 = k p ( 3 − 0.5 ) = 6.25 K p

0.20

2.45 = k p ( 4 − 0.5 ) = 12.25 K p

0.20

2

2

4.10 = k p ( 5 − 0.5 ) = 20.25 K p 2

0.202 Avg K p = 0.20 mA/V 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ iD (sat) = 0.20 (3.5 − 0.5) 2 = 1.8 mA 2 (c) iD (sat) = 0.20 (4.5 − 0.5) = 3.2 mA ______________________________________________________________________________________ 3.16

(a) (b) (c)

VSD ( sat ) = VSG + VTP VSD ( sat ) = −1 + 2 ⇒ VSD ( sat ) = 1 V VSD ( sat ) = 0 + 2 ⇒ VSD ( sat ) = 2 V VSD ( sat ) = 1 + 2 ⇒ VSD ( sat ) = 3 V k ′p W k ′p W 2 2 ⋅ (VSG + VTP ) = ⋅ ⋅ ⎡⎣VSD ( sat ) ⎤⎦ 2 L 2 L 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )(1) ⇒ I D = 0.12 mA ⎝ 2 ⎠ ID =

(a) (b)

2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )( 2 ) ⇒ I D = 0.48 mA 2 ⎝ ⎠

2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )( 3) ⇒ I D = 1.08 mA 2 ⎝ ⎠ (c) ______________________________________________________________________________________

3.17

k ′p ⎛ W ⎞ 50 ⎛ 12 ⎞ 2 ⎜ ⎟ ⇒ K p = 0.375 mA/V ⎜ ⎟= 2 ⎝ L ⎠ 2 ⎝ 0. 8 ⎠ (a) Nonsaturation 2 I D = (0.375) 2(2 − 0.5)(0.2) − (0.2) = 0.21 mA (b) Nonsaturation 2 I D = (0.375) 2(2 − 0.5)(0.8) − (0.8) = 0.66 mA (c) Nonsaturation 2 I D = (0.375) 2(2 − 0.5)(1.2) − (1.2) = 0.81 mA (d) Saturation 2 I D = (0.375)(2 − 0.5) = 0.844 mA (e) Saturation 2 I D = (0.375)(2 − 0.5) = 0.844 mA ______________________________________________________________________________________ Kp =

[

]

[

]

[

]

3.18 k ′p = μ p Cox =

μ p ∈ox t0 x

=

( 250 )( 3.9 ) (8.85 × 10−14 )

(a)

tox = 500Å ⇒ k ′p = 17.3 μ A/V

(b)

250Å ⇒ k ′p = 34.5 μ A/V 2

(c)

100Å ⇒ k ′p = 86.3 μ A/V 2

(d)

50Å ⇒ k ′p = 173 μ A/V 2

(e)

25Å ⇒ k ′p = 345 μ A/V 2

t0 x 2

=

8.629 × 10 −11 t0 x

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 3.19 Cox =

−14 ∈ox ( 3.9 ) ( 8.85 × 10 ) = = 6.90 × 10 −8 F/cm 2 t0 x 500 × 10−8

kn′ = ( μ n Cox ) = ( 675 ) ( 6.90 × 10 −8 ) ⇒ 46.6 μ A/V 2

k ′p = ( μ p Cox ) = ( 375 ) ( 6.90 × 10−8 ) ⇒ 25.9 μ A/V 2

PMOS: ID =

k ′p ⎛ W ⎞ 2 ⎜ ⎟ (VSG + VTP ) 2 ⎝ L ⎠p

2 ⎛ 0.0259 ⎞ ⎛ W ⎞ ⎛W ⎞ 0.8 = ⎜ ⎟ ⎜ ⎟ ( 5 − 0.6 ) ⇒ ⎜ ⎟ = 3.19 L 2 ⎝ ⎠⎝ ⎠p ⎝ L ⎠p

L = 4 μ m ⇒ W p = 12.8 μ m ⎛ 0.0259 ⎞ 2 Kp = ⎜ ⎟ ( 3.19 ) ⇒ K p = 41.3 μ A/V = K n ⎝ 2 ⎠ Want Kn = Kp k ′p ⎛ W ⎞ kn′ ⎛ W ⎞ ⎜ ⎟ = ⎜ ⎟ = 41.3 2 ⎝ L ⎠N 2 ⎝ L ⎠p ⎛ 46.6 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎜ ⎟ ⎜ ⎟ = 41.3 ⇒ ⎜ ⎟ = 1.77 ⎝ 2 ⎠ ⎝ L ⎠N ⎝ L ⎠N L = 4 μ m ⇒ WN = 7.09 μ m

______________________________________________________________________________________ 3.20 VGS = 2 V, I D = ( 0.2 )( 2 − 1.2 ) = 0.128 mA 2

r0 =

1 1 = ⇒ r0 = 781 kΩ λ I D ( 0.01)( 0.128 )

VGS = 4 V, I D = ( 0.2 )( 4 − 1.2 ) = 1.57 mA 1 r0 = ⇒ r = 63.7 kΩ ( 0.01)(1.57 ) 0 2

VA =

1

=

1

⇒ VA = 100 V

λ ( 0.01) ______________________________________________________________________________________ 3.21 k n′ ⎛ W ⎞ ⎛ 0.12 ⎞ 2 2 ⎟(4 )(2 − 0.5) = 0.54 mA ⎜ ⎟(VGS − VTN ) = ⎜ 2 ⎝L⎠ ⎝ 2 ⎠ 1 1 1 ro = ⇒λ = = = 0.00926 V −1 ro I DQ λ I DQ 200 × 10 3 0.54 × 10 − 3

ID =

(

Then V A =

1

)(

)

= 108 V λ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.22 VTN = VTNO + γ ⎡⎣ 2φ f + VSB − 2φ f ⎤⎦ ΔVTN = 2 = ( 0.8 ) ⎡ 2φ f + VSB − 2 ( 0.35 ) ⎤ ⎣ ⎦ 2.5 + 0.837 = 2 ( 0.35 ) + VSB ⇒ VSB = 10.4 V

______________________________________________________________________________________ 3.23

VTN = VTNo + r ⎡⎣ 2φ f + VSB − 2φ f ⎤⎦ = 0.75 + 0.6 ⎡ 2 ( 0.37 ) + 3 − 2 ( 0.37 ) ⎤ ⎣ ⎦ = 0.75 + 0.6 [1.934 − 0.860] VTN = 1.39 V VDS (sat) = 2.5 − 1.39 = 1.11 V 2 ⎛ 0.08 ⎞ Sat Region I D = (15 ) ⎜ ⎟ ( 2.5 − 1.39 ) ⎝ 2 ⎠ I D = 0.739 mA

(a)

2 ⎛ 0.08 ⎞ ⎡ Non-Sat I D = (15 ) ⎜ ⎟ 2 ( 2.5 − 1.39 )( 0.25 ) − ( 0.25 ) ⎦⎤ ⎝ 2 ⎠⎣ I D = 0.296 mA

(b) ______________________________________________________________________________________ 3.24

Ε ox = 6×10 6 V/cm

(

)(

)

(a) (i) VG = Ε ox t ox = 6 ×10 6 120 ×10 −8 = 7.2 V (ii) VG =

7.2 = 2.4 V 3

(

)(

)

(b) (i) VG = 6 ×10 6 200 ×10 −8 = 12 V 12 = 4V 3 ______________________________________________________________________________________

(ii) VG =

3.25 Want

(3)(24) = Ε ox t ox = (6 ×10 6 )t ox

t0 x = 1.2 × 10−5 cm = 1200 Angstroms

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.26 ⎛ R2 ⎞ ⎛ 18 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 3.6 V ⎝ 18 + 32 ⎠ ⎝ R1 + R2 ⎠ Assume transistor biased in saturation region V V − VGS 2 ID = S = G = K n (VGS − VTN ) RS RS 3.6 − VGS = ( 0.5 )( 2 )(VGS − 0.8 )

2

= VGS2 − 1.6VGS + 0.64

VGS2 − 0.6VGS − 2.96 = 0 VGS = ID =

0.6 ±

VG − VGS RS

( 0.6 )

2

+ 4 ( 2.96 )

⇒ VGS = 2.046 V 2 3.6 − 2.046 = ⇒ I D = 0.777 mA 2

VDS = VDD − I D ( RD + RS )

= 10 − ( 0.777 )( 4 + 2 ) ⇒ VDS = 5.34 V

VDS > VDS ( sat ) ______________________________________________________________________________________ 3.27

(a)

VGS = 4 V VDS (sat) = 4 − 0.8 = 3.2 V If Sat I D = 0.25 ( 4 − 0.8 ) = 2.56 2

VDS = 1.44 × Non-Sat 2 ⎤⎦ + VDS 4 = I D RD + VDS = K n RD ⎡⎣ 2 (VGS − VT ) VDS − VDS 2 ⎤⎦ + VDS 4 = ( 0.25 )(1) ⎡⎣ 2 ( 4 − 0.8 ) VDS − VDS 2 4 = 2.6VDS − 0.25VDS 2 − 2.6VDS + 4 = 0 0.25VDS

VDS = ID =

2.6 ± 6.76 − 4 = 1.88 V 2 ( 0.25 )

4 − 1.88 = 2.12 mA 1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(b)

Non-Sat region 2 ⎤⎦ + VDS 5 = I D RD + VDS = K n RD ⎡⎣ 2 (VGS − VT ) VDS − VDS 2 ⎤⎦ + VDS 5 = ( 0.25 )( 3) ⎡⎣ 2 ( 5 − 0.8 ) VDS − VDS 2 5 = 7.3VDS − 0.75VDS

0.75 VDS2 − 7.3VDS + 5 = 0 VDS =

7.3 ± 53.29 − 15 2 ( 0.75 )

VDS = 0.741 V 5 − 0.741 = 1.42 mA 3 ______________________________________________________________________________________ ID =

3.28 ⎛ 0.12 ⎞ 2 0.8 = ⎜ ⎟(80 )(VGS − 0.4 ) ⇒ VGS = 0.808 V 2 ⎝ ⎠ 1 VGS = ⋅ Rin ⋅ V DD R1

0.80825 =

1 (200)(1.8) ⇒ R1 = 445 k Ω R1

R1 R2 = Rin = 200 k Ω ⇒ R 2 = 363 k Ω ______________________________________________________________________________________ 3.29

(a)

VSG = VDD = 3.5

VSD ( sat ) = 3.5 − 0.8 = 2.7 V

If biased in Sat region, I D = ( 0.2 )( 3.5 − 0.8 ) = 1.46 mA

VSD = 3.5 − (1.46 )(1.2 ) = 1.75 V Biased in Non-Sat Region.

×

2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2 ⎤⎦ 3.5 = VSD + I D RD = VSD + K p RD ⎡⎣ 2 (VSG + VTP ) VSD − VSD 2 ⎤⎦ 3.5 = VSD + ( 0.2 )(1.2 ) ⎡⎣ 2 ( 3.5 − 0.8 ) VSD − VSD 2 3.5 = VSD + 1.296 VSD − 0.24 VSD 2 0.24 VSD − 2.296 VSD + 3.5 = 0

VSD =

+2.296 ± 5.272 − 3.36 use − sign VSD = 1.90 V 2 ( 0.24 )

2 I D = ( 0.2 ) ⎡ 2 ( 3.5 − 0.8 )(1.9 ) − (1.9 ) ⎤ = 0.2 [10.26 − 3.61] ⎣ ⎦ 3.5 − 1.90 = 1.33 mA ID = 1.2 I D = 1.33 mA

(b)

VSG = VDD = 5 V VSD ( sat ) = 5 − 0.8 = 4.2 V

I = ( 0.2 )( 5 − 0.8 ) = 3.53 mA, VSD < 0 If Sat Region D Non-Sat Region. 2 ⎤⎦ 5 = VSD + K p RD ⎡⎣ 2 (VSG + VTP ) VSD − VSD 2

2 ⎤⎦ 5 = VSD + ( 0.2 )( 4 ) ⎡⎣ 2 ( 5 − 0.8 ) VSD − VSD 2 5 = VSD + 6.72 VSD − 0.8 VSD 2 0.8 VSD − 7.72 VSD + 5 = 0

VSD =

7.72 ± 59.598 − 16 use − sign VSD = 0.698 V 2 ( 0.8 )

5 − 0.698 ⇒ I D = 1.08 mA 4 ______________________________________________________________________________________ ID =

3.30 ⎛ 22 ⎞ VG = ⎜ ⎟(6 ) − 3 = 1.40 V ⎝ 22 + 8 ⎠

3 = K p R S (V SG + VTP ) + V SG + VG 2

(

)

2 3 = (0.5)(0.5) V SG − 1.6V SG + 0.64 + V SG + 1.40

0.25V

2 SG

+ 0.6V SG − 1.44 = 0 ⇒ V SG = 1.483 V

I D = (0.5)(1.483 − 0.8) = 0.2332 mA V SD = 6 − (0.2332 )(0.5 + 5) = 4.72 V ______________________________________________________________________________________ 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.31 VG = 0, VSG = VS Assume saturation region 2 I D = 0.4 = K p (VSG + VTP ) 0.4 = ( 0.2 )(VS − 0.8 )

2

0.4 + 0.8 ⇒ VS = 2.21 V 0.2 VD = I D RD − 5 = ( 0.4 )( 5 ) − 5 = −3 V VSD = VS − VD = 2.21 − ( −3) ⇒ VSD = 5.21 V VS =

VSD > VSD ( sat )

______________________________________________________________________________________ 3.32 ID =

V DD − V DSQ − V RS RD

=

3.3 − 1.6 − 0.8 = 0.18 mA 5

⎛ 0.12 ⎞⎛ W ⎞ ⎛W ⎞ 2 0.18 = ⎜ ⎟⎜ ⎟(0.8 − 0.4 ) ⇒ ⎜ ⎟ = 18.75 2 L ⎝ ⎠⎝ ⎠ ⎝L⎠ 0.8 RS = = 4.44 k Ω 0.18

⎛ R2 VG = 0.8 + 0.8 = 1.6 = ⎜⎜ ⎝ R1 + R 2 Now I R = (0.05)(0.18) ⇒ 9 μ A

⎞ ⎟⎟(3.3) ⎠

3.3 = 367 k Ω 0.009 ⎛ R ⎞ Then 1.6 = ⎜⎜ 2 ⎟⎟(3.3) ⇒ R 2 = 178 k Ω and R1 = 189 k Ω ⎝ 367 ⎠ ______________________________________________________________________________________

So R1 + R 2 =

3.33 ID = V DS

0.2 = 0.2 mA 1 = 1.8 − (0.2 )(4 + 1) = 0.8 V

Now V DS (sat ) = 0.8 − 0.4 = 0.4 V VDS (sat ) = VGS − VTN ⇒ 0.4 = VGS − 0.4 ⇒ VGS = 0.8 V ID =

k n′ ⎛ W ⎞ 2 ⎜ ⎟(VGS − VTN ) 2 ⎝L⎠

⎛ 0.12 ⎞⎛ W ⎞ ⎛W ⎞ 2 0.2 = ⎜ ⎟⎜ ⎟(0.8 − 0.4 ) ⇒ ⎜ ⎟ = 20.8 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ Now VG = VGS + I D R S = 0.8 + (0.2 )(1) = 1.0 V

VG = 1 =

1 1 ⋅ Rin ⋅ V DD = (200)(1.8) ⇒ R1 = 360 k Ω R1 R1

R1 R2 = Rin = 200 k Ω ⇒ R 2 = 450 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 3.34 ⎛ 0.12 ⎞ 2 (a) 0.35 = ⎜ ⎟(50 )(VGS − 0.4 ) ⇒ VGS = 0.742 V ⎝ 2 ⎠ (b) V DS = 1.8 − (0.35)(2 ) = 1.1 V V DS (sat ) = VGS − VTN = 0.742 − 0.4 = 0.342 V V DS > V DS (sat ) ⇒ Saturation _____________________________________________________________________________________

3.35 ⎛ 6 ⎞ VG = ⎜ ⎟(10 ) − 5 = −2 V ⎝ 6 + 14 ⎠ ⎛ k′ VG = VGS + I D R S − 5 = VGS + ⎜⎜ n ⎝ 2 ⎛ 0.12 ⎞ 2 5 − 2 = VGS + ⎜ ⎟(25)(0.5) VGS 2 ⎝ ⎠

(

⎞⎛ W ⎞ 2 ⎟⎟⎜ ⎟ R S (VGS − VTN ) − 5 ⎠⎝ L ⎠ − 0.8VGS + 0.16

)

2 + 0.4VGS − 2.88 = 0 ⇒ VGS = 1.71 V Or 0.75VGS

⎛ 0.12 ⎞ 2 ID = ⎜ ⎟(25)(1.71 − 0.4 ) = 2.58 mA 2 ⎝ ⎠ V DS = 10 − (2.58)(1.2 + 0.5) = 5.62 V ______________________________________________________________________________________

3.36 ⎛W ⎞ Let ⎜ ⎟ = 20 for example, then ⎝L⎠ ⎛ 0.05 ⎞ 2 0.20 = ⎜ ⎟(20 )(V SG − 0.6 ) ⇒ V SG = 1.232 V 2 ⎝ ⎠ Then V RS = 1.232 = (0.2)R S ⇒ R S = 6.16 k Ω 6 − 1.232 − 3 = 8.84 k Ω 0.2 VG = 3 − 1.232 − 1.232 = 0.536 V RD =

IR =

6 = (0.1)(0.2 ) = 0.02 mA, ⇒ R1 + R 2 = 300 k Ω R1 + R 2

⎛ R2 ⎞ ⎛ R ⎞ ⎟⎟(6 ) − 3 = ⎜⎜ 2 ⎟⎟(6) − 3 VG = 0.536 = ⎜⎜ ⎝ 300 ⎠ ⎝ R1 + R 2 ⎠ Or R 2 = 176.8 k Ω and R1 = 123.2 k Ω ______________________________________________________________________________________

3.37 I Q = 50 = 500 (VGS − 1.2 ) ⇒ VGS = 1.516 V 2

(a) (i)

VDS = 5 − ( −1.516 ) =⇒ VDS = 6.516 V I Q = 1 = ( 0.5 )(VGS − 1.2 ) ⇒ VGS = 2.61 V 2

(ii)

VDS = 5 − ( −2.61) ⇒ VDS = 7.61 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

V = VDS = 1.516 V (i) Same as (a) GS V = VDS = 2.61 V (ii) GS ______________________________________________________________________________________ (b)

3.38

I D = K n (VGS − VTN )

2

0.25 = ( 0.2 )(VGS − 0.6 )

2

0.25 + 0.6 ⇒ VGS = 1.72 V ⇒ VS = −1.72 V 0.2 VD = 9 − ( 0.25 )( 24 ) ⇒ VD = 3 V VGS =

______________________________________________________________________________________ 3.39

(a)

RD =

5 −1 ⇒ RD = 8 K 0.5

I DQ = 0.5 = 0.25 (VGS − 1.4 ) ⇒ VGS = 2.81 V 2

RS =

−2.81 − ( −5 )

⇒ RS = 4.38 K 0.5 (b) Let RD = 8.2 K, RS = 4.3 K −VGS − ( −5 ) 2 Now = I D = 0.25 (VGS − 1.4 ) 4.3 5 − VGS = 1.075 (VGS2 − 2.8 VGS + 1.96 ) 1.075 VGS2 − 2.01 VGS − 2.89 = 0 VGS =

2.01 ± 4.04 + 12.427 ⇒ VGS = 2.82 V 2 (1.075 )

I D = 0.25 ( 2.82 − 1.4 ) ⇒ I D = 0.504 mA 2

VDS = 10 − ( 0.504 )( 8.2 + 4.3) ⇒ VDS = 3.70 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(c)

If RS = 4.3 + 10% = 4.73 K 5 − VGS = 1.18 (VGS2 − 2.8VGS + 1.96 ) 1.18 VGS2 − 2.31 VGS − 2.68 = 0 VGS =

2.31 ± 5.336 + 12.65 = 2.78 V 2 (1.18 )

I D = ( 0.25 )( 2.78 − 1.4 ) ⇒ I D = 0.476 mA 2

If Rs = 4.3 − 10% = 3.87 K 5 − VGS = ( 0.9675 ) (VGS2 − 2.8VGS + 1.96 ) 0.9675VGS2 − 1.71VGS − 3.10 = 0 VGS =

1.71 ± 2.924 + 12.0 = 2.88 V 2 ( 0.9675 )

I D = ( 0.25 )( 2.88 − 1.4 ) = 0.548 mA ______________________________________________________________________________________ 2

3.40 9 − 2 .5 ⇒ R = 65 k Ω R ⎛ k ′p ⎞⎛ W ⎞ ⎟⎜ ⎟(V SG + VTP )2 I D = ⎜⎜ ⎟⎝ L ⎠ 2 ⎝ ⎠ Now V SG = V SD = 2.5 V I D = 0.10 =

⎛ 0.05 ⎞⎛ W ⎞ ⎛W ⎞ 2 0.10 = ⎜ ⎟⎜ ⎟(2.5 − 0.7 ) ⇒ ⎜ ⎟ = 1.235 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ And W = (1.235)(0.8) = 0.988 μ m ______________________________________________________________________________________

3.41

5 = I DQ RS + VSDQ = I DQ ( 2 ) + 2.5 I DQ = 1.25 mA IR =

10 = (1.25 )( 0.1) ⇒ R1 + R2 = 80 k Ω R1 + R2

I DQ = K p (VSG + VTP )

2

1.25 = 0.5 (VSG + 1.5 ) ⇒ 2

1.25 − 1.5 = VSG 0.5

VSG = 0.0811 V VG = VS − VSG = 2.5 − 0.0811 = 2.42 V ⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎛R ⎞ 2.42 = ⎜ 2 ⎟ (10 ) − 5 ⇒ R2 = 59.4 k Ω, R1 = 20.6 k Ω ⎝ 80 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.42

(a)

RD =

VD − ( −5 ) I DQ

=

5−2 ⇒ RD = 12 K 0.25

2 ⎛ W ⎞ ⎛ k ′p ⎞ I D = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) L 2 ⎝ ⎠⎝ ⎠ 2 ⎛ 0.035 ⎞ 0.25 = (15 ) ⎜ ⎟ (VSG − 1.2 ) ⇒ VSG = 2.18 V ⎝ 2 ⎠ 5 − 2.18 RS = ⇒ RS = 11.3 K 0.25 VSD = 2.18 − ( −2 ) = 4.18 V

(b) k ′p = 35 + 5% = 36.75 μ A/V 2 5 − VSG 2 ⎛ 0.03675 ⎞ I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) = 2 11.3 ⎝ ⎠

2 − 2.4VSG + 1.44 ) = 5 − VSG 3.11(VSG 2 − 6.46VSG − 0.522 = 0 3.11VSG

VSG =

6.46 ± 41.73 + 6.49 = 2.155 V 2 ( 3.11)

5 − 2.155 = 0.252 mA 11.3 = 10 − ( 0.252 )(12 + 11.3) = 4.13 V

ID = VSD

k ′p = 35 − 5% = 33.25 μ A/V 2 5 − VSG 2 ⎛ 0.03325 ⎞ I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) = 2 11.3 ⎝ ⎠

2 2.82 (VSG − 2.4VSG + 1.44 ) = 5 − VSG 2 2.82VSG − 5.77VSG − 0.939 = 0

VSG =

5.77 ± 33.29 + 10.59 = 2.198 V 2 ( 2.82 )

5 − 2.198 = 0.248 mA 11.3 VSD = 10 − ( 0.248 )(12 + 11.3) = 4.22 V ______________________________________________________________________________________ ID =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.43 ID =

−VSD − ( −10 ) RD

⇒5=

−6 + 10 ⇒ RD = 0.8 kΩ RD

I D = K P (VSG + VTP ) ⇒ 5 = 3 (VSG − 1.75 ) 2

2

5 + 1.75 = 3.04 V ⇒ VG = −3.04 3

VSG =

⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 = −3.04 ⎝ R1 + R2 ⎠ Rin = R1 || R2 = 80 kΩ 1 ⋅ ( 80 )(10 ) = 5 − 3.04 ⇒ R1 = 408 kΩ R1 408 R2 = 80 ⇒ R2 = 99.5 kΩ 408 + R2 ______________________________________________________________________________________

3.44 (a) Both M 1 and M 2 in saturation I D1 = I D 2 k n′ k′ (4)(υ I − 0.4)2 = n (1)[0 − (− 0.6)]2 ⇒ υ I = 0.7 V 2 2 (b) (i) υ I = 0.6 V; M 1 in saturation, M 2 in nonsaturation I D1 = I D 2 k n′ k′ (4)(0.6 − 0.4)2 = n (1) 2(0 − (− 0.6))(5 − υ O ) − (5 − υ O )2 2 2 2 We find (5 − υ O ) − 1.2(5 − υ O ) + 0.16 = 0 ⇒ (5 − υ O ) = 0.153

[

]

So υ O = 4.85 V (ii) υ I = 1.5 V; M 1 in nonsaturation, M 2 in saturation I D1 = I D 2 k n′ k′ (4) 2(1.5 − 0.4)υ O − υ O2 = n (1)(0 − (− 0.6))2 2 2 2 We find υ O − 2.2υ O + 0.09 = 0 ⇒ υ O = 0.0417 V ______________________________________________________________________________________

[

]

3.45 M 1 in nonsaturation, M 2 in saturation I D1 = I D 2

[

]

k n′ ⎛ W ⎞ k n′ 2 (1)[0 − (− 0.6)]2 ⎜ ⎟ 2(3 − 0.4 )(0.025) − (0.025) = 2 ⎝ L ⎠1 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ (0.1294 ) = 0.36 ⇒ ⎜ ⎟ = 2.78 L ⎝ ⎠1 ⎝ L ⎠1 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.46 (a) Transistors matched VGS1 = VGS 2 = 2.5 V, VO = 2.5 V ⎛ 0.12 ⎞ 2 ID = ⎜ ⎟(30 )(2.5 − 0.4) = 7.938 mA ⎝ 2 ⎠ (b) I D1 = I D 2 k n′ ⎛ W ⎞ k n′ ⎛ W ⎞ 2 2 ⎜ ⎟ (VGS 1 − VTN ) = ⎜ ⎟ (VGS 2 − VTN ) 2 ⎝ L ⎠1 2 ⎝ L ⎠2 VGS 2 = 5 − VGS1 30 (VGS1 − 0.4) = (5 − VGS1 − 0.4) 15 Which yields VGS1 = 2.14 V, VGS 2 = VO = 2.86 V

Then

⎛ 0.12 ⎞ 2 ID = ⎜ ⎟(30 )(2.14 − 0.4 ) = 5.45 mA 2 ⎝ ⎠ (c) I D1 = I D 2

(15)(VGS1 − 0.4)2 = (30)(VGS 2 − 0.4)2 0.7071(VGS1 − 0.4 ) = 5 − VGS1 − 0.4

Which yields VGS1 = 2.86 V, VGS 2 = VO = 2.14 V ⎛ 0.12 ⎞ 2 ID = ⎜ ⎟(15)(2.86 − 0.4 ) = 5.45 mA ⎝ 2 ⎠ ______________________________________________________________________________________

3.47 (a) V1 = VGS 3 = 2.5 V ⎛ 0.12 ⎞⎛ W ⎞ ⎛W ⎞ 2 0.8 = ⎜ ⎟⎜ ⎟ (2.5 − 0.6 ) ⇒ ⎜ ⎟ = 3.69 ⎝ 2 ⎠⎝ L ⎠ 3 ⎝ L ⎠3 V 2 = 6 = VGS 2 + V1 = VGS 2 + 2.5 ⇒ VGS 2 = 3.5 V ⎛ 0.12 ⎞⎛ W ⎞ ⎛W ⎞ 2 0.8 = ⎜ ⎟⎜ ⎟ (3.5 − 0.6 ) ⇒ ⎜ ⎟ = 1.59 ⎝ 2 ⎠⎝ L ⎠ 2 ⎝ L ⎠2 VGS1 = 9 − V 2 = 9 − 6 = 3 V ⎛ 0.12 ⎞⎛ W ⎞ ⎛W ⎞ 2 0.8 = ⎜ ⎟⎜ ⎟ (3 − 0.6 ) ⇒ ⎜ ⎟ = 2.31 ⎝ 2 ⎠⎝ L ⎠ 1 ⎝ L ⎠1

(b) (i) k n′ = (120)(1.05) = 126 μ A/V 2

No change; V1 = 2.5 V, V 2 = 6 V

(ii) k n′ = (120)(0.95) = 114 μ A/V 2 No change; V1 = 2.5 V, V 2 = 6 V (c) k n′1 = 114 μ A/V 2 , k n′ 2 = k n′ 3 = 126 μ A/V 2 I D 2 = I D3 ⎛ 0.126 ⎞ ⎛ 0.126 ⎞ 2 2 ⎜ ⎟(1.59 )(VGS 2 − 0.6 ) = ⎜ ⎟(3.69 )(VGS 3 − 0.6 ) 2 2 ⎝ ⎠ ⎝ ⎠ Now VGS 3 = V1 , VGS 2 = V 2 − V1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ So (V 2 − V1 − 0.6) = 1.523(V1 − 0.6) And V 2 = 2.523V1 − 0.3138 Also I D1 = I D 3 ⎛ 0.114 ⎞ ⎛ 0.126 ⎞ 2 2 ⎜ ⎟(2.31)(VGS 1 − 0.6 ) = ⎜ ⎟(3.69 )(VGS 3 − 0.6) ⎝ 2 ⎠ ⎝ 2 ⎠ We have VGS1 = 9 − V 2 = 9 − (2.523V1 − 0.3138) = 9.3138 − 2.523V1

Also VGS 3 = V1

Then (0.13167)(9.3138 − 2.523V1 − 0.6) = (0.23247)(V1 − 0.6) 8.7138 − 2.523V1 = 1.3287(V1 − 0.6) Which yields V1 = 2.469 V and V 2 = 5.916 V ______________________________________________________________________________________ 2

2

3.48 M L in saturation, M D in nonsaturation I DD = I DL

[

]

⎛ k n′ ⎞⎛ W ⎞ ⎛ k′ ⎞ 2 2 ⎜⎜ ⎟⎟⎜ ⎟ 2(5 − 0.6 )(0.15) − (0.15) = ⎜⎜ n ⎟⎟(2 )(5 − 0.15 − 0.6 ) L 2 2 ⎝ ⎠⎝ ⎠ D ⎝ ⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ (1.2975) = 36.125 ⇒ ⎜ ⎟ = 27.8 ⎝ L ⎠D ⎝ L ⎠D ______________________________________________________________________________________

3.49 M L in saturation, M D in nonsaturation I DD = I DL

[

]

⎛ k n′ ⎞⎛ W ⎞ ⎛ k′ ⎞ 2 2 ⎜⎜ ⎟⎟⎜ ⎟ 2(5 − 0.6 )(0.10 ) − (0.10 ) = ⎜⎜ n ⎟⎟(2 )[0 − (− 1.2 )] ⎝ 2 ⎠⎝ L ⎠ D ⎝ 2 ⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ (0.87 ) = 2.88 ⇒ ⎜ ⎟ = 3.31 L ⎝ ⎠D ⎝ L ⎠D ______________________________________________________________________________________

3.50 (a) Transistor in nonsaturation 3 − 0.08 ID = = 0.09733 mA 30 ⎛ 0.12 ⎞⎛ W ⎞ 2 0.09733 = ⎜ ⎟⎜ ⎟ 2(2.6 − 0.4 )(0.08) − (0.08) 2 L ⎝ ⎠⎝ ⎠

[

]

⎛W ⎞ ⎛W ⎞ 0.09733 = ⎜ ⎟(0.020736 ) ⇒ ⎜ ⎟ = 4.69 L ⎝ ⎠ ⎝L⎠

[

⎛ 0.12 ⎞⎛ W ⎞ 2 (b) 0.09733 = ⎜ ⎟⎜ ⎟ 2(3 − 0.4 )(0.08) − (0.08) ⎝ 2 ⎠⎝ L ⎠

]

⎛W ⎞ ⎛W ⎞ 0.09733 = ⎜ ⎟(0.024576 ) ⇒ ⎜ ⎟ = 3.96 ⎝L⎠ ⎝L⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.51 5 = I D R D + Vγ + V DS

5 = (12)R D + 1.6 + 0.15 ⇒ R D = 271 Ω

[

]

⎛ 0.08 ⎞ W ⎛W ⎞ 2 I D = 12 = ⎜ 2(5 − 0.6 )(0.15) − (0.15) ⇒ ⎜ ⎟ = 231 ⎟ 2 L ⎝ ⎠ ⎝L⎠ ______________________________________________________________________________________

3.52

5 = V SD + I D R D + Vγ

5 = 0.20 + (15)R D + 1.6 ⇒ R D = 213 Ω

[

⎛ 0.04 ⎞⎛ W ⎞ 2 I D = 15 = ⎜ ⎟⎜ ⎟ 2(5 − 0.6)(0.20 ) − (0.20 ) ⎝ 2 ⎠⎝ L ⎠

]

⎛W ⎞ ⎛W ⎞ 15 = ⎜ ⎟(0.0344 ) ⇒ ⎜ ⎟ = 436 L ⎝ ⎠ ⎝L⎠ ______________________________________________________________________________________

3.53 5 − 0.15 = 0.097 mA 50 0.097 ⎛ 0.12 ⎞⎛ W ⎞ 2 = =⎜ ⎟⎜ ⎟ 2(5 − 0.6)(0.15) − (0.15) 2 ⎝ 2 ⎠⎝ L ⎠

(a) I R = ID

[

]

⎛W ⎞ ⎛W ⎞ ⎛W ⎞ 0.0485 = ⎜ ⎟(0.07785) ⇒ ⎜ ⎟ = ⎜ ⎟ = 0.623 L ⎝ ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2 (b) M 1 nonsaturation, M 2 cutoff IR = ID =

[

5 − VO ⎛ 0.12 ⎞ 2 =⎜ ⎟(0.623) 2(5 − 0.6 )VO − VO 50 ⎝ 2 ⎠

]

We find 1.869VO2 − 17.45VO + 5 = 0 ⇒ VO = 0.297 V ______________________________________________________________________________________ 3.54 (a) V DS 2 (sat ) = 0.5 = VGS 2 − VTN = VGS 2 − 0.4 ⇒ VGS 2 = 0.9 V ⎛ 120 ⎞⎛ W ⎞ ⎛W ⎞ 2 I Q1 = 125 = ⎜ ⎟⎜ ⎟ (0.9 − 0.4 ) ⇒ ⎜ ⎟ = 8.33 2 L ⎝ ⎠⎝ ⎠ 2 ⎝ L ⎠2 VGS 3 = VGS 2 = 0.9 V ⎛ 120 ⎞⎛ W ⎞ ⎛W ⎞ 2 I REF 1 = 225 = ⎜ ⎟⎜ ⎟ (0.9 − 0.4 ) ⇒ ⎜ ⎟ = 15 ⎝ 2 ⎠⎝ L ⎠ 3 ⎝ L ⎠3 ⎛W ⎞ M 1 and M 2 matched, so ⎜ ⎟ = 8.33 ⎝ L ⎠1 (b) VGS1 = 0.9 V ⇒ V D1 = V DS1 − VGS1 = 2 − 0.9 = 1.1 V 2.5 − 1.1 = 11.2 k Ω 0.125 ______________________________________________________________________________________ RD =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.55 (a) V SDB (sat ) = 0.8 = V SGB + VTP = V SGB − 0.4 ⇒ V SGB = 1.2 V ⎛ 50 ⎞⎛ W ⎞ ⎛W ⎞ 2 I Q 2 = 200 = ⎜ ⎟⎜ ⎟ (1.2 − 0.4 ) ⇒ ⎜ ⎟ = 12.5 L 2 ⎝ ⎠⎝ ⎠ B ⎝ L ⎠B V SGC = V SGB = 1.2 V ⎛ 50 ⎞⎛ W ⎞ ⎛W ⎞ 2 I REF 2 = 125 = ⎜ ⎟⎜ ⎟ (1.2 − 0.4 ) ⇒ ⎜ ⎟ = 7.81 ⎝ 2 ⎠⎝ L ⎠ C ⎝ L ⎠C ⎛W ⎞ M A and M B matched, so ⎜ ⎟ = 12.5 ⎝ L ⎠A (b) V SGA = 1.2 V V DA = V SGA − V SDA = 1.2 − 4 = −2.8 V

− 2.8 − (− 5) = 11 k Ω 0.20 ______________________________________________________________________________________ RD =

3.56

V DS 2 (sat ) = 0.5 = VGS 2 − VTN = VGS 2 − 0.6 ⇒ VGS 2 = 1.1 V ⎛ 0.12 ⎞⎛ W ⎞ ⎛W ⎞ 2 I Q = 0.35 = ⎜ ⎟⎜ ⎟ (1.1 − 0.6 ) ⇒ ⎜ ⎟ = 23.3 ⎝ 2 ⎠⎝ L ⎠ 2 ⎝ L ⎠2 VGS 3 = VGS 2 = 1.1 V ⎛ 0.12 ⎞⎛ W ⎞ ⎛W ⎞ 2 I REF = 0.15 = ⎜ ⎟⎜ ⎟ (1.1 − 0.6 ) ⇒ ⎜ ⎟ = 10 2 L ⎝ ⎠⎝ ⎠ 3 ⎝ L ⎠3 VGS 4 = 5 − 1.1 = 3.9 V ⎛ 0.12 ⎞⎛ W ⎞ ⎛W ⎞ 2 I REF = 0.15 = ⎜ ⎟⎜ ⎟ (3.9 − 0.6 ) ⇒ ⎜ ⎟ = 0.23 ⎝ 2 ⎠⎝ L ⎠ 4 ⎝ L ⎠4 ⎛W ⎞ M 1 and M 2 matched, so ⎜ ⎟ = 23.3 ⎝ L ⎠1 VGS1 = 1.1 V; V D1 = V DS1 − VGS1 = 3.5 − 1.1 = 2.4 V

5 − 2 .4 = 7.43 k Ω 0.35 ______________________________________________________________________________________ RD =

3.57

VDS ( sat ) = VGS − VP

VDS > VDS ( sat ) = −VP , I D = I DSS So ______________________________________________________________________________________

3.58

a.

VDS ( sat ) = VGS − VP = VGS + 3 = VDS ( sat )

VGS = 0 ⇒ I D = I DSS = 6 mA 2

⎛ V ⎞ ⎛ −1 ⎞ I D = I DSS ⎜1 − GS ⎟ = 6 ⎜ 1 − ⎟ ⇒ I D = 2.67 mA ⎝ −3 ⎠ ⎝ VP ⎠ b. 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

⎛ −2 ⎞ I D = 6 ⎜ 1 − ⎟ ⇒ I D = 0.667 mA ⎝ −3 ⎠ 2

c.

ID = 0 d. ______________________________________________________________________________________ 3.59 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ ⎛ 1 ⎞ 2.8 = I DSS ⎜1 − ⎟ ⎝ VP ⎠

2

2

⎛ 3 ⎞ 0.30 = I DSS ⎜ 1 − ⎟ V ⎝ P ⎠ ⎛ 1 ⎜1 − 2.8 ⎝ VP = 0.30 ⎛ 3 ⎜1 − ⎝ VP

2

2

⎞ ⎟ ⎠ = 9.33 2 ⎞ ⎟ ⎠

⎛ 1 ⎜1 − ⎝ VP ⎛ 3 ⎜1 − V ⎝ P

⎞ ⎟ ⎠ = 3.055 ⎞ ⎟ ⎠ 1 9.165 = 3.055 − 1− VP VP 8.165 = 2.055 ⇒ VP = 3.97 V VP 2

1 ⎞ ⎛ 2.8 = I DSS ⎜ 1 − ⎟ = I DSS ( 0.560 ) ⇒ I DSS = 5.0 mA ⎝ 3.97 ⎠ ______________________________________________________________________________________

3.60

VS = −VGS , VSD = VS − VDD

V ≥ VSD ( sat ) = VP − VGS Want SD VS − VDD ≥ VP − VGS − VGS − VDD ≥ VP − VGS ⇒ VDD ≤ −VP So

VDD ≤ −2.5 V ⎛ V ⎞ I D = 2 = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2

2

⎛ V ⎞ 2 = 6 ⎜1 − GS ⎟ ⇒ VGS = 1.06 V ⇒ VS = −1.06 V ⎝ 2.5 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.61 I D = K n (VGS − VTN )

2

18.5 = K n ( 0.35 − VTN ) 86.2 = K n ( 0.5 − VTN )

2

2

Then

( 0.35 − VTN ) 18.5 = 0.2146 = ⇒ VTN = 0.221 V 2 86.2 ( 0.50 − VTN ) 2

18.5 = K n ( 0.35 − 0.221) ⇒ K n = 1.11 mA / V 2 2

______________________________________________________________________________________ 3.62 I D = K (VGS − VTN )

2

250 = K ( 0.75 − 0.24 ) ⇒ K = 0.961 mA / V 2 2

______________________________________________________________________________________ 3.63 2

⎛ V ⎞ V V I D = I DSS ⎜ 1 − GS ⎟ = S = − GS RS RS ⎝ VP ⎠ 2

V ⎛ V ⎞ 10 ⎜ 1 − GS ⎟ = − GS −5 ⎠ 0.2 ⎝ 2 ⎛ 2V V ⎞ 2 ⎜1 + GS + GS ⎟ = −VGS 5 25 ⎠ ⎝ 2 2 9 VGS + VGS + 2 = 0 25 5 2 2VGS + 45VGS + 50 = 0 VGS =

−45 ±

ID = −

( 45 ) − 4 ( 2 )( 50 ) ⇒ VGS 2 ( 2) 2

= −1.17 V

VGS 1.17 = ⇒ I D = 5.85 mA 0.2 RS

VD = 20 − ( 5.85 )( 2 ) = 8.3 V VDS = VD − VS = 8.3 − 1.17 ⇒ VDS = 7.13 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.64 VDS = VDD − VS 8 = 10 − VS ⇒ VS = 2 V = I D RS = ( 5 ) RS ⇒ RS = 0.4 kΩ ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠

2

2

⎛ −1 ⎞ 5 = I DSS ⎜1 − ⎟ Let I DSS = 10 mA ⎝ VP ⎠ 2

⎛ −1 ⎞ 5 = 10 ⎜1 − ⎟ ⇒ VP = −3.41 V ⎝ VP ⎠ VG = VGS + VS = −1 + 2 = 1 V

⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R R R + 2 ⎠ 1 ⎝ 1 1 1 = ( 500 )(10 ) ⇒ R1 = 5 MΩ R1 5R2 = 0.5 ⇒ R2 = 0.556 MΩ 5 + R2 ______________________________________________________________________________________ 3.65 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ V ⎞ 5 = 8 ⎜ 1 − GS ⎟ ⇒ VGS = 0.838 V 4 ⎠ ⎝ VSD = VDD − I D ( RS + RD )

= 20 − ( 5 )( 0.5 + 2 ) ⇒ VSD = 7.5 V

VS = 20 − ( 5 )( 0.5 ) = 17.5 V VG = VS + VGS = 17.5 + 0.838 = 18.3 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 18.3 = (100 ) ( 20 ) ⇒ R1 = 109 kΩ R1

109 R2 = 100 ⇒ R2 = 1.21 MΩ 109 + R2 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.66 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠

2

2

⎛ V ⎞ 5 = 7 ⎜ 1 − GS ⎟ ⇒ VGS = 0.465 V 3 ⎠ ⎝ VSD = VDD − I D ( RS + RD ) 6 = 12 − ( 5 )( 0.3 + RD ) ⇒ RD = 0.9 kΩ VS = 12 − ( 5 )( 0.3) = 10.5 V VG = VS + VGS = 10.5 + 0.465 = 10.965 V ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 10.965 = ⎜ 2 ⎟ (12 ) ⇒ R2 = 91.4 kΩ ⇒ R1 = 8.6 kΩ ⎝ 100 ⎠ ______________________________________________________________________________________

3.67 ⎛ R2 ⎞ ⎛ 60 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ ( 20 ) ⇒ VG = 6 V R + R ⎝ 140 + 60 ⎠ ⎝ 1 2 ⎠ 2

⎛ V ⎞ V V − VGS I D = I DSS ⎜ 1 − GS ⎟ = S = G RS RS ⎝ VP ⎠ 2

⎛ V ⎞ (8 )( 2 ) ⎜⎜1 − GS ⎟⎟ = 6 − VGS ⎝ ( −4 ) ⎠ ⎛ V V2 ⎞ 16 ⎜ 1 + GS + GS ⎟ = 6 − VGS 2 16 ⎠ ⎝ VGS2 + 9VGS + 10 = 0 VGS =

−9 ±

(9)

2

− 4 (10 )

2

⇒ VGS = −1.30

⎛ ( −1.30 ) ⎞ I D = 8 ⎜⎜ 1 − ⎟ ⇒ I D = 3.65 mA ( −4 ) ⎟⎠ ⎝ VDS = VDD − I D ( RS + RD ) = 20 − ( 3.65 )( 2 + 2.7 ) VDS = 2.85 V 2

VDS > VDS ( sat ) = VGS − VP = −1.30 − ( −4 ) = 2.7 V (Yes) ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.68

VDS = VDD − I D ( RS + RD ) 5 = 12 − I D ( 0.5 + 1) ⇒ I D = 4.67 mA VS = I D RS = ( 4.67 )( 0.5 ) ⇒ VS = 2.33 V ⎛ R2 ⎞ ⎛ 20 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (12 ) ⇒ VG = 0.511 V ⎝ 450 + 20 ⎠ ⎝ R1 + R2 ⎠ VGS = VG − VS = 0.511 − 2.33 ⇒ VGS = −1.82 V

⎛ V ⎞ I D = I DSS ⎜1 − GS ⎟ ⎝ VP ⎠

2

⎛ ( −1.82 ) ⎞ 4.67 = 10 ⎜⎜1 − ⎟ ⇒ VP = −5.75 V VP ⎟⎠ ⎝ _____________________________________________________________________________________ 2

3.69 2

⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ , VGS = 0 ⎝ VP ⎠ I D = I DSS = 4 mA VDD − VDS 10 − 3 = ⇒ RD = 1.75 kΩ ID 4 ______________________________________________________________________________________ RD =

3.70

VSD = VDD − I D RS 10 = 20 − (1) RS ⇒ RS = 10 kΩ R1 + R2 =

VDD 20 = = 200 kΩ I 0.1

⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠

2

2

⎛ V ⎞ 1 = 2 ⎜1 − GS ⎟ ⇒ VGS = 0.586 V 2 ⎠ ⎝ VG = VS + VGS = 10 + 0.586 = 10.586 ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 10.586 = ⎜ 2 ⎟ ( 20 ) ⇒ R2 = 106 kΩ ⎝ 200 ⎠ R1 = 94 kΩ

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

3.71

VDS = VDD − I D ( RS + RD ) 2 = 3 − ( 0.040 )(10 + RD ) ⇒ RD = 15 kΩ I D = K (VGS − VTN )

2

40 = 250 (VGS − 0.20 ) ⇒ VGS = 0.60 V 2

VG = VGS + VS = 0.60 + ( 0.040 )(10 ) = 1.0 V ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 1 = ⎜ 2 ⎟ ( 3) ⇒ R2 = 50 kΩ ⎝ 150 ⎠ R1 = 100 kΩ

______________________________________________________________________________________ 3.72

VO = 0.70 V ⇒ VDS = 0.70 > VDS ( sat ) = VGS − VTN

For 0.75 − 0.15 = 0.6 Biased in the saturation region V − VDS 3 − 0.7 I D = DD = ⇒ I D = 46 μ A RD 50 I D = K (VGS − VTN ) ⇒ 46 = K ( 0.75 − 0.15 ) ⇒ K = 128 μ A / V 2 2

2

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 4 4.1 (a) (i) g m = 2

k n′ ⎛ W ⎞ ⎜ ⎟ I DQ 2 ⎝L⎠

⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 0.5 = 2 ⎜ ⎟⎜ ⎟(0.5) ⇒ ⎜ ⎟ = 2.5 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ ⎛ k ′ ⎞⎛ W ⎞ (ii) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟(VGSQ − VTN )2 ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.1 ⎞ 2 0. 5 = ⎜ ⎟(2.5)(VGSQ − 0.4 ) ⇒ VGSQ = 2.4 V ⎝ 2 ⎠

⎛ 0.1 ⎞⎛ W ⎞ ⎛W (b) (i) 0.5 = 2 ⎜ ⎟⎜ ⎟(0.15) ⇒ ⎜ L 2 ⎝ ⎠⎝ ⎠ ⎝L

⎞ ⎟ = 8.33 ⎠

⎛ 0.1 ⎞ 2 (ii) 0.15 = ⎜ ⎟(8.33)(VGSQ − 0.4 ) ⇒ VGSQ = 1.0 V 2 ⎝ ⎠ ______________________________________________________________________________________

4.2 ⎛ k ′p (a) (i) g m = 2 ⎜⎜ ⎝ 2

⎞⎛ W ⎞ ⎟⎜ ⎟ I DQ ⎟⎝ L ⎠ ⎠

⎛W ⎞ ⎛ 0.04 ⎞⎛ W ⎞ 1. 2 = 2 ⎜ ⎟⎜ ⎟(0.15) ⇒ ⎜ ⎟ = 120 2 L ⎝L⎠ ⎠⎝ ⎠ ⎝

⎛ k ′p ⎞⎛ W ⎞ 2 (ii) I DQ = ⎜⎜ ⎟⎟⎜ ⎟ V SGQ + VTP ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.04 ⎞ 2 0.15 = ⎜ ⎟(120 )(V SGQ − 0.6 ) ⇒ V SGQ = 0.85 V 2 ⎠ ⎝

(

)

⎛W ⎞ ⎛ 0.04 ⎞⎛ W ⎞ (b) (i) 1.2 = 2 ⎜ ⎟⎜ ⎟(0.50 ) ⇒ ⎜ ⎟ = 36 ⎝L⎠ ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.04 ⎞ 2 (ii) 0.50 = ⎜ ⎟(36 )(V SGQ − 0.6 ) ⇒ V SGQ = 1.43 V ⎝ 2 ⎠ ______________________________________________________________________________________

4.3 I D = K n (VGS − VTN ) (1 + λVDS ) 2

I D1 1 + λVDS1 3.4 1 + λ (10 ) = ⇒ = I D 2 1 + λVDS 2 3.0 1 + λ ( 5 )

3.4 [1 + 5λ ] = 3.0 [1 + 10λ ]

3.4 − 3.0 = λ ( 3 ⋅10 − ( 3.4 ) ⋅ 5 ) ⇒ λ = 0.0308 ΔVDS 5 = = 12.5 kΩ ΔI D 0.4 ______________________________________________________________________________________ ro =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.4 1 r0 = λ ID

1 1 = ⇒ I D = 0.833 mA λ r0 ( 0.012 )(100 ) ______________________________________________________________________________________ ID =

4.5 2 (a) I D = K n (VGS − VTN ) (1 + λV DS ) I D = I DO (1 + λV DS )

Then

0.258 (1 + 3.3λ ) = 0.250 (1 + 1.5λ )

Or 1.032(1 + 1.5λ ) = 1 + 3.3λ ⇒ λ = 0.01826 V −1 0.250 = I DO [1 + (0.01826 )(1.5)] ⇒ I DO = 0.2433 mA ro =

1

1

=

= 225 k Ω

(0.01826)(0.2433) = (0.2433)[1 + (0.01826 )(5)] = 0.2655 mA λ I DO

(b) I D ______________________________________________________________________________________ 4.6

(a) ro =

(i) ro =

(ii) (b)

1

λ ID

=

1

( 0.02 )( 0.05 )

= 1000 K

1 = 100 K 0.02 ( )( 0.5 )

ΔI D =

(i)

ΔVDS 1 = = 0.001 mA = 1.0 μ A ro 1000

ΔI D 1.0 = ⇒ 2% ID 50

ΔI D =

ΔVDS 1 = = 0.01 ⇒ 10 μ A 100 ro

ΔI D 10 = ⇒ 2% 500 (ii) I D ______________________________________________________________________________________ 4.7

I D = 1.0 mA 1 1 ro = = = 100 K λ I D ( 0.01)(1) ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.8 (a) V DD = I DQ R D + V DSQ 3.3 = I DQ (5) + 1.5 ⇒ I DQ = 0.36 mA

⎛ k ′ ⎞⎛ W ⎞ 2 I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ VGSQ − VTN ⎝ 2 ⎠⎝ L ⎠ ⎛ 0. 1 ⎞ 2 0.36 = ⎜ ⎟(40 ) VGSQ − 0.4 ⇒ VGSQ = 0.824 V 2 ⎠ ⎝

(

(

)

)

⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞ 2 (b) K n = ⎜⎜ n ⎟⎟⎜ ⎟ = ⎜ ⎟(40 ) = 2 mA/V 2 2 L ⎠ ⎠ ⎝ ⎝ ⎝ ⎠

g m = 2 K n I DQ = 2 (2)(0.36) = 1.697 mA/V ro =

1 1 = = 111.1 k Ω λ I DQ (0.025)(0.36 )

Aυ = − g m (ro R D ) = −(1.697)(111.1 5) = −8.12 ______________________________________________________________________________________ 4.9 (a) Aυ = − g m R D − 3.8 = − g m (10 ) ⇒ g m = 0.38 mA/V

⎛ k′ g m = 2 K n I DQ = 2 ⎜⎜ n ⎝ 2

⎞⎛ W ⎞ ⎟⎟⎜ ⎟ I DQ ⎠⎝ L ⎠

⎛W ⎞ ⎛ 0.1 ⎞⎛ W ⎞ 0.38 = 2 ⎜ ⎟⎜ ⎟(0.12 ) ⇒ ⎜ ⎟ = 6.02 2 L ⎝L⎠ ⎠⎝ ⎠ ⎝ (b) − 5 = − g m (10 ) ⇒ g m = 0.50 mA/V ⎛W ⎞ ⎛ 0.1 ⎞⎛ W ⎞ 0.50 = 2 ⎜ ⎟⎜ ⎟(0.12 ) ⇒ ⎜ ⎟ = 10.4 2 L ⎝L⎠ ⎠⎝ ⎠ ⎝ ______________________________________________________________________________________

4.10 (a) V DD = I DQ R D + V DSQ

5 = (0.5)R D + 3 ⇒ R D = 4 k Ω ⎛ k ′ ⎞⎛ W ⎞ 2 I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ VGSQ − VTN ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎛ 0.08 ⎞⎛ W ⎞ 2 0.5 = ⎜ ⎟⎜ ⎟(1.2 − 0.6 ) ⇒ ⎜ 2 L ⎝L ⎠⎝ ⎠ ⎝

(

)

⎞ ⎟ = 34.7 ⎠

⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.08 ⎞ (b) g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜ ⎟(34.7 )(0.5) = 1.666 mA/V ⎝ 2 ⎠ ⎝ 2 ⎠⎝ L ⎠ 1 1 ro = = = 133.3 k Ω λ I DQ (0.015)(0.5)

(c) Aυ = − g m (ro R D ) = −(1.666)(133.3 4) = −6.47

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.11 2

K n vgs2 = K n ⎡⎣Vgs sinω t ⎤⎦ = K nVgs2 sin 2ω t 1 [1 − cos 2ω t ] 2 K nVgs2 So K n vgs2 = [1 − cos 2ω t ] 2 sin 2 ω t =

K nVgs2 Ratio of signal at 2ω to that at ω :

2

⋅ cos 2ω t

2 K n (VGSQ − VTN ) Vgs ⋅ sin ω t

The coefficient of this expression is then:

Vgs

4 (VGSQ − VTN )

______________________________________________________________________________________ 4.12 0.01 =

Vgs

4 (VGSQ − VTN )

So Vgs = ( 0.01)( 4 )( 3 − 1) ⇒ Vgs = 0.08 V

______________________________________________________________________________________ 4.13 ⎛ R2 ⎞ ⎛ 60 ⎞ ⎟⎟ ⋅ V DD = ⎜ (a) VGS = ⎜⎜ ⎟(3.3) = 0.66 V R + R ⎝ 60 + 240 ⎠ 2 ⎠ ⎝ 1 ⎛ k ′ ⎞⎛ W ⎞ ⎛ 0. 1 ⎞ 2 2 I DQ = ⎜⎜ n ⎟⎟⎜ ⎟(VGS − VTN ) = ⎜ ⎟(80 )(0.66 − 0.4 ) = 0.270 mA 2 L 2 ⎠ ⎝ ⎝ ⎠⎝ ⎠ V DSQ = V DD − I DQ R D = 3.3 − (0.270 )(8) = 1.14 V

⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞ (b) g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜ ⎟(80)(0.270) = 2.078 mA/V ⎝ 2 ⎠ ⎝ 2 ⎠⎝ L ⎠ 1 1 = = 185 k Ω ro = λ I DQ (0.02 )(0.27 ) ⎛ R1 R 2 ⎞ ⎟ (c) Aυ = − g m (ro R D )⎜ ⎜R R +R ⎟ Si ⎠ ⎝ 1 2 We find ro R D = 185 8 = 7.668 k Ω

R1 R2 = 60 240 = 48 k Ω ⎛ 48 ⎞ So Aυ = −(2.078)(7.668)⎜ ⎟ = −15.3 ⎝ 48 + 2 ⎠ ______________________________________________________________________________________

4.14

Av = − g m ( r0 || RD ) −10 = − g m (100 || 5 ) ⇒ g m = 2.1 mA/V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.15 ⎛ R2 ⎞ ⎛ 175 ⎞ ⎟⎟ ⋅ V DD = ⎜ (a) VG = ⎜⎜ ⎟(5) = 2.1875 V R + R ⎝ 175 + 225 ⎠ 2 ⎠ ⎝ 1

2.1875 = VGS + I D R S = VGS + K n R S (VGS − VTN )

(

2 2.1875 = VGS + (1)(1) VGS − 1.6VGS + 0.64

or V

2 GS

2

)

− 0.6VGS − 1.5475 = 0 ⇒ VGS = 1.58 V

I DQ = K n (VGS − VTN ) = (1)(1.58 − 0.8) = 0.608 mA 2

2

V DSQ = V DD − I DQ (R S + R D ) = 5 − (0.608 )(1 + 4 ) = 1.96 V

(b) Aυ =

− g m RD 1 + g m RS

g m = 2 (1)(0.608 ) = 1.56 mA/V

− (1.56)(4) = −2.44 1 + (1.56)(1) − g m (R D R L ) − (1.56)(R D R L ) = = −0.6094(R D R L ) (c) Aυ = 1 + g m RS 1 + (1.56)(1)

Aυ =

− (0.75)(2.44) = −(0.6094)(R D R L ) ⇒ R D R L = 3.0 k Ω

4 R L = 3 ⇒ R L = 12 k Ω ______________________________________________________________________________________ 4.16 (a) V DSQ = V DD − I DQ (R S + R D )

5 = 12 − (2 )(R S + R D ) ⇒ R S + R D = 3.5 k Ω R S = 0.5 k Ω , then R D = 3 k Ω

(

I DQ = K n VGSQ − VTN

(

2 = 1.5 VGSQ − 1.2

)

2

)

2

⇒ VGSQ = 2.355 V

V G = V GSQ + I DQ R S = 2.355 + (2 )(0.5) = 3.355 V

⎛ R2 ⎞ 1 ⎟⎟ ⋅ V DD = VG = ⎜⎜ ⋅ Rin ⋅ V DD R1 ⎝ R1 + R 2 ⎠ 1 3.355 = (250)(12) ⇒ R1 = 894 k Ω R1

R1 R2 = 894 R2 = 250 ⇒ R2 = 347 k Ω (b) g m = 2 (1.5)(2 ) = 3.464 mA/V

Aυ =

− g m (R D R L )

=

− (3.464)(3 10)

= −2.93 1 + g m RS 1 + (3.464)(0.5) ______________________________________________________________________________________ 4.17 (a) From Problem 4.16; R S = 0.5 k Ω , R D = 3 k Ω , R1 = 894 k Ω , R 2 = 347 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) g m = 2 (1.5)(2 ) = 3.464 mA/V

Aυ = − g m (R D R L ) = −(3.464)(3 10) = −7.99

______________________________________________________________________________________ 4.18 (a) Av = − g m RD −15 = −2 RD ⇒ RD = 7.5 K

(b) Av = −5 =

− g m RD 1 + g m RS

− ( 2 )( 7.5 )

⇒ RS = 1 K 1 + ( 2 ) RS ______________________________________________________________________________________

4.19 Av =

− g m RD 1 + g m RS

−8 =

− g m RD 1 + g m (1)

(a) (1)

−16 = − g m RD

(2)

8=

16 ⇒ 1 + g m (1)

g m = 1 mA/V RD = 16 K

Then Av = −10 =

− (1)(16 )

1 + (1) RS

RS = 0.6 K (b) ______________________________________________________________________________________

4.20 ⎛ k ′ ⎞⎛ W ⎞ 2 (a) I DQ = I Q = ⎜⎜ n ⎟⎟⎜ ⎟(VGSQ − VTN ) ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.1 ⎞ 2 0. 5 = ⎜ ⎟(50 )(VGSQ − 0.8) ⇒ VGSQ = 1.247 V 2 ⎝ ⎠

(

)

V DSQ = V + − I DQ R D − − VGSQ = 5 − (0.5)(6 ) + 1.247 = 3.25 V

⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞ (b) g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜ ⎟(50)(0.5) = 2.236 mA/V L 2 ⎝ 2 ⎠ ⎝ ⎠⎝ ⎠ 1 1 = = 100 k Ω ro = λ I DQ (0.02 )(0.5)

Aυ = − g m (ro R D ) = −(2.236)(100 6) = −12.7

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(

)

( ) ) = −(2.236 )(100 6 6) = −6.51

(c) Aυ = − g m ro R D R L = −(2.236 ) 100 6 20 = −9.86 (d) Aυ = − g m (ro R D R L

______________________________________________________________________________________ 4.21 (a) V DSQ = 5 − (− 5) − I DQ (R S + R D )

5.5 = 10 − (0.1)(R S + R D ) ⇒ R S + R D = 45 k Ω

(

I DQ = K n VGSQ − VTN

)

2

(0.1) = (0.85)(VGSQ − 0.8)2 ⇒ VGSQ

= 1.143 V

VGS + I D R S = 5

1.143 + (0.1)R S = 5 ⇒ R S = 38.6 k Ω ⇒ R D = 6.43 k Ω

(b) g m = 2 K n I DQ = 2 (0.85)(0.1) = 0.583 mA/V ro =

1 1 = = 500 k Ω λ I DQ (0.02 )(0.1)

(

)

(

)

(c) Aυ = − g m R D ro R L = −(0.583) 6.43 500 40 = −3.19 ______________________________________________________________________________________ 4.22 (a) V DSQ = V DD − I DQ (R S + R D )

2 = 3.3 − (0.5)(R S + R D ) ⇒ R S + R D = 2.6 k Ω

(

) − (− 0.8))

I DQ = K n VGSQ − VTN

(

0.5 = 2 VGSQ

2 2

⇒ VGSQ = −0.3 V

V GSQ + I DQ R S = 0

− 0.3 + (0.5)R S = 0 ⇒ R S = 0.6 k Ω ⇒ RD = 2 k Ω

(b) g m = 2 K n I DQ = 2 (2)(0.5) = 2 mA/V Aυ =

− g m (R D R L )

=

− (2)(2 10)

= −1.52 1 + g m RS 1 + (2)(0.6 ) ______________________________________________________________________________________ 4.23 (a) V DD = I DQ R D + V DSQ + I DQ R S and VGS + I DQ R S = 0

Then V DD = K n R D (VGS − VTN ) + V DSQ − VGS 2

5 = (2)(2)(VGS + 0.8) + 2.5 − VGS Which yields 2 4VGS + 5.4VGS + 0.06 = 0 , ⇒ VGS = −0.0112 V 2

and I DQ = 2(− 0.0112 + 0.8) = 1.244 mA 2

5 = (1.244 )(2 ) + 2.5 + (1.244 )R S ⇒ R S = 8.99 Ω ≅ 9 Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) g m = 2 K n I DQ = 2 (2)(1.244) = 3.155 mA/V Aυ =

− g m (R D R L )

− (3.155)(2 2)

=

= −3.07 1 + g m RS 1 + (3.155)(0.009) ______________________________________________________________________________________ 4.24 a. 5 = I DQ RS + VSDQ + I DQ RD − 5 5 = I DQ RS + 6 + I DQ (10 ) − 5

I DQ =

1. 2. 3.

4 RS + 10

VS = VSDQ + I DQ RD − 5 = VSGQ

1 + I DQ (10 ) = VSGQ I DQ = K p (VSGQ − 2 )

2

Choose RS = 10 kΩ ⇒ I DQ =

4 = 0.20 mA 20

VSGQ = 1 + (0.2)(10) = 3 V 0.20 = K P (3 − 2) 2 ⇒ K P = 0.20 mA / V 2

b. I DQ = ( 0.20 )( 3 − 2 ) = 0.20 mA 2

( 0.2 )( 0.2 ) = 0.4 mA / V Av = − g m ( RD || RL ) = − ( 0.4 )(10 || 10 ) ⇒ Av = −2.0

g m = 2 K P I DQ = 2

c. 4 = 0.133 mA 30 = 1 + (0.133)(10) = 2.33 V

Choose RS = 20 kΩ ⇒ I DQ = VSGQ

0.133 = K p (2.33 − 2) 2 ⇒ K p = 1.22 mA / V 2 g m = 2 (1.22)(0.133) = 0.806 mA/V Av = −(0.806)(10 10) ⇒ Av = −4.03 A larger gain can be achieved. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.25 (a)

I DQ = K p (VSGQ + VTP )

2

0.25 = 0.8 (VSGQ − 0.5 )

2

VSGQ = 1.059 V 3 − 1.059 RS = ⇒ RS = 7.76 K 0.25 VD = VS − VSDQ = 1.059 − 1.5 = −0.441 V RD =

(b)

−0.441 − ( −3) 0.25

⇒ RD = 10.2 K

Av = − g m ( RD RL )

( 0.8 )( 0.25 ) = 0.8944 mA/V Av = − ( 0.8944 )(10.2 || 2 )

g m = 2 K p I DQ = 2 Av = −1.50

(c)

ΔVO = ΔI ( RD || RL ) = 0.25 (10.2 || 2 ) = 0.418 So ΔVO = 0.836 peak-to-peak

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.26

I DQ = K n (VGSQ − VTN )

2

g m = 2 K n I DQ 2.2 = 2 K n ( 6 ) ⇒ K n = 0.202 mA / V 2 6 = 0.202 ( 2.8 − VTN ) ⇒ VTN = −2.65 V 2

VDSQ = 18 − I DQ ( RS + RD ) 18 − 10 = 1.33 kΩ ⇒ RS = 1.33 − RD 6 g m ( RD RL )

RS + RD = Av = −

1 + g m RS

⎛ R ⋅1 ⎞ −2.2 ⎜ D ⎟ ⎝ RD + 1 ⎠ −1 = 1 + ( 2.2 )(1.33 − RD ) 1 + 2.93 − 2.2 RD =

2.2 RD 1 + RD

( 3.93 − 2.2 RD )(1 + RD ) = 2.2 RD 3.93 + 1.73RD − 2.2 RD2 = 2.2 RD 2.2 RD2 + 0.47 RD − 3.93 = 0 −0.47 +

RD =

VG = VGS + VS

( 0.47 ) + 4 ( 2.2 )( 3.93) ⇒ RD = 1.23 kΩ, 2 ( 2.2 ) = 2.8 + ( 6 )( 0.1) = 3.4 V 2

RS = 0.10 kΩ

1 1 ⋅ Rin ⋅ VDD = (100 )(18 ) = 3.4 ⇒ R1 = 529 kΩ R1 R1

VG =

529 R2 = 100 ⇒ R2 = 123 kΩ 529 + R2 ______________________________________________________________________________________

4.27 (a) V S = V SGQ = V SDQ + I DQ R D − 9

(

V SGQ = 5 + I DQ (4) − 9 = K p (4) V SGQ + VTP

(

V SGQ = 8 V

Or 8V

2 SGQ

2 SGQ

)

)

2

−4

− 2.4V SGQ + 1.44 − 4

− 20.2V SGQ + 7.52 = 0 ⇒ V SGQ = 2.071 V

(

I DQ = I Q = K p V SGQ + VTP

)

2

= 2(2.071 − 1.2 ) = 1.518 mA 2

(b) g m = 2 K p I DQ = 2 (2)(1.518) = 3.485 mA/V ro =

1 1 = = 22 k Ω λ I DQ (0.03)(1.518)

Aυ = − g m (R D ro ) = −(3.485)(4 22) = −11.8

(

)

(

)

(c) Aυ = − g m R D ro R L = −(3.485 ) 4 22 8 = −8.29 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.28 (a)

I DQ = K p (VSGQ + VTP )

2

0.5 = 0.25 (VSGQ + 0.8 )

2

VSGQ = 0.614 V = VS 10 − 0.614 RS = ⇒ RS = 18.8 K 0.5 VD = VS − VSDQ = 0.614 − 3 = −2.386 V RD =

−2.386 − ( −10 ) 0.5

⇒ RD = 15.2 K

(b) Av = − g m RD

( 0.25)( 0.5 ) = 0.7071 mA/V Av = − ( 0.7071)(15.2 )

g m = 2 K p I DQ = 2 Av = −10.7

______________________________________________________________________________________ 4.29

Av = − g m ( RD RL ) VDSQ = VDD − I DQ ( RS + RD )

10 = 20 − (1)( RS + RD ) ⇒ RS + RD = 10 kΩ

Let

RD = 8 kΩ, RS = 2 kΩ Aυ = −10 = − g m (8 20) g m = 1.75 mA/V = 2 K n I DQ = 2 K n (1) ⇒ K n = 0.766 mA/V 2

VS = I DQ RS = (1)( 2 ) = 2 V I DQ = K n (VGS − VTN ) ⇒ 1 = 0.766 (VGS − 2 ) ⇒ VGS = 3.14 V VG = VGS + VS = 3.14 + 2 = 5.14 2

VG =

2

1 1 ⋅ Rin ⋅ VDD ⇒ ( 200 )( 20 ) = 5.14 ⇒ R1 = 778 kΩ R1 R1

778R2 = 200 ⇒ R2 = 269 kΩ 778 + R2 ______________________________________________________________________________________ 4.30

(a) Aυ = Ro =

g m ro (5)(100) = 0.998 = 1 + g m ro 1 + (5)(100 ) 1 1 ro = 100 = 0.2 100 ⇒ R o ≅ 200 Ω gm 5

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ g m (ro R S )

(b) Aυ =

1 + g m (ro R S )

We have ro R S = 100 5 = 4.762 k Ω

(5)(4.762) = 0.960 1 + (5)(4.762) ______________________________________________________________________________________ Aυ =

4.31

Av = 0.98 =

g m ( RL ro )

1 + g m ( RL ro ) g m ro ⇒ g m ro = 49 1 + g m ro

Also 0.49 =

0.49 = 0.49 =

g m ( RL ro )

1 + g m ( RL ro ) g m ( RL ro )

⎛ Rr ⎞ gm ⎜ L o ⎟ ⎝ RL + ro ⎠ = ⎛ Rr ⎞ 1 + gm ⎜ L o ⎟ ⎝ RL + ro ⎠

RL + ro + g m ( RL ro )

( 49 )(1) 49 = 1 + ro + ( 49 )(1) 50 + ro

ro = 50 K g m = 0.98 mA/V ______________________________________________________________________________________ 4.32 (a)

Av =

( 2 )( 25) g m ro = 1 + g m ro 1 + ( 2 )( 25 )

Av = 0.98 Ro =

1 1 ro = 25 = 0.5 || 25 gm 2

Ro = 0.49 K (b) Av =

g m ( ro || RL )

1 + g m ( ro || RL )

=

2 ( 25 || 2 )

1 + 2 ( 25 || 2 )

=

2 (1.852 )

1 + 2 (1.852 )

Av = 0.787

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.33

0 − (− 1.5) = 0.15 mA 10 ⎛ k ′ ⎞⎛ W ⎞ = ⎜⎜ n ⎟⎟⎜ ⎟ VGSQ − VTN ⎝ 2 ⎠⎝ L ⎠

(a) I DQ = I DQ

(

(

⎛ 0.1 ⎞ 0.15 = ⎜ ⎟(80 ) VGSQ − 0.4 ⎝ 2 ⎠

)

2

)

2

⇒ VGSQ = 0.594 V

⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞ (b) g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜ ⎟(80)(0.15) = 1.549 mA/V ⎝ 2 ⎠ ⎝ 2 ⎠⎝ L ⎠ 1 1 ro = = = 333 k Ω λ I DQ (0.02)(0.15) We find ro R S = 333 10 = 9.708 k Ω Then Aυ = (c) R o =

g m (ro R S )

1 + g m (ro R S )

=

(1.549)(9.708) = 0.938 1 + (1.549 )(9.708)

1 1 ro R S = 333 10 = 0.6456 9.708 gm 1.549

or R o = 605 Ω ______________________________________________________________________________________ 4.34

(a) I DQ = (b) Aυ =

V DD − V DSQ RS

=

2.5 − 1.25 = 2.5 mA 0 .5

g m RS 1 + g m RS

0.85 =

g m (0.5) ⇒ g m = 11.33 mA/V 1 + g m (0.5)

⎛ k′ g m = 2 ⎜⎜ n ⎝ 2

⎞⎛ W ⎟⎟⎜ ⎠⎝ L

⎞ ⎟ I DQ ⎠

⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 11.33 = 2 ⎜ ⎟⎜ ⎟(2.5) ⇒ ⎜ ⎟ = 257 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ ⎛ k ′ ⎞⎛ W ⎞ (c) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟(VGSQ − VTN )2 ⎝ 2 ⎠⎝ L ⎠ ⎛ 0. 1 ⎞ 2 2. 5 = ⎜ ⎟(257 )(VGSQ − 0.6 ) ⇒ VGSQ = 1.041 V ⎝ 2 ⎠ V IQ = V GSQ + V O = 1.041 + 1.25 = 2.291 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.35 (a) P = I Q V DD ⇒ 2.5 = I Q (2.5) ⇒ I Q = 1 mA

(b) R o =

1 ro gm 1 1 = = 50 k Ω λ I DQ (0.02 )(1)

ro =

So 0.5 =

1 50 ⇒ g m = 1.98 mA/V gm

⎛ k′ g m = 2 ⎜⎜ n ⎝ 2

⎞⎛ W ⎞ ⎟⎟⎜ ⎟ I DQ ⎠⎝ L ⎠

⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 1.98 = 2 ⎜ ⎟⎜ ⎟(1) ⇒ ⎜ ⎟ = 19.6 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ g m ro (1.98)(50) = 0.990 (c) Aυ = = 1 + g m ro 1 + (1.98)(50 ) ⎛ 0.1 ⎞ (d) g m = 2 ⎜ ⎟(100 )(1) = 4.472 mA/V ⎝ 2 ⎠

1 1 ro = 50 = 0.2236 50 ⇒ R o = 223 Ω gm 4.472 ______________________________________________________________________________________ Ro =

4.36 ⎛ R2 ⎞ ⎛ 350 ⎞ ⎟⎟ ⋅ V DD = ⎜ (a) VG = ⎜⎜ ⎟(10 ) = 2.917 V ⎝ 350 + 850 ⎠ ⎝ R1 + R 2 ⎠ 10 = I DQ R S + V SGQ + V G ⎛ k ′p ⎞⎛ W ⎞ ⎛ 0.04 ⎞ 2 Now K p = ⎜⎜ ⎟⎟⎜ ⎟ = ⎜ ⎟(80 ) = 1.6 mA/V 2 2 L ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 So 10 = (1.6 )(4) V SGQ − 1.2 + V SGQ + 2.917

(

We find 6.4V

2 SGQ

)

− 14.36V SGQ + 2.133 = 0 ⇒ V SGQ = 2.084 V

I DQ = 1.6(2.084 − 1.2 ) = 1.25 mA 2

V SDQ = 10 − (1.25 )(4 ) = 5 V

(b) g m = 2 K p I DQ = 2 (1.6)(1.25) = 2.828 mA/V 1

ro =

λ I DQ

=

1

(0.05)(1.25)

= 16 k Ω

ro R S R L = 16 4 4 = 1.778 k Ω

Aυ =

(

g m ro R S R L

(

)

1 + g m ro R S R L

(c) Ag =

io

υi

=

)

=

(2.828)(1.778) = 0.834 1 + (2.828)(1.778)

io υ o 1 υo 1 ⋅ = ⋅ = (0.834 ) = 0.2085 mA/V υ o υ i RL υ i 4

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (d) R o =

1 1 R S ro = 4 16 = 0.3536 3.2 gm 2.828

R o = 318 Ω ______________________________________________________________________________________

4.37

(a) (i) K n =

k n′ W ⎛ 0.1 ⎞ 2 ⋅ =⎜ ⎟(20 ) = 1 mA/V 2 L ⎝ 2 ⎠

g m = 2 K n I DQ = 2 (1)(5) = 4.472 mA/V ro =

1

λ I DQ

=

1

(0.02)(5)

= 10 k Ω

ro R L = 10 4 = 2.857 k Ω Aυ =

(ii) R o =

g m (ro R L )

(4.472)(2.857 ) = 0.927 1 + g m (ro R L ) 1 + (4.472)(2.857 ) =

1 1 ro = 10 gm 4.472

R o = 219 Ω

(b) (i) g m = 2 (1)(2 ) = 2.828 mA/V

ro =

1

(0.02)(2)

= 25 k Ω

ro R L = 25 4 = 3.448 k Ω Aυ = (ii) R o =

(2.828)(3.448) = 0.907 1 + (2.828)(3.448) 1 1 ro = 25 gm 2.828

R o = 349 Ω ______________________________________________________________________________________

4.38 a. Av =

gm ( 4) g m RL ⇒ 0.95 = 1 + g m RL 1 + gm ( 4)

0.95 = 4 (1 − 0.95 ) g m ⇒ g m = 4.75 mA/V ⎛1 ⎞⎛W ⎞ g m = 2 ⎜ μ n Cox ⎟ ⎜ ⎟ I Q ⎝2 ⎠⎝ L ⎠ 4.75 = 2

( 0.030 ) ⎛⎜

W⎞ W = 47.0 ⎟ ( 4) ⇒ L ⎝L⎠

⎛1 ⎞⎛ W g m = 2 ⎜ μ n Cox ⎟⎜ ⎝2 ⎠⎝ L

⎞ ⎟ IQ ⎠

4.75 = 2 ( 0.030 )( 60 ) I Q ⇒ I Q = 3.13 mA b. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.39 I DQ = K n (VGS − VTN )

2

5 = 5 (VGS + 2 ) ⇒ VGS = −1 V ⇒ VS = −VGS = 1 V 2

I DQ =

VS − ( −5 ) RS

⇒ RS =

g m = 2 K n I DQ = 2 r0 = Av = =

1+ 5 ⇒ RS = 1.2 kΩ 5

( 5)( 5) = 10 mA / V

1 1 = = 20 kΩ λ I DQ ( 0.01)( 5 ) g m ( r0 RS RL )

1 + g m ( r0 RS RL )

(10 ) ( 20 1.2 2 ) ⇒ Av = 0.878 1 + (10 ) ( 20 1.2 2 )

1 1 || r0 || RS = || 20 ||1.2 ⇒ Ro = 91.9 Ω gm 10 ______________________________________________________________________________________ R0 =

4.40 (a) V S = I DQ R S − 5 = (5)(1) − 5 = 0 ⇒ VGS = 0 ⎛ k ′ ⎞⎛ W ⎞ 2 I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ VGSQ − VTN 2 L ⎝ ⎠⎝ ⎠ 0 . ⎛ 1 ⎞⎛ W ⎞ ⎛W ⎞ 2 5=⎜ ⎟⎜ ⎟[0 − (− 2 )] ⇒ ⎜ ⎟ = 25 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠

(

)

⎛ 0. 1 ⎞ (b) g m = 2 ⎜ ⎟(25)(5) = 5 mA/V ⎝ 2 ⎠ 1 1 = = 20 k Ω λ I DQ (0.01)(5)

ro =

ro R S = 20 1 = 0.9524 k Ω g m (ro R S )

Aυ =

1 + g m (ro R S )

=

(5)(0.9524) = 0.826 1 + (5)(0.9524 )

1 1 ro R S = 20 1 ⇒ Ro = 165 Ω gm 5

(c) Ro =

(d) ro R S R L = 20 1 2 = 0.6452 k Ω Aυ =

(

g m ro R S R L

(

)

1 + g m ro R S R L

)

=

(5)(0.6452 ) = 0.763 1 + (5)(0.6452 )

______________________________________________________________________________________ 4.41 R0 =

1 RS gm

Output resistance determined primarily by gm

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 Set = 0.2 kΩ ⇒ g m = 5 mA/V gm g m = 2 K n I DQ ⇒ 5 = 2 I DQ = K n (VGS − VTN )

( 4 ) I DQ

⇒ I DQ = 1.56 mA

2

1.56 = 4 (VGS + 2 ) VGS = −1.38 V, VS = −VGS = 1.38 V 2

RS =

1.38 − ( −5 )

⇒ RS = 4.09 kΩ 1.56 5 ( 4.09 2 ) g m ( RS RL ) = ⇒ Av = 0.870 Av = 1 + g m ( RS RL ) 1 + 5 ( 4.09 2 )

______________________________________________________________________________________ 4.42

(a) g m = 2 K p I DQ = 2 (5)(10) = 14.14 mA/V ro =

Aυ

1

=

1

= 10 k Ω

(0.01)(10) g m ro (14.14)(10) = 0.993 = = 1 + g m ro 1 + (14.14 )(10 )

(b) R o = (c) Aυ =

λ I DQ

1 1 ro = 10 = 0.07072 10 ⇒ R o = 70.2 Ω gm 14.14 g m (ro R L )

1 + g m (ro R L )

0.90 =

(14.14)(ro R L ) ⇒ (ro 1 + (14.14 )(ro R L )

R L ) = 0.6365 k Ω

10 R L = 0.6365 ⇒ R L = 680 Ω ______________________________________________________________________________________ 4.43

ΔiD = I DQ =

−1 ⋅ Δv0 RS RL

Δv0 = − I DQ ⋅ RS RL = − v0 ( min ) = −

Av = vi =

I DQ ⋅ RS RL RS + RL

I DQ RS R 1+ S RL g m ( RS RL )

1 + g m ( RS RL )

=

v0 vi

− I DQ ( RS RL ) ⎡⎣1 + g m ( RS RL ) ⎤⎦ g m ( RS RL ) I DQ

⎡1 + g m ( RS RL ) ⎤⎦ gm ⎣ ______________________________________________________________________________________ vi ( min ) = −

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.44 (a) VDD = VDSQ + I DQ RS

3 = 1.5 + ( 0.25 ) RS ⇒ RS = 6 K

VS = 1.5 V I DQ = K n (VGSQ − VTN ) 0.25 = 0.5 (VGSQ − 0.4 )

2

2

VGSQ = 1.107 V

VG = VGSQ + VS + 1.107 + 1.5 = 2.607 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = − RL − VDD R1 ⎝ R1 + R2 ⎠ 1 2.607 = ( 300 )( 3) ⇒ R1 = 345.2 K ⇒ R2 = 2291 K R1 (b) Av =

g m RS 1 + g m RS

g m = 2 K n I DQ = 2

( 0.5)( 0.25) = 0.7071 mA/V

Av =

( 0.7071)( 6 ) ⇒ Av = 0.809 1 + ( 0.7071)( 6 )

Ro =

1 1 6 = 1.414 || 6 RS = gm ( 0.7071)

Ro = 1.14 K ______________________________________________________________________________________ 4.45 Ri =

1 1 = 0.5 ⇒ g m = = 2 mA/V gm 0.5

⎛ k′ g m = 2 ⎜⎜ n ⎝ 2

⎞⎛ W ⎟⎟⎜ ⎠⎝ L

⎞ ⎟ I DQ ⎠

⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 2=2 ⎜ ⎟⎜ ⎟(0.25) ⇒ ⎜ ⎟ = 80 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ Aυ = g m R D

20 = (2 )R D ⇒ R D = 10 k Ω ______________________________________________________________________________________

4.46 (a) R o = R D = 500 Ω (b) VGSQ = 1.2 V

V DD − V DS 2.2 − (V DS (sat ) + 0.3) 2.2 − (1.2 − 0.4 + 0.3) = = RD 0.5 0.5 = 2.2 mA

I DQ = I DQ

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(

(c) I DQ = K n VGSQ − VTN

)

2

2.2 = K n (1.2 − 0.4) ⇒ K n = 3.438 mA/V 2 2

g m = 2 K n I DQ = 2 (3.438)(2.2) = 5.5 mA/V Ri =

1 1 = ⇒ Ri = 182 Ω g m 5.5

(d) Aυ = g m R D = (5.5)(0.5) = 2.75 ______________________________________________________________________________________ 4.47 ⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞ 2 K n = ⎜⎜ n ⎟⎟⎜ ⎟ = ⎜ ⎟(80 ) = 4 mA/V ⎝ 2 ⎠⎝ L ⎠ ⎝ 2 ⎠

g m = 2 K n I DQ = 2 (4)(0.5) = 2.828 mA/V Aυ = g m R D = (2.828)(4) = 11.3

Ri = 10

1 1 = 10 = 10 0.3536 ⇒ Ri = 342 Ω gm 2.828

______________________________________________________________________________________ 4.48 a. VGS + I DQ RS = 5 5 − VGS 2 I DQ = = K n (VGS − VTN ) RS

5 − VGS = (10 )( 3) (VGS2 − 2VGS + 1) 30VGS2 − 59VGS + 25 = 0

VGS =

59 ±

( 59 )

2

− 4 ( 30 )( 25 )

2 ( 30 )

⇒ VGS = 1.35 V

I DQ = ( 3)(1.35 − 1) ⇒ I DQ = 0.365 mA 2

VDSQ = 10 − ( 0.365 )( 5 + 10 ) ⇒ VDSQ = 4.53 V

b. g m = 2 K n I DQ = 2 r0 =

1

λ I DQ

=

1

( 3)( 0.365 ) ⇒ g m = 2.093 mA / V ⇒r =∞

( 0 )( 0.365 ) 0 RL ) = ( 2.093) ( 5 4 ) ⇒ Av = 4.65

A = g m ( RD c. v ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.49 a.

I DQ = K p (VSG + VTP )

2

0.75 = ( 0.5 )(VSG − 1) ⇒ VSG = 2.225 V 5 − 2.225 5 = I DQ RS + VSG ⇒ RS = ⇒ RS = 3.70 kΩ 0.75 VSDQ = 10 − I DQ ( RS + RD ) 6 = 10 − ( 0.75 )( 3.70 + RD ) ⇒ RD = 1.63 kΩ 2

b. Ri =

1 gm

g m = 2 K p I DQ = 2

( 0.5 )( 0.75) = 1.225 mA / V

1 ⇒ Ri = 0.816 kΩ 1.225 Ro = RD ⇒ Ro = 1.63 kΩ Ri =

c. ⎞ ⎞⎛ RS ⎟⎟ ⋅ ii ⎟ ⎜⎜ ⎠ ⎝ RS + [1/ g m ] ⎠ 3.70 ⎛ 1.63 ⎞ ⎛ ⎞ i0 = ⎜ ⎟⎜ ⎟ ii ⎝ 1.63 + 2 ⎠ ⎝ 3.70 + 0.816 ⎠ i0 = 0.368ii = i0 = 1.84sin ω t ( μ A ) ⎛ RD i0 = ⎜ ⎝ RD + RL

v0 = i0 RL = (1.84 )( 2 ) sin ω t ⇒ v0 = 3.68sin ω t ( mV )

______________________________________________________________________________________ 4.50 (a) VO = (V DS (sat ) + 0.25) − VGS VO = VGS − VTN + 0.25 − VGS = −0.4 + 0.25 = −0.15 V 1.8 − (− 0.15) = 0.975 k Ω 2 (b) Aυ = g m R D 6 = g m (0.975) ⇒ g m = 6.154 mA/V RD =

⎛ k′ g m = 2 ⎜⎜ n ⎝ 2

⎞⎛ W ⎟⎟⎜ ⎠⎝ L

⎞ ⎟ I DQ ⎠

⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 6.154 = 2 ⎜ ⎟⎜ ⎟(2 ) ⇒ ⎜ ⎟ = 94.7 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ ⎛ k ′ ⎞⎛ W ⎞ (c) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟(VGSQ − VTN )2 ⎝ 2 ⎠⎝ L ⎠ ⎛ 0. 1 ⎞ 2 2=⎜ ⎟(94.7 )(VGSQ − 0.4 ) ⇒ VGSQ = 1.05 V 2 ⎝ ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.51 2 (a) I DQ = I Q = K n VGSQ − VTN

(

(

2 = 4 VGSQ − 0.6

)

2

)

⇒ VGSQ = 1.307 V

V D = V DSQ − V GSQ = 3.5 − 1.307 = 2.193 V

RD =

3.3 − 2.193 = 0.554 k Ω 2

(b) g m = 2 K n I DQ = 2 (4)(2) = 5.657 mA/V Ri =

1 1 = ⇒ Ri = 177 Ω g m 5.657

(

)

(

)

(c) Aυ = g m R D R L = (5.657) 0.554 4 = 2.75 ______________________________________________________________________________________ 4.52 (a) I DQ = K p V SGQ + VTP

(

(

)

)

2

1.2 = 2.5 V SGQ − 0.8 ⇒ V SGQ = 1.493 V RS =

V + − V SGQ I DQ

2

=

3.3 − 1.493 = 1.51 k Ω 1. 2

V SDQ = 6.6 − I DQ (R S + R D ) ⇒ 3 = 6.6 − 1.2(1.51 + R D ) ⇒ R D = 1.49 k Ω

(b) g m = 2 K p I DQ = 2 (2.5)(1.2) = 3.464 mA/V

Aυ = g m (R D R L ) = (3.464)(1.49 4) = 3.76

______________________________________________________________________________________ 4.53

(a)

KD = KL

Aυ =

(W L )D (W L )L

=5

⎛W ⎞ So ⎜ ⎟ = 25 ⎝ L ⎠D From Example 4.11, (3.3 − 0.6) + (0.6)(1 + 5) = 1.05 V υ GSDt = 1+ 5 1.05 − 0.6 υ GSDQ = + 0.6 = 0.825 V 2 ⎛ k ′ ⎞⎛ W ⎞ ⎛ 0 .1 ⎞ 2 (b) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (VGSDQ − VTN )2 = ⎜ ⎟(25)(0.825 − 0.6 ) 2 L 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ D I DQ = 0.0633 mA

Now I DD = I DL

(

⎛W ⎞ ⎜ ⎟ VGSDQ − VTN ⎝ L ⎠D

(

)

2

(

⎛W ⎞ = ⎜ ⎟ VGSLQ − VTN ⎝ L ⎠L

)

2

)

25 VGSDQ − VTN = V DD − VO − VTN 1 5(0.825 − 0.6 ) = 3.3 − VO − 0.6 Or V DSDQ = VO = 1.575 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.54 (a) Neglect λ in dc analysis. Transition points: For point B, VOtB = V DD − VTNL = 5 − 0.8 = 4.2 V

For point A, I DD = I DL

( (1.2)(V

)

K nD VGSDQ − VTND GSDQ

− 0 .6

2

(

= K nL VGSLQ − VTNL

) = (0.2)[0 − (− 0.8)] 2

)

2

2

0. 2 (0.8) + 0.6 = 0.9266 V 1. 2 = VGSDQ − VTND = 0.9266 − 0.6 = 0.3266 V

So VGSDQ = Then VOtA

For point A: VOtA = 0.3266 V, V GSDQ = 0.9266 V For point B: VOtB = 4.2 V, V GSDQ = 0.9266 V (b) V GSDQ = 0.9266 V, 4.2 − 0.3266 + 0.3266 = 2.2633 V 2 2 2 = K nD VGSDQ − VTND = (1.2 )(0.9266 − 0.6) = 0.128 mA

V DSDQ =

(c) I DQ

(d) Aυ = − g mD

( (r

roD = roL =

oD

)

roL )

1 1 = = 390.6 k Ω λ I DQ (0.02 )(0.128)

g mD = 2 K nD I DQ = 2 (1.2)(0.128) = 0.7838 mA/V Aυ = −(0.7838)(390.6 390.6) = −153

______________________________________________________________________________________ 4.55 (a) VTN = 0.6 V

I D = K n (V DS − VTN )

2

0.5 = K n (1.5 − 0.6) ⇒ K n = 0.6173 mA/V 2 1 dI D = = 2 K n (V DS − VTN ) r dV DS 2

Then r =

1

2 K n (V DS − VTN )

=

1 ⇒ r = 900 Ω 2(0.6173)(1.5 − 0.6 )

(b) I D = (0.6173)(3 − 0.6) = 3.56 mA 2

1 ⇒ r = 337 Ω 2(0.6173)(3 − 0.6) ______________________________________________________________________________________ r=

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.56 a. I DQ = K nD (VGS − VTND ) = ( 0.5 ) ( 0 − ( −1) ) 2

2

I DQ = 0.5 mA I DQ = K nL (VGSL − VTNL ) = K nL (VDD − VO − VTNL ) 2

0.5 = 0.030 (10 − V0 − 1)

2

2

0.5 = 9 − V0 ⇒ V0 = 4.92 V 0.030

b.

I DD = I DL K nD (Vi − VTND ) = K nL (VDD − Vo − VTNL ) 2

2

K nD (Vi − VTND ) = VDD − Vo − VTNL K nL K nD (Vi − VTND ) K nL

Vo = VDD − VTNL − Av =

dVo K nD =− =− dVi K nL

(W / L )D (W / L )L

500 ⇒ Av = −4.08 30 ______________________________________________________________________________________ Av = −

4.57 (a)

I DQ = K L (VGSL − VTNL ) = K L (VDSL − VTNL ) 2

2

I D = ( 0.1)( 4 − 1) = 0.9 mA 2

I DQ = K D (VGSD − VTND )

2

0.9 = (1)(VGSD − 1) ⇒ VGSD = 1.95 V VGG = VGSD + VDSL = 1.95 + 4 ⇒ VGG = 5.95 V 2

b.

I DD = I DL K D (VGSD − VTND ) = K L (VGSL − VTNL ) 2

2

KD (VGG + Vi − Vo − VTND ) = Vo − VTNL KL ⎛ KD ⎞ Vo ⎜ 1 + ⎟= ⎜ K L ⎟⎠ ⎝ Av =

KD (VGG + Vi − VTND ) + VTNL KL

KD / KL dVo = ⇒ dVi 1 + K D / K L

Av =

1 1 + KL / KD

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) From Problem 4.55. 1 RLD = 2 K L (VDSL − VTNL ) 1

=

2 ( 0.1)( 4 − 1)

= 1.67 k Ω

g m = 2 K D I DQ = 2 (1)( 0.9 ) = 1.90 mA / V Av =

g m ( RLD || RL )

L

(1.90 )(1.67 || 4 ) ⇒ Av = 0.691 1 + g m ( RLD || RL ) 1 + (1.90 )(1.67 || 4 ) =

______________________________________________________________________________________ 4.58 a.

From Problem 4.57. g m ( RLD || RL ) (1.90 )(1.67 || 10 ) = Av = 1 + g m ( RLD || RL ) 1 + (1.90 )(1.67 || 10 ) Av = 0.731

b.

R0 =

1 1 1.67 = 0.526 ||1.67 RLD = 1.90 gm

R0 = 0.40 kΩ ______________________________________________________________________________________ 4.59 (a) Aυ = − g mD roD roL

(

)

roD =

1 1 = = 100 k Ω λ D I DQ (0.02 )(0.5)

roL =

1 1 = = 50 k Ω λ L I DQ (0.04 )(0.5)

roD roL = 100 50 = 33.33 k Ω

Then − 40 = − g mD (33.33) ⇒ g mD = 1.20 mA/V ⎛ k ′ ⎞⎛ W ⎞ g mD = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ ⎝ 2 ⎠⎝ L ⎠ D

( )

⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 1.20 = 2 ⎜ ⎟⎜ ⎟ (0.5) ⇒ ⎜ ⎟ = 14.4 ⎝ L ⎠D ⎝ 2 ⎠⎝ L ⎠ D

⎛ k ′p ⎞⎛ W ⎞ 2 (b) I DQ = ⎜⎜ ⎟⎟⎜ ⎟ V SGQ + VTP L 2 ⎝ ⎠⎝ ⎠ L ⎛ 0.04 ⎞ 2 0.5 = ⎜ ⎟(50 )(V SGQ − 0.4 ) ⇒ V SGQ = 1.107 V 2 ⎝ ⎠

(

)

V SGQ = V + − V B 1.107 = 2.5 − V B ⇒ V B = 1.393 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ k ′ ⎞⎛ W ⎞ (c) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (VGSDQ − VTN )2 ⎝ 2 ⎠⎝ L ⎠ D ⎛ 0.1 ⎞ 2 0.5 = ⎜ ⎟(14.4)(VGSDQ − 0.4) ⇒ VGSDQ = 1.233 V 2 ⎝ ⎠ ______________________________________________________________________________________

4.60

(a) (i) roD =

1 1 = = 100 k Ω λ D I DQ (0.04)(0.25)

roL =

1 1 = = 200 k Ω λ L I DQ (0.02 )(0.25)

roD roL = 100 200 = 66.67 k Ω Aυ = − g mD (roD roL )

− 25 = − g mD (66.67 ) ⇒ g mD = 0.375 mA/V

⎛ k ′p g mD = 2 ⎜⎜ ⎝ 2

⎞⎛ W ⎞ ⎟⎜ ⎟ I DQ ⎟⎝ L ⎠ D ⎠

⎛ 0.04 ⎞⎛ W ⎞ ⎛W ⎞ 0.375 = 2 ⎜ ⎟⎜ ⎟ (0.25) ⇒ ⎜ ⎟ = 7.03 2 L ⎝ ⎠⎝ ⎠ D ⎝ L ⎠D ⎛ k ′ ⎞⎛ W ⎞ (ii) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (V B − VTN )2 ⎝ 2 ⎠⎝ L ⎠ L ⎛ 0.1 ⎞ 2 0.25 = ⎜ ⎟(10 )(V B − 0.4 ) ⇒ V B = 1.107 V ⎝ 2 ⎠

⎛ k ′p ⎞⎛ W ⎞ 2 (iii) I DQ = ⎜⎜ ⎟⎟⎜ ⎟ V SGDQ + VTP L 2 ⎝ ⎠ D ⎝ ⎠ ⎛ 0.04 ⎞ 2 0.25 = ⎜ ⎟(7.03)(V SGDQ − 0.6) ⇒ V SGDQ = 1.933 V ⎝ 2 ⎠

(

)

(b) (i) roD =

1 1 = = 250 k Ω λ D I DQ (0.04 )(0.1)

roL =

1 1 = = 500 k Ω λ L I DQ (0.02 )(0.1)

roD roL = 250 500 = 166.7 k Ω

Aυ = − g mD (roD roL )

− 25 = − g mD (166.7 ) ⇒ g mD = 0.15 mA/V ⎛ k ′p g mD = 2 ⎜⎜ ⎝ 2

⎞⎛ W ⎞ ⎟⎜ ⎟ I DQ ⎟⎝ L ⎠ D ⎠

⎛ 0.04 ⎞⎛ W ⎞ ⎛W ⎞ 0.15 = 2 ⎜ ⎟⎜ ⎟ (0.1) ⇒ ⎜ ⎟ = 2.81 ⎝ 2 ⎠⎝ L ⎠ D ⎝ L ⎠D

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ k ′ ⎞⎛ W ⎞ (ii) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (V B − VTN )2 ⎝ 2 ⎠⎝ L ⎠ L ⎛ 0.1 ⎞ 2 0.1 = ⎜ ⎟(10 )(V B − 0.4 ) ⇒ V B = 0.874 V 2 ⎝ ⎠

⎛ k ′p ⎞⎛ W ⎞ 2 (iii) I DQ = ⎜⎜ ⎟⎟⎜ ⎟ V SGDQ + VTP L 2 ⎝ ⎠⎝ ⎠ D ⎛ 0.04 ⎞ 2 0.1 = ⎜ ⎟(2.81)(V SGDQ − 0.6 ) ⇒ V SGDQ = 1.934 V 2 ⎝ ⎠ ______________________________________________________________________________________

(

)

4.61 ⎛ 85 ⎞ K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2 ⎝ 2⎠

( 2.125)( 0.1) = 0.9220

g m1 = 2 K n1 I D1 = 2

ro1 = ro 2 =

1 1 = = 200 K λ1 I D1 ( 0.05 )( 0.1) 1

1

=

= 133.3 K

( 0.075)( 0.1) Av = − g m1 ( ro1 || ro 2 ) = − ( 0.922 )( 200 || 133.3) λ2 I D 2

Av = −73.7

______________________________________________________________________________________ 4.62 K p1 =

k p′ ⎛ w ⎞ ⎛ 40 ⎞ 2 ⎜ ⎟ = ⎜ ⎟ ( 50 ) ⇒ 1.0 mA/V 2 ⎝ L⎠ ⎝ 2 ⎠

g m1 = 2 K p1 I D1 = 2 (1)( 0.1) = 0.6325 mA/V ro1 = ro 2 =

1

λ1 I D1

=

1 = 133.3 K 0.075 ( )( 0.1)

1 1 = = 200 K λ2 I D 2 ( 0.05 )( 0.1)

Av = − g m1 ( ro1 ro 2 ) = − ( 0.6325 )(133.3 || 200 ) Av = −50.6 ______________________________________________________________________________________ 4.63 (a) I DD = I DL

K nD (VGSD − VTND ) = K nL (VGSL − VTNL ) 2

2 (V I − VO − 0.4) = VO − 0.4 0.5 2V I − 0.8 = 3VO − 0.4 ⎛2⎞ ⎛1⎞ VO = ⎜ ⎟V I − ⎜ ⎟(0.4 ) ⎝3⎠ ⎝3⎠

2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For V I = 0.8 V, VO = 0.4 V For V I = 2.5 V, V O= 1.533 V (b) I D = K nL (VGSL − VTNL ) = (0.5)(VO − 0.4) 2

2

2

2

⎡⎛ 2 ⎞ ⎡⎛ 2 ⎞ ⎤ ⎤ ⎛1⎞ I D = (0.5)⎢⎜ ⎟V I − ⎜ ⎟(0.4 ) − 0.4⎥ = (0.5)⎢⎜ ⎟V I − 0.533⎥ ⎝ 3⎠ ⎣⎝ 3 ⎠ ⎦ ⎣⎝ 3 ⎠ ⎦ For V I = 0.8 V, I D = 0 ; For V I = 2.5 V, I D = 0.642 mA (c) From (a), voltage gain = constant = 2/3 = 0.667 ______________________________________________________________________________________ 4.64 (a) V SD (sat ) = V SG + VTP = (2.5 − 1) − 0.4 = 1.1 V

V SD = V + − VO ⇒ VO (max ) = V + − V SD (sat ) = 2.5 − 1.1 = 1.4 V

⎛ k ′p ⎞⎛ W ⎞ ⎛ 0.04 ⎞ 2 2 (b) I D = ⎜⎜ ⎟⎟⎜ ⎟ (V SGL + VTP ) = ⎜ ⎟(5)(2.5 − 1 − 0.4 ) = 0.121 mA 2 L 2 ⎝ ⎠ ⎝ ⎠⎝ ⎠ L ′ k ⎛ p ⎞⎛ W ⎞ 2 (c) I D = ⎜⎜ ⎟⎟⎜ ⎟ (V SGD + VTP ) L 2 ⎝ ⎠⎝ ⎠ D ⎛ 0.04 ⎞ 2 0.121 = ⎜ ⎟(50 )(V SGD − 0.4 ) ⇒ V SGD = 0.748 V 2 ⎝ ⎠ ⎛ k ′p (d) g mD = 2 ⎜⎜ ⎝ 2

roD = roL Aυ =

⎞⎛ W ⎟⎜ ⎟⎝ L ⎠ 1 = = λID

⎛ 0.04 ⎞ ⎞ ⎟ ID = 2 ⎜ ⎟(50 )(0.121) = 0.6957 mA/V ⎠D ⎝ 2 ⎠

1 = 330.6 k Ω (0.025)(0.121)

g mD (roD roL )

1 + g mD (roD

(0.6957 )(165.3) = 0.9914 roL ) 1 + (0.6957 )(165.3) =

______________________________________________________________________________________ 4.65 (a) I DQ = K n VGSDQ − VTN

(

(

1 = 2 VGSDQ − 0.6

(

)

I DQ = K p V SGLQ + VTP

(

2

⇒ VGSDQ = 1.307 V

2

1 = 0.5 V SGLQ − 0.6

)

)

2

)

2

⇒ V SGLQ = 2.014 V

VO = 3.3 − 2.014 = 1.286 V V DSDQ = VO − V S = 1.286 − (− 1.307 ) = 2.593 V

(b) I d = I l g mDVi = g mLVo

Aυ =

Vo g mD = = Vi g mL

Kn Kp

2 =2 0. 5 ______________________________________________________________________________________

(c) Aυ =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.66 (a) ⎛ 85 ⎞ K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2 ⎝ 2⎠ g m1 = 2 K n1 I D1 = 2 ro1 = ro 2 =

1

λ1 I D1

=

( 2.125 )( 0.1) = 0.922 mA/V

1 = 200 K 0.05 ( )( 0.1)

1 1 = = 133.3 K λ2 I D 2 ( 0.075 )( 0.1)

(b) Ri1 =

1 1 = = 1.085 K g m1 0.922

Ri1 ⎛ ⎞ 1.085 ⎛ ⎞ Vgs1 = − ⎜ ⎟ Vi = − ⎜ ⎟ Vi = −0.956Vi + + R 0.050 1.085 0.050 ⎝ ⎠ ⎝ i1 ⎠ Vgs1 = + ( 0.956 )( 0.922 )( 200 )(133.3) Av = − g m1 ( ro1 ro 2 ) ⋅ Vi Av = 70.5 Ri = 0.05 +

(c)

1 1 = 0.05 + ⇒ Ri = 1.135 K 0.922 g m1

Ro ≈ ro1 ro 2 = 200 133.7 ⇒ Ro ≈ 80 K (d) ______________________________________________________________________________________ 4.67 (a) g m1 = 2 K n I D1 = 2

( 2 )( 0.1) = 0.8944 mA/V

gm2 = 2 K p I D 2 = 2

( 2 )( 0.1) = 0.8944 mA/V

1 1 = = 100 K λ I D ( 0.1)( 0.1) (b) The small-signal equivalent circuit ro1 = ro 2 =

g m1Vi +

(1)

Vsg 2 ro1

+ g m 2Vsg 2 +

Vsg 2 − Vo ro 2

=0

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (2) Vo Vo − Vsg 2 + = g m 2Vsg 2 ro ro 2 ⎛1 1 ⎞ ⎛ 1 ⎞ Vo ⎜ + + gm2 ⎟ ⎟ = Vsg 2 ⎜ r r r ⎝ o o2 ⎠ ⎝ o2 ⎠ 1 ⎞ ⎛ 1 ⎛ 1 ⎞ Vo ⎜ + + 0.8944 ⎟ ⇒ Vsg 2 = Vo ( 0.03317 ) ⎟ = Vsg 2 ⎜ 50 100 100 ⎝ ⎠ ⎝ ⎠

(1) ⎛ 1 1 ⎞ V g m1Vi + Vsg 2 ⎜ + g m 2 + ⎟ = o ro 2 ⎠ ro 2 ⎝ ro1 1 ⎞ Vo ⎛ 1 0.8944 Vi + Vo ( 0.03317 ) ⎜ + 0.8944 + ⎟= 100 ⎠ 100 ⎝ 100 0.8944 Vi = Vo ( 0.01 − 0.03033) Vo = −44 Vi

(c) For output resistance, set Vi = 0.

g m 2Vsg 2 + I x =

(1) Vsg 2

(2) (2)

ro1

Vx Vx − Vsg 2 + ro ro 2

+ g m 2Vsg 2 +

Vsg 2 − Vx ro 2

=0

⎛ 1 1 ⎞ V Vsg 2 ⎜ + g m 2 + ⎟ = x r r ro 2 o2 ⎠ ⎝ o1 1 ⎞ Vx ⎛ 1 Vsg 2 ⎜ + 0.8944 + ⎟= 100 ⎠ 100 ⎝ 100 Vsg 2 = Vx ( 0.010936 )

(1) ⎛1 1 ⎞ ⎛ 1 ⎞ I x = Vx ⎜ + ⎟ − Vsg 2 ⎜ + g m 2 ⎟ ⎝ ro ro 2 ⎠ ⎝ ro 2 ⎠ 1 ⎞ ⎛ 1 ⎛ 1 ⎞ I x = Vx ⎜ + + 0.8944 ⎟ ⎟ − Vx ( 0.010936 ) ⎜ ⎝ 50 100 ⎠ ⎝ 100 ⎠ I x = Vx ( 0.03 − 0.0098905 ) V Ro = x = 49.7 K Ix ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.68 (a) I DQ1 = K n1 VGSQ1 − VTN 1

(

(

0.2 = 0.2 VGSQ1 − 0.6

(

)

)

⇒ VGSQ1 = 1.6 V

2

I DQ 2 = K p 2 V SGQ 2 + VTP 2

(

0.5 = 1.0 V SGQ 2 − 0.6

)

2

2

)

2

⇒ V SGQ 2 = 1.307 V

0.6 = 3kΩ 0.2 = 0.6 + 2 = 2.6 V

R S1 =

V D1

5 − 2 .6 = 12 k Ω 0. 2 = V GSQ1 + 0.6 = 1.6 + 0.6 = 2.2 V

R D1 = V G1

⎛ R2 VG1 = 2.2 = ⎜⎜ ⎝ R1 + R 2 Or R1 = 909 k Ω and

⎞ 1 1 ⎟⎟(5) = (400 )(5) ⋅ Rin ⋅ (5) = R1 R1 ⎠ R1 R2 = Rin = 400 k Ω ⇒ R 2 = 714 k Ω

V S 2 = V D1 + V SGQ 2 = 2.6 + 1.307 = 3.907 V

5 − 3.907 = 2.19 k Ω 0 .5 = V S 2 − 3 = 3.907 − 3 = 0.907 V

RS 2 =

VD2

RD2 =

0.907 = 1.81 k Ω 0.5

(b) g m1 = 2 K n1 I DQ1 = 2 (0.2)(0.2) = 0.4 mA/V

g m 2 = 2 K p 2 I DQ 2 = 2 (1)(0.5) = 1.414 mA/V Aυ = (− g m1 R D1 )(− g m 2 R D 2 ) = g m1 g m 2 R D1 R D 2

Aυ = (0.4 )(1.414 )(12 )(1.81) = 12.3 ______________________________________________________________________________________

4.69 2 (a) I DQ1 = K n1 (VGS1 − VTN 1 ) 0.1 = 0.2(VGS1 − 0.6 ) ⇒ VGSQ1 = 1.307 V 2

I DQ 2 = K p 2 V SGQ 2 + VTP

)

0.25 = 1.0 V SGQ 2 − 0.6

⇒ V SGQ 2 = 1.10 V

(

(

)

2

2

V G1 = V GSQ1 + I DQ1 R S 1 = 1.307 + (0.1)(1) = 1.407 V

⎛ R2 ⎞ 1 ⎟⎟ ⋅ V DD = ⋅ Rin ⋅ V DD VG1 = ⎜⎜ R1 ⎝ R1 + R 2 ⎠ 1 1.407 = (250 )(3.3) ⇒ R1 = 586 k Ω R1

R1 R2 = Rin = 250 k Ω ⇒ R 2 = 436 k Ω

V D1 = I DQ1 R S 1 + V DSQ1 = (0.1)(1) + 1.2 = 1.3 V

R D1 =

3. 3 − 1. 3 = 20 k Ω 0.1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V S 2 = V D1 + V SGQ 2 = 1.3 + 1.1 = 2.4 V 3. 3 − 2. 4 = 3. 6 k Ω 0.25 = V S 2 − V SDQ 2 = 2.4 − 1.8 = 0.6 V

RS 2 = VD2

RD2 =

0. 6 = 2. 4 k Ω 0.25

(b) g m1 = 2 K n1 I DQ1 = 2 (0.2)(0.1) = 0.2828 mA/V

g m 2 = 2 K p 2 I DQ 2 = 2 (1)(0.25) = 1.0 mA/V

Aυ = g m1 g m 2 R D1 R D 2 = (0.2828 )(1.0 )(20 )(2.4 ) = 13.6 ______________________________________________________________________________________

4.70 ⎛ 0.04 ⎞ 2 K p1 = ⎜ ⎟(20 ) = 0.4 mA/V 2 ⎝ ⎠ ⎛ 0.1 ⎞ 2 K n2 = ⎜ ⎟(80 ) = 4.0 mA/V ⎝ 2 ⎠ 0.6 (a) R S 1 = = 6kΩ 0.1 V D1 = 1.8 − 0.6 − 1 = 0.2 V R D1 =

0.2 − (− 1.8) = 20 k Ω 0.1

I DQ1 = K p1 V SGQ1 + VTP

)

0.1 = 0.4 V SGQ1 − 0.4

⇒ V SGQ1 = 0.90 V

(

(

(

)

2

2

I DQ 2 = K n 2 VGSQ 2 − VTN

(

0.3 = 4 VGSQ 2 − 0.4

)

2

)

2

⇒ VGSQ 2 = 0.6739 V

V G1 = 1.8 − 0.6 − V SGQ1 = 1.8 − 0.6 − 0.9 = 0.3 V

⎛ R2 ⎞ ⎟⎟(3.6 ) − 1.8 VG1 = ⎜⎜ ⎝ R1 + R 2 ⎠ 1 0.3 = (200)(3.6) − 1.8 ⇒ R1 = 343 k Ω R1

R1 R2 = 200 k Ω ⇒ R 2 = 480 k Ω V D1 = 1.8 − 0.6 − 1.0 = 0.2 V V S 2 = V D1 − V GSQ 2 = 0.2 − 0.6739 = −0.4739 V RS 2 =

− 0.4739 − (− 1.8) = 4.42 k Ω 0. 3

⎛ − g m1 R D1 ⎞⎛ g m 2 R S 2 ⎟⎜ (b) Aυ = ⎜⎜ ⎟⎜ ⎝ 1 + g m1 R S1 ⎠⎝ 1 + g m 2 R S 2

⎞ ⎟ ⎟ ⎠

g m1 = 2 K p1 I DQ1 = 2 (0.4)(0.1) = 0.4 mA/V g m 2 = 2 K n 2 I DQ 2 = 2 (4)(0.3) = 2.191 mA/V Aυ =

− (0.4 )(20) (2.191)(4.42) ⋅ = −2.13 1 + (0.4 )(6) 1 + (2.191)(4.42)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 1 (c) R o = RS 2 = 4.42 = 0.4564 4.42 ⇒ Ro = 414 Ω g m2 2.191 ______________________________________________________________________________________ 4.71 (a)

I DQ1 =

10 − VGS1 2 = K n1 (VGS1 − VTN 1 ) RS 2

10 − VGS 1 = ( 4 )(10 ) (VGS2 1 − 4VGS 1 + 4 ) 40VGS2 1 − 159VGS 1 + 150 = 0

(159 ) − 4 ( 40 )(150 ) ⇒ VGS 1 = 2.435 V 2 ( 40 ) 2 I DQ1 = ( 4 )( 2.435 − 2 ) ⇒ I DQ1 = 0.757 mA VDSQ1 = 20 − ( 0.757 )(10 ) ⇒ VDSQ1 = 12.4 V VGS 1 =

159 ±

2

Also I DQ 2 = 0.757 mA VDSQ 2 = 20 − ( 0.757 )(10 + 5 ) ⇒ VDSQ 2 = 8.65 V (b) c.

g m1 = g m 2 = 2 KI DQ = 2

( 4 )( 0.757 ) ⇒ gm1 = gm 2 = 3.48 mA / V

V0 = − ( g m 2Vgs 2 ) ( RD RL ) Vgs 2 = ( − g m1Vgs1 − g m 2Vgs 2 ) ( RS1 RS 2 ) Vi = Vgs1 − Vgs 2 ⇒ Vgs1 = Vi + Vgs 2

Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) = − g m1 (Vi + Vgs 2 ) ( RS 1 RS 2 )

Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) + g m1Vgs 2 ( RS 1 Rs 2 ) = − g m1Vi ( RS1 RS 2 ) Vgs 2 = Av =

− g m1Vi ( RS 1 RS 2 )

1 + g m 2 ( RS 1 RS 2 ) + g m1 ( RS 1 RS 2 )

V0 g m1 g m 2 ( RS 1 RS 2 )( RD RL ) = Vi 1 + ( g m1 + g m 2 ) ( RS1 RS 2 )

( 3.48 ) (10 10 )( 5 2 ) Av = ⇒ Av = 2.42 1 + ( 3.48 + 3.48 ) (10 10 ) 2

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.72 a. I DQ = 3 mA

VS 1 = I DQ RS − 5 = ( 3)(1.2 ) − 5 = −1.4 V I DQ = K1 (VGS − VTN )

2

3 = 2 (VGS − 1) ⇒ VGS = 2.225 V 2

VG1 = VGS + VS 1 = 2.225 − 1.4 = 0.825 V ⎛ ⎞ R3 ⎛ R3 ⎞ VG1 = ⎜ ⎟ ( 5 ) ⇒ 0.825 = ⎜ ⎟ ( 5 ) ⇒ R3 = 82.5 kΩ ⎝ 500 ⎠ ⎝ R1 + R2 + R3 ⎠ VD1 = VS1 + VDSQ1 = −1.4 + 2.5 = 1.1 V VG 2 = VD1 + VGS = 1.1 + 2.225 = 3.325 V ⎛ R2 + R3 ⎞ ⎛ R2 + R3 ⎞ VG 2 = ⎜ ⎟ ( 5 ) ⇒ 3.325 = ⎜ ⎟ ( 5) R + R + R ⎝ 500 ⎠ 2 3 ⎠ ⎝ 1 R2 + R3 = 332.5 ⇒ R2 = 250 kΩ

R1 = 500 − 250 − 82.5 ⇒ R1 = 167.5 kΩ VD 2 = VD1 + VDSQ 2 = 1.1 + 2.5 = 3.6 V 5 − 3.6 RD = ⇒ RD = 0.467 kΩ 3 b. Av = − g m1 RD g m1 = 2 K n I DQ = 2

( 2 )( 3) = 4.90 mA / V

Av = − ( 4.90 )( 0.467 ) ⇒ Av = −2.29

______________________________________________________________________________________ 4.73 a.

VS 1 = I DQ RS − 10 = ( 5 )( 2 ) − 10 ⇒ VS 1 = 0 I DQ = K1 (VGS1 − VTN )

2

5 = 4 (VGS 1 − 1.5 ) ⇒ VGS1 = 2.618 V VG1 = VGS 1 + VS1 = 2.618 V = IR3 = ( 0.1) R3 ⇒ R3 = 26.2 kΩ 2

VD1 = VS 1 + VDSQ1 = 0 + 3.5 = 3.5 V VG 2 = VD1 + VGS = 3.5 + 2.62 = 6.12 V = ( 0.1)( R2 + R3 ) R2 + R3 = 61.2 kΩ ⇒ R2 = 35 kΩ

VD 2 = VD1 + VDSQ 2 = 3.5 + 3.5 = 7.0 V 10 − 7 RD = ⇒ RD = 0.6 kΩ 5 10 − 6.12 R1 = ⇒ R1 = 38.8 kΩ 0.1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b. Av = − g m1 RD

( 4 )( 5 ) = 8.944 mA / V

g m1 = 2 K n I DQ = 2

Av = − ( 8.944 )( 0.6 ) ⇒ Av = −5.37

______________________________________________________________________________________ 4.74 a.

I DQ

⎛ V ⎞ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

⎛ V ⎞ 4 = 6 ⎜⎜ 1 − GS ⎟⎟ ⎝ ( −3 ) ⎠

2

2

⎡ 4⎤ VGS = ( −3) ⎢1 − ⎥ ⇒ VGS = −0.551 V 6⎦ ⎣ VDSQ = VDD − I DQ RD 6 = 10 − ( 4 ) RD ⇒ RD = 1 kΩ

b.

2 I DSS ⎛ VGS ⎞ 2 ( 6 ) ⎛ −0.551 ⎞ 1− 1− = ⎟ ⇒ g m = 3.265 mA/V −3 ⎠ ( −VP ) ⎜⎝ VP ⎟⎠ 3 ⎜⎝ 1 1 = ⇒ r0 = 25 kΩ r0 = λ I DQ ( 0.01)( 4 ) gm =

c.

Aυ = − g m (ro R D ) = −(3.265)(25 1) ⇒ Aυ = −3.14

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.75

VGS + I DQ ( RS1 + RS 2 ) = 0 ⎛ V ⎞ I DQ = I DSS ⎜1 − GS ⎟ ⎝ VP ⎠

2

2

VGS

⎛ V ⎞ + I DSS ( RS1 + RS 2 ) ⎜1 − GS ⎟ = 0 ⎝ VP ⎠ 2

⎛ V ⎞ VGS + ( 2 )( 0.1 + 0.25 ) ⎜1 − GS ⎟ = 0 ⎝ VP ⎠ ⎛ 2 V2 ⎞ VGS + 0.7 ⎜ 1 − VGS + GS 2 ⎟ = 0 ⎜ ⎟ ⎝ ( −2 ) ( −2 ) ⎠ 0.175VGS2 + 1.7VGS + 0.7 = 0 VGS =

−1.7 ±

(1.7 ) − 4 ( 0.175)( 0.7 ) ⇒ VGS 2 ( 0.175 ) VGS ⎞ 2 ( 2 ) ⎛ −0.431 ⎞ 2

= −0.4314 V

2 I DSS ⎛ ⎜1 − ⎟= ⎜1 − ⎟ ⇒ g m = 1.569 mA/V −VP ⎝ −2 ⎠ VP ⎠ 2 ⎝ − g m ( RD RL ) − (1.569 ) ( 8 4 ) = ⇒ Av = −3.62 Av = 1 + g m RS1 1 + (1.569 )( 0.1)

gm =

i0 ( v0 / RL ) v0 RG ⎛ 50 ⎞ = = ⋅ = ( −3.62 ) ⎜ ⎟ ⇒ Ai = −45.2 ii ( vi / RG ) vi RL ⎝ 4⎠ ______________________________________________________________________________________ Ai =

4.76 I DSS = 4 mA 2 V VDSQ = DD = 10 V 2 VDSQ = VDD − I DQ ( RS + RD ) 10 = 20 − ( 4 )( RS + RD ) ⇒ RS + RD = 2.5 kΩ VS = 2 V = I DQ RS = 4 RS ⇒ RS = 0.5 kΩ, RD = 2.0 kΩ I DQ =

I DQ

⎛ V ⎞ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ ⎞ ⎛ V 4⎞ 4 = 8 ⎜⎜ 1 − GS ⎟⎟ ⇒ VGS = ( −4.2 ) ⎜⎜1 − ⎟⎟ ⇒ VGS = −1.23 V − 4.2 8 )⎠ ⎝ ⎠ ⎝ ( VG = VS + VGS = 2 − 1.23 ⎛ R2 ⎞ ⎛ R2 ⎞ VG = 0.77 V = ⎜ ⎟ ( 20 ) = ⎜ ⎟ ( 20 ) ⇒ R2 = 3.85 kΩ, R1 = 96.2 KΩ ⎝ 100 ⎠ ⎝ R1 + R2 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.77 a. I DSS = 5 mA 2 V 12 =6V VDSQ = DD = 2 2 12 − 6 ⇒ RS = 1.2 kΩ RS = 5 I DQ =

I DQ

⎛ V ⎞ = I DSS ⎜1 − GS ⎟ ⎝ VP ⎠

2

2

⎛ ⎛ V ⎞ 5 ⎞ 5 = 10 ⎜⎜1 − GS ⎟⎟ ⇒ VGS = ( −5 ) ⎜⎜1 − ⎟ ⇒ VGS = −1.464 V 10 ⎟⎠ ⎝ ⎝ ( −5 ) ⎠ VG = VS + VGS = 6 − 1.464 = 4.536 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R R R + ⎝ 1 2 ⎠ 1 1 4.536 = (100 )(12 ) ⇒ R1 = 265 kΩ R1 265R2 = 100 ⇒ R2 = 161 kΩ 265 + R2

b. gm = r0 =

2 I DSS ⎛ VGS 1− ( −VP ) ⎜⎝ VP

⎞ 2 (10 ) ⎛ −1.46 ⎞ ⎟= ⎜1 − ⎟ ⇒ g m = 2.83 mA/V −5 ⎠ 5 ⎝ ⎠

1 1 = = 20 kΩ λ I DQ ( 0.01)( 5 )

Av = Av =

(

g m r0 Rs RL

(

)

1 + g m r0 RS RL

)

( 2.83) ( 20 1.2 0.5 ) ⇒ Av = 0.495 1 + ( 2.83) ( 20 1.2 0.5 )

1 1 1.2 = 0.353 1.2 ⇒ R0 = 0.273 kΩ RS = 2.83 gm ______________________________________________________________________________________ R0 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.78 a. ⎛ R2 ⎞ ⎛ 110 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 5.5 V ⎝ 110 + 90 ⎠ ⎝ R1 + R2 ⎠ I DQ =

10 − (VG − VGS ) RS

10 − 5.5 + VGS

⎛ V ⎞ = I DSS ⎜1 − GS ⎟ ⎝ VP ⎠

⎛ V ⎞ = ( 2 )( 5 ) ⎜1 − GS ⎟ ⎝ 1.75 ⎠

2

2

4.5 + VGS = 10 (1 − 1.143VGS + 0.3265VGS2 ) 3.265VGs2 − 12.43VGS + 5.5 = 0 VGS =

12.43 ±

(12.43) − 4 ( 3.265)( 5.5 ) ⇒ VGS 2 ( 3.265 ) 2

= 0.511 V

2

⎛ 0.511 ⎞ I DQ = ( 2 ) ⎜1 − ⎟ ⇒ I DQ = 1.00 mA 1.75 ⎠ ⎝ VSDQ = 10 − (1.00 )( 5 ) ⇒ VSDQ = 5.0 V

b.

⎞ 2 ( 2 ) ⎛ 0.511 ⎞ ⎟= ⎜1 − ⎟ ⇒ g m = 1.618 mA/V 1.75 ⎠ ⎠ 1.75 ⎝ (1.618 ) ( 5 10 ) = ⇒ Av = 0.844 Av = 1 + g m ( RS RL ) 1 + (1.618 ) ( 5 10 )

gm =

⎛ VGS ⎜1 − VP ⎝ g m ( RS RL )

2 I DSS VP

⎛R ⎞ i0 ( v0 / RL ) = = Av ⋅ ⎜ i ⎟ ii ( vi / Ri ) ⎝ RL ⎠ Ri = R1 R2 = 90 110 = 49.5 kΩ Ai =

⎛ 49.5 ⎞ Ai = ( 0.844 ) ⎜ ⎟ ⇒ Ai = 4.18 ⎝ 10 ⎠

c.

Δid = 1.0 mA vsd = ( 3.33)(1.0 ) = 3.33 V

Maximum swing in output voltage = 6.66 V peak-to-peak ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.79 ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠

2

2 ⎛ 4⎞ ⎛ V ⎞ 4 = 8 ⎜ 1 − GS ⎟ ⇒ VGS = 4 ⎜⎜ 1 − ⎟ ⇒ VGS = 1.17 V 4 ⎠ 8 ⎟⎠ ⎝ ⎝ VSDQ = VDD − I DQ ( RS + RD ) 7.5 = 20 − 4 ( RS + RD ) ⇒ RS + RD = 3.125 kΩ

2 I DSS ⎛ VGS ⎞ 2 ( 8 ) ⎛ 1.17 ⎞ ⎜1 − ⎟= ⎜1 − ⎟ ⇒ g m = 2.83 mA/V 4 ⎝ 4 ⎠ VP ⎝ VP ⎠ RS = 3.125 − RD gm =

Av =

− g m RD 1 + g m RS

−3 (1 + g m RS ) = − g m RD

3 ⎡⎣1 + ( 2.83)( 3.125 − RD ) ⎤⎦ = ( 2.83) RD 9.844 − 2.83RD = 0.9433RD ⇒ RD = 2.61 kΩ RS = 0.516 kΩ VS = 20 − ( 4 )( 0.516 ) ⇒ VS = 17.94 V VG = VS − VGS = 17.94 − 1.17 = 16.77 V ⎛ R2 ⎞ ⎛ R2 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ ( 20 ) ⇒ R2 = 335 kΩ. R1 = 65 kΩ ⎝ 400 ⎠ ⎝ R1 + R2 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 5 5.1 (a) i E = (1 + β )i B ⇒ 1 + β =

325 = 116 ⇒ β = 115 2.8

β 115 = = 0.9914 1 + β 116 iC = i E − i B = 325 − 2.8 = 322 μ A

α=

(b) 1 + β =

1.80 = 90 ⇒ β = 89 0.020

89 = 0.9889 90 iC = 1.80 − 0.02 = 1.78 mA ______________________________________________________________________________________

α=

5.2 (a) α =

i C 726 = = 0.9918 i E 732

α 0.9918 = = 121 1 − α 1 − 0.9918 i B = i E − iC = 732 − 726 = 6 μ A β=

2.902 = 0.9801 2.961 α 0.980074 β= = = 49.19 1 − α 1 − 0.980074 i B = 2.961 − 2.902 ⇒ i B = 59 μ A ______________________________________________________________________________________

(b) α =

5.3 (a)

For

For (b) or

β = 110:

β = 180:

α=

α=

β 1+ β

=

110 = 0.99099 111

180 = 0.99448 181

0.99099 ≤ α ≤ 0.99448

I C = β I B = 110 ( 50 μ A ) ⇒ I C = 5.50 mA I C = 180 ( 50 μ A ) ⇒ I C = 9.00 mA

5.50 ≤ I C ≤ 9.0 mA so ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.4 (a) i B =

iE 1.25 = ⇒ i B = 8.28 μ A 1 + β 151

⎛ β ⎞ ⎛ 150 ⎞ ⎟⎟ ⋅ i E = ⎜ iC = ⎜⎜ ⎟(1.25) = 1.242 mA ⎝ 151 ⎠ ⎝1+ β ⎠ 150 α= = 0.9934 151 4.52 (b) i B = ⇒ i B = 55.8 μ A 81 ⎛ 80 ⎞ i C = ⎜ ⎟(4.52 ) = 4.46 mA ⎝ 81 ⎠ 80 = 0.9877 81 ______________________________________________________________________________________

α=

5.5 (a)

α

β=

0.9 0.95 0.98 0.99 0.995 0.999

9 19 49 99 199 999

α 1−α

(b)

β

α=

β 1+ β

20 0.9524 50 0.9804 100 0.9901 150 0.9934 220 0.9955 400 0.9975 ______________________________________________________________________________________ 5.6 (a) I B =

IE 1.2 = ⇒ I B = 14.8 μ A 1 + β 81

⎛ β ⎞ ⎛ 80 ⎞ ⎟⎟ ⋅ I E = ⎜ ⎟(1.2 ) = 1.185 mA I C = ⎜⎜ + 1 β ⎝ 81 ⎠ ⎝ ⎠ 80 α= = 0.9877 81 VC = 5 − (1.185)(2 ) = 2.63 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 0.8 ⇒ I B = 9.88 μ A 81 ⎛ 80 ⎞ I C = ⎜ ⎟(0.8) = 0.790 mA ⎝ 81 ⎠ α = 0.9877 VC = 5 − (0.790)(2 ) = 3.42 V

(b) I B =

1.2 ⇒ I B = 9.92 μ A 121 ⎛ 120 ⎞ IC = ⎜ ⎟(1.2 ) = 1.19 mA ⎝ 121 ⎠ 120 α= = 0.9917 121 VC = 5 − (1.19 )(2 ) = 2.62 V

(c) (i) I B =

0.8 ⇒ I B = 6.61 μ A 121 ⎛ 120 ⎞ IC = ⎜ ⎟(0.8) = 0.7934 mA ⎝ 121 ⎠ α = 0.9917 VC = 5 − (0.7934 )(2 ) = 3.41 V ______________________________________________________________________________________

(ii) I B =

5.7 ⎛V I E = I Eo exp⎜⎜ BE ⎝ VT

⎞ ⎟⎟ ⎠

⎛V 0.80 × 10 − 3 = 5 × 10 −14 exp⎜⎜ BE ⎝ VT

(

)

⎞ ⎟⎟ ⎠

⎛ 0.80 × 10 −3 ⎞ ⎟ = 0.6109 V Then V BE = (0.026) ln⎜⎜ −14 ⎟ ⎝ 5 × 10 ⎠ 0.9910 α = = 110 β= 1 − α 1 − 0.9910 I C = αI E = (0.9910)(0.80) = 0.7928 mA

IE 0.80 = ⇒ I B = 7.21 μ A 1 + β 111 VC = 5 − I C RC = 5 − (0.7928)(2) = 3.41 V ______________________________________________________________________________________ IB =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.8 0.75 ⇒ 12.3 μ A 61 ⎛ 60 ⎞ = ( 0.75 ) ⎜ ⎟ = 0.738 mA ⎝ 61 ⎠ 60 = = 0.9836 61 = I C RC − 10 = ( 0.738 )( 5 ) − 10 = −6.31 V

IB = IC

α

(a)

VC VC

(b)

1.5 ⇒ 24.6 μ A 61 ⎛ 60 ⎞ I C = (1.5 ) ⎜ ⎟ = 1.475 mA ⎝ 61 ⎠ ⎛ 60 ⎞ α = ⎜ ⎟ = 0.9836 ⎝ 61 ⎠ VC = (1.475 )( 5 ) − 10 ⇒ VC = −2.625 V IB =

(c) Yes, VC < 0 in both cases so that B-C junction is reverse biased. ______________________________________________________________________________________ 5.9 (a) VC = I C (5) − 10 − 1.2 = I C (5) − 10 ⇒ I C = 1.76 mA I 1.76 IE = C = = 1.785 mA α 0.986 I B = I E − I C = 1.785 − 1.76 ⇒ I B = 25 μ A ⎛V (b) I E = I Eo exp⎜⎜ EB ⎝ VT

⎞ ⎟⎟ ⎠

⎛V 1.785 × 10 −3 = 2 × 10 −15 exp⎜⎜ EB ⎝ VT

⎞ ⎟⎟ ⎠

⎛ 1.785 × 10 −3 ⎞ ⎟ = 0.7154 V V EB = (0.026) ln⎜⎜ −15 ⎟ ⎠ ⎝ 2 × 10 ______________________________________________________________________________________

5.10 ⎛υ iC = I S exp⎜⎜ BE ⎝ VT i C = 93.7 μ A

⎞ ⎛ 0.615 ⎞ ⎟⎟ = 5 × 10 −15 exp⎜ ⎟ ⎝ 0.026 ⎠ ⎠

(

)

93.7 = 0.7495 μ A 125 i E = (126 )(0.7495) = 94.44 μ A ______________________________________________________________________________________ iB =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.11 vEB / VT ⇒ 0.5 × 10 −3 = I Eo1e 0.650 / 0.026 Device 1: iE = I Eo1e So that I EO1 = 6.94 × 10−15 A −3 0.650 / 0.026 Device 2: 12.2 × 10 = I Eo 2 e Or I Eo 2 = 1.69 × 10−13 A

I Eo 2 1.69 × 10 −13 = ⇒ Ratio = 24.4 I Eo1 6.94 × 10 −15 ______________________________________________________________________________________ Ratio of areas =

5.12 For transistor A: ⎛ 275 × 10 −6 ⎞

⎞

⎛I

⎟ = 0.6906 V υ BE ( A) = VT ln⎜⎜ C ⎟⎟ = (0.026 ) ln⎜⎜ −16 ⎟ ⎠ ⎝ 8 × 10 ⎝ I SA ⎠

For transistor B: I SB = 4 I SA = 4 8 ×10 −16 = 3.2 × 10 −15 A

(

)

⎛ 275 × 10 −6 ⎞ ⎟ = 0.6546 V −15 ⎟ ⎝ 3.2 × 10 ⎠ ______________________________________________________________________________________

υ BE (B ) = (0.026 ) ln⎜⎜

5.13 ⎛ υ ⎞ (a) i C = I Co ⎜⎜1 + CE ⎟⎟ VA ⎠ ⎝ 2 ⎞ ⎛ 0.6 = I Co ⎜1 + ⎟ ⇒ I Co = 0.58537 mA ⎝ 80 ⎠

At υ CE = 5 V 5 ⎞ ⎛ i C = (0.58537 )⎜1 + ⎟ = 0.622 mA ⎝ 80 ⎠ Δυ CE 5−2 (b) ro = = ⇒ ro = 137 k Ω Δi C 0.621956 − 0.60 ______________________________________________________________________________________

5.14 BVC E 0 =

BVC B 0 3

β

=

60 3

100

BVC E 0 = 12.9 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.15

BVC E 0 = 56 =

220 3

β

BVC B 0 3

β

⇒3β =

220 = 3.93 56

β = 60.6 ______________________________________________________________________________________ 5.16

BVC E 0 =

BVC B 0 3

β

BVC B 0 = ( BVC E 0 ) 3 β = ( 50 ) 3 50 BVC B 0 = 184 V ______________________________________________________________________________________ 5.17

−0.7 − ( −10 )

IE =

= 1.86 mA 5 ⎛ 75 ⎞ I C = (1.86 ) ⎜ ⎟ = 1.836 mA ⎝ 76 ⎠ VC = −0.7 + 4 = 3.3 V

(a)

RC =

10 − 3.3 ⇒ RC = 3.65 K 1.836

0.5 = 0.00658 mA 76 VB = I B RB = ( 0.00658 )( 25 ) ⇒ VB = 0.164 V IB =

(b)

⎛ 75 ⎞ I C = ( 0.5 ) ⎜ ⎟ = 0.493 mA ⎝ 76 ⎠ −1 − ( −5 ) ⇒ RC = 8.11 K RC = 0.493

IE (10 ) + 0.7 + I E ( 4 ) − 8 76 7.3 = I E ( 4 + 0.132 ) ⇒ I E = 1.767 mA O=

⎛ 75 ⎞ I C = (1.767 ) ⎜ ⎟ = 1.744 mA ⎝ 76 ⎠ VCE = 8 − (1.744 )( 4 ) − ⎡⎣(1.767 )( 4 ) − 8⎤⎦ (c)

= 16 − 6.972 − 7.068 ⇒ VCE = 1.96 V ⎛I ⎞ 5 = I E (10 ) + ⎜ E ⎟ ( 20 ) + 0.7 + I E ( 2 ) = I E (10 + 0.263 + 2 ) + 0.7 ⎝ 76 ⎠ I E = 0.3506 mA ⇒ I B = 4.61 μ A VC = 5 − ( 0.3506 )(10 )

VC = 1.49 V (d) ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.18 For Fig. P5.17(a), R E = 5 + 5% = 5.25 k Ω IE =

−0.7 − ( −10 )

5.25 I C = 1.75 mA

= 1.77 mA

10 − 3.3 = 3.83 K 1.75 RE = 5 − 5% = 4.75 K

RC =

IE =

−0.7 − ( −10 )

= 1.96 mA 4.75 I C = 1.93 mA 10 − 3.3 = 3.47 K RC = 1.93 So 1.75 ≤ I C ≤ 1.93 mA 3.47 ≤ RC ≤ 3.83 K For Fig. P5.17(c), R E = 4 + 5% = 4.2 k Ω IB =

8 − 0.7 = 0.0222 mA 10 + ( 76 )( 4.2 )

I C = 1.66 mA

I E = 1.69 mA

VCE = 16 − (1.66 )( 4 ) − (1.69 )( 4.2 ) = 16 − 6.64 − 7.098 ⇒ VCE = 2.26 V RE = 4 − 5% = 3.8 K IB =

8 − 0.7 = 0.0244 I C = 1.83 mA 10 + ( 76 )( 3.8 )

I E = 1.86 mA VCE = 16 − (1.83)( 4 ) − (1.86 )( 3.8 ) = 16 − 7.32 − 7.068 VCE = 1.61 V So 1.66 ≤ I C ≤ 1.83 mA 1.61 ≤ VCE ≤ 2.26 V ______________________________________________________________________________________

5.19 (a) VCC = I C RC + VCE 2.5 − 1.1 = 0.35 mA 4 ⎛V ⎞ I C = I S exp⎜⎜ BE ⎟⎟ ⎝ VT ⎠

IC =

⎛ 0.35 × 10 −3 V BE = V BB = (0.026 ) ln⎜⎜ −16 ⎝ 5 × 10

⎞ ⎟ = 0.7091 V ⎟ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) I E =

VCC − VCE 2.5 − 1.1 = = 0.7 mA RE 2

⎛ β I C = ⎜⎜ ⎝ 1+ β

⎞ ⎛ 90 ⎞ ⎟⎟ ⋅ I E = ⎜ ⎟(0.70 ) = 0.6923 mA ⎝ 91 ⎠ ⎠ ⎛ 0.6923 × 10 −3 ⎞ ⎟ = 0.7269 V V BE = (0.026 ) ln⎜⎜ −16 ⎟ ⎠ ⎝ 5 × 10 V BB = V BE + I E R E = 0.7269 + (0.7 )(2 ) = 2.127 V ______________________________________________________________________________________

5.20 (a) I C = 0 , VCE = 2 V (b) I C = βI B = (120 )(2 ) ⇒ I C = 0.24 mA VCE = 2 − (0.24 )(4 ) = 1.04 V 1.4 − 0.7 = 0.35 mA 2 ⎛ 120 ⎞ IC = ⎜ ⎟(0.35) = 0.3471 mA ⎝ 121 ⎠ VCE = 2 − (0.3471)(4 ) − (0.35)(2 ) = −0.088 V - Not possible Transistor in Saturation VCE = 0.2 V

(c) I E =

V E = 0.7 V ⇒ VC = 0.9 V 2 − 0.9 = 0.275 mA 4 ______________________________________________________________________________________ IC =

5.21

2 − (0.7 + 0.2 ) = 0.7333 mA 1.5 ⎛ β ⎞ ⎛ 120 ⎞ ⎟⎟ ⋅ I E = ⎜ I C = ⎜⎜ ⎟(0.7333) = 0.7273 mA ⎝ 121 ⎠ ⎝ 1+ β ⎠

(a) I E =

V EC = V E = 0.9 V

(b) I C = βI B = (120 )(15) ⇒ I C = 1.8 mA - Not possible Transistor in saturation V EC = 0.2 V 2 − 0.2 = 1.2 mA 1.5 I C = I E − I B = 1.2 − 0.015 = 1.185 mA IE =

(c) Transistor cutoff I C = 0 , V EC = 2 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.22

V BB − V BE (on ) 1.3 − 0.7 ⇒ RB = = 120 k Ω RB 0.005 = β I BQ = (100 )(0.005 ) = 0.5 mA

(a) I BQ = I CQ

3 − 1.5 = 3kΩ 0.5 (b) For β = 75 , I CQ = (75 )(0.005 ) = 0.375 mA RC =

VCE = 3 − (0.375)(3) = 1.875 V For β = 125 , I CQ = (125 )(0.005 ) = 0.625 mA VCE = 3 − (0.625)(3) = 1.125 V

So 1.125 ≤ VCE ≤ 1.875 V ______________________________________________________________________________________ 5.23 (a) VB = − I B RB ⇒ I B =

−VB − ( −1) = RB 500

I B = 2.0 μ A VE = −1 − 0.7 = −1.7 V IE =

VE − ( −3) RE

=

−1.7 + 3 = 0.2708 mA 4.8

IE 0.2708 = (1 + β ) = = 135.4 ⇒ β = 134.4 IB 0.002

α=

β

⇒ α = 0.9926 1+ β I C = β I B ⇒ I C = 0.269 mA VCE = 3 − VE = 3 − ( −1.7 ) ⇒ VCE = 4.7 V

(b) 5−4 ⇒ I E = 0.5 mA 2 4 = 0.7 + I B RB + ( I B + I C ) RC − 5 I B + IC = I E IE =

I B + IC = I E 4 = 0.7 + I B (100 ) + ( 0.5 )( 8 ) − 5 I B = 0.043 ⇒

IE 0.5 = (1 + β ) = = 11.63 IB 0.043

β = 10.63, α =

β

⇒ α = 0.9140 1+ β ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.24

(V − 0.7 ) + 5 5 − VB ⎛ β ⎞ ⎛ 90 ⎞ ⎟⎟ ⋅ I E = ⎜ ⎟ ⋅ I E , IE = B , I C = ⎜⎜ 10 3 ⎝ 91 ⎠ ⎝ 1+ β ⎠ 5 − V B ⎛ 90 ⎞⎛ V B + 4.3 ⎞ Then = ⎜ ⎟⎜⎜ ⎟⎟ ⇒ V B = −2.136 V 10 3 ⎝ 91 ⎠⎝ ⎠

(a) I C =

−2.136 − 0.7 + 5 = 0.721 mA 3 = 10 − I C (10) − I E (3)

IE =

(b) VCE

⎡ ⎛ 90 ⎞ ⎤ 2 = 10 − I E ⎢3 + ⎜ ⎟(10 )⎥ = 10 − I E (12.89 ) ⎣ ⎝ 91 ⎠ ⎦ Then I E = 0.6206 mA And V B = 0.7 + (0.6206)(3) − 5 = −2.438 V ______________________________________________________________________________________

5.25 3.3 − 0.85 = 0.245 mA 10 0.85 − 0.7 IB = ⇒ IB = 3μ A 50 I C = I E − I B = 0.245 − 0.003 = 0.242 mA

(a) I E =

β=

I C 0.242 = = 80.67 I B 0.003

β

80.67 = 0.9878 1 + β 81.67 VC = (0.242 )(10 ) − 3.3 = −0.88 V V EC = 0.85 − (− 0.88) = 1.73 V (b) β = (80.67 )(1.10 ) = 88.73

α=

IB =

=

V E − 0.7 3.3 − V E ⎛ V − 0.7 ⎞ , IE = = (89.73)⎜⎜ E ⎟⎟ , ⇒ V E = 0.8371 V 50 10 ⎝ 50 ⎠

3.3 − 0.8371 = 0.2463 mA 10 ⎛ 88.73 ⎞ IC = ⎜ ⎟(0.2463) = 0.2435 mA ⎝ 89.73 ⎠ IE =

VC = (0.2435)(10 ) − 3.3 = −0.8645 V V EC = 0.8371 − (− 0.8645) = 1.70 V ______________________________________________________________________________________

5.26 5 − 0.7 ⇒ 17.2 μ A 250 I C = (120 )( 0.0172 ) = 2.064 mA VC = ( 2.064 )(1.5 ) − 5 = −1.90 V VEC = 5 − ( −1.90 ) ⇒ VEC = 6.90 V IB =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

______________________________________________________________________________________ 5.27 ⎛ 50 ⎞ I C = ⎜ ⎟ (1) = 0.98 mA ⎝ 51 ⎠ VC = I C RC − 9 = ( 0.98 )( 4.7 ) − 9 or VC = −4.39 V 1 = 0.0196 mA 51 VE = I B RB + VEB ( on ) = ( 0.0196 )( 50 ) + 0.7 or VE = 1.68 V IB =

______________________________________________________________________________________ 5.28 0.5 ⎛ 50 ⎞ I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA, I B = = 0.0098 mA 51 ⎝ 51 ⎠ VE = I B RB + VEB ( on ) = ( 0.0098 )( 50 ) + 0.7 or VE = 1.19 V VC = I C RC − 9 = ( 0.49 )( 4.7 ) − 9 = −6.70 V

Then VEC = VE − VC = 1.19 − ( −6.7 ) == 7.89 V

PQ = I CVEC + I BVEB = ( 0.49 )( 7.89 ) + ( 0.0098 )( 0.7 ) or PQ = 3.87 mW

Power Dissipated = PS = I Q ( 9 − VE ) = ( 0.5 )( 9 − 1.19 ) P = 3.91 mW Or S ______________________________________________________________________________________ 5.29 I ⇒ I E1 = I E 2 = 0.5 mA 2 ≈ 0.5 mA

I E1 = I E 2 = I C1 = I C 2

VC1 = VC 2 = 5 − ( 0.5 )( 4 ) ⇒ VC1 = VC 2 = 3 V

______________________________________________________________________________________ 5.30

RE = 0 I B =

(a)

2 − 0.7 1.3 = RB RB

⎛ 1.3 ⎞ 5 − 2 I C = ( 80 ) ⎜ ⎟ = = 0.8 ⇒ RC = 3.75 K ⎝ RB ⎠ RC RB = 130 K

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 0.8 ⎛ 81 ⎞ RE = 1 K I B = = 0.010 mA I E = 0.8 ⎜ ⎟ = 0.81 mA 80 ⎝ 80 ⎠ 2 = ( 0.010 )( RB ) + 0.7 + ( 0.81)(1) ⇒ RB = 49 K (b) (c)

5 = ( 0.8 ) RC + 2 + ( 0.81)(1) ⇒ RC = 2.74 K IB =

For part (a)

2 − 0.7 = 0.01 mA 130

I C = (120 )( 0.01) ⇒ I C = 1.20 mA VCE = 5 − (1.2 )( 3.75 ) ⇒ VCE = 0.5 V

2 = I B ( 49 ) + 0.7 + (121) I B (1) For part (b) I B = 0.00765 mA, I E = 0.925 mA, I C = 0.918 mA VCE = 5 − ( 0.918 )( 2.74 ) − ( 0.925 )(1) ⇒ VCE = 1.56 V Including RE result in smaller changes in Q-point values. ______________________________________________________________________________________ 5.31

(a) RC =

VCC − VCEQ I CQ

=

9 − 4.5 = 18 k Ω 0.25

0.25 ⇒ I BQ = 3.125 μ A 80 9 − 0.7 RB = ⇒ R B = 2.656 M Ω 0.003125 (b) I CQ = (120 )(0.003125 ) = 0.375 mA I BQ =

V CEQ = 9 − (0.375 )(18 ) = 2.25 V

______________________________________________________________________________________ 5.32 (a) I C = I E = 0 , VC = 6 V 0.9 − 0.7 ⎛ 150 ⎞ = 0.2 mA, I C = ⎜ ⎟(0.2 ) = 0.1987 mA 1 ⎝ 151 ⎠ VC = 6 − (0.1987 )(10 ) = 4.013 V

(b) I E =

1.5 − 0.7 = 0.8 mA 1 Transistor in saturation VC = 1.5 − 0.7 + 0.2 = 1 V

(c) I E =

6 −1 = 0.5 mA 10 2.2 − 0.7 = 1.5 mA (d) I E = 1 VC = 2.2 − 0.7 + 0.2 = 1.7 V IC =

6 − 1 .7 = 0.43 mA 10 ______________________________________________________________________________________ IC =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.33

VBB = 0. ⎛ RL ⎞ ⎛ 10 ⎞ Cutoff V0 = ⎜ ⎟ VCC = ⎜ ⎟ (5) ⎝ 10 + 5 ⎠ ⎝ RC + RL ⎠ V0 = 3.33 V

a.

VBB = 1 V 1 − 0.7 ⇒ 6 μA 50 I C = β I B = ( 75 )( 6 ) ⇒ I C = 0.45 mA IB =

5 − V0 V = IC + 0 5 10 ⎛1 1 ⎞ 1 − 0.45 = V0 ⎜ + ⎟ ⇒ V0 = 1.83 V ⎝ 5 10 ⎠

b.

V0 = VCE ( sat ) = 0.2 V c. Transistor in saturation ______________________________________________________________________________________

5.34

β = 100

(a)

⎛ 100 ⎞ I Q = 0.1 mA I C = ⎜ ⎟ ( 0.1) = 0.0990 mA ⎝ 101 ⎠ VO = 5 − ( 0.099 )( 5 ) ⇒ VO = 4.505 V

(i)

⎛ 100 ⎞ I Q = 0.5 mA I C = ⎜ ⎟ ( 0.5 ) = 0.495 mA ⎝ 101 ⎠ VO = 5 − ( 0.495 )( 5 ) ⇒ VO = 2.525 V

(ii)

I Q = 2 mA Transistor is in saturation

(iii) (b)

VO = −VBE ( sat ) + VCE ( sat ) = −0.7 + 0.2 ⇒ VO = −0.5 V

β = 150

⎛ 150 ⎞ I Q = 0.1 mA I C = ⎜ ⎟ ( 0.1) = 0.09934 mA ⎝ 151 ⎠ VO = 5 − ( 0.09934 )( 5) ⇒ VO = 4.503 V

(i)

% change =

4.503 − 4.505 × 100% = −0.044% 4.503

⎛ 150 ⎞ I Q = 0.5 mA I C = ⎜ ⎟ ( 0.5 ) = 0.4967 mA ⎝ 151 ⎠ VO = 5 − ( 0.4967 )( 5 ) ⇒ VO = 2.517 V

(ii)

% change =

2.517 − 2.525 × 100% = −0.32% 2.525

I Q = 2 mA Transistor in saturation

V = −8.5 V No change (iii) o ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

5.35 5−4 = 0.20 mA 5 ⎛ 121 ⎞ IQ = I E = ⎜ ⎟(0.20 ) = 0.2017 mA ⎝ 120 ⎠

(a) I C =

5−2 = 0.60 mA 5 ⎛ 121 ⎞ IQ = I E = ⎜ ⎟(0.60 ) = 0.605 mA ⎝ 120 ⎠

(b) I C =

5−0 = 1.0 mA 5 ⎛ 121 ⎞ IQ = I E = ⎜ ⎟(1.0 ) = 1.008 mA ⎝ 120 ⎠ ______________________________________________________________________________________

(c) I C =

5.36

For

I Q = 0,

then

PQ = 0

⎛ 50 ⎞ I Q = 0.5 mA, I C = ⎜ ⎟ ( 0.5) = 0.49 mA ⎝ 51 ⎠ For 0.5 IB = = 0.0098 mA, VB = 0.490 V , VE = 1.19 V 51 VC = ( 0.49 )( 4.7 ) − 9 = −6.70 V ⇒ VEC = 7.89 V P ≅ I CVEC = ( 0.49 )( 7.89 ) ⇒ P = 3.87 mW For For

I Q = 1.0 mA,

Using the same calculations as above, we find P = 5.95 mW I Q = 1.5 mA, P = 6.26 mW

For

I Q = 2 mA, P = 4.80 mW

For

I Q = 2.5 mA, P = 1.57 mW

For

I Q = 3 mA,

Transistor is in saturation. 0.7 + I B ( 50 ) = 0.2 + I C ( 4.7 ) − 9 I E = IQ = I B + IC ⇒ I B = 3 − IC

Then, 0.7 + ( 3 − I C )( 50 ) = 0.2 + I C ( 4.7 ) − 9 Which yields I C = 2.916 mA and I B = 0.084 mA P = I BVEB + I CVEC = ( 0.084 )( 0.7 ) + ( 2.916 )( 0.2 )

or P = 0.642 mW ______________________________________________________________________________________ 5.37 IE =

VEE − VEB ( on ) RE

=

9 − 0.7 ⇒ I E = 2.075 mA 4

I C = α I E = ( 0.9920 ) ( 2.075 ) ⇒ I C = 2.06 mA VBC + I C RC = VCC

VBC = 9 − ( 2.06 ) ( 2.2 ) ⇒ VBC = 4.47 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 5.38 12 − 6 2.727 = 2.727 mA, I B = = 0.03409 mA 2.2 80 0.7 − (− 12 ) = = 0.127 mA 100 = I B + I R 2 = 0.1611 mA

(a) I C = I R2

I R1

V I = I R1 R1 + 0.7 = (0.1611)(15) + 0.7 = 3.12 V

12 − 9 = 1.364 mA, I B = 0.01705 mA 2.2 = 0.01705 + 0.127 = 0.14405 mA

(b) For V CEQ = 9 V, I C = I R1

V I = (0.14405)(15) + 0.7 = 2.86 V

12 − 3 = 4.0909 mA, I B = 0.05114 mA 2.2 = 0.05114 + 0.127 = 0.1781 mA

For V CEQ = 3 V, I C = I R1

V I = (0.1781)(15) + 0.7 = 3.37 V

So 2.86 ≤ V I ≤ 3.37 V ______________________________________________________________________________________ 5.39 For VCE = 4.5 5 − 4.5 = 0.5 mA I CQ = 1 0.5 I BQ = = 0.02 mA 25 0.7 − ( −5 ) IR2 = = 0.057 mA 100 I R1 = I R 2 + I BQ = 0.057 + 0.02 = 0.077 mA V1 = I R1 R1 + VBE ( on ) = ( 0.077 )(15 ) + 0.7 = 1.86 V For VCE = 1.0 5 −1 I CQ = = 4 mA 1 4 I BQ = = 0.16 mA 25 I R 2 = 0.057 mA I R1 = I R 2 + I BQ = 0.057 + 0.16 = 0.217 mA V1 = ( 0.217 )(15 ) + 0.7 ⇒ 3.96 V

So

1.86 ≤ V1 ≤ 3.96 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

______________________________________________________________________________________ 5.40 5 − 2.5 =5K 0.5 0.5 IB = = 0.00417 mA 120 5 − 0.7 RB = = 1032 K 0.00417

RC =

(a)

Choose RC = 5.1 K RB = 1 MΩ (b) For RB = 1 MΩ + 10% = 1.1 M, RC = 5.1 k + 10% = 5.61 K

5 − 0.7 = 3.91 μ A ⇒ I CQ = 0.469 mA 1.1 = 2.37 V

I BQ = VCEQ

RB = 1 MΩ + 10% = 1.1M, RC = 5.1 K − 10% = 4.59 K I BQ = 3.91 μ A ⇒ I CQ = 0.469 mA VCEQ = 2.85 V

RB = 1 MΩ − 10% = 0.90 MΩ RC = 5.1 k + 10% = 5.61 K 5 − 0.7 I BQ = = 4.78 μ A ⇒ I C = 0.573 mA 0.90 VCEQ = 1.78 V RB = 1 MΩ − 10% = 0.90 MΩ I BQ = 4.78 μ A ⇒ I C = 0.573 mA VCEQ = 2.37 V

RC = 5.1 k − 10% = 4.59 K

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

______________________________________________________________________________________ 5.41 VE 2 = 5 − VBE 2

VE1 = 5 − VBE1

VO = VE 2 − VE1 = ( 5 − VBE 2 ) − ( 5 − VBE1 ) VO = VBE1 − VBE 2

⎛I ⎞ We have VBE1 = VE ln ⎜ E1 ⎟ ⎝ I EO ⎠ ⎛I ⎞ VBE 2 = VT ln ⎜ E 2 ⎟ ⎝ I EO ⎠ ⎡ ⎛I VO = VT ⎢ln ⎜ E1 ⎣ ⎝ I EO

⎞ ⎛ IE2 ⎟ − ln ⎜ ⎠ ⎝ I EO

⎞⎤ ⎟⎥ ⎠⎦

⎛I ⎞ ⎛ 10 ⎞ VO = VT ln ⎜ E1 ⎟ = VT ln ⎜ I ⎟ ⎝ I ⎠ ⎝ IE2 ⎠ kT ln (10 ) e ______________________________________________________________________________________ VO =

5.42 5−4 0.25 = 0.25 mA, I B = = 0.002083 mA 120 4 V I = (0.002083)(200 ) + 0.7 = 1.117 V

(a) (i) I C =

⎛ 121 ⎞ (ii) I C = 0.25 mA, I E = ⎜ ⎟(0.25) = 0.252 mA ⎝ 120 ⎠ V I = (0.002083)(200 ) + 0.7 + (0.252)(1) = 1.369 V

5 − 2.5 0.625 = 0.625 mA, I B = = 0.005208 mA 120 4 V I = (0.005208)(200 ) + 0.7 = 1.742 V

(b) (i) I C =

⎛ 121 ⎞ (ii) I E = ⎜ ⎟(0.625) = 0.6302 mA ⎝ 120 ⎠ V I = 1.742 + (0.6302 )(1) = 2.372 V (c) Transistor biased in saturation 3.5 = I B (200 ) + 0.7 + I E (1) V − 0.2 5 − VO , IE = O IC = 4 1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V − 0.2 ⎛ 5 − VO ⎞ I B = I E − IC = O − ⎜⎜ ⎟⎟ = VO (1.25) − 1.45 1 ⎝ 4 ⎠ Then 3.5 = [VO (1.25) − 1.45](200 ) + 0.7 + (VO − 0.2 ) VO = 1.167 V ______________________________________________________________________________________ 5.43

For 4.3 ≤ VI ≤ 5 Q is cutoff I C = 0 VO = 0 If Q reaches saturation, VO = 4.8 4.8 = 1.2 mA 4 5 − 0.7 − VI 1.2 IB = = 0.015 = ⇒ VI = 1.6 80 180 So VI ≤ 1.6, VO = 4.8 IC =

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.44

For VI ≥ 4.3, Q is off and VO = 0 ⎛ 101 ⎞ 5=⎜ ⎟ I C (1) + 0.2 + I C ( 4 ) ⇒ I C = 0.958 mA ⎝ 100 ⎠ When transistor enters saturation, (a)

VO = 3.832 V I B = 0.00958 mA

⎛ 101 ⎞ 5=⎜ ⎟ ( 0.958 )(1) + 0.7 + ( 0.00958 )(180 ) + VI ⎝ 100 ⎠ VI = 5 − 0.7 − 0.9676 − 1.7244 ⇒ VI = 1.61 V For VI = 0, transistor in saturation 5 = I E (1) + 0.2 + I C ( 4 ) ⇒ 5 = I C (1) + I B (1) + 0.2 + I C ( 4 ) 5 = I E (1) + 0.7 + I B (180 ) 5 = I C (1) + I B (1) + 0.7 + I B (180 ) I E = IC + I B 4.8 = 5 I C + I B (1) 4.3 = 1I C + 181I B I B = 4.8 − 5 I C 4.3 = I C + (181)( 4.8 − 5 I C ) 904 I C = 864.5 I C = 0.956 mA VO = 3.825 V

______________________________________________________________________________________ 5.45 IC =

VCC − VCE (sat ) 5 − 0.2 = = 24 mA 0.2 RC

IC 24 = 20 ⇒ I B = = 1.2 mA IB 20

V I − V BE (on ) 5 − 0.7 ⇒ RB = = 3.58 k Ω RB 1.2 ______________________________________________________________________________________ IB =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.46 (a) V I = 0 , ⇒ I B = I C = I E = 0, VO = 0 (b) V I = 2.5 V, V I = I B R B + V BE (on ) + I E R E , and I E = (1 + β )I B

Then I B =

V I − V BE (on ) 2.5 − 0.7 = ⇒ I B = 50.7 μ A R B + (1 + β )R E 10 + (51)(0.5)

I C = (50 )(0.0507 ) = 2.535 mA, I E = (51)(0.0507 ) = 2.586 mA VO = I E R E = (2.586 )(0.5) = 1.293 V (c) V I = 5 V, Transistor in saturation 2.8 = 5.6 mA 0 .5 V − V BE (on ) − VO 5 − 0.7 − 2.8 = = 0.15 mA IB = I 10 RB

VO = 2.8 V, I E =

I C = I E − I B = 5.6 − 0.15 = 5.45 mA ______________________________________________________________________________________

5.47 IC =

VO 8.8 = = 17.6 mA RC 0.5

IC 17.6 = 25 ⇒ I B = = 0.704 mA IB 25

9 − V EB (on ) − V I 9 − 0.7 − 5 ⇒ RB = = 4.69 k Ω RB 0.704 ______________________________________________________________________________________ IB =

5.48 3 − 1.6 0.7 = 0.7 mA, I BQ = ⇒ I BQ = 5.833 μ A 120 2 V − V BE (on ) 1 − 0 .7 R B = BB = = 51.4 k Ω I BQ 0.005833

(a) I CQ =

(b) ΔVO = 3.3 − 0.2 = 2.8 V, peak-to-peak ΔVO 3 − 1.6 = = Aυ = −4.67 (c) ΔV I 0.7 − 1.0 ΔVO (max ) 2.8 (d) ΔV I (max ) = = = 0.6 V 4.667 Aυ So υ i = 0.6 V, peak-to-peak ______________________________________________________________________________________ 5.49 I BQ =

I CQ

β

=

0.15 ⇒ I BQ = 1.25 μ A 120

⎛1+ β ⎞ ⎛ 121 ⎞ ⎟⎟ = (0.15)⎜ I EQ = I CQ ⎜⎜ ⎟ = 0.15125 mA β ⎝ 120 ⎠ ⎝ ⎠ We have RTH = 200 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VTH = I BQ RTH + V BE (on ) + I EQ R E = (0.00125 )(200 ) + 0.7 + (0.15125 )(2 ) = 1.2525 V ⎛ R2 ⎞ 1 ⎟⎟ ⋅ VCC = VTH = ⎜⎜ ⋅ RTH ⋅ VCC R + R R 2 ⎠ 1 ⎝ 1 1 So 1.2525 = (200)(2.5) R1 ⇒ R1 = 399 k Ω and R 2 = 401 k Ω VCEQ = 2.5 − I CQ R C − I EQ R E = 2.5 − (0.15 )(6 ) − (0.15125 )(2 ) = 1.30 V

______________________________________________________________________________________ 5.50 I CQ 0.20 ⎛ 1+ β ⎞ ⎛ 151 ⎞ ⎟⎟ ⋅ I CQ = ⎜ = ⇒ I BQ = 1.33 μ A I EQ = ⎜⎜ ⎟(0.20 ) = 0.2013 mA, I BQ = β 150 ⎝ 150 ⎠ ⎝ β ⎠ V CC = I CQ R C + V CEQ + I EQ R E 2.5 = (0.20 )RC + 1.5 + (0.2013)(1) ⇒ RC = 4 k Ω

VTH = I BQ RTH + V BE (on ) + I EQ R E = (0.00133 )(120 ) + 0.7 + (0.2013 )(1) = 1.061 V

VTH =

1 1 ⋅ RTH ⋅ VCC ⇒ 1.061 = (120)(2.5) R1 R1

So R1 = 283 k Ω and R 2 = 208 k Ω ______________________________________________________________________________________ 5.51

RTH = R1 R2 = 20 15 = 8.57 k Ω ⎛ R2 VTH = ⎜⎜ ⎝ R1 + R 2 VCC

⎞ ⎛ 15 ⎞ ⎟⎟ ⋅ VCC = ⎜ ⎟(10 ) = 4.29 V ⎝ 15 + 20 ⎠ ⎠ I EQ = I EQ R E + V EB (on ) + ⋅ RTH + VTH 1+ β

⎛ 8.57 ⎞ 10 = I EQ (1) + 0.7 + I EQ ⎜ ⎟ + 4.29 ⎝ 101 ⎠ 10 − 0.7 − 4.29 5.01 = ⇒ I EQ = 4.62 mA Then I EQ = 8.57 1.085 1+ 101 I EQ ⎛ 4.62 ⎞ VB = ⋅ RTH + VTH = ⎜ ⎟(8.57 ) + 4.29 ⇒ V B = 4.68 V 1+ β ⎝ 101 ⎠ ______________________________________________________________________________________

5.52 (a)

RTH = 42 58 = 24.36 k Ω ⎛ 42 ⎞ VTH = ⎜ ⎟(24 ) = 10.08 V ⎝ 100 ⎠ 10.08 − 0.7 9.38 I BQ = = ⇒ I BQ = 7.30 μ A 24.36 + (126)(10) 1284.36 I CQ = 0.913 mA, I EQ = 0.9202 mA V CEQ = 14.8 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(b) R1 + 5% = 60.9, R2 + 5% = 44.1 RTH = 25.58 K 10.08 − 0.7 9.38 I BQ = = ⇒ 7.30 μ A 25.58 + 126 (10 ) 1285.58 I CQ = 0.912 mA VCEQ = 14.81

I EQ = 0.919

R1 + 5% = 60.9, R2 − 5% = 39.90 RTH = 24.11 K 9.50 − 0.7 8.8 I BQ = = = 6.85 μ A 24.11 + (126 )(10 ) 1284.11 I CQ = 0.857 mA VCEQ = 15.37 V

VTH = 10.08

VTH = 9.50

I EQ = 0.8635 mA

R1 − 5% = 55.1 K R2 + 5% = 44.1 K 10.67 − 0.7 9.97 I BQ = = = 7.76 μ A 24.50 + 1260 1284.5 I CQ = 0.970 mA I EQ = 0.978 mA VCEQ = 14.22 V

R1 − 5% = 55.1 K R2 − 5% = 39.90 10.08 − 0.7 9.38 I BQ = = = 7.31 μ A 23.14 + 1260 1283.14 I CQ = 0.914 mA I EQ = 0.9211 mA VCEQ = 14.79 V

RTH = 24.50 K

RTH = 23.14 K

VTH = 10.67 V

VTH = 10.08

So we have 0.857 ≤ I CQ ≤ 0.970 mA 14.22 ≤ VCEQ ≤ 15.37 V

______________________________________________________________________________________ 5.53 (a) RTH = R1 R2 = 96 24 = 19.2 k Ω ⎛ R2 ⎞ ⎛ 24 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH = ⎜⎜ ⎟(9 ) = 1.80 V ⎝ 24 + 96 ⎠ ⎝ R1 + R 2 ⎠ V − V BE (on ) 1.80 − 0.7 I BQ = TH = ⇒ I BQ = 10.98 μ A RTH + (1 + β )R E 19.2 + (81)(1)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I CQ = β I BQ = (80 )(0.01098 ) = 0.8782 mA, I EQ = (81)(0.01098 ) = 0.8892 mA V CEQ = 9 − (0.8782 )(5.25 ) − (0.8892 )(1) = 3.50 V

1.80 − 0.7 ⇒ I BQ = 7.846 μ A 19.2 + (121)(1) = (120 )(0.007846 ) = 0.9415 mA, I EQ = (121)(0.007846 ) = 0.9494 mA

(b) I BQ = I CQ

V CEQ = 9 − (0.9415 )(5.25 ) − (0.9494 )(1) = 3.108 V

⎛ 0.9415 − 0.8782 ⎞ For I CQ : ⎜ ⎟ × 100% = 7.21% 0.8782 ⎝ ⎠ ⎛ 3.108 − 3.500 ⎞ For VCEQ : ⎜ ⎟ × 100% = −11.2% 3.500 ⎝ ⎠ ______________________________________________________________________________________

5.54 (a) I CQ ≅ I EQ = 0.4 mA 3 3 RC = ⇒ RC = 7.5 k Ω; RE = ⇒ RE = 7.5 k Ω 0.4 0.4 9 R1 + R2 ≅ = 112.5 k Ω ( 0.2 )( 0.4 ) ⎛ R2 ⎞ VTH = ⎜ ⎟ (VCC ) = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE ⎝ R1 + R2 ⎠ (112.5 − R2 ) R2 RR 0.4 , I BQ = = 0.004 mA RTH = 1 2 = 112.5 100 R1 + R2 ⎡ (112.5 − R2 ) R2 ⎤ ⎛ 9 ⎞ R2 ⎜ ⎥ + 0.7 + (101)( 0.004 )( 7.5 ) ⎟ = ( 0.004 ) ⎢ 112.5 ⎝ 112.5 ⎠ ⎣ ⎦

We obtain

R2 ( 0.08) = 0.004R2 − 3.56 × 10−5 R22 + 3.73

R = 48 k Ω ⇒ R1 = 64.5 k Ω From this quadratic, we find 2 (b) Standard resistor values: Set R E = RC = 7.5 k Ω and R1 = 62 k Ω , R 2 = 47 k Ω Now RTH = R1 R 2 = 62 47 = 26.7 k Ω ⎛ R2 ⎞ ⎛ 47 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ ( 9 ) = 3.88 V ⎝ 47 + 62 ⎠ ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

So I BQ =

3.88 − 0.7 = 0.00406 mA 26.7 + (101)( 7.5 )

Then I CQ = 0.406 mA VRC = VRE = ( 0.406 )( 7.5 ) = 3.05 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

5.55 (a)

RTH = R1 R2 = 12 2 = 1.714 k Ω ⎛ R2 VTH = ⎜⎜ ⎝ R1 + R 2

(b)

⎞ ⎛ 2⎞ ⎟⎟(10 ) − 5 = ⎜ ⎟(10 ) − 5 ⇒ VTH = −3.571 V ⎝ 14 ⎠ ⎠

VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5

−3.57 = I BQ (1.714 ) + 0.7 + (101) I BQ ( 0.5 ) − 5 I BQ =

5 − 0.7 − 3.571 0.729 = ⇒ 13.96 μ A 1.714 + (101)( 0.5 ) 52.21

I CQ = 1.396 mA, I EQ = 1.410 mA

VCEQ = 10 − (1.396 )( 5 ) − (1.41)( 0.5 ) ⇒ VCEQ = 2.32 V

(d)

RE = 0.5 + 5% = 0.525 K I BQ =

RC = 5 + 5% = 5.25 K

0.729 ⇒ 13.32 μ A 1.714 + (101)( 0.525 )

I CQ = 1.332 mA

I EQ = 1.345 mA

VCEQ = 10 − (1.332 )( 5.25 ) − (1.345 )( 0.525 ) = 10 − 6.993 − 0.7061 ⇒ VCEQ = 2.30 V RE = 0.5 + 5% = 0.525 K I CQ = 1.332 mA

RC = 5 − 5% = 4.75 K

I EQ = 1.345 mA

VCEQ = 10 − (1.332 )( 4.75 ) − (1.345 )( 0.525 ) = 10 − 6.327 − 0.7061 ⇒ VCEQ = 2.97 V RE = 0.5 − 5% = 0.475 K I BQ =

RC = 5 + 5% = 5.25 K

0.729 ⇒ 14.67 μ A 1.714 + (101)( 0.475 )

I CQ = 1.467 mA

I EQ = 1.482 mA

VCEQ = 10 − (1.467 )( 5.25 ) − (1.482 )( 0.475 ) = 10 − 7.70175 − 0.70395 ⇒ VCEQ = 1.59 V RE = 0.5 − 5% = 0.475 K I CQ = 1.467 mA

RC = 5 − 5% = 4.75 K

I EQ = 1.482 mA

VCEQ = 10 − (1.467 )( 4.75 ) − (1.482 )( 0.475 ) = 10 − 6.96825 − 0.70395 ⇒ VCEQ = 2.33 V

______________________________________________________________________________________ 5.56 (a) RTH = R1 R2 = 40 40 = 20 k Ω ⎛ R2 VTH = ⎜⎜ ⎝ R1 + R 2

⎞ + ⎛ 40 ⎞ ⎟⎟ ⋅ V = ⎜ ⎟(2.5) = 1.25 V ⎝ 40 + 40 ⎠ ⎠

V + = I EQ R E + V EB (on ) + I BQ RTH + VTH

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V + − V EB (on ) − VTH 2.5 − 0.7 − 1.25 = ⇒ I BQ = 6.57 μ A RTH + (1 + β )R E 20 + (91)(0.7 ) = 0.5914 mA, I EQ = 0.5980 mA

So I BQ = I CQ

V ECQ = 2.5 − (0.5914 )(1.6 ) − (0.5980 )(0.7 ) = 1.135 V

2.5 − 0.7 − 1.25 ⇒ I BQ = 4.375 μ A 20 + (151)(0.7 ) = 0.6563 mA, I EQ = 0.6607 mA

(b) I BQ = I CQ

V ECQ = 2.5 − (0.6563 )(1.6 ) − (0.6607 )(0.7 ) = 0.9874 V

⎛ 0.6563 − 0.5914 ⎞ For I CQ : ⎜ ⎟ × 100% = 10.97% 0.5914 ⎝ ⎠ ⎛ 0.9874 − 1.135 ⎞ For V ECQ : ⎜ ⎟ × 100% = −13.0% 1.135 ⎝ ⎠ ______________________________________________________________________________________

5.57 (a)

RTH = 36 68 = 23.5 k Ω ⎛ 36 ⎞ VTH = ⎜ ⎟(10 ) = 3.46 V ⎝ 36 + 68 ⎠ 3.46 − 0.7 I BQ = = 0.00178 mA 23.5 + (51)(30) I CQ = 0.0888 mA, I EQ = 0.0906 mA

VCEQ = 10 − (0.0888 )(42 ) − (0.0906 )(30 ) ⇒ V CEQ = 3.55 V

(b) R1 = 22.7, R2 = 12 K, RC = 14 K, RE = 10 K RTH = 7.85 k I BQ =

VTH = 3.46

3.46 − 0.7 = 0.00533 mA 7.85 + ( 51)(10 )

I CQ = 0.266 mA

I EQ = 0.272 mA

VCE = 10 − ( 0.266 )(14 ) − ( 0.272 )(10 ) VCE = 3.56 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.58 (a) ⎛ 68 ⎞ RTH = 36 68 = 23.5 k Ω ; VTH = ⎜ ⎟(10 ) − 5 = 1.54 V ⎝ 36 + 68 ⎠ 5 = (51)I BQ (30 ) + 0.7 + I BQ (23.5) + 1.54

I BQ =

2.76 = 1.78 μA ⇒ I CQ = 0.0888 mA 1553.5 I EQ = 0.0906 mA

VECQ = 10 − ( 0.0906 )( 30 ) − ( 0.0888 )( 42 ) = 10 − 2.718 − 3.7296 ⇒ VECQ = 3.55 V (b)

RTH = 12 22.7 = 7.85 k Ω VTH = 1.54 V, R E = 10 k Ω , RC = 14 k Ω 5 = (51)I BQ (10 ) + 0.7 + I BQ (7.85 ) + 1.54 2.76 ⇒ 5.33 μ A, I CQ = 0.266 mA, I EQ = 0.272 mA 517.85 = 10 − (0.272 )(10 ) − (0.266 )(14 ) = 3.56 V

I BQ = V ECQ

______________________________________________________________________________________ 5.59 (a)

RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.5 ) = 5.05 k Ω VTH = I BQ =

Then

1 ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE R1 I CQ

β

=

0.8 = 0.008 mA 100

1 ( 5.05)(10 ) = ( 0.008)( 5.05) + 0.7 + (101)( 0.008 )( 0.5) R1

or R1 = 44.1 k Ω,

44.1R2 = 5.05 ⇒ R2 = 5.70 k Ω 44.1 + R2

⎛ 101 ⎞ Now I EQ = ⎜ ⎟ ( 0.8 ) = 0.808 mA ⎝ 100 ⎠ VCC = I CQ RC + VCEQ + I EQ RE 10 = ( 0.8 ) RC + 5 + ( 0.808 )( 0.5 ) RC = 5.75 k Ω

(b)

For

75 ≤ β ≤ 150

⎛ R2 ⎞ ⎛ 5.7 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 1.145 V ⎝ 5.7 + 44.1 ⎠ ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

β = 75, I BQ = For Then

1.145 − 0.7 = 0.0103 mA 5.05 + ( 76 )( 0.5 )

I CQ = ( 75 )( 0.0103) = 0.775 mA

β = 150, I BQ = For Then

1.145 − 0.7 = 0.00552 mA 5.05 + (151)( 0.5 )

I CQ = 0.829 mA

% Change =

ΔI CQ I CQ

=

0.829 − 0.775 × 100% ⇒ % Change = 6.75% 0.80

For RE = 1 k Ω RTH = ( 0.1)(101)(1) = 10.1 k Ω 1 1 VTH = ⋅ RTH ⋅ VCC = (10.1)(10 ) = ( 0.008 )(10.1) + 0.7 + (101)( 0.008 )(1) R1 R1

(c)

which yields R1 = 63.6 k Ω

And

63.6 R2 = 10.1 ⇒ R2 = 12.0 k Ω 63.6 + R2

⎛ R2 ⎞ ⎛ 12 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 1.587 V R1 + R2 ⎠ 12 + 63.6 ⎠ ⎝ ⎝ Now 1.587 − 0.7 β = 75, I BQ = = 0.0103 mA 10.1 + ( 76 )(1) For I = 0.773 mA So CQ 1.587 − 0.7 β = 150, I BQ = = 0.00551 mA 10.1 + (151)(1) For I = 0.826 mA Then CQ ΔI CQ 0.826 − 0.773 = × 100% ⇒ % Change = 6.63% % Change = 0.8 I CQ

______________________________________________________________________________________ 5.60

VCC ≅ I CQ ( RC + RE ) + VCEQ

10 = ( 0.8 )( RC + RE ) + 5 ⇒ RC + RE = 6.25 k Ω

Let RE = 0.875 k Ω

R = ( 0.1)(121)( 0.875 ) = 10.6 k Ω Then, for bias stable TH 0.8 I BQ = = 0.00667 mA 120 1 (10.6 )(10 ) = ( 0.00667 )(10.6 ) + 0.7 + (121)( 0.00667 )( 0.875 ) R1

71.8R2 = 10.6 ⇒ R2 = 12.4 k Ω 71.8 + R2 R = 71.8 k Ω 1 So and

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10 = 0.119 mA + 12.4 71.8 Then This is close to the design specification. ______________________________________________________________________________________ IR ≅

5.61 I CQ ≈ I EQ ⇒ VCEQ

= VCC − I CQ ( RC + RE )

6 = 12 − I CQ ( 2 + 0.2 )

I CQ = 2.73 mA,

I BQ = 0.0218 mA

VCEQ = 6 V

VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E − 6

⎛ R2 ⎞ ⎟⎟(12) − 6 , RTH = R1 R 2 VTH = ⎜⎜ ⎝ R1 + R 2 ⎠ Bias stable ⇒ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(126 )( 0.2 ) = 2.52 kΩ ⎛ 1⎞ VTH = ⎜ ⎟ ( RTH )(12 ) − 6 ⎝ R1 ⎠ 1 ( 2.52 )(12 ) − 6 = ( 0.0218)( 2.52 ) + 0.7 + (126 )( 0.0218)( 0.2 ) − 6 R1

1 ( 30.24 ) = 0.7549 + 0.5494 R1 R1 = 23.2 kΩ,

23.2R 2 = 2.52 23.2 + R 2

R2 = 2.83 kΩ

______________________________________________________________________________________ 5.62 (a) RTH = (0.1)(1 + β )R E = (0.1)(121)(0.2 ) = 2.42 k Ω ⎡ ⎛ 1+ β VCEQ = 6 − I CQ ⎢ RC + ⎜⎜ ⎝ β ⎣ 2.8 = 6 − I CQ (2.202 ) ⇒ I CQ

⎞ ⎤ ⎟⎟ R E ⎥ = 6 − I CQ (2.202 ) ⎠ ⎦ = 1.453 mA

Then I EQ = 1.465 mA, I BQ = 12.11 μ A

VTH = I BQ RTH + V BE (on ) + I EQ R E − 3 = (0.01211)(2.42 ) + 0.7 + (1.465 )(0.2 ) − 3

Then VTH = −1.978 V =

1 1 ⋅ RTH (6) − 3 = (2.42)(6) − 3 R1 R1

Which yields R1 = 14.2 k Ω and R 2 = 2.92 k Ω (b) For R1 = (1.05)(14.2) = 14.91 k Ω R 2 = (0.95)(2.92 ) = 2.774 k Ω

RTH = R1 R 2 = 2.34 k Ω 2.774 ⎛ ⎞ VTH = ⎜ ⎟(6) − 3 = −2.059 V ⎝ 2.774 + 14.91 ⎠ VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E − 3

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3 − 2.059 − 0.7 ⇒ I BQ = 9.08 μ A So I BQ = 2.34 + (121)(0.2) I CQ = 1.090 mA, I EQ = 1.099 mA V CEQ = 6 − (1.09 )(2 ) − (1.099 )(0.2 ) = 3.60 V

For R1 = (0.95)(14.2 ) = 13.49 k Ω R 2 = (1.05)(2.92 ) = 3.066 k Ω

RTH = R1 R 2 = 2.50 k Ω 3.066 ⎛ ⎞ VTH = ⎜ ⎟(6 ) − 3 = −1.889 V ⎝ 3.066 + 13.49 ⎠ 3 − 1.889 − 0.7 I BQ = ⇒ I BQ = 15.39 μ A 2.50 + (121)(0.2) I CQ = 1.847 mA, I EQ = 1.863 mA

V CEQ = 6 − (1.847 )(2 ) − (1.863)(0.2 ) = 1.933 V

So 1.09 ≤ I CQ ≤ 1.847 mA 1.933 ≤ VCEQ ≤ 3.60 V

______________________________________________________________________________________ 5.63

Let

VCEQ ≅ VCC − I CQ ( RC + RE ) 5 = 12 − 3 ( RC + RE ) ⇒ RC + RE = 2.33 k Ω

RE = 0.333 k Ω

R = 2 kΩ and C β = 100

Nominal value of RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.333) = 3.36 kΩ 3 I BQ = = 0.03 mA 100 1 1 VTH = ⋅ RTH ⋅ (12 ) − 6 = ( 3.36 )(12 ) − 6 R1 R1

Then VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6

1 ( 3.36 )(12 ) − 6 = ( 0.03)( 3.36 ) + 0.7 + (101)( 0.03)( 0.333) − 6 R1 which yields R1 = 22.3 k Ω and R2 = 3.96 k Ω ⎛ R2 ⎞ ⎛ 3.96 ⎞ Now VTH = ⎜ ⎟ (12 ) − 6 = ⎜ ⎟ (12 ) − 6 or VTH = −4.19 V ⎝ 3.96 + 22.3 ⎠ ⎝ R1 + R2 ⎠ For β = 75, VTH = I BQ RTH + VBE ( on) + (1 + β ) I BQ RE − 6 VTH + 6 − 0.7 −4.19 + 6 − 0.7 I BQ = = = 0.0387 mA ⇒ I C = 2.90 mA RTH + (1 + β ) RE 3.36 + ( 76 )( 0.333)

β = 150, I BQ = For

−4.19 + 6 − 0.7 = 0.0207 mA 3.36 + (151)( 0.333)

I = 3.10 mA Then C Specifications are met. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.64 ⎡ ⎛ 1+ β ⎞ ⎤ ⎟⎟ R E ⎥ (a) V + = V ECQ + I CQ ⎢ RC + ⎜⎜ ⎝ β ⎠ ⎦ ⎣ 3.3 = 1.5 + I CQ (3.022 ) ⇒ I CQ = 0.5956 mA I BQ = 6.618 μ A, I EQ = 0.6022 mA

V + = I EQ R E + V EB (on ) + I BQ RTH + VTH

3.3 = (0.6022 )(2 ) + 0.7 + (0.006618)(2.4 ) + VTH

So VTH = 1.380 V =

1 1 ⋅ RTH ⋅ V + = (2.4)(3.3) R1 R1

Which yields R1 = 5.74 k Ω and R 2 = 4.12 k Ω

V + − V EB (on ) − VTH 3.3 − 0.7 − 1.38 = ⇒ I BQ = 4.61 μ A RTH + (1 + β )R E 2.4 + (131)(2 ) = 0.60 mA, I EQ = 0.6045 mA

(b) I BQ = I CQ

V ECQ = 3.3 − (0.60 )(1) − (0.6045 )(2 ) = 1.49 V

______________________________________________________________________________________ 5.65

I CQ = 4.8 mA → I EQ = 4.84 mA VCEQ = VCC − I CQ RC − I EQ RE 6 = 18 − ( 4.8 )( 2 ) − ( 4.84 ) RE ⇒ RE = 0.496 kΩ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 0.496 ) = 6.0 kΩ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE I BQ = 0.040 mA 1 1 VTH = ⋅ RTH ⋅VCC = ( 6.0 )(18 ) R1 R1 1 ( 6.0 )(18 ) = ( 0.04 )( 6.0 ) + 0.70 + (121)( 0.04 )( 0.496 ) R1 1 (108) = 3.34 R1 R1 = 32.3 kΩ,

32.3 R2 = 6.0 32.3 + R2

R2 = 7.37 kΩ

______________________________________________________________________________________ 5.66

For I EQ ≅ I CQ , RC + R E =

VCC − VCEQ I CQ

=

2.5 − 1.6 = 4.5 k Ω 0.2

So R E = 0.5 k Ω For β = 100

RTH = (0.1)(101)(0.5) = 5.05 k Ω 0.2 ⇒ I BQ = 2 μ A, I EQ = 0.202 mA 100 = I BQ RTH + V BE (on ) + I EQ R E = (0.002 )(5.05 ) + 0.7 + (0.202 )(0.5) = 0.8111 V

I BQ = VTH

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VTH =

1 1 ⋅ RTH ⋅ VCC ⇒ 0.8111 = (5.05)(2.5) R1 R1

So R1 = 15.6 k Ω and R 2 = 7.47 k Ω For β = 80 ,

VTH − V BE (on ) 0.8111 − 0.7 = ⇒ I BQ = 2.439 μ A, I CQ = 0.1951 mA RTH + (1 + β )R E 5.05 + (81)(0.5) For β = 120 , I BQ =

0.8111 − 0.7 ⇒ I BQ = 1.695 μ A, I CQ = 0.2034 mA 5.05 + (121)(0.5) Design is valid ______________________________________________________________________________________ I BQ =

5.67 I CQ = 1 mA → I EQ = 1.017 mA VCEQ = VCC − I CQ RC − I EQ RE

5 = 15 − (1)( 5 ) − (1.017 ) RE ⇒ RE = 4.92 kΩ Bias stable: RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)( 4.92 ) = 30.0 kΩ 1 = 0.0167 mA 60 1 = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE R1

I BQ = VTH

1 ( 30.0 )(15 ) = ( 0.0167 )( 30.0 ) + 0.70 + (1.017 )( 4.92 ) R1 1 ( 448.5) = 6.197 R1 R1 = 72.5 kΩ,

72.5 R2 = 30.0 72.5 + R2

R2 = 51.2 kΩ

β = 45 Check: For ⎛ 51.2 ⎞ VTH = ⎜ ⎟ (15 ) = 6.21V ⎝ 51.2 + 72.5 ⎠ V − VBE ( on ) 6.21 − 0.7 I BQ = TH = = 0.0215 mA RTH + (1 + β ) RE 30 + ( 46 )( 4.92 ) I CQ = 0.967 mA,

Check: For I BQ =

β = 75

ΔI C = 3.27% IC

6.21 − 0.7 = 0.0136 mA 30.0 + ( 76 )( 4.92 )

ΔI C = 2.31% IC Design criterion is satisfied. ______________________________________________________________________________________ I CQ = 1.023 mA,

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.68 (a)

VCC ≅ I CQ ( RC + RE ) + VCEQ

3 = ( 0.1)( 5 RE + RE ) + 1.4 ⇒ RE = 2.67 k Ω 100 = 0.833 μ A 120 = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2.67 ) = 32.3 k Ω

RC = 13.3 k Ω, I BQ = RTH

VTH =

1 1 ⋅ RTH ⋅ VCC = ( 32.3)( 3) R1 R1

= I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

= ( 0.000833)( 32.3) + 0.7 + (121)( 0.000833 )( 2.67 ) which gives R1 = 97.3 k Ω, and R2 = 48.4 k Ω

(b) IR ≅

3 3 = ⇒ 20.6 μ A R1 + R2 97.3 + 48.4

I CQ = 100 μ A

P = ( I CQ + I R ) VCC = (100 + 20.6 )( 3) or P = 362 μW

______________________________________________________________________________________ 5.69 IE =

5 − VE 5 = = 1.67 mA RE 3

RTH = R1 || R2 = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 3) = 30.3 kΩ ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ ( 4 ) − 2 = ⋅ RTH ⋅ ( 4 ) − 2 R1 ⎝ R1 + R2 ⎠ I EQ = 0.0165 mA I BQ = 1+ β 5 = I EQ RE + VEB ( on ) + I B RTH + VTH 5 = (1.67 )( 3) + 0.7 + ( 0.0165 )( 30.3) + 0.80 =

1 ( 30.3)( 4 ) − 2 R1

1 ( 30.3)( 4 ) ⇒ R1 = 152 kΩ R1

152 R2 = 30.3 ⇒ R2 = 37.8 kΩ 152 + R2 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.70 a.

RTH = R1 R2 = 10 20 = 6.67 k Ω ⎛ R2 VTH = ⎜⎜ ⎝ R1 + R 2

b.

⎞ ⎛ 20 ⎞ ⎟⎟(10) − 5 = ⎜ ⎟(10 ) − 5 = 1.67 V ⎝ 20 + 10 ⎠ ⎠

10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH I BQ =

10 − 0.7 − 1.67 7.63 = ⇒ I BQ = 0.0593 mA 6.67 + ( 61)( 2 ) 128.7

I CQ = 3.56 mA, I EQ =3.62 mA VE = 10 − I EQ RE = 10 − ( 3.62 )( 2 ) VE = 2.76 V VC = I CQ RC − 10 = ( 3.56 )( 2.2 ) − 10 VC = −2.17 V

______________________________________________________________________________________ 5.71

V + − V − ≅ I CQ ( RC + RE ) + VECQ

20 = ( 0.5 )( RC + RE ) + 8 ⇒ ( RC + RE ) = 24 k Ω

Let Let

RE = 10 k Ω

RC = 14 k Ω

then

β = 60 from previous problem. RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)(10 )

Or RTH = 61 k Ω 0.5 I BQ = = 0.00833 mA 60 ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10 ) − 5 = ⋅ RTH ⋅10 − 5 R1 ⎝ R1 + R2 ⎠

Now 10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 10 = ( 61)( 0.00833) (10 ) + 0.7 + ( 0.00833)( 61) +

1 ( 61)(10 ) − 5 R1

Then R1 = 70.0 k Ω and R2 = 474 kΩ 10 10 = ⇒ 18.4 μ A R1 + R2 70 + 474 40 μ A So the current limit is met. ______________________________________________________________________________________ IR ≅

5.72 ⎡ ⎛ 1+ β (a) V ECQ = V + − V − − I CQ ⎢ RC + ⎜⎜ ⎝ β ⎣

(

)

⎤ ⎞ ⎟⎟ ⋅ R E ⎥ ⎠ ⎦

⎡ ⎛ 81 ⎞ ⎤ 2.7 = 5 − (0.15)⎢ RC + ⎜ ⎟(2 )⎥ ⇒ RC = 13.3 k Ω ⎝ 80 ⎠ ⎦ ⎣ I EQ = 0.1519 mA, I BQ = 1.875 μ A RTH = (0.1)(81)(2 ) = 16.2 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V + = I EQ R E + V EB (on ) + I BQ RTH + VTH

2.5 = (0.1519 )(2 ) + 0.7 + (0.001875)(16.2) + VTH ⇒ VTH = 1.466 V

VTH =

1 1 ⋅ RTH (5) − 2.5 ⇒ 1.466 = (16.2)(5) − 2.5 R1 R1

So R1 = 20.4 k Ω and R 2 = 78.7 k Ω (b) For β = 60 V + − VTH − V EB (on ) 2.5 − 1.466 − 0.7 = ⇒ I BQ = 2.417 μ A RTH + (1 + β )R E 16.2 + (61)(2 ) = 0.145 mA , I EQ = 0.1474 mA

I BQ = I CQ

V ECQ = 5 − (0.145 )(13.3) − (0.1474 )(2 ) = 2.777 V

For β = 100

2.5 − 1.466 − 0.7 ⇒ I BQ = 1.531 μ A 16.2 + (101)(2) = 0.1531 mA, I EQ = 0.1546 mA

I BQ = I CQ

V ECQ = 5 − (0.1531)(13.3) − (0.1546 )(2 ) = 2.655 V

⎛ 0.1531 − 0.145 ⎞ For I CQ : ⎜ ⎟ × 100% = 5.4% 0.15 ⎝ ⎠ ⎛ 2.655 − 2.777 ⎞ For V ECQ : ⎜ ⎟ × 100% = −4.52% 2.70 ⎝ ⎠ ____________________________________________________________________________________

5.73 a. RTH = 500 500 70 = 250 70 = 54.7 k Ω

5 − VTH 3 − VTH VTH − (− 5) + = 500 500 70 5 3 5 1 1 ⎞ ⎛ 1 + − = VTH ⎜ + + ⎟ ⇒ −0.0554 = VTH (0.0183) 500 500 70 ⎝ 500 500 70 ⎠

VTH = −3.03 V

b. I BQ = =

VTH − VBE ( on ) − ( −5 ) RTH + (1 + β ) RE

−3.03 − 0.7 + 5 54.7 + (101)( 5 )

I BQ = 0.00227 mA I CQ = 0.227 mA, I EQ = 0.229

VCEQ = 20 − ( 0.227 )( 50 ) − ( 0.229 )( 5 ) VCEQ = 7.51 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.74 RE ≅

VE 1.5 = = 1.87 k Ω I CQ 0.8

10 = (0.8)RC + 4 + 1.5 ⇒ RC = 5.63 k Ω

RTH = (0.1)(1 + β )R E = (0.1)(121)(1.87 ) = 22.6 k Ω I BQ =

0.8 ⇒ I BQ = 6.67 μ A 120

VTH = I BQ RTH + V BE (on ) + I EQ R E =

1 ⋅ RTH ⋅ VCC R1

(0.00667 )(22.6) + 0.7 + (0.807 )(1.87 ) =

1 (22.6)(10) R1

which yields R1 = 95.8 k Ω and R 2 = 29.6 k Ω ______________________________________________________________________________________ 5.75

I CQ = 50 μ A, I BQ = 0.625 μ A, I EQ = 50.6 μ A

(a) 1 = 19.8 K 0.0506 5 = ( 0.050 ) RC + 5 + ( 0.0506 )(19.8 ) − 5 RE =

RC = 80 K

RTH = R1 R 2 , Design bias stable circuit

RTH = (0.1)(51)(19.8) = 101 k Ω

⎛ R2 ⎞ 1 ⎟⎟(10 ) − 5 = ⋅ RTH ⋅ (10 ) − 5 VTH = ⎜⎜ R1 ⎝ R1 + R 2 ⎠ 1 So (101) (10 ) − 5 = I BQ (101) + 0.7 + ( 0.0506 )(19.8) − 5 R1

1 (1010 ) = 0.0631 + 0.7 + 1 R1 R1 = 573 K

573 R2 = 101 573 + R2

R2 = 123 K (b) RTH = 101 K, VTH = −3.23 V

VTH = I BQ RTH + 0.7 + (121)(19.8 ) I BQ − 5

1.07 = I BQ (101 + 2395.8 ) ⇒ I BQ = 0.429 μ A I CQ = 0.0514 mA, I EQ = 0.0519 mA

VCEQ = 10 − ( 0.0514 )( 80 ) − ( 0.0519 )(19.8 ) = 10 − 4.11 − 1.03 ⇒ VCEQ = 4.86 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.76

(a) R E ≅ V CEQ

0.7 0.5 = 1.4 k Ω , I BQ = ⇒ I BQ = 4.167 μ A 0.5 120 = 6 − I CQ R C − 0.7

2.5 = 6 − (0.5)RC − 0.7 ⇒ RC = 5.6 k Ω RTH = (0.1)(121)(1.4 ) = 16.9 k Ω VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E − 3

VTH = (0.004167 )(16.9 ) + 0.7 + (121)(0.004167 )(1.4 ) − 3 = −1.5237 V 1 1 ⋅ RTH (6 ) − 3 ⇒ −1.5237 = (16.9)(6) − 3 R1 R1

VTH =

which yields R1 = 68.7 k Ω and R 2 = 22.4 k Ω (b) For standard resistor values: Let R E = 1.5 k Ω , RC = 5.6 k Ω , R1 = 68 k Ω , R 2 = 22 k Ω

RTH = R1 R 2 = 68 22 = 16.62 k Ω ⎛ R2 ⎞ ⎛ 22 ⎞ ⎟⎟(6) − 3 = ⎜ VTH = ⎜⎜ ⎟(6 ) − 3 = −1.533 V R + R ⎝ 22 + 68 ⎠ 2 ⎠ ⎝ 1 3 + VTH − V BE (on ) 3 − 1.533 − 0.7 I BQ = = ⇒ I BQ = 3.87 μ A RTH + (1 + β )R E 16.62 + (121)(1.5) I CQ = 0.4646 mA, I EQ = 0.4684 mA V CEQ = 6 − (0.4646 )(5.6 ) − (0.4684 )(1.5) = 2.70 V

______________________________________________________________________________________ 5.77 0.7 6−3 = 7 k Ω , RC + R E = = 30 k Ω , ⇒ RC = 23 k Ω 0.1 0.1 = (0.1)(1 + β )R E = (0.1)(111)(7 ) = 77.7 k Ω

(a) R E ≅ RTH

I BQ =

0 .1 ⇒ I BQ = 0.909 μ A, I EQ = 0.1009 mA 110

V + = I EQ R E + V EB (on ) + I BQ RTH + VTH

3 = (0.1009 )(7 ) + 0.7 + (0.000909 )(77.7 ) + VTH , ⇒ VTH = 1.523 V

VTH =

1 1 ⋅ RTH (6 ) − 3 ⇒ 1.523 = (77.7 )(6) − 3 R1 R1

which yields R1 = 103 k Ω and R 2 = 316 k Ω

V + − V EB (on ) − VTH 3 − 0.7 − 1.523 = ⇒ I BQ = 0.685 μ A RTH + (1 + β )R E 77.7 + (151)(7 ) = 0.1027 mA, I EQ = 0.1034 mA

I BQ = I CQ

V ECQ = 6 − (0.1027 )(23) − (0.1034 )(7 ) = 2.914 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.78 (a) V ECQ = 18 − I CQ R C − V RE 6 = 18 − (1.2 )RC − 1.5 ⇒ RC = 8.75 k Ω

1.2 ⎛ 76 ⎞ I EQ = ⎜ ⎟(1.2 ) = 1.216 mA, I BQ = ⇒ I BQ = 16 μ A 75 75 ⎝ ⎠ 1.5 RE = = 1.234 k Ω 1.216 RTH = (0.1)(1 + β )R E = (0.1)(76 )(1.234 ) = 9.375 k Ω

V + = I EQ R E + V EB (on ) + I BQ RTH + VTH

9 = (1.216 )(1.234 ) + 0.7 + (0.016 )(9.375) + VTH ⇒ VTH = 6.65 V

VTH =

1 (RTH )(18) − 9 ⇒ 6.65 = 1 (9.375)(18) − 9 R1 R1

which yields R1 = 10.78 k Ω and R 2 = 71.8 k Ω Set RC = 9.1 k Ω , R E = 1.2 k Ω , R1 = 11 k Ω , R 2 = 68 k Ω

RTH = R1 R2 = 11 68 = 9.47 k Ω ⎛ R2 ⎞ ⎛ 68 ⎞ ⎟⎟(18) − 9 = ⎜ VTH = ⎜⎜ ⎟(18) − 9 = 6.494 V ⎝ 68 + 11 ⎠ ⎝ R1 + R 2 ⎠ V + − V EB (on ) − VTH 9 − 0.7 − 6.494 I BQ = = ⇒ I BQ = 17.94 μ A RTH + (1 + β )R E 9.47 + (76 )(1.2 ) I CQ = 1.345 mA, I EQ = 1.363 mA

V ECQ = 18 − (1.345 )(9.1) − (1.363)(1.2 ) = 4.12 V

______________________________________________________________________________________ 5.79

RTH = R1 R2 = 100 40 = 28.6 k Ω ⎛ R2 ⎞ ⎛ 40 ⎞ ⎟⎟(10 ) = ⎜ VTH = ⎜⎜ ⎟(10 ) = 2.86 V R + R ⎝ 40 + 100 ⎠ 2 ⎠ ⎝ 1 V − V BE (on ) 2.86 − 0.7 = I B1 = TH RTH + (1 + β )R E1 28.6 + (121)(1) I B1 = 0.0144 mA, I C1 = 1.73 mA, I E1 = 1.75 mA 10 − VB 2 = I C1 + I B 2 3 VB 2 − VBE ( on ) − ( −10 ) IE2 = 5 10 − VB 2 VB 2 − 0.7 + 10 = I C1 + 3 (121)( 5)

⎛1 ⎞ 10 9.3 1 − 1.73 − = VB 2 ⎜⎜ + ⎟⎟ 3 605 ⎝ 3 (121)( 5 ) ⎠

1.588 = VB 2 ( 0.335) ⇒ VB 2 = 4.74 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.74 − 0.7 − ( −10 ) IE2 = ⇒ I E 2 = 2.808 mA 5 I B 2 = 0.0232 mA I C 2 = 2.785 mA VCEQ1 = 4.74 − (1.75 ) (1) ⇒ VCEQ1 = 2.99 V VCEQ 2 = 10 − ( 4.74 − 0.7 ) ⇒ VCEQ 2 = 5.96 V

______________________________________________________________________________________ 5.80

VE1 = −0.7 I R1 = VE 2

−0.7 − ( −5 )

= 0.215 mA 20 = −0.7 − 0.7 = −1.4

IE2 =

−1.4 − ( −5 ) 1

⇒ I E 2 = 3.6 mA I B 2 = 0.0444 mA I C 2 = 3.56 mA

I E1 = I R1 + I B 2 = 0.215 + 0.0444 I E1 = 0.259 mA I B1 = 0.00320 mA I C1 = 0.256 mA

______________________________________________________________________________________ 5.81

V B1 = V RE + V BE (on ) = 0.5 + 0.7 = 1.2 V 1.2 = 60 k Ω 0.020 0.5 RE = = 2.5 k Ω 0.2 V B 2 = V BE (on ) + VCE + V RE = 0.7 + 1.2 + 0.5 = 2.4 V R3 =

R2 =

V B 2 − V B1 2.4 − 1.2 = = 60 k Ω I R2 0.020

R1 =

V + − V B 2 5 − 2.4 = = 130 k Ω 0.020 I R1

VC 2 = 2VCE + V RE = 2(1.2) + 0.5 = 2.9 V

V + − VC 2 5 − 2.9 = = 10.5 k Ω 0.20 I C2 ______________________________________________________________________________________ RC =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.82 RTH = 40 80 = 26.67 k Ω ⎛ 40 ⎞ VTH = ⎜ ⎟(9 ) = 3 V ⎝ 40 + 80 ⎠ VTH − V BE (on ) 3 − 0 .7 = ⇒ I B1 = 8.56 μ A I B1 = RTH + (1 + β n )R E1 26.67 + (121)(2)

I C1 = 1.027 mA, I E1 = 1.036 mA 9 − V C1 + I B 2 = I C1 2 9 = I E 2 (0.1) + V EB (on ) + VC1 ⇒ I B 2 =

So

9 − 0.7 − VC1 8.3 − VC1 = 1 + β p (0.1) 8.1

(

)

9 − VC1 8.3 − VC1 + = 1.027 ⇒ VC1 = 7.214 V 2 8.1 8.3 − 7.214 I B2 = = 0.134 mA 8.1 I C 2 = 10.73 mA, I E 2 = 10.86 mA

V E1 = I E1 R E1 = (1.036 )(2 ) = 2.072 V

VCE1 = VC1 − V E1 = 7.214 − 2.072 = 5.14 V

V EC 2 = 9 − (10.86 )(0.1) − (10.73)(0.2 ) = 5.77 V ______________________________________________________________________________________

5.83

RTH = R1 R 2 = 50 100 = 33.3 k Ω ⎛ R2 ⎞ ⎛ 100 ⎞ ⎟⎟(10 ) − 5 = ⎜ VTH = ⎜⎜ ⎟(10 ) − 5 = 1.67 V ⎝ 100 + 50 ⎠ ⎝ R1 + R 2 ⎠ 5 = I E1 R E1 + V EB (on ) + I B1 RTH + VTH ⎛ 101 ⎞ I E1 = ⎜ ⎟ ( 0.8 ) = 0.808 mA ⎝ 100 ⎠ I B1 = 0.008 mA

5 = ( 0.808 ) RE1 + 0.7 + ( 0.008 )( 33.3) + 1.67

RE1 = 2.93 kΩ

VE1 = 5 − ( 0.808 )( 2.93) = 2.63 V VC1 = VE1 − VECQ1 = 2.63 − 3.5 = −0.87 V VE 2 = −0.87 − 0.70 = −1.57 V IE2 =

−1.57 − ( −5 ) RE 2

= 0.808 ⇒ RE 2 = 4.25 kΩ

VCEQ 2 = 4 ⇒ VC 2 = −1.57 + 4 = 2.43 V 5 − 2.43 RC 2 = ⇒ RC 2 = 3.21 kΩ 0.8

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I RC1 = I C1 − I B 2 = 0.8 − 0.008 = 0.792 mA −0.87 − ( −5 )

⇒ RC1 = 5.21 kΩ 0.792 ______________________________________________________________________________________ RC1 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 6 6.1 I CQ

(a) (i) g m =

VT

β VT

rπ =

ro =

I CQ

= =

0.5 = 19.23 mA/V 0.026

(180)(0.026) = 9.36 k Ω 0.5

V A 150 = = 300 k Ω I CQ 0.5

2 = 76.92 mA/V 0.026 (180)(0.026) = 2.34 k Ω rπ = 2 150 ro = = 75 k Ω 2 0.25 (b) (i) g m = = 9.615 mA/V 0.026 (80)(0.026) = 8.32 k Ω rπ = 0.25 100 ro = = 400 k Ω 0.25 0.08 (ii) g m = = 3.077 mA/V 0.026 (80)(0.026) = 26 k Ω rπ = 0.08 100 ro = = 1250 k Ω 0.08 ______________________________________________________________________________________

(ii) g m =

6.2

(a) g m = rπ = ro =

I CQ VT

β VT I CQ

⇒ I CQ = (95)(0.026 ) = 2.47 mA

=

(125)(0.026) = 1.32 k Ω 2.47

VA 200 = = 81 k Ω I CQ 2.47

(b) I CQ = g mVT = (120)(0.026) = 3.12 mA

β = g m rπ = (120)(1.2) = 144

______________________________________________________________________________________ 6.3

0.8 1.2 = 30.77 mA/V; g m = = 46.15 mA/V 0.026 0.026 So 30.77 ≤ g m ≤ 46.15 mA/V gm =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ rπ =

(180)(0.026) = 5.85 k Ω ;

rπ =

(90)(0.026) = 1.95 k Ω

0.8 1.2 So 1.95 ≤ rπ ≤ 5.85 k Ω ______________________________________________________________________________________ 6.4

(

)

⎛ 0.12 ⎞ −3 (a) iC = I CQ + g mυ be = 0.12 + ⎜ ⎟ 5 × 10 sin ω t 0 . 026 ⎝ ⎠ iC = 0.12 + 0.0231 sin ω t (mA) υ CE = VCEQ − i C RC = [3.3 − (0.12)(15)] − (15)(0.0231 sin ω t )

υ CE = 1.5 − 0.346 sin ω t (V) υ − 0.346 = −69.2 (b) Aυ = ce = 0.005 υ be ______________________________________________________________________________________ 6.5

(a) I CQ =

β (V BB − V BE (on )) RB

=

(120)(1.10 − 0.7 ) = 0.436 mA 110

0.436 = 16.78 mA/V 0.026 (120)(0.026) = 7.15 k Ω = 0.436 80 = = 183 k Ω 0.436 ⎞ ⎛ rπ 7.15 ⎞ ⎟ = −(16.78)(4 183)⎛⎜ = − g m (RC ro )⎜⎜ ⎟ ⎟ + 110 ⎠ r R + 7 . 15 ⎝ B ⎠ ⎝ π = −4.0

gm = rπ ro

(b) Aυ Aυ

0.5 sin (100t ) = −0.125 sin (100t ) (V) Aυ −4 ______________________________________________________________________________________

(c) υ s =

υo

=

6.6

a. rπ = 5.4 =

β VT I CQ

=

(120 )( 0.026 ) I CQ

⇒ I CQ = 0.578 mA

1 1 VCEQ = VCC = ( 5 ) = 2.5 V 2 2 VCEQ = VCC − I CQ RC ⇒ 2.5 = 5.0 − ( 0.578 ) RC ⇒ RC = 4.33 kΩ 0.578 = 0.00482 mA 120 VBB = I BQ RB + VBE ( on ) = ( 0.00482 )( 25 ) + 0.70 ⇒ VBB = 0.820 V I BQ =

I CQ

β

=

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b. rπ = gm = r0 =

β VT I CQ I CQ VT

= =

(120 )( 0.026 ) 0.578

= 5.40 kΩ

0.578 = 22.2 mA/V 0.026

VA 100 = = 173 kΩ I CQ 0.578

⎛ r ⎞ V0 = − g m ( r0 RC ) Vπ , Vπ = ⎜ π ⎟ VS r R + B ⎠ ⎝ π β ( r0 RC ) ⎛ r ⎞ Av = − g m ⎜ π ⎟ ( r0 RC ) = − rπ + RB ⎝ rπ + RB ⎠ Av = −

(120 ) ⎡⎣173

4.33⎤⎦

(120 )( 4.22 )

⇒ Av = −16.7 5.40 + 25 30.4 ______________________________________________________________________________________

6.7

rπ =

=−

(120)(0.026) = 6.24 k Ω

0.5 (a) Ri = R B + rπ = 50 + 6.24 = 56.24 k Ω (b) Ri = R B rπ = 100 6.24 = 5.87 k Ω (c) Ri = r π = 6.24 k Ω ______________________________________________________________________________________ 6.8

ro =

VA 80 = = 400 k Ω I CQ 0.2

(a) R o = RC ro = 4 400 = 3.96 k Ω (b) R o = RC ro = 10 400 = 9.76 k Ω (c) R o = RC R L ro = 10 5 400 = 3.333 400 = 3.31 k Ω ______________________________________________________________________________________ 6.9

10 − 4 = 1.5 mA 4 1.5 = 0.015 mA I BQ = 100 (100 )( 0.026 ) = 1.73 K rπ = 1.5 5sin ω t ( mV ) v = 2.89sin ω t ( μ A ) ib = be = 1.73 kΩ rπ I CQ =

So

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

i B (t ) = I BQ + i b (t ) = 15 + 2.89 sin ω t ( μ A) iC1 (t ) = β i B (t ) = 1.5 + 0.289 sin ω t (mA)

υ C1 (t ) = 10 − iC1 (t )RC = 10 − [1.5 + 0.289 sin ω t ](4 ) (V) υ C1 (t ) = 4 − 1.156 sin ω t (V) υ (t ) − 1.156 Aυ = c = ⇒ Aυ = −231 υ be (t ) 0.005 ______________________________________________________________________________________ 6.10 a. 1 VECQ = VCC = 5 V 2 VECQ = 10 − I CQ RC ⇒ 5 = 10 − ( 0.5 ) RC ⇒ RC = 10 kΩ 0.5 = 0.005 β 100 VEB ( on ) + I BQ RB = VBB = ( 0.70 ) + ( 0.005 )( 50 ) ⇒ VBB = 0.95 V I BQ =

I CQ

=

gm =

I CQ

=

b.

rπ = r0 =

VT

β VT I CQ

0.5 ⇒ g m = 19.2 mA/V 0.026

=

(100 )( 0.026 ) 0.5

⇒ rπ = 5.2 kΩ

VA ∞ = ⇒ r0 = ∞ I CQ 0.5

β RC

Av = −

(100 )(10 ) ⇒ A

v = −18.1 5.2 + 50 rπ + RB c. ______________________________________________________________________________________

6.11

=−

vo = 1.2sin ω t ( V ) iC ( t ) RC + vo = 0 ⇒ iC ( t ) = iC ( t ) = −0.60sin ω t ( mA ) ib ( t ) =

iC ( t )

β

−1.2sin ω t 2

= −6sin ω t ( μ A )

vbe ( t ) = ib ( t ) ⋅ rπ g m rπ = β 100 rπ = =2K 50 vbe ( t ) = −12sin ω t ( mV )

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.12 a. I CQ ≈ I EQ VCEQ = 5 = 10 − I CQ ( RC + RE ) = 10 − I CQ (1.2 + 0.2) I CQ = 3.57 mA 3.57 = 0.0238 mA 150 R1 R 2 = RTH = (0.1)(1 + β )R E I BQ =

= (0.1)(151)(0.2) = 3.02 k Ω

VTH =

1 ⋅ RTH ⋅ (10) − 5 R1

VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5 1 (3.02)(10) − 5 = ( 0.0238)(3.02) + 0.7 + (151)(0.0238)(0.2) − 5 R1 1 ( 30.2 ) = 1.50 ⇒ R1 = 20.1 k Ω R1 20.1R2 = 3.02 ⇒ R2 = 3.55 kΩ 20.1 + R2

b. rπ =

(150)(0.026) = 1.09 k Ω

3.57 3.57 gm = = 137 mA/V 0.026

− β RC − (150 )(1.2 ) = ⇒ Aυ = −5.75 rπ + (1 + β )R E 1.09 + (151)(0.2) ______________________________________________________________________________________ Aυ =

6.13 (a) RTH = R1 R 2 = 33 50 = 19.88 k Ω ⎛ R2 ⎞ ⎛ 50 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH = ⎜⎜ ⎟(3.3) = 1.988 V ⎝ 50 + 33 ⎠ ⎝ R1 + R 2 ⎠ VCC = I BQ (1 + β )R E + V EB (on ) + I BQ RTH + VTH 3.3 − 0.7 − 1.988 = 0.005063 mA 19.88 + (101)(1) = 0.506 mA; I EQ = 0.511 mA

Then I BQ = I CQ

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V ECQ = VCC − I CQ RC − I EQ R E = 3.3 − (0.506 )(2 ) − (0.511)(1) V ECQ = 1.78 V

(b) rπ = Aυ =

β VT I CQ

=

(100)(0.026) = 5.14 k Ω 0.506

− β RC − (100 )(2 ) = = −1.884 rπ + (1 + β )R E 5.14 + (101)(1)

(c) R1 = (1.05)(33) = 34.65 k Ω R 2 = (0.95)(50 ) = 47.5 k Ω

RTH = R1 R 2 = 34.65 47.5 = 20.03 k Ω 47.5 ⎛ ⎞ VTH = ⎜ ⎟(3.3) = 1.908 V ⎝ 47.5 + 34.65 ⎠

⎛ 3.3 − 0.7 − 1.908 ⎞ ⎟⎟ = 0.5718 mA I CQ = (100)⎜⎜ ⎝ 20.03 + (101)(1) ⎠ (100 )(0.026) = 4.547 k Ω rπ = 0.5718 − (100 )(2) = −1.895 Aυ = 4.547 + (101)(1) Also R1 = (0.95)(33) = 31.35 k Ω R 2 = (1.05)(50 ) = 52.5 k Ω RTH = 31.35 52.5 = 19.63 k Ω

52.5 ⎞ ⎛ VTH = ⎜ ⎟(3.3) = 2.066 V 52 . 5 31 . 35 + ⎠ ⎝ ⎛ 3.3 − 0.7 − 2.066 ⎞ ⎟⎟ = 0.4427 mA I CQ = (100)⎜⎜ ⎝ 19.63 + (101)(1) ⎠ rπ =

(100)(0.026) = 5.873 k Ω

0.4427 − (100 )(2 ) = −1.871 Aυ = 5.873 + (101)(1)

So 1.871 ≤ Aυ ≤ 1.895 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.14 (a) VCC = ⎜⎜ 1+ β ⎟⎟ I CQ RE + VECQ + I CQ RC ⎛ ⎝

⎞

β

⎠

⎛ 101 ⎞ 12 = ⎜ ⎟ I CQ (1) + 6 + I CQ ( 2 ) ⎝ 100 ⎠ so that I CQ = 1.99 mA 1.99 = 0.0199 mA 100 = ( 0.1)(1 + β ) RE = ( 0.1)(101)(1) = 10.1 k Ω

I BQ = RTH

⎛ R2 ⎞ 1 1 VTH = ⎜ ⎟ VCC = ⋅ RTH ⋅ VCC = (10.1)(12 ) R R R R + 2 ⎠ 1 1 ⎝ 1 VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 121.2 12 = (101)( 0.0199)(1) + 0.7 + ( 0.0199)(10.1) + R1 which yields R1 = 13.3 k Ω and R2 = 41.6 k Ω − β RC − (100 )(2 ) = ⇒ Aυ = −1.95 rπ + (1 + β )R E 1.31 + (101)(1) ______________________________________________________________________________________

(b)

Aυ =

6.15 I CQ = 0.25 mA, I EQ = 0.2525 mA I BQ = 0.0025 mA I BQ RB + VBE ( on ) + I EQ ( RS + RE ) − 5 = 0 ( 0.0025)( 50 ) + 0.7 + ( 0.2525)( 0.1 + RE ) = 5 RE = 16.4 kΩ

VE = − ( 0.0025 )( 50 ) − 0.7 = −0.825 V VC = VCEQ + VE = 3 − 0.825 = 2.175 V 5 − 2.175 ⇒ RC = 11.3 kΩ RC = 0.25 − β RC Av = rπ + (1 + β ) RS rπ =

(100 )( 0.026 )

= 10.4 kΩ 0.25 − (100 )(11.3) Av = ⇒ Av = −55.1 10.4 + (101)( 0.1)

Ri = RB ⎣⎡ rπ + (1 + β ) RS ⎦⎤ = 50 ⎣⎡10.4 + (101)( 0.1) ⎦⎤ Ri = 50 20.5 ⇒ Ri = 14.5 kΩ

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.16

(a) I CQ ≅

VCC − VCEQ RC + R E

=

9 − 5.20 = 0.905 mA 2.2 + 2

I BQ ≅ 0.00754 mA; I EQ ≅ 0.9123 mA

RTH = (0.1)(1 + β )R E = (0.1)(121)(2) = 24.2 k Ω VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E

= (0.00754 )(24.2 ) + 0.7 + (0.9123)(2 ) = 2.707 V

Now VTH = 2.707 =

1 1 ⋅ RTH ⋅ VCC = (24.2)(9) R1 R1

Or R1 = 80.5 k Ω and then R 2 = 34.6 k Ω

(120)(0.026) = 3.448 k Ω

(b) rπ =

0.905 0.905 gm = = 34.81 mA/V 0.026 100 ro = = 110 k Ω 0.905

(

) (

Vπ = i s R1 R 2 rπ = i s 80.5 34.6 3.448 Vπ = i s (3.02 )

(

υ o = − g mVπ ro RC R L Then R m =

υo is

)

) (

)

= − g m (3.02) 110 2.2 1 = −(34.81)(3.02 )(110 0.6875)

Or R m = −71.8 V/mA (c) For β = 100 , 2.707 − 0.7 = 0.008873 mA 24.2 + (101)(2) = 0.8873 mA

I BQ = I CQ

(100)(0.026) = 2.93 k Ω ,

0.8873 100 gm = = 34.13 mA/V, ro = = 113 k Ω 0.8873 0.026 0.8873 Vπ = i s (RTH rπ ) = i s (24.2 2.93) = i s (2.614 )

rπ =

υo

= −(34.13)(2.614 )(113 0.6875) = −61.0 V/mA is For β = 150 ,

Rm =

⎛ 2.707 − 0.7 ⎞ ⎟⎟ = 0.923 mA I CQ = (150 )⎜⎜ ⎝ 24.2 + (151)(2 ) ⎠ rπ = 4.225 k Ω , g m = 35.5 mA/V, ro = 108 k Ω

Vπ = i s (24.2 4.225) = i s (3.597 ) Rm =

υo is

= −(35.5)(3.597 )(108 0.6875) = −87.2 V/mA

So 61.0 ≤ R m ≤ 87.2 V/mA ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.17 0.8 = 0.009877 mA, I CQ = 0.790 mA 81 0.2 = = 20.25 k Ω 0.009877 − 3 − (− 5) = = 2.53 k Ω 0.79 0.79 100 = = 30.38 mA/V, ro = = 127 k Ω 0.026 0.79 RC = 127 2.53 = 2.48 k Ω

(a) (i) I BQ = RB

RC (ii) g m ro

⎛ 2.48 ⎞ i o = − g mυ s ⎜ ⎟ ⎝ 2.48 + 4 ⎠ i ⎛ 2.48 ⎞ G f = o = −(30.38)⎜ ⎟ = −11.63 mA/V υs ⎝ 2.48 + 4 ⎠ 0.8 = 0.00661 mA, I CQ = 0.7934 mA 121 0.2 RB = = 30.3 k Ω 0.00661 − 3 − (− 5) RC = = 2.52 k Ω 0.7934 0.7934 80 (ii) g m = = 30.52 mA/V, ro = = 101 k Ω 0.026 0.7934

(b) (i) I BQ =

ro RC = 101 2.52 = 2.459 k Ω

⎛ 2.459 ⎞ G f = −(30.52)⎜ ⎟ = −11.62 mA/V ⎝ 2.459 + 4 ⎠ ______________________________________________________________________________________ 6.18 0.25 = 0.00208 mA, I EQ = 0.252 mA 120 5 = (0.00208)(2.5) + 0.7 + (0.252 )R E ⇒ R E = 17.0 k Ω

(a) (i) I BQ =

VCEQ = 3 = 10 − (0.252 )(17 ) − (0.25)RC ⇒ RC = 10.9 k Ω

(120)(0.026) = 12.48 k Ω 0.25 = 9.615 mA/V, rπ = 0.026 0.25 ⎛ rπ ⎞ 12.48 ⎞ ⎟ = −(9.615)(10.9 5)⎛⎜ Aυ = − g m (RC R L )⎜⎜ ⎟ = −27.5 ⎟ ⎝ 12.48 + 2.5 ⎠ ⎝ rπ + R S ⎠

(ii) g m =

(

)

(iii) υ o = −(27.49 ) 5 × 10 −3 sin ω t = −0.137 sin ω t (V) (b) (i) 5 = 0.7 + (0.252 )R E ⇒ R E = 17.1 k Ω VCEQ = 3 = 10 − (0.252)(17.06) − (0.25)RC ⇒ RC = 10.8 k Ω (ii) g m = 9.615 mA/V, rπ = 12.48 k Ω

Aυ = − g m (RC R L ) = −(9.615)(10.8 5) = −32.9

(

)

(iii) υ o = −(32.86 ) 5 × 10 −3 sin ω t = −0.164 sin ω t (V) ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.19 5 − 0.7 = 0.005292 mA, ⇒ I CQ = 0.4234 mA 2.5 + (81)(10 ) (80)(0.026) = 4.91 k Ω 0.4234 gm = = 16.28 mA/V, rπ = 0.026 0.4234 ⎛ rπ ⎞ 4.91 ⎞ ⎟ = −(16.28)(5 5)⎛⎜ Aυ = − g m (RC R L )⎜⎜ ⎟ = −26.97 ⎟ + 4 . 91 + 2.5 ⎠ r R ⎝ S ⎠ ⎝ π υ Aυ io = o = υ s RL RL

(a) (i) I BQ =

or

Gf =

io

υs

=

(

− 26.97 = −5.39 mA/V 5

)

(ii) υ o = −(26.97 ) 4 × 10 −3 sin ω t = −0.108 sin ω t (V)

(

i o = − 5.39 × 10

−3

)(4 ×10

−3

)

sin ω t ⇒ −21.6 sin ω t ( μ A)

⎛ ⎞ 5 − 0.7 ⎟⎟ = 0.4256 mA (b) (i) I CQ = (120 )⎜⎜ ⎝ 2.5 + (121)(10 ) ⎠ (120)(0.026) = 7.33 k Ω 0.4256 gm = = 16.37 mA/V, rπ = 0.026 0.4256 ⎛ 7.33 ⎞ Aυ = −(16.37 )(5 5)⎜ ⎟ = −30.5 ⎝ 7.33 + 2.5 ⎠

Gf =

−30.5 = −6.1 mA/V 5

(

)

(ii) υ o = −(30.5) 4 × 10 −3 sin ω t = −0.122 sin ω t (V) i o = −24.4 sin ω t ( μ A) ______________________________________________________________________________________ 6.20 RTH = R1 R2 = 27 15 = 9.64 K ⎛ R2 ⎞ ⎛ 15 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 9 ) = 3.214 V ⎝ 15 + 27 ⎠ ⎝ R1 + R2 ⎠ V − VBE ( on ) 3.214 − 0.7 2.514 I BQ = TH = = RTH + (1 + β ) RE 9.64 + (101)(1.2 ) 130.84

I BQ = 0.0192 mA I CQ = 1.9214 mA gm =

ro =

(100 )( 0.026 ) 1.92 = 73.9 mA/V rπ = = 1.35 K 0.026 1.92

100 = 52.1 K 1.92

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ r R Vπ = ⎜ π TH ⎜r R +R S ⎝ π TH = 1.35 9.64 = 1.184 K

(

Vo = − g mVπ r0 RC RL rπ RTH

⎛ 1.184 ⎞ Vπ = ⎜ ⎟ VS ⎝ 1.184 + 10 ⎠ = 0.1059VS

)

(

Av = − ( 73.9 ) ( 0.1059 ) 52.1 2.2 2

⎞ ⎟⎟ VS ⎠

)

= − ( 73.9 ) ( 0.1059 ) ( 52.1 1.0476 ) = − ( 73.9 ) ( 0.1059 ) (1.027 ) Av = −8.04 ⎛ ro RC − g mVπ ⎜⎜ I ⎝ ro RC + RL AI = o = Vπ IS RTH rπ

⎞ ⎟⎟ ⎠

⎛ ro RC AI = − g m ( RTH rπ ) ⎜ ⎜r R +R L ⎝ o C

⎞ ⎟⎟ ⎠

ro RC = 52.1 2.2 = 2.11 K RTH rπ = 9.64 1.35 = 1.184 K ⎛ 2.11 ⎞ AI = − ( 73.9 )(1.184 ) ⎜ ⎟ ⎝ 2.11 + 2 ⎠ AI = −44.9 Ri = RTH rπ = 9.64 1.35 Ri = 1.184 K

______________________________________________________________________________________ 6.21 a. 0.35 = 0.00347 mA 101 V B = − I B R B = −(0.00347 )(10 ) ⇒ V B = −0.0347 V I E = 0.35 mA, I B =

V E = V B − V BE (on ) ⇒ V E = −0.735 V

b. VC = VCEQ + V E = 3.5 − 0.735 = 2.77 V ⎛ β ⎞ ⎛ 100 ⎞ ⎟⎟ ⋅ I E = ⎜ I C = ⎜⎜ ⎟(0.35) = 0.347 mA ⎝ 101 ⎠ ⎝ 1+ β ⎠ V + − VC 5 − 2.77 RC = = ⇒ RC = 6.43 k Ω IC 0.347

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) ⎛ R B rπ ⎞ ⎟(R r ) Aυ = − g m ⎜ ⎜R r +R ⎟ C o S ⎠ ⎝ B π 0.347 100 gm = = 13.3 mA/V, ro = = 288 k Ω 0.026 0.347 (100)(0.026) = 7.49 k Ω rπ = 0.347 R B rπ = 10 7.49 = 4.28 k Ω ⎛ 4.28 ⎞ Aυ = −(13.3)⎜ ⎟(6.43 288) ⇒ Aυ = −81.7 ⎝ 4.28 + 0.1 ⎠

d. ⎛ R B rπ ⎞ ⎟(R r ) Aυ = − g m ⎜ ⎜R r +R ⎟ C o S ⎠ ⎝ B π R B rπ = 10 7.49 = 4.28 k Ω ⎛ 4.28 ⎞ Aυ = −(13.3)⎜ ⎟(6.43 288) ⇒ Aυ = −74.9 ⎝ 4.28 + 0.5 ⎠ ______________________________________________________________________________________

6.22 a. RTH = R1 R2 = 6 1.5 = 1.2 kΩ ⎛ R2 ⎞ + ⎛ 1.5 ⎞ VTH = ⎜ ⎟V = ⎜ ⎟ ( 5 ) = 1.0 V ⎝ 1.5 + 6 ⎠ ⎝ R1 + R2 ⎠ V − VBE ( on ) 1.0 − 0.7 = = 0.0155 mA I BQ = TH RTH + (1 + β ) RE 1.2 + (181)( 0.1)

I CQ = 2.80 mA, I EQ = 2.81 VCEQ = V + − I CQ RC − I EQ RE = 5 − ( 2.8 )(1) − ( 2.81)( 0.1) ⇒ VCEQ = 1.92 V

b. rπ =

(180)(0.026) = 1.67 k Ω

2.80 2.80 = 108 mA/V gm = 0.026 (c)

⎛ R1 R 2 rπ ⎞ ⎟(R R ) Aυ = g m ⎜ ⎜R R r +R ⎟ C L S ⎠ ⎝ 1 2 π R1 R 2 rπ = 6 1.5 1.67 = 0.698 k Ω ⎛ 0.698 ⎞ Aυ = (108)⎜ ⎟(1 1.2) = 45.8 ⎝ 0.698 + 0.2 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.23 a.

9 = I EQ RE + VEB ( on ) + I BQ RS 0.75 I EQ = 0.75 mA, I BQ = = 0.00926 mA 81 I CQ = 0.741 mA

9 = ( 0.75 ) RE + 0.7 + ( 0.00926 )( 2 ) ⇒ RE = 11.0 kΩ

b.

VE = 9 − ( 0.75 )(11) = 0.75 V VC = VE − VECQ = 0.75 − 7 = −6.25 V RC =

VC − ( −9 ) I CQ

=

9 − 6.25 ⇒ RC = 3.71 kΩ 0.741

c.

(

⎛ rπ ⎞ ⎟ RC R L ro Aυ = − g m ⎜⎜ ⎟ ⎝ rπ + R S ⎠ (80)(0.026) = 2.81 k Ω rπ = 0.741 80 = 108 k Ω ro = 0.741 −80 Aυ = 3.71 10 108 2.81 + 2 Aυ = −43.9

(

)

)

d. Ri = R S + rπ = 2 + 2.81 ⇒ Ri = 4.81 k Ω ______________________________________________________________________________________

6.24 ⎡ 4 − 0.7 ⎤ (a) For β = 80 , I CQ = (80 )⎢ ⎥ = 0.6439 mA ⎣ 5 + (81)(5) ⎦ (80)(0.026) = 3.23 k Ω , r = 50 = 77.7 k Ω 0.6439 gm = = 24.77 mA/V, rπ = o 0.026 0.6439 0.6439 (3.23 5) rπ R B ( ) ⋅υ s = Vπ = (3.23 5) + 1 ⋅υ s = 0.6624 υ s rπ R B + R S Aυ = − g m ⋅

Vπ

υs

(

)

(

⎡ 4 − 0.7 ⎤ For β = 120 , I CQ = (120)⎢ ⎥ = 0.6492 mA ⎣ 5 + (121)(5) ⎦ g m = 24.97 mA/V, rπ = 4.806 k Ω , ro = 77 k Ω Vπ =

)

⋅ ro RC R L = −(24.77 )(0.6624) 77.7 4 4 = −32

(4.806 5) ( ) (4.806 5) + 1 ⋅υ = 0.710 υ s

(

)

s

Aυ = −(24.97 )(0.710 ) 77 4 4 = −34.6

So 32 ≤ Aυ ≤ 34.6

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) Ri = R B rπ For β = 80 , Ri = 5 3.23 = 1.96 k Ω For β = 120 , Ri = 5 4.806 = 2.45 k Ω So 1.96 ≤ Ri ≤ 2.45 k Ω (c) R o = RC ro For β = 80 , R o = 4 77.7 = 3.804 k Ω For β = 120 , R o = 4 77 = 3.802 k Ω So 3.802 ≤ R o ≤ 3.804 k Ω ______________________________________________________________________________________ 6.25

Assume an npn transistor with β = 100 and V A = ∞ . Let VCC = 10 V. 0.5 = 50 0.01 I CQ = 1 mA

Av =

Bias at and let RE = 1 k Ω For a bias stable circuit RTH = (0.1)(1 + β )R E = (0.1)(101)(1) = 10.1 k Ω VTH =

1 1 ⋅ RTH ⋅ VCC = (10.1)(10) = 101 R1 R1 R1

1 = 0.01 mA 100 = I BQ RTH + V BE (on ) + (1 + β )I BQ R E

I BQ = VTH

101 = (0.01)(10.1) + 0.7 + (101)(0.01)(1) R1

which yields R1 = 55.8 k Ω , R 2 = 12.3 k Ω Now (100)(0.026) = 2.6 k Ω rπ = 1 1 gm = = 38.46 mA/V 0.026

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V o = − g m Vπ R C ⎞ ⎛ R1 R 2 rπ ⎟ ⋅ V = ⎛⎜ 10.1 2.6 ⎞⎟ ⋅ V = 0.674V where Vπ = ⎜ s ⎜ R R r + R ⎟ s ⎜ 10.1 2.6 + 1 ⎟ s S ⎠ ⎝ ⎠ ⎝ 1 2 π V Then Av = o = − ( 0.674 ) g m RC = − ( 0.674 )( 38.46 ) RC = −50 Vs which yields RC = 1.93 k Ω

With this RC, the dc bias is OK. Finish Design, Set RC = 2 k Ω , R E = 1 k Ω R1 = 56 k Ω , R 2 = 12 k Ω RTH = R1 R2 = 9.88 K ⎛ R2 ⎞ ⎛ 12 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 1.765 V ⎝ 12 + 56 ⎠ ⎝ R1 + R2 ⎠ 1.765 − 0.7 = 9.60 μ A I BQ = 9.88 + (101)(1) I CQ = 0.9605 mA rπ =

(100 )( 0.026 )

0.9605 RTH rπ = 2.125 K

= 2.707 K

gm =

0.9605 = 36.94 0.026

⎛ RTH rπ ⎞ ⎟ ⋅ V = ⎛ 2.125 ⎞ ⋅ V = (0.680 ) ⋅ Vi Vπ = ⎜ ⎜ R r + R ⎟ i ⎜⎝ 2.125 + 1 ⎟⎠ i S ⎠ ⎝ TH π ( ) Aυ = − 0.680 g m RC = −(0.680 )(36.94 )(2) = −50.2 Design specification met. ______________________________________________________________________________________

6.26 a. I BQ =

6 − 0.7 = 0.0169 mA 10 + (101)( 3)

I CQ = 1.69 mA, I EQ = 1.71 mA VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3) VCEQ = 5.38 V

b. 1.69 = 65 mA/V 0.026 (100)(0.026) = 1.54 k Ω , r = ∞ rπ = o 1.69

gm =

(c) Av =

− β ( RC RL )

⋅

RB Rib

rπ + (1 + β ) RE RB Rib + RS

Rib = rπ + (1 + β ) RE = 1.54 + (101)(3) = 304.5 k Ω RB Rib = 10 304.5 = 9.68 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then Aυ =

− (100 )(6.8 6.8) ⎛ 9.68 ⎞ ⋅⎜ ⎟ = −1.06 1.54 + (101)(3) ⎝ 9.68 + 0.5 ⎠

⎛ RC i o = ⎜⎜ ⎝ RC + R L

⎞ ⎟(− β i b ) ⎟ ⎠

⎛ RB ib = ⎜⎜ R + r + ⎝ B π (1 + β )R E

⎞ ⎟ ⋅ is ⎟ ⎠

⎛ RC ⎞⎛ ⎞ RB ⎟⎜ ⎟ Ai = −(β )⎜⎜ ⎟⎜ R + r + (1 + β )R ⎟ R R + L ⎠⎝ B π E ⎠ ⎝ C ⎞ 10 ⎛ 6.8 ⎞⎛ ⎟⎟ ⇒ Ai = −1.59 = −(100 )⎜ ⎟⎜⎜ ( )( ) 6 . 8 6 . 8 10 1 . 54 101 3 + + + ⎝ ⎠⎝ ⎠

(d) (e)

Ris = RS + RB Rib = 0.5 + 10 304.5 = 10.2 k Ω

Aυ =

− β (RC R L )

rπ + (1 + β )R E

=

− (100 )(6.8 6.8) 1.54 + (101)(3)

Aυ = −1.12 Ai = −1.59 , same as part (c) ______________________________________________________________________________________

6.27

r=

υ ce 1 = g mυ ce g m

So re = rπ

1 ro gm

______________________________________________________________________________________ 6.28

(a) Set

RC = 12.5 ⇒ RC = 12.5 R E RE

Set VCEQ = 1.5 ≅ 3.3 − I CQ (RC + R E ) = 3.3 − I CQ (13.5R E ) Set I CQ = 0.1 mA, ⇒ R E = 1.33 k Ω and RC = 16.7 k Ω We find I BQ = 0.833 μ A

Set RTH = (0.1)(1 + β )R E = (0.1)(121)(1.33) = 16.1 k Ω VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E = (0.000833)(16.1) + 0.7 + (121)(0.000833)(1.33)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VTH = 0.8475 =

1 1 ⋅ RTH ⋅ VCC = (16.1)(3.3) R1 R1

So R1 = 62.7 k Ω , R 2 = 21.7 k Ω Actual gain: (120)(0.026) = 31.2 k Ω rπ = 0.1 Rib = rπ + (1 + β )R E = 31.2 + (121)(1.33) = 192.1 k Ω Ri = RTH Rib = 16.1 192.1 = 14.85 k Ω

Then Aυ =

− (120 )(16.7 ) ⎛ 14.85 ⎞ ⋅⎜ ⎟ = −10.4 31.2 + (121)(1.33) ⎝ 14.85 + 0.1 ⎠

(b) Ris = R S + Ri = 0.1 + 14.85 = 14.95 k Ω R o = RC = 16.7 k Ω ______________________________________________________________________________________ 6.29 100 = 20. Need a voltage gain of 5

Assume a sign inversion from a common-emitter is not important. Use the configuration for Figure 6.31. Let RS = 0. Need an input resistance of 5 × 10 −3 = 25 × 10 3 ⇒ Ri = 25 k Ω 0.2 × 10 − 6 Ri = RTH Rib , Let RTH = 50 k Ω , then Rib = 50 k Ω Ri =

Rib = rπ + (1 + β )R E ≅ (1 + β )R E

Rib 50 = = 0.495 k Ω 1 + β 101 = 10 V, I CQ = 0.2 mA

For β = 100 , R E ≅ Let R E = 0.5 k Ω , VCC

0.2 = 0.002 mA 100 = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

Then I BQ = VTH

1 1 ⋅ RTH ⋅ VCC = ( 50)(10) = ( 0.002)(50) + 0.7 + (101)( 0.002)( 0.5) R1 R1 which yields R1 = 555 k Ω and R2 = 55 k Ω Now Av =

(100)( 0.026) − β RC = 13 k Ω , rπ = rπ + (1 + β ) RE 0.2

So −20 =

− (100 ) RC

13 + (101)( 0.5)

⇒ RC = 12.7 k Ω

I R = ( 0.2 )(12.7 ) = 2.54 V. So dc biasing is OK.] [Note: CQ C ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.30

Set

RC = 15 ⇒ RC = 15 R E RE

Set RC = 5 k Ω , then R E = 0.333 k Ω Set I CQ = 0.5 mA, then V ECQ ≅ 5 − (0.5)(5.333) = 2.33 V

rπ = 4.68 k Ω , Rib = rπ + (1 + β )R E = 4.68 + (91)(0.333) = 35 k Ω

Set Ri = 22 k Ω = RTH Rib = RTH 35 ⇒ RTH ≅ 60 k Ω Now I BQ = 0.00556 mA, I EQ = 0.506 mA

VCC = I EQ R E + V EB (on ) + I BQ RTH + VTH

5 = (0.506 )(0.333) + 0.7 + (0.00556 )(60 ) + VTH

VTH = 3.798 =

1 1 ⋅ RTH ⋅ VCC = (60)(5) R1 R1

So that R1 = 79 k Ω and R 2 = 249 k Ω ______________________________________________________________________________________ 6.31

β = 120 Let I CQ = 0.35 mA, I EQ = 0.353 mA I BQ = 0.00292 mA Let RE = 2 kΩ. For VCEQ = 4 V ⇒ 10 = 4 + ( 0.35) RC + ( 0.353)( 2 ) RC = 15.1 kΩ, rπ = Av =

− β ( RC RL ) rπ

(120 )( 0.026 )

=−

= 8.91 kΩ 0.35 (120 ) (15.1 10 ) 8.91

Av = −81.0

For bias stable circuit: R1 R2 = RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2 ) = 24.2 kΩ ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10) − 5 = ⋅ RTH ⋅ (10 ) − 5 R1 ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 ( 24.2 )(10 ) − 5 = ( 0.00292 )( 24.2 ) + 0.7 + (121)( 0.00292 )( 2 ) − 5 R1 1 ( 242 ) = 1.477, R1 = 164 kΩ R1 164 R2 = 24.2 ⇒ R2 = 28.4 kΩ 164 + R2 10 = 0.052, 0.35 + 0.052 = 0.402 mA 164 + 28.4 So bias current specification is met. ______________________________________________________________________________________

6.32 RTH = R1 R 2 = 33 50 = 19.88 k Ω

⎛ R2 ⎞ ⎛ 50 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH = ⎜⎜ ⎟(3.3) = 1.988 V ⎝ 50 + 33 ⎠ ⎝ R1 + R 2 ⎠ 3.3 = (101)I BQ (1) + 0.7 + I BQ (19.88) + 1.988

So I BQ = 0.005063 mA, I CQ = 0.5063 mA, I EQ = 0.5114 mA V ECQ = 3.3 − (0.5114)(1) − (0.5063)(2 ) = 1.776 V

Then ΔV EC = 1.776 − 0.5 = 1.276 V, or ΔV EC = 3 − 1.776 = 1.224 V So ΔV EC = 2(1.224 ) = 2.448 V peak-to-peak ______________________________________________________________________________________ 6.33 I BQ =

5 − 0.7 = 0.00315 mA 50 + (101)( 0.1 + 12.9 )

I CQ = 0.315 mA, I EQ = 0.319 mA VCEQ = ( 5 + 5 ) − ( 0.315 )( 6 ) − ( 0.319 )(13) VCEQ = 3.96 V

1 Δv 6.1 eC For ΔiC = 0.315 − 0.05 = 0.265 ⇒ ΔvEC = 1.62

ΔiC = −

vEC ( min ) = 3.96 − 1.62 = 2.34

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Output signal swing determined by current: = 3.24 V peak-to-peak

Max. output swing ______________________________________________________________________________________ 6.34 (a) 5 = (81)I BQ (10) + 0.7 + I BQ (2.5)

So that I BQ = 0.005292 mA, I CQ = 0.4234 mA, I EQ = 0.4287 mA Now V ECQ = 10 − (0.4234 )(5) − (0.4287 )(10 ) = 3.60 V

ΔV EC = ΔI C (RC R L ) = ΔI C (2.5)

For ΔI C = 0.4234 ⇒ ΔV EC = 1.06 V So that ΔV EC = 2(1.06) = 2.12 V peak-to-peak (b) ΔI C = 2(0.4234 ) = 0.847 mA peak-to-peak ______________________________________________________________________________________ 6.35

I EQ = 0.8 mA, I CQ = 0.790 mA, I BQ = 0.009877 mA

V E = 0.7 + (0.009877 )(20 ) = 0.898 V

VC = (0.79 )(2.5) − 5 = −3.025 V Then V ECQ = V E − VC = 3.923 V

ΔV EC = ΔI C (RC R L ) = ΔI C (2.5 4 ) = ΔI C (1.538)

For ΔI C = 0.79 − 0.08 = 0.71 mA, then ΔV EC = (0.71)(1.538) = 1.09 V So, ΔI C = 2(0.71) = 1.42 mA peak-to-peak, ⎛ 2.5 ⎞ Δi o = ⎜ ⎟ΔI C = 0.546 mA peak-to-peak ⎝ 2.5 + 4 ⎠ ______________________________________________________________________________________ 6.36 I BQ =

6 − 0.7 = 0.0169 mA 10 + (101)( 3)

I CQ = 1.69 mA, I EQ = 1.71 mA VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3) VCEQ = 5.38 V

Δi c = −

1 Δυ ce 6.4

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For υ ce (min ) = 1 V, Δυ ce = 5.38 − 1 = 4.38 V ⇒ Δi c =

4.38 = 0.684 mA 6.4

Output swing limited by voltage: Δvce = Max. swing in output voltage = 8.76 V peak-to-peak 1 ΔiC ⇒ Δi0 = 0.342 mA 2 or Δi0 = 0.684 mA (peak-to-peak) ______________________________________________________________________________________ Δi0 =

6.37

ro =

100 I CQ

Neglect ro as 1st approx.: dc load line

VCE = 9 − I C ( 3.4 )

ΔI C = I CQ − 0.1 ΔVCE = VCEQ − 1

Also ΔVCE = ΔI C ( RC RL ) = ΔI C (1.05 ) Or VCEQ − 1 = ( I CQ − 0.1) (1.05 )

Substituting the expression for the dc load line. ⎡⎣9 − I CQ ( 3.4 ) − 1⎤⎦ = ( I CQ − 0.1) (1.05 ) 8.105 = I CQ ( 4.45 ) ⇒ I CQ = 1.821 mA

VCEQ = 2.81 V 1.821 = 0.01821 100 = ( 0.1)(101)(1.2 ) = 12.12 K

I BQ = RTH

VTH =

1 1 ⋅ RTH ⋅ VCC = (12.12 ) ( 9 ) = ( 0.01821) (12.12 ) + 0.7 + (101)( 0.01821)(1.2 ) R1 R1

= 0.2207 + 0.7 + 2.20705 R1 = 34.9 K R2 = 18.6 K 34.9 R2 = 12.12 34.9 + R2 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.38 dc load line

For maximum symmetrical swing ΔiC = I CQ − 0.25 ΔvCE = VCEQ − 0.5 and ΔiC = I CQ − 0.25 =

1 ⋅ | ΔvCE | 0.545 kΩ

VCEQ − 0.5

0.545 VCEQ = 5 − I CQ (1.1)

0.545 ( I CQ − 0.25 ) = ⎡⎣5 − I CQ (1.1) ⎤⎦ − 0.5

( 0.545 + 1.1) I CQ = 5 − 0.5 + 0.136 I CQ = 2.82 mA,

I BQ = 0.0157 mA

RTH = R1 R 2 = (0.1)(1 + β )R E

= (0.1)(181)(0.1) = 1.81 k Ω

VTH =

1 ⋅ RTH ⋅ V + = I BQ RTH + V BE (on ) + (1 + β )I BQ R E R1

1 (1.81)( 5) = ( 0.0157 )(1.81) + 0.7 + (181)( 0.0157 )( 0.1) R1 1 ( 9.05) = 1.013 ⇒ R1 = 8.93 kΩ R1 8.93R2 = 1.81 ⇒ R2 =2.27 kΩ 8.93 + R2

______________________________________________________________________________________ 6.39

I CQ = 0.647 mA, VCEQ = 10 − (0.647 )(9) = 4.18 V Δi c = I CQ = 0.647 mA

So Δυ ce = Δi c (4 4 ) = (0.647 )(2 ) = 1.294 V

Voltage swing is well within the voltage specification. Then Δυ ce = 2(1.294) = 2.59 V peak-to-peak. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.40

(a) Aυ =

(1 + β )R E rπ + (1 + β )R E

0.92 =

(121)(0.5) ⇒ r π rπ + (121)(0.5)

= 5.261 =

β VT I CQ

=

(120)(0.026) ⇒ I I CQ

CQ

= 0.593 mA

VA 20 = = 33.7 k Ω I CQ 0.593

(b) ro = Aυ =

(1 + β )(ro R E ) rπ + (1 + β )(ro R E )

ro R E = 33.7 0.5 = 0.4927 k Ω

(121)(0.4927 ) = 0.919 5.261 + (121)(0.4927 )

Then Aυ =

(c) For (a), R o = R E

rπ 5.261 = 0.5 = 0.5 0.04348 ⇒ R o = 40 Ω 1+ β 121

rπ = 0.04 33.7 ⇒ R o = 39.95 Ω 1+ β ______________________________________________________________________________________

For (b), Ro = R E ro

6.41 (a) Rib = rπ + (1 + β )R E

(1 + β )R E (1 + β )R E = rπ + (1 + β )R E Rib ( A )(R ) (0.95)(50) = υ ib = = 0.586 k Ω

Aυ = RE

1+ β 81 50 = rπ + (81)(0.586 ) ⇒ rπ = 2.5 k Ω rπ =

β VT I CQ

(b) R o = R E

⇒ I CQ =

(80)(0.026) = 0.832 mA 2.5

rπ 2.5 = 0.586 = 0.586 0.03086 1+ β 81

R o = 29.3 Ω ______________________________________________________________________________________

6.42 (a) g m = 76.92 mA/V, rπ = 1.04 k Ω , ro = 25 k Ω (1 + β )ro (81)(25) = = 0.999 Aυ = R S + rπ + (1 + β )ro 1 + 1.04 + (81)(25)

Ri = R S + rπ + (1 + β )ro = 1 + 1.04 + (81)(25) ⇒ Ri = 2.027 M Ω

Ro = ro

rπ + R S ⎛ 1.04 + 1 ⎞ = 25 ⎜ ⎟ = 25 0.0252 ⇒ R o = 25.2 Ω 1+ β ⎝ 81 ⎠

(b) rπ = 10.4 k Ω , ro = 250 k Ω Aυ =

(81)(250) = 0.9994 1 + 10.4 + (81)(250)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Ri = 1 + 10.4 + (81)(250) ⇒ Ri = 20.26 M Ω ⎛ 10.4 + 1 ⎞ R o = 250 ⎜ ⎟ = 250 0.1407 ⇒ R o = 141 Ω ⎝ 81 ⎠ ______________________________________________________________________________________

6.43 rπ + R S 1+ β r + 0.5 0.015 = π ⇒ rπ = 1.315 k Ω 121 β VT (120)(0.026) = 2.37 mA rπ = ⇒ I CQ = I CQ 1.315

(a) R o =

50 = 21.07 k Ω 2.373 (1 + β )ro (121)(21.07 ) = = 0.9993 Aυ = R S + rπ + (1 + β )ro 0.5 + 1.315 + (121)(21.07 ) ______________________________________________________________________________________ (b) ro =

6.44 a. RTH = R1 R 2 = 10 10 = 5 k Ω ⎛ R2 ⎞ ⎛ 10 ⎞ ⎟⎟(18) − 9 = ⎜ VTH = ⎜⎜ ⎟(18) − 9 = 0 ⎝ 10 + 10 ⎠ ⎝ R1 + R 2 ⎠ 0 − 0.7 − (− 9 ) I BQ = = 0.0869 mA 5 + (181)(0.5) I CQ = 15.6 mA, I EQ = 15.7 mA VCEQ = 18 − (15.7 )(0.5) = 10.1 V

b.

c. rπ = Aυ =

(180)(0.026) = 0.30 k Ω 15.6

(1 + β )(R E R L ) ⎛⎜ R1 R2 Rib ⎞⎟ ⋅ rπ + (1 + β )(R E R L ) ⎜ R1 R 2 Rib + R S ⎟ ⎝ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Rib = rπ + (1 + β )(R E R L ) = 0.30 + (181)(0.5 0.3) = 34.2 k Ω R1 R 2 Rib = 5 34.2 = 4.36 k Ω

(181)(0.5 0.3) ⎛ 4.36 ⎞ ⋅⎜ ⎟ ⇒ Aυ 0.3 + (181)(0.5 0.3) ⎝ 4.36 + 1 ⎠

Aυ =

d.

= 0.806

Rib = rπ + (1 + β )(R E R L )

Rib = 0.30 + (181)(0.188) = 34.3 k Ω ⎛ rπ + R1 R 2 R S ⎞ ⎟ = 0.5 ⎛⎜ 0.3 + 5 1 ⎞⎟ ⇒ R = 6.18 Ω Ro = R E ⎜ o ⎜ ⎟ ⎜ 181 ⎟ 1+ β ⎝ ⎠ ⎝ ⎠ ______________________________________________________________________________________

6.45 a. RTH = R1 R 2 = 10 10 = 5 k Ω

⎛ R2 ⎞ ⎟⎟(− 10 ) = −5 V VTH = ⎜⎜ ⎝ R1 + R 2 ⎠ VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E − 10 − 5 − 0.7 − (− 10 ) = 0.0174 mA 5 + (121)(2) = 2.09 mA, I EQ = 2.11 mA

I BQ =

I CQ

VCEQ = 10 − (2.09 )(1) − (2.11)(2 ) = 3.69 V

b.

c. rπ =

(120 )( 0.026 )

= 1.49 kΩ 2.09 (1 + β ) ( RE RL ) ⎛ R1 R2 Rib ⎞ Av = ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎜⎝ R1 R2 Rib + RS ⎟⎠ Rib = rπ + (1 + β ) ( RE RL ) = 1.49 + (121) ( 2 2) Rib = 122.5 k Ω, R1 R2 Rib = 5 122.5 = 4.80 k Ω Av =

(121) ( 2 2) ⎛ 4.80 ⎞ ⋅⎜ ⎟ ⇒ Av 1.49 + (121) ( 2 2) ⎝ 4.80 + 5 ⎠

= 0.484

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ d.

Rib = rπ + (1 + β )(R E R L )

Rib = 1.49 + (121)(2 2) ⇒ Rib = 122 k Ω ⎛ rπ + R1 R 2 R S ⎞ ⎟ = 2 ⎛⎜ 1.49 + 5 5 ⎞⎟ ⇒ R = 32.4 Ω Ro = R E ⎜ o ⎜ ⎟ ⎜ 121 ⎟ 1+ β ⎝ ⎠ ⎝ ⎠ ______________________________________________________________________________________

6.46 (a) RTH = R1 R 2 = 585 135 = 109.7 k Ω ⎛ R2 ⎞ ⎛ 135 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH = ⎜⎜ ⎟(3.3) = 0.61875 V ⎝ 135 + 585 ⎠ ⎝ R1 + R 2 ⎠ 3.3 = (1 + β )I BQ R E + V EB (on ) + I BQ RTH + VTH 3.3 − 0.7 − 0.61875 = 0.001649 mA 109.7 + (91)(12 ) Then I CQ = 0.1484 mA, I EQ = 0.150 mA I BQ =

V ECQ = 3.3 − (0.15)(12 ) = 1.5 V

(c) rπ = Aυ =

(90)(0.026) = 15.77 k Ω , 0.1484

ro =

(1 + β )(ro R E R L ) rπ + (1 + β )(ro R E R L )

60 = 404 k Ω 0.1484

ro R E R L = 404 12 4 = 2.978 k Ω

(91)(2.978) = 0.945 15.77 + (91)(2.978) = rπ + (1 + β )(ro R E R L ) = 15.77 + (91)(2.978) = 286.8 k Ω

Aυ = Rib

⎛ RTH Ai = (1 + β )⎜⎜ ⎝ RTH + Rib (d) Rib = 286.8 k Ω R o = R E ro

⎞⎛ ro R E ⎟⎜ ⎟⎜ r R + R L ⎠⎝ o E

⎞ 109.7 ⎞⎛⎜ 404 12 ⎞⎟ ⎟ = (91)⎛⎜ = 18.7 ⎟ ⎟ ⎝ 109.7 + 286.8 ⎠⎜⎝ 404 12 + 4 ⎟⎠ ⎠

rπ 15.77 = 12 404 ⇒ R o = 171 Ω 1+ β 91

______________________________________________________________________________________ 6.47 (a) 0.5 = 0.00617 mA 81 VB = I BQ RB = ( 0.00617 )(10 ) ⇒ VB = 0.0617 V

I BQ =

VE = VB + 0.7 ⇒ VE = 0.7617 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) ⎛ 80 ⎞ I CQ = ( 0.5 ) ⎜ ⎟ = 0.494 mA ⎝ 81 ⎠ I CQ 0.494 = ⇒ g m = 19 mA / V gm = 0.026 VT rπ = ro =

β VT I CQ

=

(80 )( 0.026 ) 0.494

⇒ rπ = 4.21 k Ω

VA 150 = ⇒ ro = 304 k Ω I CQ 0.494

(c)

For R S = 0 ⎛V ⎞ V o = −⎜⎜ π + g mVπ ⎟⎟(R L ro ) ⎝ rπ ⎠ −V o so that Vπ = ⎛ 1+ β ⎞ ⎜ ⎟ ⎜ r ⎟(R L ro ) ⎝ π ⎠ Now V s + Vπ = V o V s = V o − Vπ = V o +

We find Aυ =

Vo ⎛ 1+ β ⎜ ⎜ r ⎝ π

⎞ ⎟(R L ro ) ⎟ ⎠

(1 + β )(R L ro ) (81)(0.5 304) Vo = = V s rπ + (1 + β )(R L ro ) 4.21 + (81)(0.5 304)

Aυ = 0.906

Rib = rπ + (1 + β )(R L ro ) ≅ 4.21 + (81)(0.5) = 44.7 k Ω ⎛ RB I b = ⎜⎜ ⎝ R B + Rib

⎛ ro ⎞ ⎞ ⎟ ⋅ I s and I o = ⎜ ⎟ ⎜ r + R ⎟(1 + β )I b ⎟ L ⎠ ⎝ o ⎠

Then ⎛ RB ⎞⎛ ro ⎞ Io = (1 + β ) ⎜ ⎟⎜ ⎟ Is ⎝ RB + Rib ⎠⎝ ro + RL ⎠ ⎛ 10 ⎞ Ai ≅ ( 81) ⎜ ⎟ (1) ⇒ Ai = 14.8 ⎝ 10 + 44.7 ⎠ Ai =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (d) ⎛ R B Rib ⎞ ⎛ 10 44.7 ⎞ ⎟ ⋅Vs = ⎜ ⎟ ⋅ V = (0.803)V s V s′ = ⎜ ⎜R R +R ⎟ ⎜ 10 44.7 + 2 ⎟ s S ⎠ ⎝ B ib ⎝ ⎠ ( )( ) Then Aυ = 0.803 0.906 ⇒ Aυ = 0.728 Ai = 14.8 (unchanged) ______________________________________________________________________________________

6.48 (a) I CQ = 1.98 mA, rπ =

(100)(0.026) = 1.313 k Ω 1.98

VA 100 = = 50.5 k Ω I CQ 1.98

ro =

rπ + R S 1.31 + 10 ro = 50.5 ⇒ R o = 112 Ω 1+ β 101 (b) From Equation (6.68) (1 + β ) ( ro RL ) 100 Av = ro = = 50.5 K 1.98 rπ + (1 + β ) ( ro RL ) Ro =

(i) RL = 0.5 K

(101) ( 50.5 0.5) 1.31 + (101) ( 50.5 0.5 ) (101)( 0.4951) Av = ⇒ Av = 0.974 1.31 + (101)( 0.4951) Av =

(ii) RL = 5 K Av =

ro RL = 50.5 5 = 4.5495

(101)( 4.55) ⇒ Av = 0.997 1.31 + (101)( 4.55 )

______________________________________________________________________________________ 6.49

(a)

5 − 0.7 = I EQ = 1.303 mA, I CQ = 1.29 mA 3.3 V ECQ = 0.7 − (− 5) = 5.7 V

(b) rπ = Aυ =

(110)(0.026) = 2.217 k Ω , 1.29

(1 + β )(ro R E R L ) rπ + (1 + β )(ro R E R L )

ro =

50 = 38.76 k Ω 1.29

ro R E R L = 38.76 3.3 1 = 0.7525 k Ω

(111)(0.7525) = 0.974 2.217 + (111)(0.7525) = rπ + (1 + β )(ro R E R L ) = 2.217 + (111)(0.7525) = 85.7 k Ω

Aυ = Rib

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ rπ 2.217 R E ro = 3.3 38.76 ⇒ R o = 19.8 Ω 1+ β 111 υ (t ) 2.8 sin ω t (c) i s (t ) = s = ⇒ i s (t ) = 32.7 sin ω t ( μ A) Rib 85.7 Ro =

⎛ R E ro ⎞ ⎟ ⋅ i (t ) = (111)⎛⎜ 3.041 ⎞⎟(32.7 sin ω t ) i o (t ) = (1 + β )⎜ ⎜R r +R ⎟ s ⎝ 3.041 + 1 ⎠ L ⎠ ⎝ E o or i o (t ) = 2.73 sin ω t (mA) υ o (t ) = i o (t )R L = 2.73 sin ω t (V) υ eb (t ) = −i s (t )rπ = −(32.7 sin ω t )(2.217 ) υ eb (t ) = −72.5 sin ω t (mV) ______________________________________________________________________________________

6.50 a. I EQ = 1 mA , VCEQ = VCC − I EQ RE

5 = 10 − (1)( RE ) ⇒ RE = 5 kΩ

1 = 0.0099 mA 101 10 = I BQ RB + VBE ( on ) + I EQ RE

I BQ =

10 = ( 0.0099 ) RB + 0.7 + (1)( 5 ) ⇒ RB = 434 kΩ

b.

rπ =

(100 )( 0.026 ) 0.99 (1 + β ) RE

= 2.63 kΩ

(101)( 5 ) v0 = = = 0.995 vb rπ + (1 + β ) RE 2.63 + (101)( 5 ) ⇒ vb =

v0 4 = ⇒ vb = 4.02 V peak-to-peak at base 0.995 0.995

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Rib = rπ + (1 + β ) RE = 508 kΩ RB Rib = 434 508 = 234 kΩ vb =

RB Rib RB Rib + RS

⋅ vS =

vb = 0.997vS ⇒ vS =

c.

234vS 234 = vS 234 + 0.7 234.7

4.02 ⇒ vS = 4.03 V peak-to-peak 0.997

Rib = rπ + (1 + β ) ( RE RL )

Rib = 2.63 + (101) ( 5 1) = 86.8 kΩ RB Rib = 434 86.8 = 72.3 kΩ

⎛

72.3 ⎞ ⎟ ⋅υ s = 0.99υ s = (0.99 )(4.03) 72 . ⎝ 3 + 0.7 ⎠

υb = ⎜

υ b = 3.99 V peak-to-peak

(1 + β )(R E R L ) ⋅υ b rπ + (1 + β )(R E R L ) (101)(0.833) ⋅ (3.99) = 2.63 + (101)(0.833)

υo =

υ o = 3.87 V peak-to-peak ______________________________________________________________________________________ 6.51 RTH = R1 R 2 = 40 60 = 24 k Ω ⎛ 60 ⎞ VTH = ⎜ ⎟(10 ) = 6 V ⎝ 60 + 40 ⎠ 6 − 0.7 = 0.0131 mA For β = 75 , I BQ = 24 + (76 )(5) I CQ = 0.984 mA

For β = 150 , I BQ =

6 − 0.7 = 0.00680 mA 24 + (151)(5)

I CQ = 1.02 mA

For β = 75 , rπ =

(75)(0.026) = 1.98 k Ω

0.984 (150)(0.026) = 3.82 k Ω For β = 150 , rπ = 1.02 For β = 75 , Rib = rπ + (1 + β )(R E R L ) = 65.3 k Ω

For β = 150 , Rib = 130 k Ω Aυ =

R1 R 2 Rib (1 + β )(R E R L ) ⋅ rπ + (1 + β )(R E R L ) R1 R 2 Rib + R S

For β = 75 , R1 R 2 Rib = 40 60 65.3 = 17.5 k Ω Aυ =

(76)(0.833) ⋅ 17.5 = 0.789 1.98 + (76 )(0.833) 17.5 + 4

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For β = 150 , R1 R 2 Rib = 40 60 130 = 20.3 k Ω Aυ =

(151)(0.833) ⋅ 20.3 = 0.811 3.82 + (151)(0.833) 20.3 + 4

So 0.789 ≤ Av ≤ 0.811

⎛ RE ⎞ ⎛ RTH ⎞ Ai = (1 + β ) ⎜ ⎟ ⎟⎜ ⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠ β = 75 24 ⎞ ⎛ 5 ⎞⎛ Ai = ( 76 ) ⎜ ⎟⎜ ⎟ ⇒ Ai = 17.0 ⎝ 5 + 1 ⎠ ⎝ 24 + 65.3 ⎠ β = 150 ⎛ 5 ⎞ ⎛ 24 ⎞ Ai = (151) ⎜ ⎟ ⎜ ⎟ ⇒ Ai = 19.6 ⎝ 6 ⎠ ⎝ 24 + 130 ⎠ 17.0 ≤ Ai ≤ 19.6

______________________________________________________________________________________ 6.52 (a) ⎛ I ⎞ 9 = ⎜ E ⎟ (100 ) + VBE ( on ) + I E RE ⎝ 1+ β ⎠ 9 − 0.7 IE = ⎛ 100 ⎞ ⎜ ⎟ + RE ⎝ 1+ β ⎠ 8.3 = 2.803 mA ⎛ 100 ⎞ ⎜ ⎟ +1 ⎝ 51 ⎠ 8.3 = 5.543 mA β = 200 I E = ⎛ 100 ⎞ + 1 ⎜ ⎟ ⎝ 201 ⎠

β = 50 I E =

2.80 ≤ I E ≤ 5.54 mA VE = I E RE , β = 50, VE = 2.80 V

β = 200, VE = 5.54 V β = 50, I CQ = 2.748 mA, rπ = 0.473 K (b) β = 200, I CQ = 5.515 mA, rπ = 0.943 K

[

]

Ri = R B rπ + (1 + β )(R E R L )

β = 50 , Ri = 100 [0.473 + (51)(1 1)] = 100 25.97 = 20.6 k Ω

β = 200 , Ri = 100 [0.943 + (201)(1 1)] = 100 101.4 = 50.3 k Ω From Fig. (6.68)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(1 + β ) ( RE RL ) ⎛ Ri ⎞ ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎝ Ri + RS ⎠ ( 51) (1 1) ⎛ 20.6 ⎞ = ⋅⎜ ⎟ 0.473 + ( 51) (1 1) ⎝ 20.6 + 10 ⎠

Av =

β = 50 ⇒ Av = 0.661 β = 200 ⇒ Av =

( 201) (1 1) ⎛ 50.3 ⎞ ⎜ ⎟ 0.943 + ( 201) (1 1) ⎝ 50.3 + 10 ⎠

Av = 0.826

______________________________________________________________________________________ 6.53

Vo = (1 + β ) I b RL Vs rπ + (1 + β ) RL

Ib =

so Av =

(1 + β ) RL rπ + (1 + β ) RL

For β = 100, RL = 0.5 k Ω rπ =

(100 )( 0.026 ) 0.5

Then Av ( min ) =

= 5.2 k Ω

(101)( 0.5 ) = 0.9066 5.2 + (101)( 0.5 )

Then β = 180, RL = 500 k Ω rπ =

(180 )( 0.026 ) 0.5

Then Av ( max ) =

= 9.36 k Ω

(181)( 500 ) = 0.9999 9.36 + (181)( 500 )

______________________________________________________________________________________ 6.54 (a) VCEQ = VCC − I EQ R E

Let VCEQ = 2.5 V, then 2.5 = 5 − I EQ (0.5) ⇒ I EQ ≅ I CQ = 5 mA, I BQ = 0.04167 mA rπ =

(120)(0.026) = 0.624 k Ω

5 Rib = rπ + (1 + β )(R E R L ) = 0.624 + (121)(0.5 0.5) = 30.87 k Ω ⎛ RTH Ai = (1 + β )⎜⎜ ⎝ RTH + Rib

⎞⎛ R E ⎟⎜⎜ ⎟ R +R L ⎠⎝ E

⎞ ⎟⎟ ⎠

⎛ ⎞⎛ 0.5 ⎞ RTH ⎟⎟⎜ 10 = (121)⎜⎜ ⎟ ⇒ RTH = 6.113 k Ω ⎝ RTH + 30.87 ⎠⎝ 0.5 + 0.5 ⎠ VTH = I BQ RTH + V BE (on ) + I EQ R E = (0.04167 )(6.113) + 0.7 + (5)(0.5) = 3.455 V VTH = 3.455 =

1 1 ⋅ RTH ⋅ VCC = (6.113)(5) R1 R1

So R1 = 8.85 k Ω and R 2 = 19.8 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Ro = R E

rπ 0.624 = 0.5 ⇒ Ro = 5.10 Ω 1+ β 121

⎛ RTH ⎞⎛ R E ⎞ ⎟⎜⎜ ⎟⎟ (b) Ai = (1 + β )⎜⎜ ⎟ ⎝ RTH + Rib ⎠⎝ R E + R L ⎠ Rib = rπ + (1 + β )(R E R L ) = 0.624 + (121)(0.5 2 ) = 49.02 k Ω

6.113 ⎛ ⎞⎛ 0.5 ⎞ Ai = (121)⎜ ⎟⎜ ⎟ = 2.68 ⎝ 6.113 + 49.02 ⎠⎝ 0.5 + 2 ⎠ ______________________________________________________________________________________ 6.55

Ri = RTH Rib where Rib = rπ + (1 + β ) RE 5 − 3.5 VCEQ = 3.5, I CQ = 0.75 mA 2 (120 )( 0.026 ) rπ = = 4.16 k Ω 0.75 Rib = 4.16 + (121) ( 2 ) = 246 k Ω Then Ri = 120 = RTH 246 ⇒ RTH = 234 k Ω 0.75 I BQ = = 0.00625 mA 120 VTH = I BQ RTH + VBE ( on ) + (1+ β ) I BQ RE 1 1 ⋅ RTH ⋅ VCC = ( 234)(5) = ( 0.00625) ( 234) + 0.7 + (121)( 0.00625 )( 2 ) R1 R1 which yields R1 = 318 k Ω and R2 = 886 k Ω

______________________________________________________________________________________ 6.56 a. Let RE = 24 Ω and VCEQ = 12 VCC = 12 V ⇒ I EQ = I CQ = 0.493 A, I BQ = 6.58 mA rπ =

( 75)( 0.026 ) 0.493

= 3.96 Ω

⎛ RE ⎞ I 0 = (1 + β ) I b ⎜ ⎟ ⎝ RE + RL ⎠ ⎛ RTH ⎞ Ib = I S ⎜ ⎟ ⎝ RTH + Rib ⎠

12 = 0.5 A 24

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Rib = rπ + (1 + β )(R E R L )

= 3.96 + (76 )(24 8) ⇒ Rib = 460 Ω ⎛ RE ⎞ ⎛ RTH ⎞ I0 = (1 + β ) ⎜ ⎟ ⎟⎜ IS ⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠ ⎞ ⎛ 24 ⎞ ⎛ RTH 8 = ( 76) ⎜ ⎟⎜ ⎟ ⎝ 24 + 8 ⎠ ⎝ RTH + 460 ⎠

Ai =

0.140 =

RTH ⇒ RTH = 74.9 Ω (Minimum value) RTH + 460

dc analysis: 1 VTH = ⋅ RTH ⋅ VCC R1

= I BQ RTH + VBE ( on ) + I EQ RE 1 ( 74.9 )( 24 ) = ( 0.00658)( 74.9 ) + 0.70 + ( 0.5)( 24 ) R1 = 13.19 R1 = 136 Ω,

136 R2 = 74.9 ⇒ R2 = 167 Ω 136 + R2

b.

1 ΔiC = − Δvce 6 For ΔiC = 0.493 ⇒ Δvce = ( 0.493)( 6 ) ⇒ Max. swing in output voltage for this design = 5.92 V peak-to-peak

c. rπ 3.96 24 = 0.0521 24 ⇒ R0 = 52 mΩ RE = 1+ β 76 ______________________________________________________________________________________ R0 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.57 The output of the emitter follower is ⎛ RL vo = ⎜ ⎝ RL + Ro

⎞ ⎟ ⋅ vTH ⎠

For vO to be within 5% for a range of RL , we have RL ( min )

RL ( min ) + Ro

= ( 0.95 )

RL ( max )

RL ( max ) + Ro

4 10 which yields Ro = 0.364 k Ω = ( 0.95 ) 4 + Ro 10 + Ro

⎛ rπ + R1 R 2 R S ⎞ ⎟R r We have Ro = ⎜ ⎟ E o ⎜ 1+ β ⎠ ⎝ The first term dominates Let R1 R 2 R S ≅ R S , then rπ + R S r +4 ⇒ 0.364 = π 1+ β 1+ β rπ β VT 4 4 or 0.364 = + = + 1 + β 1 + β I CQ (1 + β ) 1 + β Ro ≅

0.364 ≅

VT 4 + I CQ 1 + β

4 1+ β

V 4 4 Ro ≅ 0.32 = T = 0.044 = 0.0305. I CQ The factor is in the range of 91 to 131 We can set I CQ = 0.08125 mA. I CQ = 0.15 mA,

Or

To take into account other factors, set

0.15 I BQ = = 0.00136 mA 110 5 RE = = 33.3 k Ω VCEQ ≅ 5 V , 0.15 For set

Design a bias stable circuit. ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10) − 5 = ( RTH )(10) − 5 + R R R ⎝ 1 2 ⎠ 1 RTH = ( 0.1)(1 + β ) RE = ( 0.1)(111)(33.3) = 370 k Ω VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 ( 370 )(10 ) − 5 = ( 0.00136 )( 370 ) + 0.7 + (111)( 0.00136 )( 33.3) − 5 R 1 which yields R1 = 594 k Ω and R2 = 981 k Ω

So

(1 + β ) ( RE RL ) ⎛ RTH Rib ⎞ ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎝ RTH Rib + RS ⎠ β VT Rib = rπ + (1 + β ) ( RE RL ) and rπ = I CQ

Now Av =

For β = 90, RL = 4 k Ω, rπ = 15.6 k Ω , Rib = 340.6 k Ω

(91)(33.3 4) 370 340.6 ⋅ ⇒ Aυ 15.6 + (91)(33.3 4 ) 370 340.6 + 4

Aυ =

= 0.9332

For β = 90, RL = 10 k Ω

Rib = 715.4 k Ω Aυ =

(91)(33.3 10) 370 715.4 ⋅ ⇒ Aυ 15.6 + (91)(33.3 10) 370 715.4 + 4

= 0.9625

For β = 130, RL = 4 k Ω rπ = 22.5 k Ω , Rib = 490 k Ω Aυ =

(131)(33.3 4) 370 490 ⋅ ⇒ Aυ 22.5 + (131)(33.3 4 ) 370 490 + 4

= 0.9360

For β = 130, RL = 10 k Ω Rib = 1030 k Ω Aυ =

(131)(33.3 10) 370 1030 ⋅ ⇒ Aυ 22.5 + (131)(33.3 10 ) 370 1030 + 4

= 0.9645

Now vO ( min ) = Av ( min ) .vS = 3.73sin ω t vO ( max ) = Av ( max ) .vS = 3.86sin ω t ΔvO = 3.5% vO ______________________________________________________________________________________

6.58

PAVG = iL2 ( rms ) RL ⇒ 1 = iL2 ( rms )(12 ) so iL ( rms ) = 0.289 A ⇒ iL ( peak ) = 2 ( 0.289 ) iL ( peak ) = 0.409 A

vL ( peak ) = iL ( peak ) ⋅ RL = ( 0.409 )(12 ) = 4.91 V Need a gain of

4.91 = 0.982 5

With RS = 10 k Ω, we will not be able to meet this voltage gain requirement. Need to insert a buffer or an op-amp voltage follower (see Chapter 9) between RS and CC1 .

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 (12 − ( −12 ) ) = 8 V 3 = ( 0.5 ) RE + 8 ⇒ RE = 32 Ω

Set I EQ = 0.5 A, VCEQ = 24 = I EQ RE + VCEQ

50 ( 0.5 ) = 0.49 A 51 β VT ( 50 )( 0.026 ) rπ = = = 2.65 Ω I CQ 0.49

Let β = 50, I CQ =

Rib = rπ + (1 + β ) ( RE RL ) = 2.65 + ( 51) ( 32 12 ) Rib = 448 Ω Av =

(1 + β ) ( RE RL ) ( 51) ( 32 12 ) = = 0.994 rπ + (1 + β ) ( RE RL ) 2.65 + ( 51) ( 32 12 )

So gain requirement has been met. 0.49 I BQ = = 0.0098 A = 9.8 mA 50 24 Let I R ≅ ≅ 10 I B = 98 mA R1 + R2 So that R1 + R2 = 245 Ω R2 VTH = ( 24 ) − 12 = I BQ RTH + VBE ( on ) + I EQ RE − 12 R1 + R2

( 0.0098) R1 R2 ⎛ R2 ⎞ + 0.7 + ( 0.5 )( 32 ) ⎜ 245 ⎟ ( 24 ) = 245 ⎝ ⎠ Now R1 = 245 − R2 So we obtain 4 × 10−5 R22 + 0.0882 R2 − 16.7 = 0 which yields R2 = 175 Ω and R1 = 70 Ω ______________________________________________________________________________________ 6.59 1 = 38.46 mA/V 0.026 Aυ = g m RC = (38.46 )(2 ) = 76.9

(a) g m =

β 120 = = 0.9917 1 + β 121 r (120)(0.026) = 3.12 k Ω (c) Ri = π , rπ = 1+ β 1 3.12 Ri = ⇒ Ri = 25.8 Ω 121 (d) Ro = RC = 2 k Ω ______________________________________________________________________________________ (b) Ai =

6.60

⎛ 80 ⎞ (a) I CQ = ⎜ ⎟(2 ) = 1.975 mA ⎝ 81 ⎠ (80)(0.026) = 1.053 k Ω 1.975 = 75.97 mA/V, rπ = gm = 0.026 1.975

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Aυ = g m (RC R L ) = (75.97 )(1.5 2.5) = 71.2 ⎛ β ⎞⎛ RC ⎞ ⎛ 80 ⎞⎛ 1.5 ⎞ ⎟ = ⎜ ⎟⎜ ⎟⎟⎜⎜ (b) Ai = ⎜⎜ ⎟ = 0.370 ⎟ ⎝ 1 + β ⎠⎝ RC + R L ⎠ ⎝ 81 ⎠⎝ 1.5 + 2.5 ⎠ r 1.053 (c) Ri = π = ⇒ Ri = 13 Ω 1+ β 81 80 = 40.5 k Ω (d) ro = 1.975 (i) R oc = ro = 40.5 k Ω R o = ro RC R L = 40.5 1.5 2.5 = 0.916 k Ω

(ii)

______________________________________________________________________________________ 6.61

⎛ 110 ⎞ (a) I CQ = ⎜ ⎟(0.5) = 0.4955 mA ⎝ 111 ⎠ (110)(0.026) = 5.77 k Ω 0.4955 = 19.06 mA/V, rπ = gm = 0.026 0.4955 ⎛ R ⎞⎡ r ⎤ ⎛ 4 ⎞ ⎡ 5.77 ⎤ Aυ = g m ⎜⎜ C ⎟⎟ ⎢ π R S ⎥ = (19.06 )⎜ ⎟ ⎢ 1⎥ = (76.24 ) 0.05198 1 ⎝ 1 ⎠ ⎣ 111 ⎦ ⎦ ⎝ R S ⎠ ⎣1 + β Aυ = 3.77

[

β

]

110 = 0.991 111 r 5.77 (c) Ri = R S + π = 1 + = 1.052 k Ω 1+ β 111 (d) Ro = RC = 4 k Ω ______________________________________________________________________________________

(b) Ai =

1+ β

=

6.62

0.7 = 0.25 mA 2.8 ≅ 1.50 − 0.25 = 1.25 mA

(a) I R 2 = I CQ

VC = (0.25)(5 + 2.8) = 1.95 V = VCEQ

(120)(0.026) = 2.5 k Ω 1.25 = 48.08 mA/V, rπ = 0.026 1.25 Aυ = g m (R1 R L ) = (48.08)(5 10 ) = 160.3

(b) g m =

⎛ β ⎞⎛ R1 ⎞ ⎛ 120 ⎞⎛ 5 ⎞ ⎟⎟ = ⎜ ⎟⎟⎜⎜ (c) Ai = ⎜⎜ ⎟⎜ ⎟ = 0.3306 ⎝ 1 + β ⎠⎝ R1 + R L ⎠ ⎝ 121 ⎠⎝ 5 + 10 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.63 (a) 10 − 0.7 = 0.93 mA 10 = 0.921 mA

I EQ = I CQ

VECQ = 20 − ( 0.93)(10 ) − ( 0.921)( 5 ) VECQ = 6.10 V

(b) 0.921 = 35.42 mA/V 0.026 Av = g m ( RC RL ) = ( 35.42 ) ( 5 50 )

gm =

Av = 161

______________________________________________________________________________________ 6.64

(a)

I EQ = 0.93 mA, I CQ = 0.921 mA VECQ = 6.10 V

0.921 = 35.42 mA/V rπ = 2.82 K 0.026 (b) From Eq. 6.90 gm =

Av = g m = Av =

(R

RL ) ⎡ rπ ⎤ ⎢1 + β RE RS ⎥ RS ⎣ ⎦

C

( 35.42 ) ( 50 5 ) ⎡ 2.82 0.1

⎤ ⎢ 101 10 0.1⎥ ⎣ ⎦

( 35.42 )( 4.545) 0.1

[ 0.0218]

Av = 35.1

______________________________________________________________________________________ 6.65

(a)

I EQ = 1 mA, I CQ = 0.9917 mA VC = 5 − ( 0.9917 )( 2 ) = 3.017 V VE = −0.7 V VCEQ = 3.72 V

(b)

Av = g m ( RC RL ) 0.9917 = 38.14 mA/V 0.026 Av = ( 38.14 ) ( 2 10 ) ⇒ Av = 63.6

gm =

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.66 a. Emitter current

I EQ = I CC = 0.5 mA 0.5 = 0.00495 mA 101 VE = I EQ RE = ( 0.5 )(1) ⇒ VE = 0.5 V

I BQ =

VB = VE + VBE ( on ) = 0.5 + 0.7 ⇒ VB = 1.20 V VC = VB + I BQ RB = 1.20 + ( 0.00495 )(100 ) ⇒ VC = 1.7 V

b. rπ =

(100 )( 0.026 ) = 5.25 kΩ (100 )( 0.00495 ) (100 )( 0.00495 )

gm =

= 19.0 mA/V

0.026

V o = − g mVπ (R B R L )

Vπ =

− (R E Rie )

R E Rie + R S

⋅ V s = −(0.4971)V s

V o = (19 )(0.4971)V s (100 1) Aυ = 9.37

c.

IX =

VX VX + − g mVπ , Vπ = −V X R E rπ

IX 1 1 1 = = + + gm VX Ri R E rπ

or Ri = R E rπ

1 1 = 1 5.253 = 0.84 0.05252 gm 19

Ri = 49.4 Ω ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.67 (a) RTH = R1 R 2 = 150 50 = 37.5 k Ω ⎛ R2 ⎞ ⎛ 50 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH = ⎜⎜ ⎟(9) = 2.25 V R R + ⎝ 50 + 150 ⎠ 2 ⎠ ⎝ 1 VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E 2.25 − 0.7 = 0.00373 mA, I CQ = 0.4663 mA, I EQ = 0.470 mA 37.5 + (126)(3) = 9 − (0.4663)(6 ) − (0.470 )(3) = 4.79 V

or I BQ = VCEQ

(125)(0.026) = 6.97 k Ω 0.4663 = 17.93 mA/V, rπ = 0.026 0.4663 ⎛ R S R E ⎞⎛ β ⎞ υ ⎟⎜ ⎟(RC R L ) Rm = o = ⎜ i s ⎜⎝ R S R E + Rie ⎟⎠⎜⎝ 1 + β ⎟⎠ We find R S R E = 100 3 = 2.913 k Ω

(b) g m =

Rie =

rπ + RTH 6.97 + 37.5 = = 0.3529 k Ω 1+ β 126

2.913 ⎞⎛ 125 ⎞ ⎛ Rm = ⎜ ⎟(6 4 ) = 2.12 V/mA ⎟⎜ ⎝ 2.913 + 0.3529 ⎠⎝ 126 ⎠

(

) (

)

(c) υ s = i s R S R E Rie = i s 100 3 0.3529 = i s (0.3148) So i s =

υs 0.3148

Then Aυ =

υ o υ o is 2.12 = ⋅ = = 6.73 υ s i s υ s 0.3148

______________________________________________________________________________________ 6.68 (a) VCEQ ≅ VCC − I CQ (RC + R E )

Let VCEQ = 2.5 V and I CQ = 0.25 mA

Then 2.5 = 5 − (0.25)(RC + 0.5) ⇒ RC = 9.5 k Ω ⎛ β ⎝ 1+ β

υ o = i c (RC R L ) = ⎜⎜ ie =

υs Rie

Aυ =

=

υs ⎛ rπ + RTH ⎜⎜ ⎝ 1+ β

υo ⎛ β =⎜ υ s ⎜⎝ rπ + RTH

Now rπ =

⎞ ⎟⎟ ⋅ i e ⋅ (RC R L ) ⎠

⎞ ⎟⎟ ⎠ ⎞ ⎟(RC R L ) ⎟ ⎠

(100)(0.026) = 10.4 k Ω 0.25

⎛ 100 Aυ = 25 = ⎜⎜ ⎝ 10.4 + RTH Also I BQ = 0.0025 mA

⎞ 530.2 ⎟⎟(9.5 12 ) = ⇒ RTH = 10.81 k Ω 10.4 + RTH ⎠

VTH = I BQ RTH + V BE (on ) + (1 + β )I BQ R E = (0.0025)(10.81) + 0.7 + (101)(0.0025)(0.5)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VTH = 0.853275 =

1 1 ⋅ RTH ⋅ VCC = (10.81)(5) R1 R1

We obtain R1 = 63.3 k Ω and R 2 = 13 k Ω (b) From part (a), I CQ = 0.25 mA, VCEQ = 2.5 V

(c) Aυ = g m (RC R L )

0.25 = 9.615 mA/V 0.026 Aυ = (9.615)(9.5 12 ) = 51 gm =

______________________________________________________________________________________ 6.69 (a) ⎛ 60 ⎞ I CQ = ⎜ ⎟ (1) ⇒ I CQ = 0.984 mA ⎝ 61 ⎠ ⎛ 1⎞ VCEQ = I BQ RB + VBE ( on ) = ⎜ ⎟ (100 ) + 0.7 ⎝ 61 ⎠ VCEQ = 2.34 V

(b) Av = g m

(R

RL ) ⎡ rπ ⎤ RS ⎥ ⎢ RS ⎣1 + β ⎦

B

0.984 = 37.85 mA/V 0.026 rπ = 1.59 K

gm =

( 37.85) (100 2 ) ⎡1.59

⎤ 0.05⎥ ⎢ 0.05 ⎣ 61 ⎦ = 1484 ⎡⎣ 0.0261 0.05⎤⎦

Av =

Av = 25.4 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.70

i s ( peak ) = 2.5 μ A, V o ( peak ) = 5 mV

So we need R m =

υo is

=

5 × 10 −3 = 2 × 10 3 Ω ⇒ R m = 2 k Ω −6 2.5 × 10

We have ⎛ RS RE ⎞ Vo ⎛ β ⎞ =⎜ ⎟⎟ ⎟ ( RC RL ) ⎜⎜ Is ⎝ 1+ β ⎠ ⎝ RS RE + Rie ⎠ Let RC = 4 k Ω, RL = 5 k Ω, RE = 2 k Ω Now β = 120, so we have ⎛ RS RE ⎛ 120 ⎞ 2=⎜ ⎟ ( 4 5 ) ⎜⎜ ⎝ 121 ⎠ ⎝ RS RE + Rie RS RE Then = 0.9075 RS RE + Re

⎞ ⎛ RS RE ⎞ ⎟⎟ = 2.204 ⎜⎜ ⎟⎟ ⎠ ⎝ RS RE + Rie ⎠

RS RE = 50 2 = 1.923 k Ω, so that Rie = 0.196 k Ω Assume VCEQ = 3 V VCC ≅ I CQ ( RC + RE ) + VCEQ 5 = I CQ ( 4 + 2 ) + 3 ⇒ I CQ = 0.333 mA

(120 )( 0.026 )

rπ =

= 9.37 k Ω 0.333 r + RTH 9.37 + RTH ⇒ 0.196 = Rie = π 1+ β 121 which yields RTH = 14.35 k Ω

Now VTH = I BQ RTH + VBE ( on ) + I EQ RE 1 ⎛ 121 ⎞ = 0.00833 mA, I EQ = ⎜ ⎟ (1) = 1.008 mA 120 ⎝ 120 ⎠ 1 1 = ⋅ RTH ⋅ VCC = (14.35 )( 5 ) = ( 0.00833)(14.35 ) + 0.7 + (1.008 )( 2 ) R1 R1

I BQ = VTH

which yields R1 = 25.3 k Ω and R2 = 33.2 k Ω ______________________________________________________________________________________

6.71 a. 20 − 0.7 = 1.93 mA 10 = 1.91 mA

I EQ = I CQ

VECQ = VCC + VEB ( on ) − I C RC = 25 + 0.7 − (1.91)( 6.5 ) ⇒ VECQ = 13.3 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b.

Neglect effect hoe Assume 2.45 ≤ hie ≤ 3.7 kΩ 80 ≤ h fe ≤ 120

(

)

V o = h fe I b (RC R L ) ⎛ RE ⎞ hie ⎟⋅ Is , I e = ⎜⎜ ⎟ 1 + h fe ⎝ R E + Rie ⎠ ⎛ I ⎞ Vs Ib = ⎜ e ⎟ , Is = ⎜ 1 + h fe ⎟ R S + (R E Rie ) ⎝ ⎠ ⎛ h fe ⎞ ⎞ ⎛ ⎞⎛ 1 ⎟(R R )⎜ R E ⎟⎜ ⎟ Aυ = ⎜ ⎜ 1 + h fe ⎟ C L ⎜ R E + Rie ⎟⎜ R + R R ⎟ S E ie ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ hie = 3.7 kΩ, h fe = 120 High gain device: 3.7 Rie = = 0.0306 k Ω 121 R E Rie = 10 0.0306 = 0.0305 k Ω

Rie =

10 1 ⎛ 120 ⎞ ⎛ ⎞⎛ ⎞ Aυ = ⎜ ⎟(6.5 5)⎜ ⎟⎜ ⎟ ⇒ Aυ = 2.711 + + 121 10 0 . 0306 1 0 . 0305 ⎝ ⎠ ⎝ ⎠⎝ ⎠ h = 2.45 kΩ, h = 80

fe Low gain device: ie 2.45 Rie = = 0.03025 k Ω 81 R E Rie = 10 0.03025 = 0.0302 k Ω

10 1 ⎛ 80 ⎞ ⎛ ⎞⎛ ⎞ Aυ = ⎜ ⎟(6.5 5)⎜ ⎟⎜ ⎟ ⇒ Aυ = 2.70 ⎝ 81 ⎠ ⎝ 10 + 0.03025 ⎠⎝ 1 + 0.0302 ⎠ 2.70 ≤ Av ≤ 2.71

So A υ ≅ constant

c. Ri = R E Rie

We found

0.0302 ≤ Ri ≤ 0.0305 kΩ

h , R = RC = 6.5 kΩ Neglecting oe o ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.72 a.

Small-signal voltage gain Aυ = g m (RC R L ) ⇒ 25 = g m (RC 1)

For V ECQ = 3 V, VC = −V ECQ + V EB (on ) = −3 + 0.7 = −2.3 V VCC − I CQ RC + VC = 0 ⇒ I CQ =

5 − 2.3 2.7 = RC RC

For I CQ = 1 mA, RC = 2.7 k Ω 1 = 38.5 mA/V 0.026 Aυ = (38.5)(2.7 1) = 28.1

gm =

Design criterion satisfied and

VECQ

satisfied.

⎛ 101 ⎞ IE = ⎜ ⎟ (1) = 1.01 mA ⎝ 100 ⎠ VEE = I E RE + VEB ( on ) ⇒ RE =

b. rπ =

β VT I CQ

=

(100)( 0.026) 1

5 − 0.7 ⇒ RE = 4.26 kΩ 1.01

⇒ rπ = 2.6 kΩ, g m = 38.5 mA/V, ro = ∞

______________________________________________________________________________________ 6.73

1 = 38.46 mA/V 0.026 (120)(0.026) = 3.12 k Ω , r = (80)(0.026) = 2.08 k Ω rπ 1 = π2 1 1 ⎞ ⎛ r (1 + β 1 )⎜⎜ R E π 2 ⎟⎟ 1 β2 ⎠ + V ⎝ Aυ1 = o1 = Vi ⎞ ⎛ r rπ 1 + (1 + β 1 )⎜ R E π 2 ⎟ ⎜ 1 + β 2 ⎟⎠ ⎝

(a) g m1 = g m 2 =

We find R E Then Aυ1 = (b) Aυ 2 =

rπ 2 2.08 =1 = 0.02504 k Ω 1+ β 2 81

(121)(0.02504) = 0.4927 3.12 + (121)(0.02504 )

Vo 2 = g m 2 RC = (38.46 )(4) = 153.8 Vo1

(c) Aυ = Aυ1 ⋅ Aυ 2 = (0.4927 )(153.8) = 75.8 ______________________________________________________________________________________ 6.74

0.5 2 = 19.23 mA/V, g m 2 = = 76.92 mA/V 0.026 0.026 (100)(0.026) = 5.2 k Ω , r = (100)(0.026) = 1.3 k Ω rπ 1 = π2 0.5 2

(a) g m1 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Vo1 = − g m1 (RC1 Rib 2 ) Vi

Aυ1 =

where Rib 2 = rπ 2 + (1 + β 2 )R E 2 = 1.3 + (101)(4 ) = 405.3 k Ω Aυ1 = −(19.23)(4 405.3) = −76.17

(b) Aυ 2 =

Vo 2 (1 + β 2 )(R E 2 ) (101)(4) = 0.9968 = = Vo1 rπ 2 + (1 + β 2 )R E 2 1.3 + (101)(4)

(c) Aυ = Aυ1 ⋅ Aυ 2 = (− 76.17 )(0.9968) = −75.93 ______________________________________________________________________________________ 6.75 a. ⎛ R2 ⎞ ⎛ 20 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH 1 = ⎜⎜ ⎟(10 ) = 2.0 V R + R ⎝ 20 + 80 ⎠ 2 ⎠ ⎝ 1 RTH 1 = R1 R 2 = 20 80 = 16 k Ω 2 − 0.7 = 0.0111 mA 16 + (101)(1) 1.11 I C1 = 1.11 mA; g m1 = = 42.74 mA/V 0.026 (100)(0.026) = 2.34 k Ω rπ 1 = 1.11 ∞ ro1 = =∞ 1.11 ⎛ R4 ⎞ 15 ⎞ ⎟ ⋅ VCC = ⎛⎜ VTH 2 = ⎜⎜ ⎟(10 ) = 1.50 V ⎟ ⎝ 15 + 85 ⎠ ⎝ R3 + R 4 ⎠ RTH 2 = R3 R 4 = 15 85 = 12.75 k Ω I B1 =

IB2 =

1.50 − 0.70 = 0.01265 mA 12.75 + (101)( 0.5 )

I C 2 = 1.265 mA ⇒ g m 2 = rπ 2 =

(100 )( 0.026 ) 1.26

1.265 ⇒ g m2 = 48.65 mA/V 0.026

⇒ rπ 2 = 2.06 kΩ r02 = ∞

b.

Av1 = − g m1 RC1 = − ( 42.7 )( 2 ) ⇒ Av1 = −85.48 Av 2 = − g m 2 ( RC 2 RL ) = − ( 48.5) ( 4 4) ⇒ Av 2 = −97.3

c.

Input resistance of 2nd stage Ri 2 = R3 R 4 rπ 2 = 15 85 2.06 = 12.75 2.06 Ri 2 = 1.773 k Ω

Aυ′1 = − g m1 (RC1 Ri 2 ) = −(42.7 )(2 1.773) Aυ′1 = −40.17

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Overall gain: Av = ( −40.17 )( −97.3) ⇒ Av = 3909 If we had Av1 ⋅ Av 2 = ( −85.48 )( −97.3) = 8317 Loading effect reduces overall gain ______________________________________________________________________________________

6.76 a. ⎛ R2 ⎞ ⎛ 12.7 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH 1 = ⎜⎜ ⎟(12 ) = 1.905 V ⎝ 12.7 + 67.3 ⎠ ⎝ R1 + R 2 ⎠ RTH 1 = R1 R 2 = 12.7 67.3 = 10.68 k Ω 1.905 − 0.7 = 0.00477 mA 10.68 + (121)(2 ) = 0.572 mA

I B1 = I C1

0.572 = 22 mA/V 0.026 (120)(0.026) = 5.45 k Ω rπ 1 = 0.572 ∞ ro1 = =∞ 0.572 ⎛ R4 ⎞ 45 ⎞ ⎟ ⋅ VCC = ⎛⎜ VTH 2 = ⎜⎜ ⎟(12 ) = 9.0 V ⎟ R + R 45 ⎝ + 15 ⎠ 4 ⎠ ⎝ 3 RTH 2 = R3 R 4 = 15 45 = 11.25 k Ω g m1 =

9.0 − 0.7 = 0.0405 mA 11.25 + (121)(1.6) = 4.86 mA

I B2 = I C2

4.86 = 187 mA/V 0.026 (120)(0.026) = 0.642 k Ω , r = ∞ = o2 4.86

g m2 = rπ 2

b. I E1 = 0.577 mA VCEQ1 = 12 − ( 0.572 ) (10 ) − ( 0.577 ) ( 2 ) ⇒ VCEQ1 = 5.13 V I E 2 = 4.90 VCEQ 2 = 12 − ( 4.90 ) (1.6 ) ⇒ VCEQ 2 = 4.16 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Ri 2 = R3 R 4 Rib

Rib = rπ 2 + (1 + β )(R E 2 R L )

= 0.642 + (121)(1.6 0.25) = 26.8 k Ω

Ri 2 = 15 45 26.8 = 7.92 k Ω c.

Aυ1 = − g m1 (RC1 Ri 2 ) = −(22 )(10 7.92 ) ⇒ Aυ1 = −97.2 Aυ 2 =

(1 + β )(R E 2 R L ) (121)(0.216) = rπ 2 + (1 + β )(R E 2 R L ) 0.642 + (121)(0.216)

Aυ 2 = 0.976 Overall gain, Aυ = −(97.2)(0.976 ) = −94.9 d. RiS = R1 R 2 rπ 1 = 67.3 12.7 5.45 = 3.61 k Ω Ro =

rπ 2 + R S RE 2 1+ β

where R S = R3 R3 RC1 = 15 45 10 = 5.29 k Ω Ro =

0.642 + 5.29 1.6 = 0.049 1.6 ⇒ R o = 47.6 Ω 121

e. −1 ⋅ Δvce , ΔiC = 4.86 0.216 kΩ Δvce = ( 4.86 )( 0.216 ) = 1.05 V Max. output voltage swing = 2.10 V peak-to-peak ΔiC =

______________________________________________________________________________________ 6.77

5 − 2(0.7 ) 0.7 = 72 mA, I R1 = = 1.4 mA 0.050 0.5 70.6 ⎛ 80 ⎞ = 72 − 1.4 = 70.6 mA, I C 2 = ⎜ ⎟(70.6) = 69.73 mA, I B 2 = = 0.8716 mA 81 ⎝ 81 ⎠

(a) I R 2 = I E2

⎛ 120 ⎞ I E1 = 1.4 + 0.8716 = 2.2716 mA, I C1 = ⎜ ⎟(2.2716) = 2.253 mA ⎝ 121 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

2.253 69.73 = 86.65 mA/V, g m 2 = = 2681.9 mA/V 0.026 0.026 (120)(0.026) = 1.385 k Ω , r = (80)(0.026) = 0.02983 k Ω rπ 1 = π2 2.253 69.73 V s = V π 1 + Vπ 2 + V o

g m1 =

⎡⎛ V ⎤ ⎞ V o = ⎢⎜⎜ π 1 + g m1Vπ 1 ⎟⎟ + g m 2Vπ 2 ⎥ (0.05) ⎢⎣⎝ rπ 1 ⎥⎦ ⎠ ⎛V ⎞ ⎛ 1 ⎞ Vπ 2 = ⎜⎜ π 1 + g m1Vπ 1 ⎟⎟(0.5 rπ 2 ) = Vπ 1 ⎜ + 86.65 ⎟(0.5 0.02983) = Vπ 1 (2.4595) r 1 . 385 ⎝ ⎠ ⎝ π1 ⎠ ⎡ ⎛ 1 ⎤ ⎞ V o = ⎢Vπ 1 ⎜ + 86.65 ⎟ + (2681.9 )Vπ 1 (2.4595)⎥ (0.05) = Vπ 1 (334.175) 1 . 385 ⎠ ⎣ ⎝ ⎦ V s = Vπ 1 + Vπ 2 + V o = Vπ 1 + Vπ 1 (2.4595) + Vo So Vπ 1 = (V s − Vo )(0.28906 ) And V o = (334.175)(V s − Vo )(0.28906 ) = 96.596(V s − Vo ) V Aυ = o = 0.990 Vs

(c) For Rib : We have V o = (0.989754)V s Vπ 1 = (V s − V o )(0.28906 ) = V s (1 − 0.98975)(0.28906 ) = V s (0.0029618) V Vs rπ 1 Rib = s = = ⇒ Rib = 467.6 k Ω I s ⎛Vπ 1 ⎞ 0.0029618 ⎜ rπ 1 ⎟⎠ ⎝

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For R o :

(1) Vπ 1 + Vπ 2 + V x = 0 ⎛V ⎞ V (2) I x + g m 2Vπ 2 + ⎜⎜ π 1 + g m1Vπ 1 ⎟⎟ = x ⎝ rπ 1 ⎠ 0.05 We had Vπ 2 = Vπ 1 (2.4595) (1) Vπ 1 + Vπ 1 (2.4595) + V x = 0 ⇒ Vπ 1 = −V x (0.28906 ) ⎛ 1 ⎞ V (2) I x + g m 2Vπ 1 (2.4595) + Vπ 1 ⎜⎜ + g m1 ⎟⎟ = x r ⎝ π1 ⎠ 0.05 ⎡ ⎛ 1 ⎞⎤ V I x + Vπ 1 ⎢(2681.9 )(2.4595) + ⎜ + 86.65 ⎟⎥ = x ⎝ 1.385 ⎠⎦ 0.05 ⎣ V I x − V x (0.28906 )[6683.5] = x 0.05 V R o = x = 0.512 Ω Ix ______________________________________________________________________________________

6.78 a. RTH = R1 R2 = 335 125 = 91.0 kΩ ⎛ R2 ⎞ VTH = ⎜ ⎟ VCC ⎝ R1 + R2 ⎠ ⎛ 125 ⎞ =⎜ ⎟ (10 ) = 2.717 V ⎝ 125 + 335 ⎠ VTH = I B1 RTH + VBE1 + VBE 2 + I E 2 RE 2 I E 2 = (1 + β ) I E1 = (1 + β ) I B1 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.717 − 1.40 I B1 = ⇒ I B1 = 0.128 μΑ 2 91.0 + (101) (1) I C1 = 12.8 μΑ I C 2 = β I E1 = β (1 + β ) I B1 = (100 )(101)( 0.128 μΑ ) I C 2 = 1.29 mΑ, I E 2 = 1.31 mΑ I RC = I C 2 + I C1 = 1.29 + 0.0128 = 1.30 mΑ

VC = 10 − I RC RC = 10 − (1.30 )( 2.2 ) = 7.14 V VE = I E 2 RE 2 = (1.30 )(1) = 1.30 V VCE 2 = 7.14 − 1.30 = 5.84 V VCE1 = VCE 2 − VBE 2 = 5.84 − 0.7 VCE1 = 5.14 V Summary: I C1 = 12.8 μΑ I C 2 = 1.29 mΑ

VCE1 = 5.14 V

VCE 2 = 5.84 V

b. 0.0128 = 0.492 mΑ / V 0.026 1.292 = = 49.7 mΑ / V 0.026

g m1 = gm2

V0 = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC VS = Vπ 1 + Vπ 2 , Vπ 1 = VS − Vπ 2 ⎛V ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2 r ⎝ π1 ⎠ ⎛1+ β ⎞ Vπ 2 = Vπ 1 ⎜ ⎟ rπ 2 ⎝ rπ 1 ⎠ V0 = − ⎡⎣ g m1 (VS − Vπ 2 ) + g m 2Vπ 2 ⎤⎦ RC V0 = − ⎡⎣ g m1VS + ( g m 2 − g m1 ) Vπ 2 ⎤⎦ RC ⎛r ⎞ Vπ 2 = (VS − Vπ 2 )(1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ ⎡ ⎛ r ⎞⎤ ⎛r ⎞ Vπ 2 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥ = VS (1 + β ) ⎜ π 2 ⎟ r ⎝ π 1 ⎠⎦ ⎝ rπ 1 ⎠ ⎣

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎧ ⎛r ⎞⎫ VS (1 + β ) ⎜ π 2 ⎟ ⎪ ⎪ ⎪ ⎝ rπ 1 ⎠ ⎪ R V0 = − ⎨ g m1VS + ( g m 2 − g m1 ) ⋅ ⎬ C ⎛r ⎞ ⎪ 1 + (1 + β ) ⎜ π 2 ⎟ ⎪ ⎪ ⎝ rπ 1 ⎠ ⎪⎭ ⎩ V Av = 0 VS 2.01 ⎞ ⎫ ⎧ ( 49.7 − 0.492 )(101) ⎛⎜ ⎟⎪ ⎪⎪ ⎝ 203 ⎠ ⎪ 2.2 = − ⎨( 0.492 ) + ⎬ ⎛ 2.01 ⎞ ⎪ ⎪ 1 + (101) ⎜ ⎟ ⎪⎩ ⎪⎭ ⎝ 203 ⎠ Av = −55.2

c. Ris = R1 R2 Rib Rib = rπ 1 + (1 + β ) rπ 2 = 203 + (101)( 2.01) = 406 kΩ Ris = 91 406 = 74.3 kΩ = Ris R0 = RC = 2.2 kΩ

______________________________________________________________________________________ 6.79 I E1 = I Bias + I B 2 = I Bias +

I C2

β2

⎛ β ⎞⎡ I ⎤ I C1 = ⎜⎜ 1 ⎟⎟ ⎢ I Bias + C 2 ⎥ β2 ⎦ ⎝ 1 + β 1 ⎠⎣

1⎤ ⎛ 120 ⎞ ⎡ (a) I C 2 = I Bias = 1 mA, I C1 = ⎜ ⎟ ⎢1 + ⎥ = 1.004 mA ⎝ 121 ⎠ ⎣ 80 ⎦ g m1 = 38.62 mA/V, rπ 1 = 3.108 k Ω , ro1 = 79.68 k Ω g m 2 = 38.46 mA/V, rπ 2 = 2.08 k Ω , ro 2 = 50 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V V − Vπ 2 (1) I x = x + g m 2Vπ 2 + x + g m1V π 1 ro 2 ro1 ⎡V V − Vπ 2 ⎤ (2) Vπ 2 = ⎢ π 1 + g m1Vπ 1 + x ⎥ ⋅ rπ 2 ro1 ⎦ ⎣ rπ 1 (3) Vπ 1 + Vπ 2 = 0 ⇒ Vπ 1 = −Vπ 2 V V − Vπ 2 − g m1Vπ 2 Then (1) I x = x + g m 2Vπ 2 + x ro 2 ro1 ⎛ 1 ⎛ ⎞ 1 ⎞ 1 ⎟ + Vπ 2 ⎜ g m 2 − I x = V x ⎜⎜ + − g m1 ⎟⎟ ⎟ ⎜ ro1 ⎝ ro 2 ro1 ⎠ ⎝ ⎠ ⎛ 1 ⎛r 1 ⎞ ⎟ ⋅ rπ 2 + V x ⎜ π 2 (2) Vπ 2 = −Vπ 2 ⎜⎜ + g m1 + ⎟ ⎜r r r o1 ⎠ ⎝ π1 ⎝ o1 ⎡ ⎛ 1 ⎤ ⎛ rπ ⎞ 1 ⎞ ⎟ ⋅ rπ 2 ⎥ = V x ⎜ 2 ⎟ + g m1 + Vπ 2 ⎢1 + ⎜⎜ ⎟ ⎜r ⎟ ro1 ⎠ ⎢⎣ ⎝ rπ 1 ⎥⎦ ⎝ o1 ⎠

⎞ ⎟ ⎟ ⎠

⎡ ⎛ 1 ⎤ 1 ⎞ ⎛ 2.08 ⎞ + 38.62 + Now (2) Vπ 2 ⎢1 + ⎜ ⎟(2.08)⎥ = V x ⎜ ⎟ 3 . 108 79 . 68 ⎠ ⎝ 79.68 ⎠ ⎣ ⎝ ⎦ Vπ 2 = V x (0.00031825) 1 ⎞ 1 ⎛ 1 ⎛ ⎞ Then (1) I x = V x ⎜ + − 38.62 ⎟ = V x (0.32495) ⎟ + V x (0.00031825)⎜ 38.46 − 79.68 ⎝ 50 79.68 ⎠ ⎝ ⎠ V R o = x = 30.77 k Ω Ix

1 ⎞ ⎛ 120 ⎞⎛ (b) I C 2 = 1 mA, I Bias = 0.2 mA, I C1 = ⎜ ⎟⎜ 0.2 + ⎟ = 0.2107 mA 121 80 ⎝ ⎠⎝ ⎠ g m1 = 8.104 mA/V, rπ 1 = 14.81 k Ω , ro1 = 379.7 k Ω g m 2 = 38.46 mA/V, rπ 2 = 2.08 k Ω , ro 2 = 50 k Ω ⎡ ⎛ 1 ⎤ 1 ⎞ ⎛ 2.08 ⎞ + 8.104 + Now (2) Vπ 2 ⎢1 + ⎜ ⎟(2.08)⎥ = V x ⎜ ⎟ 379.7 ⎠ ⎝ 379.7 ⎠ ⎣ ⎝ 14.81 ⎦ Vπ 2 = V x (0.0003043) 1 ⎞ 1 ⎛ 1 ⎛ ⎞ (1) I x = V x ⎜ + − 8.104 ⎟ = V x (0.031867 ) ⎟ + V x (0.0003043)⎜ 38.46 − 379.7 ⎝ 50 379.7 ⎠ ⎝ ⎠ Vx = 31.38 k Ω Ro = Ix ⎛ 120 ⎞⎛ 2 ⎞ (c) I C 2 = 2 mA, I Bias = 0 , I C1 = ⎜ ⎟⎜ ⎟ = 0.02479 mA ⎝ 121 ⎠⎝ 80 ⎠ g m1 = 0.9536 mA/V, rπ 1 = 125.9 k Ω , ro1 = 3327 k Ω g m 2 = 76.92 mA/V, rπ 2 = 1.04 k Ω , ro 2 = 25 k Ω ⎡ ⎛ 1 ⎤ 1 ⎞ ⎛ 1.04 ⎞ + 0.9536 + Now (2) Vπ 2 ⎢1 + ⎜ ⎟(1.04 )⎥ = V x ⎜ ⎟ 125 . 9 3327 ⎠ ⎝ 3327 ⎠ ⎣ ⎝ ⎦ Vπ 2 = V x (0.00015627 )

1 ⎞ 1 ⎛ 1 ⎛ ⎞ (1) I x = V x ⎜ + − 0.9536 ⎟ = V x (0.05217 ) ⎟ + V x (0.00015627)⎜ 76.92 − 3327 ⎝ 25 3327 ⎠ ⎝ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V R o = x = 19.17 k Ω Ix ______________________________________________________________________________________

6.80 (a) RTH = R1 R 2 = 250 75 = 57.69 k Ω ⎛ R2 ⎞ ⎛ 75 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH = ⎜⎜ ⎟(5) = 1.1538 V R + R ⎝ 75 + 250 ⎠ 2 ⎠ ⎝ 1 V − V BE (on ) 1.1538 − 0.7 = = 0.003483 mA I BQ = TH RTH + (1 + β )R E 57.69 + (121)(0.6)

⇒ I CQ = 0.418 mA, I EQ = 0.4214 mA

VCEQ = 5 − (0.418)(5.6 ) − (0.4214 )(0.6 ) = 2.406 V

For Q : PQ = I CQ VCEQ = (0.418)(2.406 ) = 1.01 mW 2 RC = (0.418) (5.6) = 0.978 mW For RC : PRC = I CQ 2

2 R E = (0.4214 ) (0.6) = 0.107 mW For R E : PRE = I EQ 2

(b) Rib = rπ + (1 + β )R E rπ =

(120)(0.026) = 7.464 k Ω

0.418 Rib = 7.464 + (121)(0.6) = 80.06 k Ω Ri = RTH Rib = 57.69 80.06 = 33.53 k Ω

RiS = R S + Ri = 0.5 + 33.53 = 34.03 k Ω

⎛ RTH ⎞ ⎛ RTH ⎞⎛ υ s ⎞ ⎟ ⋅ is = ⎜ ⎟⎜ ⎟ ; ib = ⎜⎜ ⎟ ⎜ ⎟⎜ ⎟ RiS ⎝ RTH + Rib ⎠ ⎝ RTH + Rib ⎠⎝ RiS ⎠ ⎛ RTH ⎛ RTH ⎞⎛ υ s ⎞ ⎞⎛ υ s ⎞ ⎟⎜ ⎟⎜ ⎟ ; i e = (1 + β )⎜ ⎟ i c = β ⎜⎜ ⎜ ⎟ ⎟⎜ ⎜ ⎟ ⎟ ⎝ RTH + Rib ⎠⎝ RiS ⎠ ⎝ RTH + Rib ⎠⎝ RiS ⎠ 57.69 ⎛ ⎞⎛ 0.1 sin ω t ⎞ Now i c = (120)⎜ ⎟ = 0.1477 sin ω t (mA) ⎟⎜ ⎝ 57.69 + 80.06 ⎠⎝ 34.03 ⎠ is =

υs

i e = 0.1489 sin ω t (mA)

1 2 1 2 I c RC = 0.978 + (0.1477 ) (5.6) = 1.039 mW 2 2 1 1 2 2 R E + I e2 R E = 0.107 + (0.1489) (0.6) = 0.1137 mW For R E : PRE = I EQ 2 2 1 1 For Q : PQ = I CQ VCEQ − I c2 RC − I e2 R E = 1.01 − 0.0611 − 0.00665 = 0.942 mW 2 2 ______________________________________________________________________________________ 2 RC + For RC : PRC = I CQ

6.81 (a) I BQ R B + V BE (on ) + (1 + β )I BQ R E + V − = 0 5 − 0.7 = 0.007363 mA; I CQ = 0.8836 mA, I EQ = 0.8909 mA 100 + (121)(4) = 10 − (0.8836)(4 ) − (0.8909)(4) = 2.902 V

I BQ =

VCEQ

For Q : PQ = I CQ V CEQ= (0.8836)(2.902 ) = 2.564 mW

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2 For RC : PRC = I CQ RC = (0.8836) (4 ) = 3.123 mW 2

2 R E = (0.8909 ) (4 ) = 3.175 mW For R E : PRE = I EQ 2

(b) ΔVCE = ΔI C (RC ) = (0.8836)(4 ) = 3.534 V - Not possible ΔVCE = 2.902 − 0.5 = 2.402 V 2.402 = 0.6005 mA 4 1 1 2 2 p RC = (ΔI C ) RC = (0.6005) (4 ) = 0.721 mW 2 2 ______________________________________________________________________________________

So ΔI C =

6.82 a. I BQ =

10 − 0.7 = 0.00596 mA 50 + (151)(10 )

I CQ = 0.894 mA, I EQ = 0.90 mA

VECQ = 20 − ( 0.894 )( 5 ) − ( 0.90 )(10 ) ⇒ VECQ = 6.53 V PQ ≅ I CQVECQ = ( 0.894 )( 6.53) ⇒ PQ = 5.84 mW 2 PRC ≅ I CQ RC = ( 0.894 ) ( 5 ) ⇒ PRC = 4.0 mW 2

2 PRE ≅ I EQ RE = ( 0.90 ) (10 ) ⇒ PRE = 8.1 mW 2

b.

−1 ⋅ Δvec 1.43 kΩ ΔiC = 0.894 ⇒ Δvec = ( 0.894 )(1.43) = 1.28 V ΔiC =

⎛ 5 ⎞ Δi0 = ⎜ ⎟ ΔiC = 0.639 mA ⎝5+2⎠ 1 2 PRL = ( 0.639 ) ( 2 ) ⇒ PRL = 0.408 mW 2 1 2 PRC = ⋅ ( 0.894 − 0.639 ) ( 5 ) ⇒ PRC = 0.163 mW 2 PRE = 0 PQ = 5.84 − 0.408 − 0.163 ⇒ PQ = 5.27 mW

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

6.83 I BQ =

10 − 0.70 = 0.00838 mA 100 + (101)(10 )

I CQ = 0.838 mA, I EQ = 0.846 mA

VCEQ = 20 − ( 0.838 )(10 ) − ( 0.846 )(10 ) ⇒ VCEQ = 3.16 V

100 = 119 kΩ 0.838 Neglecting base currents: a. RL = 1 kΩ −1 −1 slope = = 10 1 119 0.902 kΩ r0 =

−1 ⋅ ΔVce 0.902 kΩ ΔiC = 0.838 ⇒ ΔVce = ( 0.902 )( 0.838 ) = 0.756 V ΔiC =

1 ( 0.756 ) ⇒ PRL = 0.286 mW PRL = 2 1 2

b. RL = 10 kΩ

slope =

−1 −1 = 10 10 119 4.80

For ΔiC = 0.838 ⇒ Δvce = ( 0.838 )( 4.80 ) = 4.02 1 ( 3.16 ) ⇒ PRL = 0.499 mW 2 10 2

Max. swing determined by voltage PRL =

______________________________________________________________________________________ 6.84 a. I BQ =

10 − 0.7 = 0.00838 mA 100 + (101)(10 )

I CQ = 0.838 mA, I EQ = 0.846 mA

VCEQ = 20 − ( 0.838 )(10 ) − ( 0.846 )(10 ) ⇒ VCEQ = 3.16 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ PQ ≅ I CQVCEQ = ( 0.838 )( 3.16 ) ⇒ PQ = 2.65 mW 2 PRC ≅ I CQ RC = ( 0.838 ) (10 ) ⇒ PRC = 7.02 mW 2

b.

−1 ⋅ Δvce 0.909 kΩ For ΔiC = 0.838 ⇒ Δvce = ( 0.909 )( 0.838 ) = 0.762 V

ΔiC =

⎛ RC ⎞ ⎛ 10 ⎞ Δi0 = ⎜ ⎟ ΔiC = ⎜ ⎟ ΔiC = 0.762 mA ⎝ 10 + 1 ⎠ ⎝ RC + RL ⎠ 1 2 PRL = ( 0.762 ) (1) ⇒ PRL = 0.290 mW 2 1 2 PRC = ⋅ ( 0.838 − 0.762 ) (10 ) ⇒ PRC = 0.0289 mW 2 PQ = 2.65 − 0.290 − 0.0289 ⇒ PQ = 2.33 mW

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 7 7.1 a. T ( s) = T (s) =

V0 ( s ) Vi ( s )

=

1/ ( sC1 )

⎡⎣1/ ( sC1 ) ⎤⎦ + R1

1 1 + sR1C1

b.

fH =

1 1 = ⇒ f H = 159 Hz 3 2π R1C1 2π (10 )(10−6 )

c. V0 ( s ) = Vi ( s ) ⋅

1 1 + sR1C1

For a step function

Vi ( s ) =

1 s

K K2 1 1 V0 ( s ) = ⋅ = 1+ s 1 + sR1C1 s 1 + sR1C1 = =

K1 (1 + sR1C1 ) + K 2 s s (1 + sR1C1 )

K1 + s ( K1 R1C1 + K 2 ) s (1 + sR1C1 )

K 2 = − K1 R1C1 and K1 = 1 − R1C1 1 V0 ( s ) = + s 1 + sR1C1 1 1 = − 1 s +s R1C1 v0 ( t ) = 1 − e − t / R1C1

______________________________________________________________________________________ 7.2 a.

T (s) = T (s) =

V0 ( s ) Vi ( s )

=

R2 R2 + ⎡⎣1/ ( sC2 ) ⎤⎦

sR2 C2 1 + sR2 C2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b.

fL =

1 1 = ⇒ f L = 1.59 Hz 2π R2 C2 2π (104 )(10 × 10−6 )

c. V0 ( s ) = Vi ( s ) ⋅

sR2 C2 1 + sR2 C2

Vi ( s ) =

1 s

V0 ( s ) =

R2 C2 1 = 1 + sR2 C2 s + 1 R2 C2

v0 ( t ) = e − t / R2C2

______________________________________________________________________________________ 7.3

V (a) T (s ) = o = Vi

R2 R2

1 sC 2

1 + R1 sC 2

⎛ 1 ⎞ ⎟⎟ R 2 ⎜⎜ R2 1 ⎝ sC 2 ⎠ Now R 2 = = 1 sC 2 1 + sR 2 C 2 R2 + sC 2

R2 1 + sR 2 C 2 R2 = Then T (s ) = R2 R1 + R 2 + sR1 R 2 C 2 + R1 1 + sR 2 C 2 ⎛ R2 T (s ) = ⎜⎜ ⎝ R1 + R 2

⎞ 1 ⎟⎟ ⋅ ( 1 + s R 1 R 2 )C 2 ⎠

(b) τ = (R1 R 2 )C 2 = (10 20 )× 10 3 × 10 × 10 −6 ⇒ τ = 66.7 ms 1 = 2.39 Hz 2πτ 2π 66.7 × 10 −3 ______________________________________________________________________________________

(c)

f =

1

=

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 7.4

a.

τ S = (Ri + R P )C S = (30 + 10)×10 3 × (10 ×10 −6 ) ⇒ τ S = 0.40 s

τ P = (Ri R P )C P = (30 10)× 10 3 × (50 × 10 −12 ) ⇒ τ P = 0.375 μ s b. fL = fH =

1 2π τ S 1 2π τ P

=

1 ⇒ f L = 0.398 Hz 2π (0.4)

=

1 ⇒ f H = 424 kHz 2π 0.375 × 10 − 6

(

)

At midband. CS → short, CP → open Vo = I i ( Ri RP ) T ( s ) = Ri R P = 30 10 ⇒ T ( s ) = 7.5 k Ω

c.

______________________________________________________________________________________ 7.5

(a)

Vo R2 20 = = = 0.667 Vi R 2 + R1 20 + 10

(b)

Vo =1 Vi

(c) T (s ) =

Vo (s ) = Vi (s )

R2 R 2 + R1

R2

= 1 sC1

R2 +

R1 1 + sR1C1

⎛ R2 ⎞ (1 + sR1C1 ) R 2 (1 + sR1C1 ) ⎟⎟ ⋅ = ⎜⎜ R1 + R 2 + sR1 R 2 C1 ⎝ R1 + R 2 ⎠ 1 + s (R1 R 2 )C1 R2 We have K = , τ A = R1C1 , τ B = R1 R2 C1 R1 + R 2 ______________________________________________________________________________________ T (s ) =

[ (

]

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 7.6

a. T (s ) =

RP

Vo (s ) = Vi (s )

RP

1 sC P

⎛ 1 1 + ⎜⎜ R S + sC P ⎝ sC S 1 RP ⋅ sC P RP 1 = = 1 sC P 1 + sR P C P RP + sC P RP

⎞ ⎟ ⎟ ⎠

Then RP

T (s) =

⎛ 1 ⎞ RP + ⎜ RS + ⎟ (1 + sRP CP ) sCS ⎠ ⎝ RP = RP CP 1 + + sRS RP CP RP + RS + CS sCS

⎛ RP T (s) = ⎜ ⎝ RP + RS

⎤⎞ ⎞ ⎛ ⎡ sRP RS RP C 1 ⋅ P+ + ⋅ CP ⎥ ⎟ ⎟ × ⎜⎜1/ ⎢1 + ⎟ + + + R R C s R R C R R ( ) P S S S P S S P ⎠ ⎝ ⎣⎢ ⎦⎥ ⎠

b. −11 ⎛ ⎤⎞ 1 ⎛ 10 ⎞ ⎜ ⎡ 10 10 3 −11 ⎟ ⎥ + × ⋅ 5 10 10 T (s) = ⎜ s ( ) ⎟ × 1/ ⎢1 + ⋅ −6 + s ( 2 × 104 ) ⋅10 −6 ⎥⎦ ⎟ ⎝ 10 + 10 ⎠ ⎜ ⎢⎣ 20 10 ⎝ ⎠ 1 1 ≅ ⋅ 1 2 1+ + s ( 5 × 10−8 ) s ( 0.02 )

s = jω T ( jω ) =

1 ⋅ 2

1 ⎡ ⎤ 1 1 + j ⎢ω ( 5 × 10 −8 ) − ⎥ ω ( 0.02 ) ⎦ ⎣ 1 1 For ω L = = = 50 ( RS + RR ) CS ( 2 ×104 )(10−6 )

T ( jω ) =

1 ⋅ 2

1

⎡ ⎤ 1 1 + j ⎢( 50 ) ( 5 × 10 −8 ) − ⎥ 50 0.02 ( )( ) ⎣ ⎦ 1 1 1 1 ≈ ⋅ ⇒ T ( jω ) = ⋅ 2 1− j 2 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For

ωH =

(R

T ( jω ) =

S

1 1 = = 2 × 10 7 −11 3 R P )C P 5 × 10 10

(

1 ⋅ 2

)(

)

1 ⎡ ⎤ 1 1 + j ⎢ 2 ×10 7 5 ×10 −8 − ⎥ 7 2 × 10 (0.02) ⎦ ⎣ 1 1 1 1 T ( jω ) ≅ ⋅ ⇒ T ( jω ) = ⋅ 2 1+ j 2 2

(

)(

In each case, T ( jω ) = c.

)

(

)

RP R 2 S + RP

1

⋅

RS = RP = 10 kΩ, CS = CP = 0.1 μ F T (s) =

⎛ ⎡ ⎤⎞ 1 ⎜ ⎢ 1 1 1 ⋅ 1/ 1 + ⋅ + + s ( 5 × 103 )(10−7 ) ⎥ ⎟ ⎥ ⎟⎟ 2 ⎜⎜ ⎢ 2 1 s 2 × 104 (10−7 ) ⎦⎠ ⎝ ⎣

(

)

s = jω T ( jω ) =

1 ⋅ 2

1

⎡ ⎤ 1 1 ⎥ + j ⎢ω ( 5 × 10−4 ) − −3 2 ω ( 2 × 10 ) ⎥⎦ ⎢⎣ 1 For ω = = 500 4 2 10 × ( )(10−7 )

T ( jω ) =

1+

1 ⋅ 2

1

⎡ ⎤ 1 ⎥ 1.5 + j ⎢( 500 ) ( 5 × 10−4 ) − −3 ⎢⎣ ( 500 ) ( 2 ×10 ) ⎥⎦ 1 1 = ⋅ ⇒ T ( jω ) = 0.298 2 1.5 − j ( 0.75 )

For ω =

1

( 5 ×103 )(10−7 )

= 2 × 103

⎧ ⎡ ⎤ ⎞ ⎫⎪ 1 ⎪ ⎛⎜ 1 ⎥ ⎟⎬ ⋅ ⎨1/ 1.5 + j ⎢( 2 × 103 )( 5 × 10−4 ) − −3 3 2 ⎪ ⎜ 2 × 10 )( 2 × 10 ) ⎥⎦ ⎟ ⎪ ⎢⎣ ( ⎝ ⎠⎭ ⎩ 1 1 = ⋅ ⇒ T ( jω ) = 0.298 2 1.5 + j ( 0.75 )

T ( jω ) =

In each case, T ( jω )
> τ C1 and f = Then

f3− dB ( CC 2 ) ⇒ CC1

1 2π τ

so f 3− dB (C C 2 ) C ⇒ C

L μ L dominates frequency response. c. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 8 8.1 (b) (i) R D = PD , max

24 = 6Ω 4 = (12 )(2 ) = 24 W

(ii) PD , max = 30 = (20 )I DQ ⇒ I DQ = 1.5 A I D , max = 2(1.5) = 3 A

RD =

(c) (i) I D , max

40 = 13.3 Ω 3 = 4A

(ii) I D , max = 3 A ______________________________________________________________________________________ 8.2 (a) PQ , max = VCEQ ⋅ I CQ ⎛ 24 ⎞ 25 = ⎜ ⎟ ⋅ I CQ ⇒ I CQ = 2.083 A ⎝ 2 ⎠ 24 − 12 RL = = 5.76 Ω 2.083 2.083 I BQ = = 0.03472 A 60 24 − 0.7 RB = = 671 Ω 0.03472 β VT (60 )(0.026 ) (b) rπ = = = 0.7489 Ω I CQ 2.083 Ib =

Vp rπ

=

12 mV = 16.02 mA 0.7489

I c = βI b = (60 )(0.01602 ) = 0.9614 A 1 2 1 2 I c RC = (0.9614 ) (5.76 ) = 2.66 W 2 2 For the transistor, PQ = 25 − 2.66 = 22.34 W Pavg =

______________________________________________________________________________________ 8.3 30 = 25 Ω 1. 2 I CQ 0.6 I BQ = = = 0.0075 A 80 β V − V BE (on ) 30 − 0.7 R B = CC = ⇒ R B = 3.91 k Ω I BQ 0.0075

(a) R L =

PQ , max = I CQ V CEQ = (0.6 )(15 ) = 9 W

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ VCC − VCEQ ⎞ ⎟ ⋅ VCEQ (b) PQ , max = I CQ VCEQ = ⎜⎜ ⎟ RL ⎝ ⎠ 1 VCEQ = VCC 2 1 ⎛ VCC ⎞⎛ V CC ⎞ 22.36 5= ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⇒ V CC = 22.36 V; I C , max = = 0.8944 A 25 25 ⎝ 2 ⎠⎝ 2 ⎠ 1 (ΔI C )2 R L = 1 (0.6)2 (25) = 4.5 W 2 2 0.8944 1 2 For (b): ΔI C = I CQ = = 0.4472 A; PL = (0.4472) (25) = 2.5 W 2 2 ______________________________________________________________________________________

(c) For (a): ΔI C = 0.6 A; PL =

8.4

Point (b): Maximum power delivered to load. Point (a): Will obtain maximum signal current output. Point (c): Will obtain maximum signal voltage output. ______________________________________________________________________________________ 8.5 a.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b. VGG = 5 V, I D = 0.25 ( 5 − 4 ) = 0.25 A, VD S = 37.5 V, P = 9.375 W 2

VGG = 6 V, I D = 0.25 ( 6 − 4 ) = 1.0 A, VD S = 30 V, P = 30 W 2

VGG = 7 V, I D = 0.25 ( 7 − 4 ) = 2.25 A, VD S = 17.5 V, P = 39.375 W 2

VGG = 8 V, I D

= 0.25 ⎡⎣ 2 ( 8 − 4 ) VD S − VD2 S ⎤⎦ =

40 − VD S

10 I D = 3.71 A, P = 10.8 W VGG = 9 V, I D

⇒ VD S = 2.92

= 0.25 ⎡⎣ 2 ( 9 − 4 )VD S − VD2S ⎤⎦ =

40 − VD S

⇒ VD S = 1.88 V 10 I D = 3.81 A, P = 7.16 W V = 7 V, P = 39.375 W > PD ,max = 35 W c. Yes, at GG ______________________________________________________________________________________

8.6 a. VDD = 25 V 2 50 − 25 = = 1.25 A 20

Set VDSQ = I DQ

I DQ = K n (VGS − VTN )

2

1.25 + 4 = VGS = 6.5 V 0.2 ⎛ R2 ⎞ VGS = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ Let R1 + R2 = 100 kΩ ⎛ R ⎞ 6.5 = ⎜ 2 ⎟ ( 50 ) ⇒ R2 = 13 kΩ ⎝ 100 ⎠ R1 = 87 kΩ

b. c.

PD = I DQVDSQ = (1.25 )( 25 ) ⇒ PD = 31.25 W I D ,max = 2 I DQ ⇒ I D ,max = 2.5 A VDS ,max = VDD ⇒ VDS ,max = 50 V PD ,max = 31.25 W

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ d. V0 = g m RL Vi g m = 2 K n I DQ = 2

( 0.2 )(1.25) = 1 A / V

V0 = (1)( 20 )( 0.5 ) = 10 V 1 V02 1 (10 ) ⋅ = ⋅ ⇒ PL = 2.5 W 2 RL 2 20 2

PL =

PQ = 31.25 − 2.5 ⇒ PQ = 28.75 W

______________________________________________________________________________________ 8.7 (a)

(b)

PD = PD ,max − ( Slope ) (T j − 25 )

At PD = 0, T j ,max =

60 + 25 ⇒ T j ,max = 145°C 0.5

T j ,max − Tcase

145 − 25 ⇒ θ dev − amb = 2°C/W 60 or (c) ______________________________________________________________________________________ PD ,max =

8.8

PD ,rated =

θ dev − amb

θ dev − amb =

T j ,max − Tamb

θ dev − case

or θ dev − case =

T j ,max − Tamb PD ,rated

150 − 25 = 2.5°C/W 50 Then Tdev − Tamb = PD (θ dev − case + θ case − amb ) 150 − 25 = PD ( 2.5 + θ case − amb ) ⇒ 125 = PD ( 2.5 + θ case − amb ) =

______________________________________________________________________________________ 8.9 (a) T j , max − Tamb = PT (θ dev − case + θ snk − amb + θ case − snk ) 120 − 25 = PT (1.5 + 2.8 + 0.6) ⇒ PT = 19.39 W

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) Tcase = 25 + (19.39 )(0.6 + 2.8) = 90.9°C (c) Tsnk = 25 + (19.39)(2.8) = 79.3°C ______________________________________________________________________________________ 8.10 (a) T j , max − Tamb = P (θ dev − case + θ case − amb )

150 − 25 = 30(2.8 + θ case − amb ) ⇒ θ case − amb = 1.37°C / W (b) T j , max = 25 + 20(2.8 + 1.37 ) = 108°C

______________________________________________________________________________________ 8.11 (a) 150 − 25 = PT (3.8 + 1.5 + 4 ) ⇒ PT = 13.4 W (b) P = I CQ V CEQ 13.4 = (3) ⋅ V CEQ ⇒ V CEQ = 4.48 V V CC = V CE , max = 2V CEQ = 8.96 V

______________________________________________________________________________________ 8.12

η=

PL PS

PS = VCC ⋅ I Q ⎛V ⎞ PL = VP ⋅ I P = ⎜ CC ⎟ ( I Q ) ⎝ 2 ⎠ 1 ⋅ VCC ⋅ I Q η= 2 ⇒ η = 50% VCC ⋅ I Q

______________________________________________________________________________________ 8.13

(a) Aυ =

(1 + β )R L rπ + (1 + β )R L

=

(1 + β )R L

βVT IC

+ (1 + β )R L

We have (1 + β ) ≅ β I C RL RL RL Aυ = = = V 1 I C R L + VT RL + RL + T gm IC 8

(b) (i) 0.9 =

8+

1 gm

⇒ g m = 1.125 A/V, I C = 29.25 mA

8

(ii) 0.95 =

8+

1 gm 8

⇒ g m = 2.375 A/V, I C = 61.75 mA

⇒ g m = 41.54 A/V, I C = 108 mA 1 8+ gm ______________________________________________________________________________________

(iii) 0.997 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.14 2 2 1 Vp 1 Vp ⋅ ⇒ 0.5 = ⋅ ⇒ V p = 2.828 V 2 RL 2 8 V p 2.828 = = 0.3536 A Ip = 8 RL (b) For V o = −V p = −2.828 V

(a) PL =

I L = 0.3536 = (0.9)I O ⇒ I O = 0.393 A

______________________________________________________________________________________ 8.15 1.6 = 0.20 A 8 I C ≅ 0.2 + 0.25 = 0.45 A

(a) VO = 1.6 V, I L =

0.45 = 17.31 A/V 0.026 8 Aυ = = 0.9928 1 8+ 17.31 (b) VO = 0 , I L = 0 , so I C = 0.25 A gm =

0.25 = 9.615 A/V 0.026 8 Aυ = = 0.9872 1 8+ 9.615 (c) VO = −1.6 V, I L = −0.2 A, I C ≅ 0.25 − 0.2 = 0.05 A gm =

0.05 = 1.923 A/V 0.026 8 Aυ = = 0.939 1 8+ 1.923 ______________________________________________________________________________________ gm =

8.16

vo ( max ) = 4.8 V iC 3 = iC 2 = vI = vo + 0.7

−0.7 − ( −5 ) 1

iL ( max ) = −4.3 mA =

so − 3.6 ≤ vI ≤ 5.5 V vo ( min ) = −4.3 V

= 4.3 mA vS ( min ) 1

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.17 I D 3 = K (VGS 3 − VTN ) = 2

0 − VGS 3 − ( −5 ) R

12 (VGS 3 − 0.5 ) = 5 − VGS 3 2

2VGS2 3 − 11VGS − 2 = 0 VGS 3 =

11 ±

(11)

2

+ 4 (12 )( 2 )

2 (12 )

VGS 3 = VGS 2 = 1.072 V I D 3 = I D 2 = 12 (1.072 − 0.5 ) = 3.93 mA VDS 2 ( sat ) = VGS 2 − VTN = 1.072 − 0.5 = 0.572 V 2

vo ( min ) : i2 ( max ) = −3.93 = vI ( min ) = vo ( min ) + VTN vI ( min ) = −3.43 V

V0 ( min )

1 = −3.93 + 0.5

⇒ V0 ( min ) = −3.93 V

vo ( max ) = 5 − VDS ( sat ) = 5 − 0.572 vo ( max ) = 4.43 V I D1 ( max ) = 3.93 +

4.43 = 8.36 mA 1

I D1 = 8.36 = 12 (VGS 1 − 0.5 ) ⇒ VGS 1 = 1.33 V vI ( max ) = vo + VGS1 = 4.43 + 1.33 ⇒ vI ( max ) = 5.76 V 2

______________________________________________________________________________________ 8.18

(a) For υ O = −12 + 0.7 = −11.7 V, I Q =

11.3 + 50 = 615 mA 0.02

⎛ 2⎞ 2 ⎞ ⎛ I REF = ⎜⎜1 + ⎟⎟ ⋅ I Q = ⎜1 + ⎟(615) = 645.75 mA 40 β ⎝ ⎠ ⎝ ⎠ 0 − 0.7 − (− 12 ) R= ⇒ R = 17.5 Ω 0.6475 11.3 For υ O = 12 − 0.7 = +11.3 V, i L = = 565 mA 0.02 i E1 (max ) = I Q + i L = 615 + 565 ⇒ i E1 (max ) = 1.18 A 2 2 1 VO 1 (11.3) ⋅ = ⋅ = 3.19 W 2 RL 2 20 PS = I Q (24 ) = (0.615 )(24 ) = 14.76 W

(b) PL =

PL 3.19 = × 100% = 21.6% PS 14.76 ______________________________________________________________________________________

Define η =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.19 −20 (a) VO = −20 V, i L = = −0.10 A 200 I Q = 0.10 + i E1 (min ) = 0.10 + 0.02 = 0.12 A ⎛ 2⎞ 2 ⎞ ⎛ I REF = ⎜⎜1 + ⎟⎟ ⋅ I Q = ⎜1 + ⎟(0.12 ) = 0.1248 A ⎝ 50 ⎠ ⎝ β⎠ 0 − 0.7 − (− 24 ) R= = 187 Ω 0.1248 (b) PQ1 = I Q VCE1 = (0.12 )(24 ) = 2.88 W

P = I Q VCE 2 + I REF (24 ) = (0.12 )(24 ) + (0.1248 )(24 ) = 5.88 W

(c) PL =

( )

2 1 VO 1 20 2 ⋅ = ⋅ =1W 2 R L 2 200

1 × 100% = 11.4% 2.88 + 5.88 ______________________________________________________________________________________

η=

8.20

I D1 = K n (VGS − VTN ) = 12 ( 0 − ( −1.8) ) 2

2

I D1 = 38.9 mA (a) For RL = ∞

vo ( max ) = 4.8 V

VDS ( sat ) = VGS − VTN = 1.8 V vo ( min ) = −5 + 1.8 = −3.2 V vI = vo + 0.7 ⇒ −2.5 ≤ vI ≤ 5.5 V (b)

For

RL = 500 Ω vo ( max ) = 4.8 V

vo < 0, vo ( min ) = −3.2 V

For

I 2′ =

vo −3.2 = = −6.4 mA RL 0.5

−2.5 ≤ vI ≤ 5.5 V

(c)

For vo = −2V , I 2′ ( max ) = −38.9 mA R2 ( min ) =

−2 ⇒ RL ( min ) = 51.4 Ω −38.9

1 v2 1 ( 2) ⇒ PL = 38.9 mW PL = ⋅ o = ⋅ 2 RL 2 51.4 2

38.9 = 10% 389 ______________________________________________________________________________________ PL = 10 ( 38.9 ) = 389 mW % =

8.21

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ + V 2 (V ) PL = P = RL RL

2

1 (V ) 1 (V ) , V − = −V + + ⋅ PS = ⋅ 2 RL 2 RL + 2

− 2

(V ) =

+ 2

So PS

η=

RL

PL

⇒ η = 100% PS ______________________________________________________________________________________

8.22 (a) As maximum conversion efficiency

η=

π 4

,

VP = 0.785 VCC

⎛4⎞ So V p ( max ) = ( 0.785 )( 5 ) ⎜ ⎟ ⎝π ⎠ V p ( max ) = 5 V

(b)

Maximum power dissipation occurs when V2 Pθ ( max ) = 2CC π RL 2=

( 5)

Vp =

2VCC

π

=

2 ( 5)

π

= 3.183 V

2

π 2 RL

⇒ RL = 1.27 Ω

(c) ______________________________________________________________________________________ 8.23 P=

(a) (b)

2 1 Vp ⋅ 2 RL

2 1 Vp ⋅ ⇒ V p = 49 V ⇒ V + = 52 V, V − = −52 V 2 24 VP 49 IP = = = 2.04 A RL 24

50 =

η=

π VP ⋅

4 VCC

=

π ⎛ 49 ⎞ ⎜ ⎟ 4 ⎝ 52 ⎠

η = 74.0% (c) ______________________________________________________________________________________

8.24

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) VDS ≥ VDS ( sat ) = VGS − VTN = VGS VDS = 10 − Vo ( max ) and I D = I L = K n (VGS ) Vo ( max ) RL VGS =

= K n (VGS )

2

Vo ( max ) RL ⋅ K n Vo ( max )

So 10 − Vo ( max ) = ⎡⎣10 − V0 ( max ) ⎤⎦ = 2

RL ⋅ K n

=

2

V0 ( max ) =

VGS

( 5 )( 0.4 )

V0 ( max )

V02 ( max ) − 20.5V0 ( max ) + 100 = 0 20.5 ±

Vo ( max )

V0 ( max )

100 − 20V0 ( max ) + V02 ( max ) =

iL =

2

( 20.5 )

2

2

− 4 (100 )

2

⇒ V0 ( max ) = 8 V

8 ⇒ iL = 1.6 mA 5 i 1.6 = L = = 2 V ⇒ VI = 10 V 0.4 Kn

b. 1 (8) ⋅ = 6.4 mW 2 5 20 (1.6 ) PS = = 10.2 mW 2

PL =

π

η=

PL

=

6.4 ⇒ η = 62.7% 10.2

PS ______________________________________________________________________________________ 8.25 (b) υ I = υ GSn + υ o

Also υ o = i L R L = i dn R L = KR L (υ GSn ) , but υ GSn = υ I − υ o 2

So υ o = KR L (υ I − υ o )

2

⎛ dυ dυ o = 2 KR L (υ I − υ o )⎜⎜1 − o dυ I ⎝ dυ I

⎞ ⎟⎟ ⎠

2 KR L (υ I − υ o ) dυ o = , also υ I − υ o = dυ I 1 + 2 KR L (υ I − υ o )

Then

2 KR L ⋅ υ o dυ o = Aυ = dυ I 1 + 2 KR L ⋅ υ o

υo KR L

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ KR L =

We find

(4)(0.05) = 0.4472

(i)

For υ o = 0 , Aυ = 0

(ii)

For υ o = 1 V, Aυ =

2(0.4472)(1) = 0.472 1 + 2(0.4472)(1)

For υ o = 10 V, Aυ =

2(0.4472 ) 10

= 0.739 1 + 2(0.4472 ) 10 ______________________________________________________________________________________

(iii)

8.26 ⎛ 10 −3 ⎛i ⎞ ⎛V ⎞ (a) iCn = I S exp⎜⎜ BE ⎟⎟ ⇒ V BE = VT ln⎜⎜ Cn ⎟⎟ = (0.026) ln⎜⎜ −15 ⎝ VT ⎠ ⎝ 2 × 10 ⎝ IS ⎠ V BB = 2V BE = 1.40077 V PQ = i C ⋅υ CE = (1)(5) = 5 mW

⎞ ⎟ = 0.7004 V ⎟ ⎠

3.5 = 3.5 mA ≅ i L 1 ⎛ 3.5 × 10 −3 ⎞ ⎟ = 0.732957 V = (0.026) ln⎜⎜ −15 ⎟ ⎠ ⎝ 2 × 10 = 1.40077 − 0.732957 = 0.66781 V

(b) For υ o = −3.5 V, iCp ≅

υ EB υ BE

(

)

⎛ 0.66781 ⎞ i Cn = 2 × 10 −15 exp⎜ ⎟ ⇒ i Cn = 0.2857 mA ⎝ 0.026 ⎠ Then iCp ≅ 0.2857 + 3.5 = 3.7857 mA

⎛ 3.7857 × 10 −3 ⎞ ⎟ = 0.734997 V −15 ⎟ ⎠ ⎝ 2 × 10 = 1.40077 − 0.734997 = 0.66577 V

υ EB = (0.026 ) ln⎜⎜ υ BE

(

)

⎛ 0.66577 ⎞ i Cn = 2 × 10 −15 exp⎜ ⎟ ⇒ i Cn = 0.2642 mA ⎝ 0.026 ⎠ i Cp = 3.5 + 0.2642 = 3.764 mA

V BB = −3.5 − 0.735 + 0.7004 = −3.535 V 2 2 = i L2 R L = (3.5) (1) = 12.25 mW

υ I = υ o − υ EB +

For R L : PRL For Q n : PQn = i Cnυ CEn = (0.2642 )[5 − (− 3.5)] = 2.25 mW

For Q p : PQp = i Cpυ ECp = (3.764 )[− 3.5 − (− 5)] = 5.65 mW

______________________________________________________________________________________ 8.27 2 (a) (i) i Dn = K n (υ GSn − VTN ) V BB = 2 V, ⇒ V BB = 4 V 2 = (1)(12 ) = 12 mW

1 = 4(υ GSn − 1.5) ⇒ υ GSn = 2

(ii) P = i Dn ⋅υ DSn

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V − υ DS (sat ) υ 2 (b) (i) i D = K [υ DS (sat )] = o = DD RL RL

KR L [υ DS (sat )] + υ DS (sat ) − V DD = 0 Now KR L = (4 )(1) = 4 2

We have 4[υ DS (sat )] + υ DS (sat ) − 12 = 0 ⇒ υ DS (sat ) = 1.612 V υ o (max ) = 12 − 1.612 = 10.39 V (ii) i Dn = i L = 10.39 mA i Dp = 0 2

υ GSn = 3.112 V −V BB + υ GSn + υ o = −2 + 3.112 + 10.39 = 11.5 V 2 2 (iii) For R L : PRL = i L2 R L = (10.39) (1) = 108 mW For M n : PMn = i Dnυ DSn = (10.39)[12 − 10.39] = 16.7 mW For M p : PMp = i Dpυ SDp = 0

υI =

______________________________________________________________________________________ 8.28 a.

v0 = 24 V ⇒ iL =

24 ⇒ iL ≈ iN = 3 A 8

3 ⇒ iBn = 73.2 mA 41 For iD = 25 mA ⇒ iR1 = 25 + 73.2 = 98.2 mA ⎛i ⎞ 3 ⎛ ⎞ VBE = VT ln ⎜ N ⎟ = ( 0.026 ) ln ⎜ −12 ⎟ I × 6 10 ⎝ ⎠ ⎝ S⎠ iBn =

= 0.7004 V Then 98.2 =

30 − ( 24 + 0.7 ) R1

⇒ R1 =

5.3 ⇒ R1 = 53.97 Ω 98.2

⎛ 25 × 10−3 ⎞ = 0.5759 V VD = ( 0.026 ) ln ⎜ −12 ⎟ ⎝ 6 × 10 ⎠ VEB = 2VD − VBE = 2 ( 0.5759 ) − 0.7004 = 0.4514 V ⎛V ⎞ ⎛ 0.4514 ⎞ iP = I S exp ⎜ EB ⎟ = ( 6 × 10−12 ) exp ⎜ ⎟ ⇒ iP = 0.208 mA V ⎝ 0.026 ⎠ ⎝ T ⎠ b. Neglecting base current iD ≈

30 − 0.6 30 − 0.6 = ⇒ iD ≈ 545 mA R1 53.97

⎛ 0.545 VD = ( 0.026 ) ln ⎜ −12 ⎝ 6 × 10

⎞ ⎟ = 0.656 V ⎠

Approximation for iD is okay.

⇒ iN = iP = 545 mA Diodes and transistors matched ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.29 (a) I D1 = K1 (VGS 1 − VTN ) VGS1 =

2

5 +1 = 2 V 5

I D 3 = K 3 (VGS 3 − VTN )

2

200 = K 3 ( 2 − 1) ⇒ K n3 = K p 4 = 200 μ A / V 2 2

(b) vI + VSG 4 + VGS 3 − VGS1 = vO For vo large, iL = i1 = K n1 (VGS1 − VTN ) VGS 1 =

iL + VTN = K n1

⎛ So vI + 2 + 2 − ⎜ ⎜ ⎝

2

vo + VTN RL K n1

⎞ vo + 1⎟ = v0 ( 0.5 )( 5 ) ⎟⎠

v0 −3 2.5 dv 1 dv dvI 1 =1= 0 + ⋅ ⋅ 0 dvI dvI 2 2.5v0 dvI

vI = v0 +

⎡ ⎤ 1 ⎢1 + ⎥ ⎢⎣ 2 2.5v0 ⎥⎦ vO = 5 V :

1=

For

dv0 dvI

⎤ dv dv0 ⎡ dv 1 ⎢1 + ⎥ = 0 (1.1414 ) ⇒ 0 = 0.876 dvI ⎢ 2 2.5 ( 5 ) ⎥ dvI dvI ⎣ ⎦ ______________________________________________________________________________________ 1=

8.30 vO = vI +

VBB − VGS and VGS = 2

vO ≈ 0, I Dn = I DQ + iL = I DQ +

For

vO = vI + Then

I Dn + VTN Kn vO RL

I DQ + ( vO / RL ) VBB − VTN − 2 Kn

vO ≅ vI +

vO = vI + or

I DQ VBB 1 v − VTN − ⋅ 1+ ⋅ O 2 2 I DQ RL Kn

For vO small, ⎡ 1 I DQ I DQ V 1 ⎤ vO ⎢1 + ⋅ ⋅ ⎥ = vI + BB − VTN − 2 2 K I R Kn ⎢⎣ n DQ L ⎥ ⎦

I DQ v VBB − VTN − ⋅ 1+ O 2 Kn I DQ RL

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Now dvO 1 = = 0.95 dvI ⎡ 1 I DQ 1 ⎤ ⋅ ⎢1 + ⋅ ⎥ ⎢⎣ 2 K n I DQ RL ⎥⎦

So

1 I DQ 1 1 ⋅ ⋅ = − 1 = 0.0526 2 K n I DQ RL 0.95

1

RL = 0.1 k Ω, then

K n I DQ

For Or

= 0.01052

K n I DQ = 95.1

g = 2 K n I DQ = 190 mA/V We can write m This is the required transconductance for the output transistor. This implies a very large transistor. ______________________________________________________________________________________

8.31 (a) RTH = R1 R2 = 14 10 = 5.833 k Ω ⎛ R2 ⎞ ⎛ 10 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH = ⎜⎜ ⎟(12 ) = 5 V ⎝ 10 + 14 ⎠ ⎝ R1 + R 2 ⎠ 5 − 0.7 I BQ = = 0.5619 mA, I CQ = 50.57 mA 5.833 + (91)(0.02) V 12 (b) R L = CC = ⇒ R L = 237 Ω I CQ 50.57 1 (11) ⋅ = 255 mW 2 0.237 (d) PS = I CQVCC = (50.57)(12) = 607 mW

(c) PL (max ) =

2

255 × 100% = 42% 607 ______________________________________________________________________________________

η=

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.32 V 15 I CQ = CC = = 15 mA 1 RL I BQ =

15 = 0.15 mA 100

(15) 1 V2 PL ( max ) = ⋅ CC = ⇒ PL ( max ) = 112.5 mW 2 RL 2 (1) 2

Let RTH = 10 kΩ

VTH = I BQ RTH + VBE + (1 + β ) I BQ RE = ( 0.15 )(10 ) + 0.7 + (101)( 0.15 )( 0.1) VTH = 3.715 =

1 1 ⋅ RTH ⋅ VCC = ⋅ (10 )(15 ) R1 R1

R1 = 40.4 kΩ R2 = 13.3 kΩ

______________________________________________________________________________________ 8.33 (a) RTH = R1 R 2 = 2.3 1.75 = 0.9938 k Ω ⎛ R2 ⎞ ⎛ 1.75 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH = ⎜⎜ ⎟(12 ) = 5.185 V R R + ⎝ 1.75 + 2.3 ⎠ 2 ⎠ ⎝ 1 5.185 − 0.7 I BQ = = 2.473 mA, I CQ = 98.91 mA 0.9938 + (41)(0.02) V 12 (b) Want R L′ = CC = = 121.3 Ω = a 2 R L = a 2 (8) ⇒ a = 3.89 I CQ 0.09891

(9) = 333.9 mW 1 ⋅ 2 (0.1213) 2

(c) PL =

PS = I CQVCC = (98.91)(12) ⇒ 1.187 W 0.3339 × 100% = 28.1% 1.187 ______________________________________________________________________________________

(d) η =

8.34 a.

b.

Assuming the maximum power is being delivered, then 36 9 Vo′ ( peak ) = 36 V ⇒ Vo = = 9 V ⇒ Vrms = ⇒ Vrms = 6.36 V 4 2 36 Vo = ⇒ Vo = 25.5 V 2

I rms = c.

Secondary

Primary d.

IP =

PL 2 = ⇒ I rms = 0.314 A Vrms 6.36

0.314 ⇒ I P = 78.6 mA 4

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ PS = I CQ .VCC = ( 0.15 )( 36 ) = 5.4 W 2 η= ⇒ η = 37% 5.4 ______________________________________________________________________________________

8.35 a.

⎛V ve = ⎜ π + g mVπ ⎝ rπ

⎞ ⎟ RE′ = Vπ ⎠

⎛1 ⎞ ⎜ + g m ⎟ RE′ ⎝ rπ ⎠

⎛ 1+ β = Vπ ⎜ ⎝ rπ vi = Vπ + ve ⇒ Vπ = vi − ve

⎞ ⎟ RE′ ⎠

⎛ 1+ β ⎞ ve = (vi − ve ) ⎜ ⎟ RE′ ⎝ rπ ⎠ 1+ β ⋅ RE′ 2 ⎛n ⎞ (1 + β ) RE′ ve rπ v = = = e where RE′ = ⎜ 1 ⎟ RL vi 1 + 1 + β ⋅ R′ rπ + (1 + β ) RE′ vi ⎝ n2 ⎠ E rπ ⎛n ⎞ ve so ve − v0 ⎜ 1 ⎟ ⎛ n1 ⎞ ⎝ n2 ⎠ ⎜ ⎟ ⎝ n2 ⎠ (1 + β ) RE′ v 1 so 0 = ⋅ vi ⎛ n1 ⎞ rπ + (1 + β ) RE′ ⎜ ⎟ ⎝ n2 ⎠ v0 =

b. I 1 1 n1 2 PL = ⋅ I P2 RL , a = , I CQ = P so PL = .a 2 I CQ RL 2 2 n2 a PS = I CQ .VCC For η = 50% : 1 2 2 ⋅ a I CQ RL a 2 I R V VCC VCC PL CQ L so a 2 = = 0.5 = 2 = = ⇒ a 2 = CC 5 2VCC I CQ ⋅ VCC I CQ ⋅ RL ( 0.1)( 50 ) PS c. R0 =

49 ( 0.026 ) rπ β VT = = ⇒ R0 = 0.255 Ω 1 + β (1 + β ) I CQ ( 50 )( 0.1)

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.36 a. With a 10:1 transformer ratio, we need a current gain of 8 through the transistor.

⎛ R1 R2 ie = (1 + β ) ib and ib = ⎜ ⎜R R +R ib ⎝ 1 2

⎞ ⎛ R1 R2 ie = 8 = (1 + β ) ⎜ ⎟⎟ ii ⎜R R +R ib ⎝ 1 2 ⎠ so we need ii Rib = rπ + (1 + β ) RL′ ≈ (1 + β ) RL′ = (101)( 0.8 ) = 80.8

⎞ ⎟⎟ ⎠ where

⎛ ⎞ R1 R2 Then 8 = (101) ⎜ ⎜ R R + 80.8 ⎟⎟ ⎝ 1 2 ⎠ R1 R2 = 0.0792 or R1 R2 = 6.95 kΩ R1 R2 + 80.8 Set

2VCC V 12 = RL′ ⇒ I CQ = CC = = 15 mA 2 I CQ RL′ 0.8 15 = 0.15 mA 100 = I BQ RTH + VBE

I BQ = VTH

1 ⋅ RTH ⋅ VCC = I BQ RTH + VBE R1 1 ( 6.95)(12 ) = ( 0.15)( 6.95) + 0.7 ⇒ R1 = 47.9 kΩ then R2 = 8.13 kΩ R1

b. I e = 0.9 I CQ = 13.5 mA =

IL ⇒ I L = 135 mA a

1 2 ( 0.135) ( 8) ⇒ PL = 72.9 mW 2 PS = VCC I CQ = (12 )(15 ) ⇒ PS = 180 mW PL =

η=

PL

⇒ η = 40.5% PS ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.37 a. VP = 2 RL PL VP = 2 ( 8 )( 2 ) = 5.66 V = peak output voltage IP =

VP 5.66 = = 0.708 A = peak output current RL 8

Set Ve = 0.9VCC = aVP to minimize distortion Then a =

( 0.9 )(18) 5.66

⇒ a = 2.86

b.

1 ⎛ I P ⎞ 1 ⎛ 0.708 ⎞ ⎜ ⎟ ⇒ I CQ = 0.275 A ⎜ ⎟= 0.9 ⎝ a ⎠ 0.9 ⎝ 2.86 ⎠ Then PQ = VCC I Q = (18 )( 0.275) ⇒ PQ = 4.95 W Power rating of transistor Now I CQ =

______________________________________________________________________________________ 8.38 a.

Need a current gain of 8 through the transistor.

⎛ R1 R2 ib = 8 = (1 + β ) ⎜ ⎜R R +R ii ib ⎝ 1 2

⎞ ⎟⎟ ⎠ where Rib ≈ (1 + β )( 0.9 ) = 90.9 kΩ

⎞ R1 R2 8 ⎛ =⎜ ⎟ = 0.0792 or R1 R2 = 7.82 kΩ ⎜ 101 ⎝ R1 R2 + 90.9 ⎟⎠ Set

2VCC 12 = 0.9 kΩ ⇒ I CQ = = 13.3 mA 2 I CQ 0.9

13.3 = 0.133 mA 100 1 Then ( 7.82 )(12 ) = ( 0.133)( 7.82 ) + 0.7 ⇒ R1 = 53.9 kΩ and R 2 = 9.15 kΩ R1 I BQ =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b. I I e = ( 0.9 ) I CQ = 12 mA = L ⇒ I L = 120 mA a 1 2 PL = ( 0.12 ) ( 8 ) ⇒ PL = 57.6 mW 2 PS = VCC I CQ = (12 )(13.3) ⇒ PS = 159.6 mW PL 57.6 = ⇒ η = 36.1% PS 159.6 ______________________________________________________________________________________

η=

8.39 ⎛ 10 −3 ⎞ ⎞ ⎛I ⎟ = 1.473 V (a) V BB = 2VT ln⎜⎜ Bias ⎟⎟ = 2(0.026) ln⎜⎜ −16 ⎟ ⎝ 5 × 10 ⎠ ⎝ I SD ⎠ ⎛ 1.473 2 ⎞ (b) I CQ = I SQ exp⎜ ⎟ ⇒ I CQ = 14 mA ⎝ 0.026 ⎠ ______________________________________________________________________________________

8.40 −3 ⎛ I CQ ⎞ ⎛ ⎞ ⎟ = (0.026) ln⎜ 4 × 10 ⎟ = 0.73643 = V BEn = V EBp = V D (a) V BE = VT ln⎜ ⎜ 2 × 10 −15 ⎟ ⎜ I SQ ⎟ ⎝ ⎠ ⎠ ⎝ ⎛V ⎞ ⎛ 0.73643 ⎞ I Bias = I SD exp⎜⎜ D ⎟⎟ = 4 × 10 −16 exp⎜ ⎟ ⇒ I Bias = 0.8 mA V ⎝ 0.026 ⎠ ⎝ T ⎠

(

)

(b) V BB = 2V D = 1.473 V (c) υ I = −V EBp = −0.7364 V ______________________________________________________________________________________ 8.41 ⎛ 0.5 × 10 −3 ⎞ ⎟ = 0.76025 V (a) V D1 = (0.026 ) ln⎜⎜ −16 ⎟ ⎠ ⎝ 10 ⎛ 0.5 × 10 −3 ⎞ ⎟ = 0.72421 V V D 2 = (0.026) ln⎜⎜ −16 ⎟ ⎠ ⎝ 4 × 10 V BB = V D1 + V D 2 = 1.48446 V ⎛υ (b) iCn = i Cp = I SQn exp⎜⎜ BEn ⎝ VT I SQp I SQn

⎛υ exp⎜⎜ BEn ⎝ VT = ⎛ υ EBp exp⎜⎜ ⎝ VT

⎛υ ⎞ ⎞ ⎟⎟ exp⎜⎜ BEn ⎟⎟ ⎝ VT ⎠ ⎠ = ⎛ V − υ BEn ⎞ ⎟ exp⎜⎜ BB ⎟ VT ⎝ ⎠

⎛ I SQp 2υ BEn − V BB = ln⎜ ⎜ I SQn VT ⎝

υ BEn =

⎛υ ⎞ ⎟⎟ = I SQp exp⎜ EBp ⎜ V ⎠ ⎝ T

⎞ ⎟ ⎟ ⎠

⎛ 2υ − V BB = exp⎜⎜ BEn VT ⎞ ⎝ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠

⎞ ⎟ ⎟ ⎠

⎛ I SQp 1⎡ ⎢V BB + VT ln⎜ ⎜ I SQn 2 ⎣⎢ ⎝

−15 ⎞⎤ 1 ⎡ ⎛ ⎟⎥ = ⎢1.48446 + (0.026 ) ln⎜ 1.6 × 10 − 16 ⎜ ⎟⎥ 2 ⎢ ⎝ 8 × 10 ⎣ ⎠⎦

⎞⎤ ⎟⎥ = 0.75124 V ⎟ ⎠⎦⎥

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ υ EBp = V BB − υ BEn = 0.73322 V ⎛υ ⎞ ⎛ 0.75124 ⎞ (c) I CQ = I SQn exp⎜⎜ BEn ⎟⎟ = 8 × 10 −16 exp⎜ ⎟ V ⎝ 0.026 ⎠ ⎝ T ⎠ ⇒ i Cn = i Cp = 2.828 mA

(

)

(d) υ I = −υ EBp = −0.73322 V ______________________________________________________________________________________ 8.42 a.

All transistors are matched. ⎛1+ β ⎞ iC 3 mA = iE1 + iB 3 = ⎜ ⎟ iC + β ⎝ β ⎠ ⎛ 61 1 ⎞ 3 = ⎜ + ⎟ iC ⇒ iC = 2.90 mA ⎝ 60 60 ⎠

b. R = 200 Ω. For vo = 6 V , let L 6 = 0.03 A = 30 mA ≅ iE 3 200 30 iB 3 = = 0.492 mA 61 iE1 = 3 − 0.492 = 2.508 mA 2.508 iB1 = ⇒ iB1 = 41.11 μ A 61 3 iE 2 ≅ 3 mA ⇒ iB 2 = ⇒ 49.18 μ A 61 iI = iB 2 − iB1 = 49.18 − 41.11 ⇒ iI = 8.07 μ A io =

Current gain Ai = VBE 3

30 × 10−3 ⇒ Ai = 3.72 × 103 8.07 × 10−6 ⎛i ⎞ ⎛ 30 × 10−3 ⎞ = VT ln ⎜ E 3 ⎟ = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠

VBE 3 = 0.6453 V ⎛i ⎞ ⎛ 2.508 × 10−3 ⎞ VEB1 = VT ln ⎜ E1 ⎟ = ( 0.026 ) ln ⎜ ⎟ −13 ⎝ 5 × 10 ⎠ ⎝ IS ⎠ VEB1 = 0.5807 V vI = v0 + VBE 3 − VEB1 = 6 + 0.6453 − 0.5807 vI = 6.0646 V Voltage gain v 6 Av = 0 = ⇒ Av = 0.989 vI 6.0646 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.43 1 i0 = 1 A, I B 3 ≅ ⇒ 20 mA 50 a. For ⎡10 − ( v0,max + VBE 3 ) ⎤ 10 − VEB1 = 2⎢ − 20 ⎥ R1 R1 ⎢⎣ ⎦⎥ We can then write 10 − VBE 2vo ,max = + 40 R1 R1 If, for simplicity, we assume VEB1 = VBE 3 = 0.7 V, then 9.3 2 ( 4 ) = + 40 R = R2 = 32.5 Ω v = 4 V, R R1 which yields 1 If we assume 0,max then 1 9.3 I E1 = ⇒ I E1 = 0.286 A = I E 2 v = 0, 32.5 I b. For

I = I E 4 = 2.86 A I = 10 I S 1,2 , Since S 3,4 then E 3 c. We can write

⎧ rπ 1 ⎫ ⎪ rπ 3 + R1 ⎪ 1 + β1 ⎪ 1⎪ R0 = ⎨ ⎬ 2⎪ 1 + β3 ⎪ ⎪ ⎪ ⎩ ⎭ ( 50 )( 0.026 ) βV Now rπ 3 = 3 T = = 0.4545 Ω IC3 2.86 rπ 1 =

β1VT I C1

=

(120 )( 0.026 ) 0.286

= 10.91 Ω

So 10.91 ⎫ ⎧ 0.4545 + 32.5 1 ⎪⎪ 121 ⎪⎪ R0 = ⎨ ⎬ 2⎪ 51 ⎪ ⎪⎩ ⎪⎭ 10.91 32.5 = 32.5 0.0902 = 0.0900 121 1 ⎧ 0.4545 + 0.0900 ⎫ ⎨ ⎬ or R 0 = 0.00534 Ω 51 2⎩ ⎭ ______________________________________________________________________________________ Then R0 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.44 1 Ri = rπ 1 + (1 + β ) ⎡ R1 ( rπ 3 + (1 + β ) 2 RL ) ⎤ ⎣ ⎦ 2

{

}

iC1 ≈ 7.2 mA and iC 3 ≈ 7.2 mA Then rπ =

( 60 )( 0.026 ) 7.2

{

= 0.217 kΩ

}

1 0.217 + ( 61) ⎡ 2 ( 0.217 + ( 61)( 0.2 ) ) ⎤ ⎣ ⎦ 2 1 = 0.217 + 61 ⎡⎣ 2 12.4 ⎤⎦ or Ri = 52.6 kΩ 2 ______________________________________________________________________________________ So Ri =

{

}

8.45

I DQ

(b) For M n 3 ; υ GSn3 =

K n3

+ VTN =

5 +1 = 2 V 5

υ SGp1 = υ GSn 3 = 2 V I DQ1 = 2(2 − 1) = 2 mA 2

R1 = R 2 =

(c) I DQ1 = I DQ 2

10 − 2 = 4 kΩ 2 = 2 mA

(d) For υ o = 3.5 V, Assume M p 4 cutoff, so I DQ 4 = 0 I Dn3 = i o =

υo RL

=

3.5 = 23.33 mA 0.15

23.33 + 1 = 3.160 V 5 10 − (υ o + υ GSn3 ) 10 − (3.16 + 3.5) = I R1 = = = 0.835 mA R1 4

υ GSn 3 = I Dp1

0.835 + 1 = 1.646 V 2 υ I = υ o + υ GSn 3 − υ SGp1 = 3.5 + 3.160 − 1.646 = 5.014 V

υ SGp1 =

υ I = υ GSn 2 + K n 2 R2 (υ GSn2 − VTN )2 + V −

(

)

2 15.014 = υ GSn2 + 8 υ GSn 2 − 2υ GSn 2 + 1

2 or 8υ GSn 2 − 15υ GSn 2 − 7.014 = 0 ⇒ υ GSn 2 = 2.2625 V

I Dn 2 = 2(2.2625 − 1) = 3.188 mA υ G 4 = υ I − υ GSn 2 = 5.014 − 2.2625 = 2.75 V υ SGp 4 = υ o − υ G 4 = 3.5 − 2.75 = 0.75 V, ⇒ M p 4 cutoff 2

PL =

υ o2

=

(3.5)2

= 81.7 mW 0.15 RL ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.46 For υ I = −1.5 V and υ O = 0 , ⇒ υ SG 2 = 1.5 V = υ GS1 ⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 2 i D1 = i D 2 = 0.5 = ⎜ ⎟⎜ ⎟ (1.5 − 0.8) ⇒ ⎜ ⎟ = 20.4 ⎝ 2 ⎠⎝ L ⎠ 1 ⎝ L ⎠1 ⎛ 0.04 ⎞⎛ W ⎞ ⎛W ⎞ 2 0. 5 = ⎜ ⎟⎜ ⎟ (1.5 − 0.8) ⇒ ⎜ ⎟ = 51.0 2 L ⎝ ⎠⎝ ⎠ 2 ⎝ L ⎠2 ⎛ 0.04 ⎞⎛ W ⎞ ⎛W ⎞ 2 ⎟⎜ ⎟ (1.5 − 0.8) ⇒ ⎜ ⎟ = 20.4 ⎝ 2 ⎠⎝ L ⎠ 4 ⎝ L ⎠4

υ SG 4 = 1.5 V, 0.2 = ⎜

⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 2 0. 2 = ⎜ ⎟⎜ ⎟ (1.5 − 0.8) ⇒ ⎜ ⎟ = 8.16 ⎝ 2 ⎠⎝ L ⎠ 3 ⎝ L ⎠3 ______________________________________________________________________________________

8.47

For v0 = 0 I Q = I C 3 + I C 2 + I E1 ⎛ 1 + βn I B3 = I E 2 = ⎜ ⎝ βn I C 3 = (1 + β n ) I C 2

⎞ IC 3 ⎟ IC 2 = βn ⎠

⎛ β I B 2 = I C1 = ⎜ P ⎝ 1+ βP

⎞ IC 2 ⎟ I E1 = βn ⎠

⎛ β IC 2 = β n ⎜ P ⎝ 1+ βP

⎞ ⎟ I E1 ⎠

⎛ β I C 3 = (1 + β n ) β n ⎜ P ⎜1+ β p ⎝

⎞ ⎟⎟ I E1 ⎠

⎛ β ⎞ ⎛ β ⎞ I Q = (1 + β n ) β n ⎜ P ⎟ I E1 + β n ⎜ P ⎟ I E1 + I E1 ⎝ 1+ βP ⎠ ⎝ 1+ βP ⎠ ⎛ 10 ⎞ ⎛ 10 ⎞ = ( 51)( 50 ) ⎜ ⎟ I E1 + ( 50 ) ⎜ ⎟ I E1 + I E1 ⎝ 11 ⎠ ⎝ 11 ⎠ I Q = 2318.18I E1 + 45.45 I E1 + I E1 I E1 = 1.692 μ A ⇒ I C1 = 1.534 μ A ⎛ 10 ⎞ I C 2 = ( 50 ) ⎜ ⎟ (1.692 ) ⇒ I C 2 = 76.9 μ A ⎝ 11 ⎠ ⎛ 10 ⎞ I C 3 = ( 51)( 50 ) ⎜ ⎟ (1.692 ) ⇒ I C 3 = 3.92 mA ⎝ 11 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Because of rπ 1 and Z, neglect effect of r . Then neglecting r , r and r , we find 0

01

02

03

VX I X = g m 3Vπ 3 + g m 2Vπ 2 + g m1Vπ 1 + rπ 1 + Z Now ⎛ r ⎞ Vπ 1 = ⎜ π 1 ⎟ VX , Vπ 2 ≅ g m1Vπ 1rπ 2 ⎝ rπ 1 + Z ⎠ and Vπ 3 = ( g m1Vπ 1 + g m 2Vπ 2 ) rπ 3

= ⎡⎣ g m1Vπ 1 + g m 2 ( g m1Vπ 1rπ 2 ) ⎤⎦ rπ 3

⎛ r ⎞ Vπ 3 = ⎜ π 1 ⎟ [ g m1 + g m1 g m 2 rπ 2 ] rπ 3 ⋅ VX ⎝ rπ 1 + Z ⎠ ( β + β1 β 2 ) rπ 3 ⋅ VX Vπ 3 = 1 rπ 1 + Z ⎛ r ⎞ ⎛ βr ⎞ and Vπ 2 = g m1 ⎜ π 1 ⎟ rπ 2VX = ⎜ 1 π 2 ⎟ VX ⎝ rπ 1 + Z ⎠ ⎝ rπ 1 + Z ⎠ ( β + β1 β 2 ) β 3 VX ββ β1 ⋅ V X + 1 2 ⋅ VX + ⋅ VX + Then I X = 1 rπ 1 + Z rπ 1 + Z rπ 1 + Z rπ 1 + Z Then R0 =

rπ 1 + Z VX = I X 1 + β1 + β1 β 2 + ( β1 + β1 β 2 ) β 3

(10 )( 0.026 )

rπ 1 =

1.534 Z = 25 kΩ

= 0.169 MΩ

Then R0 =

169 + 25 1 + (10 ) + (10 )( 50 ) + ⎡⎣10 + (10 )( 50 ) ⎤⎦ ( 50 )

194 = 0.00746 kΩ or Ro = 7.46 Ω 26, 011 ______________________________________________________________________________________ R0 =

8.48 a

Neglect base currents. ⎛I ⎞ VBB = 2VD = 2VT ln ⎜ Bias ⎟ ⎝ IS ⎠ ⎛ 5 × 10−3 ⎞ = 2 ( 0.026 ) ln ⎜ ⇒ VBB = 1.281 V −13 ⎟ ⎝ 10 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VBE1 + VEB 3 = VBB I E1 = I E 3 + I C 2 ⎛ β ⎞ I B 2 = IC 3 = ⎜ P ⎟ I E 3 ⎝ 1+ βP ⎠ ⎛ β ⎞ IC 2 = β n I B 2 = β n ⎜ P ⎟ I E 3 ⎝ 1+ βP ⎠ ⎛ βP I E1 = I E 3 + β n ⎜ ⎝ 1+ βP

⎞ ⎟ IE3 ⎠

⎡ ⎛ βP I E1 = I E 3 ⎢1 + β n ⎜ ⎝ 1+ βP ⎣

⎞⎤ ⎟⎥ ⎠⎦

⎡ ⎛ 1+ βn ⎞ ⎛ 1+ βP ⎞ ⎛ βP ⎜ ⎟ I C1 = ⎜ ⎟ I C 3 ⎢1 + β n ⎜ ⎝ βP ⎠ ⎝ 1+ βP ⎝ βn ⎠ ⎣

⎞⎤ ⎟⎥ ⎠⎦

⎡I ⎤ ⎡I ⎤ VBE1 = VT ln ⎢ C1 ⎥ , VEB 3 = VT ln ⎢ C 3 ⎥ ⎣ IS ⎦ ⎣ IS ⎦

(1.01) IC1 = ⎛⎜

21 ⎞ ⎟ IC 3 ⎝ 20 ⎠

I C1

⎡ ⎛ 20 ⎞ ⎤ ⎢1 + (100 ) ⎜ 21 ⎟ ⎥ ⎝ ⎠⎦ ⎣

⎡ 21 ⎤ = I C 3 ⎢ + 100 ⎥ = 101.05 I C 3 20 ⎣ ⎦ = 100.05 I C 3

⎛ 100.05I C 3 ⎞ ⎛ IC 3 VT ln ⎜ ⎟ + VT ln ⎜ IS ⎝ ⎠ ⎝ IS 2 ⎛ 100.05I C 3 ⎞ VT ln ⎜ ⎟ = VBB I S2 ⎝ ⎠ 2 ⎛V ⎞ 100.05 I C 3 = exp ⎜ BB ⎟ 2 IS ⎝ VT ⎠

⎞ ⎟ = VBB ⎠

⎛V ⎞ exp ⎜ BB ⎟ = 0.4995 mA = I C3 100.05 ⎝ VT ⎠ Then I E 3 = 0.5245 mA IC 3 =

IS

Now I C1 = 100.05I C 3 = 49.97 mA = I C1 ⎛ 20 ⎞ I C 2 = (100 ) ⎜ ⎟ ( 0.5245 ) = 49.95 mA = I C 2 ⎝ 21 ⎠ ⎛I ⎞ ⎛ 49.97 × 10−3 ⎞ VBE1 = VT ln ⎜ C1 ⎟ = 0.026 ln ⎜ ⎟ −13 ⎝ 10 ⎠ ⎝ IS ⎠ = 0.70037 ⎛I ⎞ ⎛ 0.4995 × 10−3 ⎞ VEB 3 = VT ln ⎜ C 3 ⎟ = 0.026 ln ⎜ ⎟ 10−13 ⎝ ⎠ ⎝ IS ⎠ = 0.58062 Note: VBE1 + VEB 3 = 0.70037 + 0.58062 = 1.28099 = VBB

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b. 10 v0 = 10 V ⇒ iE1 ≈ = 0.10 A = iC1 100 100 iB1 = = 1 mA 100 ⎛ 4 × 10−3 ⎞ VBB = 2 ( 0.026 ) ln ⎜ = 1.2694 V −13 ⎟ ⎝ 10 ⎠ ⎛ 0.1 ⎞ VBE1 = ( 0.026 ) ln ⎜ −13 ⎟ = 0.7184 ⎝ 10 ⎠ VEB 3 = 1.2694 − 0.7184 = 0.55099 V ⎛ 0.55099 ⎞ I C 3 = 10−13 exp ⎜ ⎟ = 0.1598 mA ⎝ 0.026 ⎠ V02 (10 ) = ⇒ PL = 1 W RL 100 2

PL =

PQ1 = iC1 ⋅ vCE1 = ( 0.1)(12 − 10 ) ⇒ PQ1 = 0.2 W PQ 3 = iC 3 ⋅ vEC 3 = ( 0.1598 ) (10 − [ 0.7 − 12]) ⇒ PQ 3 = 3.40 mW iC 2 = (100 )( iC 3 ) = (100 )( 0.1598 ) = 15.98 mA PQ 2 = iC 2 ⋅ vCE 2 = (15.98) (10 − [ −12]) ⇒ PQ 2 = 0.352 W ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.49 a. ⎛ 10 × 10−3 ⎞ VBB = 3 ( 0.026 ) ln ⎜ ⇒ VBB = 1.74195 V −12 ⎟ ⎝ 2 × 10 ⎠ VBE1 + VBE 2 + VEB 3 = VBB I C1 ≈

IC 2

βn

, IC 3 ≈

IC 2

β n2

⎛I ⎞ ⎛I ⎞ ⎛I ⎞ VT ln ⎜ C1 ⎟ + VT ln ⎜ C 2 ⎟ + VT ln ⎜ C 3 ⎟ = VBB ⎝ IS ⎠ ⎝ IS ⎠ ⎝ IS ⎠ ⎡ I3 ⎤ VT ln ⎢ 3C 2 3 ⎥ = VBB ⎣ βn IS ⎦ ⎛V ⎞ I C 2 = β n I S 3 exp ⎜ BB ⎟ ⎝ VT ⎠ ⎛ 1.74195 ⎞ = ( 20 ) ( 20 × 10−12 ) 3 exp ⎜ ⎟ ⎝ 0.026 ⎠ I C 2 = 0.20 A, I C1 ≈ 10 mA, I C 3 ≈ 0.5 mA ⎛ 10 × 10−3 ⎞ VBE1 = ( 0.026 ) ln ⎜ ⇒ VBE1 = 0.58065 V −12 ⎟ ⎝ 2 × 10 ⎠ ⎛ 0.2 ⎞ VBE 2 = ( 0.026 ) ln ⎜ ⇒ VBE 2 = 0.6585 V −12 ⎟ ⎝ 2 × 10 ⎠ ⎛ 0.5 × 10−3 ⎞ VEB 3 = ( 0.026 ) ln ⎜ ⇒ VEB 3 = 0.50276 V −12 ⎟ ⎝ 2 × 10 ⎠

b. 1 V2 1 V2 PL = 10 W= ⋅ 0 = ⋅ 0 ⇒ V0 ( max ) = 20 V 2 RL 2 20

For

v0 ( max )

PL =

:

( 20 ) v = ⇒ PL = 20 W 20 RL 2 0

2

20 = −1 A 20 iC 5 + iC 4 + iE 3 = −io ( max ) = 1 A i0 ( max ) = −

⎛ 1+ β p ⎜⎜ ⎝ βp

iC 5 +

iC 5 ⎛ β n ⎞ iC 4 ⋅⎜ ⎟+ βn ⎝ 1+ βn ⎠ βn

iC 5 +

iC 5 ⎛ β n ⎞ iC 5 ⎛ β n ⎜ ⎟+ ⎜ βn ⎝ 1+ βn ⎠ βn ⎝ 1+ βn

⎞ ⎟⎟ = 1 ⎠

⎞⎡ 1 ⎟⎢ ⎠ ⎢⎣ β n

⎛ 1+ β p ⎜⎜ ⎝ βp

⎡ 1 ⎛ 20 ⎞ ⎤ ⎛ 1 ⎞ ⎡ 1 ⎛ 6 ⎞ ⎤ iC 5 ⎢1 + ⎜ ⎟ ⎥ + ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ = 1 20 ⎝ 21 ⎠ ⎦ ⎝ 21 ⎠ ⎣ 20 ⎝ 5 ⎠ ⎦ ⎣

⎞⎤ ⎟⎟ ⎥ = 1 ⎠ ⎥⎦

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ iC 5 (1.05048) = 1 iC 5 = 0.952 A iC 4 = 0.0453 A iE 3 = 0.00272 A ⎛5⎞ iC 3 = 0.00272 ⎜ ⎟ ⎝6⎠ = 0.002267 A ⎛ 2.267 × 10 −3 ⎞ VEB 3 = ( 0.026 ) ln ⎜ ⎟ = 0.54206 V −12 ⎝ 2 × 10 ⎠ VBE1 + VBE 2 = 1.74195 − 0.54206 = 1.19989 ⎛ I ⎞ ⎛I ⎞ VT ln ⎜ C 2 ⎟ + VT ln ⎜ C 2 ⎟ = 1.19989 ⎝ βn IS ⎠ ⎝ IS ⎠

⎛ 1.19989 ⎞ iC 2 = β n ⋅ I S exp ⎜ ⎟ ⎝ 0.026 ⎠ = 20 (18.83) mA iC 2 = 93.9 mA iC 2 ⎛ β n ⎞ 93.9 = 4.47 mA ⎜ ⎟= β n ⎝ 1 + β n ⎠ 21 = I C 2 ( 24 − ( −20 ) ) = ( 0.0939 ) ( 44 ) = 4.13 W

iC1 = PQ 2

PQ 5 = ( 0.952 ) ( −10 − ( −24 ) ) = 13.3 W

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 9 9.1 (a)

vO = Ad ( v2 − v1 )

(

)

1 = Ad 10−3 − ( −10 −3 ) ⇒ Ad = 500

(b)

1 = 500 ( v2 − 10 −3 ) = 1 + 0.5 = 500v2 v2 = 3 mV

(c)

(d) (e)

5 = 500 (1 − v1 ) ⇒ 500v1 = 495 v1 = 0.990 V vO = 0

− 3 = 500 ( v2 − ( −0.5 ) ) −250 − 3 = 500v2 v2 = −0.506 V

______________________________________________________________________________________ 9.2

(a) υ 2 =

υO Aod

=

−2 = −2 × 10 − 4 V 10 4

⎛

1 ⎞ ⎟ ⋅υ I ⎝ 1 + 2000 ⎠

υ2 = ⎜

⎛ 1 ⎞ − 2 × 10 − 4 = ⎜ ⎟ ⋅υ I ⇒ υ I = −0.4002 V ⎝ 2001 ⎠ 1 ⎛ ⎞ (b) υ 2 = ⎜ ⎟ ⋅υ I ⎝ 1 + 2000 ⎠ ⎛ 1 ⎞ −3 ⎟(2 ) = 0.9995 × 10 V ⎝ 2001 ⎠

υ2 = ⎜

υ O = 1 = Aod υ 2 = Aod (0.9995 ×10 −3 ) ⇒ Aod = 1000.5 ______________________________________________________________________________________ 9.3 (a) υ O = Aod (υ 2 − υ1 ) = 5 ×10 3 (2.0000 − 2.0010) = −5 V (b) υ O = Aod (υ 2 − υ1 )

(

(

)

)

− 3.000 = 2 × 10 4 (3.0025 − υ1 ) ⇒ υ1 = 3.00265 V (c) υ O = Aod (υ 2 − υ1 )

1.80 = Aod (0.01 − (− 0.01))× 10 −3 ⇒ Aod = 9 ×10 4 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.4

⎛ Ri ⎞ vid = ⎜ ⎟ vI ⎝ Ri + 25 ⎠ ⎛ Ri ⎞ 0.790 = ⎜ ⎟ ( 0.80 ) ⎝ Ri + 25 ⎠ 0.9875 ( Ri + 25 ) = Ri 24.6875 = 0.0125 Ri Ri = 1975 K ______________________________________________________________________________________ 9.5

(a) Aυ =

− R 2 − 200 = = −10 R1 20

−120 = −3 40 −40 (c) Aυ = = −1 40 ______________________________________________________________________________________

(b) Aυ =

9.6

200 ⎫ = −10 ⎪ 20 ⎪ and ⎬ for each case ⎪ Ri = 20 kΩ ⎪ ⎭ ______________________________________________________________________________________ Av = −

9.7

a. 100 = −10 10 Ri = R1 = 10 kΩ Av = −

b. Aυ = −

100 100

= −5 10 Ri = R1 = 10 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ c. 100 = −5 10 + 10 Ri = 10 + 10 = 20 K ______________________________________________________________________________________ Av = −

9.8 0 −υO 4 ⇒ R2 = ⇒ R 2 = 200 k Ω R2 20 × 10 − 6

(a) i 2 =

− R2 − 200 ⇒ −12 = ⇒ R1 = 16.67 k Ω R1 R1

Aυ =

(b) i 2 =

0 −υO 0 − 1.5 = ⇒ i1 = i 2 = −7.5 μ A R2 200 × 10 3

υO

+ 1.5 ⇒ υ I = −0.125 V − 12 Aυ ______________________________________________________________________________________

υI =

=

9.9 Av = −

(a) (b) (c) (d) (e)

R2 R1

Av = −10 Av = −1 Av = −0.20 Av = −10 Av = −2

Av = −1 (f) ______________________________________________________________________________________

9.10

(a) − 3 =

− R 2 − 200 = ⇒ R 2 = 200 k Ω , R1 = 66.67 k Ω R1 R1

(b) − 8 =

− R 2 − 200 = ⇒ R 2 = 200 k Ω , R1 = 25 k Ω R1 R1

(c) − 20 =

− R 2 − 200 = ⇒ R 2 = 200 k Ω , R1 = 10 k Ω R1 R1

− R2 − R2 = ⇒ R 2 = 100 k Ω , R1 = 200 k Ω R1 200 ______________________________________________________________________________________

(d) − 0.5 =

9.11

υI

0.25 ⇒ R1 = 5 k Ω R1 50 × 10 − 6 − R2 − R2 Aυ = ⇒ −6.5 = ⇒ R 2 = 32.5 k Ω R1 5

(a) i1 =

⇒ R1 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

υO

(b) υ I =

Aυ

=

−4 = 0.6154 V − 6.5

4 = 0.123 mA 32.5 ______________________________________________________________________________________ i1 = i 2 =

9.12

(a) Aυ =

− R2 R1

− R2 ⇒ R1 = 25 k Ω , R 2 = 500 k Ω 25 −1000 ⇒ R1 = 50 k Ω , R 2 = 1 M Ω (b) − 20 = R1 − 20 =

(c) For (a), i1 =

υI R1

=

− 0.2 ⇒ i1 = −8 μ A 25

υI

− 0.2 = ⇒ i1 = −4 μ A R1 50 ______________________________________________________________________________________

For (b), i1 =

9.13 a. Av =

⎛R ⎞ R2 1.05 R2 ⇒ = 1.105 ⎜ 2 ⎟ R1 0.95 R1 ⎝ R1 ⎠

⎛R ⎞ 0.95 R2 = 0.905 ⎜ 2 ⎟ 1.05 R1 ⎝ R1 ⎠ Deviation in gain is +10.5% and − 9.5%

b.

Av ⇒

⎛R ⎞ ⎛R ⎞ 1.01R2 0.99 R2 = 1.02 ⎜ 2 ⎟ ⇒ = 0.98 ⎜ 2 ⎟ 0.99 R1 1.01R1 ⎝ R1 ⎠ ⎝ R1 ⎠

Deviation in gain = ±2% ______________________________________________________________________________________ 9.14

(a) (i) υ O = (ii) i 2 = iL =

− R2 − 15 ⋅υ I = (− 0.20) = 3 V R1 1 0 −υO − 3 = = −0.20 mA R2 15

υO RL

=

3 = 0.75 mA 4

i O + i 2 = i L ⇒ i O = 0.75 − (− 0.20 ) = 0.95 mA

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ −15 (0.05) = −0.75 V 1 0 − (− 0.75) (ii) i 2 = = 0.05 mA 15 −0.75 iL = = −0.1875 mA 4 i O = i L − i 2 = −0.1875 − 0.05 = −0.2375 mA

(b) (i) υ O =

−15 (8 sin ω t ) (mV) ⇒ υ O = −0.12 sin ω t (V) 1 0.12 sin ω t (ii) i 2 = ⇒ i 2 = 8 sin ω t ( μ A) 15 −0.12 sin ω t iL = ⇒ i L = −30 sin ω t ( μ A) 4 iO = i L − i 2 = −38 sin ω t ( μ A) ______________________________________________________________________________________

(c) (i) υ O =

9.15 Av = −

R2 R1 + R5

Av = −30 ± 2.5% ⇒ 29.25 ≤ Av ≤ 30.75 So

R2 R2 = 29.25 and = 30.75 R1 + 2 R1 + 1

We have 29.25 ( R1 + 2 ) = 30.75 ( R1 + 1) Which yields R1 = 18.5 k Ω and R2 = 599.6 kΩ For vI = 25 mV , then 0.731 ≤ vO ≤ 0.769 V

______________________________________________________________________________________ 9.16

υ O1 =

− R2 − 80 ⋅υ I = (− 0.15) = 1.2 V R1 10

υO =

− R4 − 100 ⋅υ O1 = (1.2) = −6 V R3 20

i1 = i 2 = i3 = i 4 =

υI R1

υ O1 R3

=

− (0.15) ⇒ i1 = i 2 = −15 μ A 10

=

1.2 ⇒ i 3 = i 4 = 60 μ A 20

At υ O1 : i 2 + i O1 = i 3 ⇒ i O1 = 60 − (− 15) = 75 μ A; Out of Op-Amp At υ O : iO 2 = i 4 = 60 μ A; Into Op-Amp ______________________________________________________________________________________ 9.17

υ O ⎛ − R 2 ⎞⎛ − R 4 ⎞ ⎟ = 100 ⎟⎜ =⎜ υ I ⎜⎝ R1 ⎟⎠⎜⎝ R3 ⎟⎠ For υ I = 50 mV, υ O = (100 )(0.05) = 5 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ If i 4 = 50 μ A, R 4 = Set R1 = 10 k Ω Then

5 ⇒ R 4 = 100 k Ω 50 × 10 − 6

υO R ⎛ R ⎞⎛ 100 ⎞ ⎟ ⇒ 2 = 10 = 100 = ⎜⎜ 2 ⎟⎟⎜⎜ ⎟ R3 υI ⎝ 10 ⎠⎝ R3 ⎠

Set R 2 = 100 k Ω , R3 = 10 k Ω ______________________________________________________________________________________ 9.18 ⎛ − R 2 ⎞⎛ − R 4 ⎟⎟⎜ Aυ = ⎜⎜ ⎜ ⎝ R1 ⎠⎝ R3

⎞⎛ − R6 ⎟⎜ ⎟⎜ R ⎠⎝ 5

For υ O = 6 V, set i 6 =

υO R6

⎞ ⎟ = −300 ⎟ ⎠

⇒ 60 μ A =

6 ⇒ R 6 = 100 k Ω R6

Set R 6 = 200 k Ω so that i 6 = 30 μ A Set R1 = 20 k Ω ⎛ R ⎞⎛ R ⎞⎛ 200 ⎞ ⎟ Now 300 = ⎜⎜ 2 ⎟⎟⎜⎜ 4 ⎟⎟⎜⎜ ⎟ ⎝ 20 ⎠⎝ R3 ⎠⎝ R5 ⎠ For example, set R 2 = 100 k Ω and R5 = 20 k Ω ⎛ R4 ⎞ ⎛ 100 ⎞⎛⎜ R 4 ⎞⎟⎛ 200 ⎞ ⎟ Then 300 = ⎜ ⎜ ⎟ = 50⎜⎜ ⎟⎜ ⎟ ⎟ ⎝ 20 ⎠⎝ R3 ⎠⎝ 20 ⎠ ⎝ R3 ⎠ R Or 4 = 6 , set R3 = 20 k Ω and R 4 = 120 k Ω R3 ______________________________________________________________________________________

9.19 − R2 ⎛ 22 ⎞ ⋅υ I = −⎜ ⎟(− 0.40 ) = 8.8 V R1 ⎝ 1 ⎠ − R2 1 1 (b) Aυ = = −(22 ) ⋅ ⋅ = −21.8993 R1 ⎡ 1 ⎡ ⎤ 1 ⎛ R 2 ⎞⎤ + 23 ( ) 1 ⎟⎥ ⎜⎜1 + ⎢1 + ⎢ 5 × 10 3 ⎥ ⎣ ⎦ R1 ⎟⎠⎥⎦ ⎢⎣ Aod ⎝ υ O = (− 21.8993)(− 0.40) = 8.7597 V

(a) υ O =

(c) Aυ = −(0.998)(22 ) = −21.956 − 21.956 = −(22) ⋅

1

⇒ Aod = 1.1477 × 10 4

⎡ ⎤ 1 (23)⎥ ⎢1 + A od ⎣ ⎦ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.20 (a) R 1 Av = − 2 ⋅ R1 ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ 100 1 =− ⋅ 1 25 ⎡ ⎤ ⎢⎣1 + 5 × 103 ( 5 ) ⎥⎦ Av = −3.9960 (b) (c) (d)

vO = −3.9960 (1.00 ) ⇒ vO = −3.9960 V 4 − 3.9960 × 100% = 0.10% 4

vO = Aod ( v2 − v1 ) = − Aod v1 v1 = −

− ( −3.9960 ) vO = Aod 5 × 10+3

v1 = 0.7992 mV

______________________________________________________________________________________ 9.21 1 −100 = −9.98431 ⋅ 10 ⎡ 1 ⎛ 100 ⎞⎤ ⎜1 + ⎟⎥ ⎢1 + 3 10 ⎠⎦ ⎣ 7 × 10 ⎝ υ 7 υI = O = = −0.7011 V Aυ − 9.9843

(a) Aυ =

−υ O −7 = ⇒ υ 1 = −1 mV Aυ 7 × 10 3 −υO − (− 5) = = = 2.5 × 10 4 υ1 0.2 × 10 −3

υ1 = (b) Aod

Aυ = (− 10) ⋅

1 = −9.9956 1 ⎡ ⎤ ⎢1 + 2.5 × 10 4 (11)⎥ ⎣ ⎦ υ −5 υI = O = = 0.50022 V Aυ − 9.9956 ______________________________________________________________________________________ 9.22

(a) Aυ =

− R2 R1

⎛ R3 R3 ⎞ − 50 ⎛ 50 50 ⎞ ⎜⎜1 + ⎟⎟ = + ⎟ = −60 + ⎜1 + 5 50 ⎠ ⎝ R 4 R 2 ⎠ 10 ⎝

⎛ 50 50 ⎞ (b) (i) − 100 = −5⎜⎜1 + + ⎟⎟ ⇒ R 4 = 2.78 k Ω ⎝ R 4 50 ⎠ ⎛ 50 50 ⎞ (ii) − 150 = −5⎜⎜1 + + ⎟⎟ ⇒ R 4 = 1.79 k Ω ⎝ R 4 50 ⎠ _____________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.23 a. Av = −

R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + R1 ⎝ R4 R2 ⎠

R1 = 500 kΩ 80 =

R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + 500 ⎝ R4 R2 ⎠

Set R2 = R3 = 500 kΩ ⎛ 500 ⎞ 500 + 1⎟ = 2 + ⇒ R4 = 6.41 kΩ 80 = 1⎜ 1 + R4 R4 ⎝ ⎠

b. For vI = −0.05 V −0.05 i1 = i2 = ⇒ i1 = i2 = −0.1 μ A 500 kΩ v X = −i2 R2 = − ( −0.1× 10−6 )( 500 × 103 ) = 0.05 i4 = −

vX 0.05 =− ⇒ i4 = −7.80 μ A 6.41 R4

i3 = i2 + i4 = −0.1 − 7.80 ⇒ i3 = −7.90 μ A

______________________________________________________________________________________ 9.24 (a)

Av = −1000 =

− R2 −500 = R1 R1

R1 = 0.5 K (b) ⎛ R3 R3 ⎞ + ⎟ ⎜1 + ⎝ R4 R2 ⎠ −250 ⎛ 500 500 ⎞ −1250 −1000 = + ⎜1 + ⎟= R1 ⎝ 250 250 ⎠ R1 Av =

− R2 R1

R1 = 1.25 K ______________________________________________________________________________________ 9.25

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ v i1 = I = i2 R ⎛v ⎞ v A = −i2 R = − ⎜ I ⎟ R = −vI ⎝R⎠ v A vI i3 = − = R R v v 2v 2v i4 = i2 + i3 = − A − A = − A = I R R R R ⎛ 2v ⎞ vB = v A − i4 R = −vI − ⎜ I ⎟ ( R ) = −3vI ⎝ R ⎠ ( −3vI ) 3vI v i5 = − B = − = R R R 2vI 3vI 5vI i6 = i4 + i5 = + = R R R v0 ⎛ 5vI ⎞ v0 = vB − i6 R = −3vI − ⎜ ⎟ R ⇒ = −8 vI ⎝ R ⎠ From Figure 9.12 ⇒ Av = −3

______________________________________________________________________________________ 9.26 − R2 − 200 1 1 ⋅ = ⋅ = −9.9978 R1 ⎡ 20 ⎡ 1 ⎛ 200 ⎞⎤ 1 ⎛ R 2 ⎞⎤ + + ⎜⎜1 + ⎟⎥ ⎜1 ⎟⎥ ⎢1 + ⎢1 4 20 ⎠⎦ R1 ⎟⎠⎥⎦ ⎣ 5 × 10 ⎝ ⎢⎣ Aod ⎝ υ − 4.80 = 0.4801056 V (b) υ I = O = Aυ − 9.9978

(a) Aυ =

(c) υ1 =

− υ O − (− 4.80 ) = ⇒ υ 1 = 96 μ V Aod 5 × 10 4

4.801056 − 4.8 × 100% = 0.022% 4.8 ______________________________________________________________________________________

(d)

9.27

(a) Aυ =

(i)

− R2 1 1 ⋅ = (− 1) ⋅ = −0.9992 R1 ⎡ 1 ⎡ ⎤ 1 ⎛ R 2 ⎞⎤ ⎜⎜1 + ⎟⎥ ⎢1 + ⎢1 + 2.5 × 10 3 (2 )⎥ ⎣ ⎦ R1 ⎟⎠⎦⎥ ⎣⎢ Aod ⎝ υ O = (− 0.9992 )(− 0.8) = 0.7993605 V

0.8 − 0.7993605 × 100% ≅ 0.08% 0.8 1 = −0.990099 (b) Aυ = (− 1) ⋅ 1 ⎡ ⎤ ⎢1 + 2 × 10 2 (2 )⎥ ⎣ ⎦ (i) υ O = (− 0.990099 )(− 0.8) = 0.79208 V

(ii)

0.8 − 0.79208 × 100% = 0.99% 0.9 ______________________________________________________________________________________

(ii)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.28 vl v v R = i2 = − O ⇒ O = − 2 R1 R1 R2 vl

ii =

(a) (b)

i2 = i1 =

⎞ vl v 1 ⎛ R2 = i3 + O = i3 + ⎜ − ⋅ vl ⎟ R1 RL RL ⎝ R1 ⎠

vl ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎝ RL ⎠ ______________________________________________________________________________________ Then i3 =

9.29 ⎛ R3 R1 ⎞ + ⎛ 0.1 1 ⎞ VX .max = ⎜ 10 ⇒ VX .max = 0.09008 V ⋅V = ⎜ ⎜ R R + R ⎟⎟ ⎜ 0.1 1 + 10 ⎟⎟ ( ) 4 ⎠ ⎝ 3 1 ⎝ ⎠ R vO = 2 ⋅ VX .max R1 10 =

R2 R ( 0.09008 ) ⇒ 2 = 111 R1 R1

So R2 = 111 k Ω

______________________________________________________________________________________ 9.30

υO = −

RF R R 120 120 120 ⋅υ I 1 − F ⋅ υ I 2 − F ⋅υ I 3 = − ⋅υ I 1 − ⋅υ I 2 − ⋅υ I 3 40 20 60 R1 R2 R3

υ O = −3υ I 1 − 6υ I 2 − 2υ I 3 (a) υ O = −3(− 0.25) − 6(0.10 ) − 2(1.5) = −2.85 V (b) 0.5 = −3(υ I 1 ) − 6(0.25) − 2(− 1.2) ⇒ υ I 1 = 0.133 V ______________________________________________________________________________________ 9.31 (a) υ O = −2.5(1.2υ I 1 + 2.5υ I 2 + 0.25υ I 3 ) = −3υ I 1 − 6.25υ I 2 − 0.625υ I 3 RF R R = 3 , F = 6.25 , F = 0.625 R1 R2 R3

Then

R3 is the largest resistor, so set R3 = 400 k Ω Then R F = 250 k Ω , R1 = 83.3 k Ω , R 2 = 40 k Ω

(b) υ O = −3(− 1) − 6.25(0.25) − 0.625(2) = 0.1875 V

υO

0.1875 ⇒ i F = 0.75 μ A RF 250 ______________________________________________________________________________________ iF =

9.32

=

υ O = −2(υ I 1 + 3υ I 2 ) = −2υ I 1 − 6υ I 2 Then

RF R = 2, F = 6 R1 R2

For υ I 1 = −1 V, υ I 2 = −0.5 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then υ O = −2(− 1) − 6(− 0.5) = 5 V Set i F = 80 μ A =

υO RF

=

5 ⇒ R F = 62.5 k Ω RF

Then R1 = 31.25 k Ω , R 2 = 10.42 k Ω ______________________________________________________________________________________ 9.33

vI 1 = ( 0.05 ) 2 sin ( 2π ft ) = 0.0707 sin ( 2π ft ) 1 1 ⇒ 1 ms vI 2 ⇒ T2 = ⇒ 10 ms 100 103 R R 10 10 vO = − F ⋅ vI 1 − F ⋅ vI 2 = − ⋅ vI 1 − ⋅ vI 2 1 5 R1 R2 f = 1 kHz ⇒ T =

vO = − (10 ) ( 0.0707 sin ( 2π ft ) ) − ( 2 )( ±1 V ) vO = −0.707 sin ( 2π ft ) − ( ±2 V )

______________________________________________________________________________________ 9.34

υO = −

RF R ⋅υ I 1 − F ⋅υ I 2 R1 R2

− 0.5 sin ω t = −

100 (0.004 + 0.125 sin ω t ) − 100 (− 0.006) R1 R2

Set − 0.5 sin ω t = − We have 0 = −

100 (0.125 sin ω t ) ⇒ R1 = 25 k Ω R1

100 (0.004) − 100 (− 0.006) R1 R2

0.4 0.6 + ⇒ R 2 = 37.5 k Ω 25 R 2 ______________________________________________________________________________________ 0=−

9.35 1 ⎡υ ⎤ (a) υ O = −2 ⎢ I 1 + 2υ I 2 + υ I 3 ⎥ = − υ I 1 − 4υ I 2 − 2υ I 3 2 ⎣ 4 ⎦ R R 1 R Then F = , F = 4 , F = 2 R1 2 R 2 R3

Set R1 = 250 k Ω , Then R F = 125 k Ω , R 2 = 31.25 k Ω , R3 = 62.5 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) For υ I 1 = −2 V, υ I 2 = 0 , υ I 3 = −1 V 1 (− 2) − 4(0) − 2(− 1) = 3 V 2 = 2 V, υ I 2 = 0.5 V, υ I 3 = 0

υO = − For υ I 1

1 (2) − 4(0.5) − 2(0) = −3 V 2 Then −3 ≤ υ O ≤ +3 V

υO = −

υO

3 max = ⇒ i F max = 24 μ A RF 125 ______________________________________________________________________________________ iF

=

max

9.36 RF R R ⋅υ I 1 − F ⋅ υ I 2 − F ⋅υ I 3 R1 R2 R3

υO = −

− 6 sin ω t = −

We have −

RF R R (2 + 2 sin ω t ) − F (0.5 sin ω t ) − F (− 4) R1 R2 R3

RF R (2) + F (4) = 0 ⇒ R3 = 2 R1 R1 R3

Also − 6 = −

RF R (2) − F (0.5) R1 R2

For υ O = 6 V, i F For υ I 1

max

max

= 4 V, i1

= 120 μ A =

max

6 ⇒ R F = 50 k Ω RF

= 120 μ A =

4 ⇒ R1 = 33.33 k Ω and R3 = 2 R1 = 66.66 k Ω R1

(50)(2) + (50)(0.5) = 100 + 25 =

100 25 + ⇒ R 2 = 8.33 k Ω R1 R2 R1 R 2 33.33 R 2 ______________________________________________________________________________________

Now 6 =

9.37 a.

v0 = −

RF R R R ⋅ a3 ( −5 ) − F ⋅ a2 ( −5 ) − F ⋅ a1 ( −5 ) − F ⋅ a0 ( −5 ) R3 R2 R1 R0

RF ⎡ a3 a2 a1 a0 ⎤ + + + ( 5) 10 ⎢⎣ 2 4 8 16 ⎥⎦ R 1 v0 = 2.5 = F ⋅ ⋅ 5 ⇒ RF = 10 kΩ 10 2

So v0 =

b. c.

v0 =

10 1 ⋅ ⋅ 5 ⇒ v0 = 0.3125 V 10 16

v0 =

10 ⎡ 1 1 1 1 ⎤ + + + ( 5) ⇒ v0 = 4.6875 V 10 ⎢⎣ 2 4 8 16 ⎥⎦

i.

ii. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.38 ⎛ 20 ⎞ ⎛ − 10 ⎞⎛ − 20 ⎞ (a) υ O = ⎜ ⎟ ⋅υ I 1 − ⎜ ⎟ ⋅υ I 2 = 200υ I 1 − 20υ I 2 ⎟⎜ ⎝ 1 ⎠ ⎝ 1 ⎠⎝ 1 ⎠ (b) υ O = (200 )(5) − (20 )(− 25 − 50 sin ω t ) = 1000 + 500 + 1000 sin ω t (mV) υ O = 1.5 + 1.0 sin ω t (V) (c) For the 20 k Ω resistor: 2.5 i max = ⇒ i max = 0.125 mA 20 For the 10 k Ω resistor: 50 mV ⎛ 10 ⎞ υ O1 = ⎜ ⎟(5) = 50 mV, i max = ⇒ i max = 5 μ A 10kΩ ⎝1⎠ ______________________________________________________________________________________

9.39 For one-input

υ1 = −

υO Aod

υ −υO υ I 1 − υ1 υ1 = + 1 R1 RF R 2 R3 ⎡1 1 1 ⎤ υO = υ1 ⎢ + + ⎥− R1 ⎣⎢ R1 R 2 R3 R F ⎥⎦ R F υ ⎡1 1 1 ⎤ υO =− O ⎢ + + ⎥− Aod ⎣⎢ R1 R 2 R3 R F ⎦⎥ R F

υ I1

⎡ 1 1 1 ⎛⎜ 1 1 = −υ O ⎢ + + + ⎜ ⎢⎣ Aod R F R F Aod ⎝ R1 R 2 R3 ⎤ υ ⎡ 1 RF 1 ⎥ =− O ⎢ +1+ ⋅ R F ⎢ Aod Aod R1 R 2 R3 ⎥ ⎣ ⎦ RF 1 where υO = − ⋅υ I 1 ⋅ R1 ⎡ 1 ⎛ R F ⎞⎤ ⎟⎟⎥ ⎜⎜1 + ⎢1 + ⎣⎢ Aod ⎝ R P ⎠⎦⎥

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

R P = R1 R2 R3

⎛R ⎞ R R −1 ⋅ ⎜⎜ F ⋅υ I 1 + F ⋅υ I 2 + F ⋅υ I 3 ⎟⎟ R2 R3 1 ⎛ R F ⎞ ⎝ R1 ⎠ ⎟⎟ ⎜⎜1 + 1+ Aod ⎝ R P ⎠ ______________________________________________________________________________________

Therefore, for three-inputs υ O =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.40 ⎛ R ⎞ ⎛ 150 ⎞ (a) Aυ = ⎜⎜1 + 2 ⎟⎟ = ⎜1 + ⎟ = 11 R1 ⎠ ⎝ 15 ⎠ ⎝ ⎛ 150 ⎞ (b) Aυ = ⎜1 + ⎟=4 50 ⎠ ⎝ ⎛ 20 ⎞ (c) Aυ = ⎜1 + ⎟ = 1.4 ⎝ 50 ⎠ ⎛ 20 ⎞ (d) Aυ = ⎜1 + ⎟ = 2 ⎝ 20 ⎠ ______________________________________________________________________________________

9.41 ⎛ R ⎞ R (a) Aυ = 15 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 14 R1 ⎠ R1 ⎝ For υ O = −7.5 V ⇒ υ1 = −0.5 V

i = 120 μ A =

7.5 − 0.5 ⇒ R 2 = 58.33 k Ω , R1 = 4.17 k Ω R2

(b) υ O = (15)(0.25) = 3.75 V

0.25 ⇒ i1 = i 2 = 60 μ A 4.17 ______________________________________________________________________________________ i1 = i 2 =

9.42 ⎛ R ⎞ (a) Aυ = ⎜⎜1 + 2 ⎟⎟ R1 ⎠ ⎝ ⎛ R ⎞ R 3 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 2 , Set R 2 = 290 k Ω , R1 = 145 k Ω R1 ⎠ R1 ⎝ ⎛ R ⎞ R (b) 9 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 8 , Set R 2 = 290 k Ω , R1 = 36.25 k Ω R R1 1 ⎠ ⎝ ⎛ R ⎞ R (c) 30 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 29 , Set R 2 = 290 k Ω , R1 = 10 k Ω R R1 1 ⎠ ⎝ ⎛ R ⎞ R (d) 1 = ⎜⎜1 + 2 ⎟⎟ ⇒ 2 = 0 , Set R 2 = 0 , R1 = 290 k Ω R1 ⎠ R1 ⎝ ______________________________________________________________________________________

9.43

⎛ 1 ⎞ vB = ⎜ ⎟ vI ⎝ 1 + 500 ⎠

⎛ 1 ⎞ v0 = Aod ⎜ ⎟ vi ⎝ 501 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

a.

⎛ 1 ⎞ 2.5 = Aod ⎜ ⎟ ( 5 ) ⇒ Aod = 250.5 ⎝ 501 ⎠

⎛ 1 ⎞ v0 = 5000 ⎜ ⎟ ( 5 ) ⇒ v0 = 49.9 V ⎝ 501 ⎠ b. ______________________________________________________________________________________ 9.44 ⎛ 50 ⎞ ⎡⎛ 20 ⎞ ⎛ 40 ⎞ ⎤ v0 = ⎜ 1 + ⎟ ⎢⎜ ⎟ vI 2 + ⎜ ⎟ vI 1 ⎥ + ⎝ 50 ⎠ ⎣⎝ 20 40 ⎠ ⎝ 20 + 40 ⎠ ⎦ v0 = 1.33vI 1 + 0.667vI 2

______________________________________________________________________________________ 9.45 (a)

vI 1 − v2 vI 2 − v2 v2 + = 20 40 10 ⎛ 100 ⎞ vO = ⎜ 1 + ⎟ v2 = 3v2 50 ⎠ ⎝ Now 2vI 1 − 2v2 + vI 2 − v2 = 4v2 ⎛v ⎞ 2vI 1 + vI 2 = 7v2 = 7 ⎜ o ⎟ ⎝3⎠ 6 3 So vO = ⋅ vI 1 + ⋅ vI 2 7 7

vO = (b)

6 3 ( 0.2 ) + ⎛⎜ ⎞⎟ ( 0.3) ⇒ vO = 0.3 V 7 ⎝7⎠

⎛6⎞ ⎛ 3⎞ vO = ⎜ ⎟ ( 0.25) + ⎜ ⎟ ( −0.4 ) ⇒ vO = 42.86 mV ⎝7⎠ ⎝7⎠ (c) ______________________________________________________________________________________ 9.46 ⎛ R4 (a) υ 2 = ⎜⎜ ⎝ R3 + R 4 ⎛

R ⎞

⎞ ⎟ ⋅υ I ⎟ ⎠ ⎛

R ⎞

υ O = ⎜⎜1 + 2 ⎟⎟ ⋅υ 2 = ⎜⎜1 + 2 ⎟⎟ ⋅ R1 ⎠ R1 ⎠ ⎛ ⎝ ⎝

1

R ⎜⎜1 + 3 ⎝ R4

⎞ ⎟⎟ ⎠

⋅υ I

⎛ R ⎞ ⎛ R ⎞ R 1 (b) 6 = ⎜⎜1 + 2 ⎟⎟ ⋅ ⇒ ⎜⎜1 + 2 ⎟⎟ = 9 ⇒ 2 = 8 R1 ⎠ ⎛ 25 ⎞ R1 ⎠ R1 ⎝ ⎝ ⎜1 + ⎟ ⎝ 50 ⎠ Set R 2 = 200 k Ω , R1 = 25 k Ω ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.47 (a) vO ⎛ 50 x ⎞ = ⎜⎜ 1 + ⎟ vI ⎝ (1 − x ) 50 ⎟⎠ vO ⎛ x ⎞ 1− x + x = ⎜1 + ⎟= vI ⎝ 1 − x ⎠ 1− x vO 1 = vI 1 − x 1 ≤ Av ≤ ∞

Av =

(b) (c) If x = 1, gain goes to infinity. ______________________________________________________________________________________ 9.48 ⎛υ ⎞ (a) υ X = ⎜⎜ I ⎟⎟(2 R ) + υ I = 3υ I ⎝ R⎠ υ X −υ I υ X υ X −υO + + =0 2R R 2R 1 1 ⎞ υ I υO ⎛ 1 = υX ⎜ + + ⎟− ⎝ 2R R 2R ⎠ 2R 2R

υ ⎛2⎞ υ 3υ I ⎜ ⎟ − I = O ⎝ R ⎠ 2R 2R

υO = 11 υI (b) For υ I = 0.25 V, ⇒ υ O = 2.75 V (c) R = 30 k Ω , υ I = −0.15 V so

0.15 ⇒ 5μ A 30 For R 2 : i = 5 μ A

For R1 : i =

υ X = 3υ I = −0.45 V 0.45 ⇒ 15 μ A 30 υ O = (11)(− 0.15) = −1.65 V

For R 4 : i =

1.65 − 0.45 ⇒ 20 μ A 60 ______________________________________________________________________________________

For R3 : i =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.49

(a) (b)

vO =1 vI

From Exercise TYU9.7 ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ vO ⎝ = vI ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ A R1 ⎠ ⎦ od ⎝ ⎣ But R2 = 0, R1 = ∞ vO v 1 1 = = ⇒ O = 0.999993 1 vI 1 + 1 vI 1+ Aod 1.5 × 105 Want

vO 1 = 0.990 = ⇒ Aod = 99 1 vI 1+ Aod

(b) ______________________________________________________________________________________ 9.50 (a) υ O = Aod (υ 2 − υ1 ) = Aod (υ I − υ O ) υ O (1 + Aod ) = Aod υ I υ 1 1 Aυ = O = = = 0.9524 1 1 υI 1+ 1+ Aod 20

(b) Aυ =

1

= 0.995 1 200 1 = 0.9995 (c) Aυ = 1 1+ 2000 1 = 0.99995 (d) Aυ = 1 1+ 20000 ______________________________________________________________________________________ 1+

9.51

(a) Aυ1 = Aυ 2 =

υ O1 ⎛ R 2 ⎞ ⎟ = ⎜1 + υ I ⎜⎝ R1 ⎟⎠ ⎛ R ⎞ υ O2 = −⎜⎜1 + 2 ⎟⎟ R1 ⎠ υI ⎝

⎛ 60 ⎞ (b) υ O1 = ⎜1 + ⎟(− 0.5) = −2 V ⎝ 20 ⎠ ⎛ ⎝

υ O 2 = −⎜ 1 +

60 ⎞ ⎟(− 0.5) = 2 V 20 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ R ⎞ ⎛ 60 ⎞ (c) υ O1 − υ O 2 = 2⎜⎜1 + 2 ⎟⎟ ⋅υ I = 2⎜1 + ⎟(0.8) = 6.4 V R ⎝ 20 ⎠ 1 ⎠ ⎝ ______________________________________________________________________________________

9.52

iL = (a) (b)

vI R1

vO1 = iL RL + vI = iL RL + iL R1

vOI ( max ) ≅ 10 V = iL (1 + 9 ) = 10iL

So iL ( max ) ≅ 1 mA

Then vI ( max ) ≅ iL R1 = (1)( 9 ) ⇒ vI ( max ) ≅ 9 V

______________________________________________________________________________________ 9.53 ⎛ 20 ⎞ (a) υ O = ⎜ ⎟ ⋅υ I = (0.3333)υ I ⎝ 20 + 40 ⎠ (i) υ O = 1 V (ii) υ O = −1.67 V ⎛ 20 ⎞ (b) υ O = ⎜ ⎟ ⋅υ I = (0.3333)υ I ⎝ 20 + 40 ⎠ (i) υ O = 1 V (ii) υ O = −1.67 V ⎛ 10 ⎞⎛ 6 ⎞ (c) υ O = ⎜1 + ⎟⎜ ⎟ ⋅υ I = (0.2222 )υ I ⎝ 10 ⎠⎝ 6 + 48 ⎠ (i) υ O = 0.667 V (ii) υ O = −1.111 V ______________________________________________________________________________________

9.54 a.

Rin =

v −v v1 and 1 0 = i1 and v0 = − Aod v1 i1 RF

So i1 =

v1 − ( − Aod v1 )

Then Rin =

RF

=

v1 RF = i1 1 + Aod

v1 (1 + Aod ) RF

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b. ⎛ RS ⎞ RF i1 = ⎜ ⋅ i1 ⎟ iS and v0 = − Aod ⋅ 1 + Aod ⎝ RS + Rin ⎠ ⎛ A ⎞⎛ RS ⎞ So v0 = − RF ⎜ od ⎟⎜ ⎟ iS ⎝ 1 + Aod ⎠⎝ RS + Rin ⎠ RF 10 = = 0.009990 Rin = 1 + Aod 1001 ⎞ RS ⎛ 1000 ⎞ ⎛ v0 = − RF ⎜ ⎟ iS ⎟⎜ ⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠ ⎞ RS ⎛ 1000 ⎞ ⎛ Want ⎜ ⎟ ≤ 0.990 ⎟⎜ ⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠ which yields RS ≥ 1.099 kΩ

______________________________________________________________________________________ 9.55

vO = iC RF , 0 ≤ iC ≤ 8 mA

For vO ( max ) = 8 V, Then RF = 1 kΩ

______________________________________________________________________________________ 9.56 vI 10 ⇒ R = 10 kΩ so 1 = R R In the ideal op-amp, R1 has no influence. ⎛ R ⎞ v0 = ⎜ 1 + 2 ⎟ vI R⎠ ⎝ Output voltage: i=

v0 must remain within the bias voltages of the op-amp; the larger the R2, the smaller the range of input voltage vI in which the output is valid. ______________________________________________________________________________________ 9.57

(a) i L = Set

−υ I − (− 5) ; R2 = =1k Ω 5 R2 RF 1 = ; For example, set R1 = R F = 10 k Ω , R3 = 1 k Ω R1 R3 R 2

(b) υ L = (5)(0.2) = 1 V = υ 1 i1 = i 2 =

υ I − υ1 R1

=

− 5 −1 = −0.6 mA 10

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

υ O = υ1 − i 2 R F = 1 − (− 0.6)(10) = 7 V υO −υ L 7 −1 i3 =

R3

=

1

= 6 mA

υL

1 = = 1 mA R2 1

i4 =

For the op-amp: iO + i 2 = i 3 ⇒ i O = i 3 − i 2 = 6 − (− 0.6 ) = 6.6 mA ______________________________________________________________________________________ 9.58 (a)

i1 = i2 and i2 =

vx + iD , vx = −i2 RF R2

⎛R Then i1 = −i1 ⎜ F ⎝ R2

⎞ ⎟ + iD ⎠

⎛ R ⎞ Or iD = i1 ⎜1 + F ⎟ R2 ⎠ ⎝ (b) R1 =

vI 5 = ⇒ R1 = 5 k Ω i1 1

⎛ R ⎞ R 12 = (1) ⎜ 1 + F ⎟ ⇒ F = 11 R2 ⎠ R2 ⎝ For example, R2 = 5 k Ω, RF = 55 k Ω

______________________________________________________________________________________ 9.59 IX =

(1) (2)

VX VX − vO + R2 R3

VX VX − vO + =0 R1 RF

⎛ R ⎞ From (2) vO = VX ⎜1 + F ⎟ R1 ⎠ ⎝ ⎛ 1 ⎛ R ⎞ 1 ⎞ 1 Then (1) I X = VX ⎜ + ⎟ − ⋅ VX ⎜1 + F ⎟ R1 ⎠ ⎝ ⎝ R2 R3 ⎠ R3 IX R R 1 1 1 1 1 = = + − − F = − F VX R0 R2 R3 R3 R1 R3 R2 R1 R3 =

R1 R3 − R2 RF R1 R2 R3

or Ro =

R1 R2 R3 R1 R3 − R2 RF

RF 1 = ⇒ R2 RF = R1 R3 then Ro = ∞, which corresponds to an ideal current source. R1 R3 R2 ______________________________________________________________________________________ Note: If

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.60 (a) Rid = R1 + R3 = 30 k Ω ; R1 = R3 = 15 k Ω R2 R4 = = 15 ⇒ R 2 = R 4 = 225 k Ω R1 R3

(b) υ O = i L R L = (0.25)(10) = 2.5 V

υ I 2 − υ I1 =

υO Ad

=

2.5 = 0.1667 V 15

(c) υ O = Ad (υ I 2 − υ I 1 ) = 15(1.2 − 1.5) = −4.5 V iL =

(d) υ O

υO RL

=

− 4.5 = −0.45 mA 10

= (0.5)(10) = 5 V

υ I 2 − υ I1 =

υO Ad

=

5 = 0.333 V 15

υ I 1 = 2 − 0.333 = 1.667 V ______________________________________________________________________________________ 9.61

(a)

R2 R4 = = 40 ; Set R 2 = R 4 = 250 k Ω , R1 = R3 = 6.25 k Ω R1 R3

(b)

R2 R4 = = 25 ; Set R 2 = R 4 = 250 k Ω , R1 = R3 = 10 k Ω R1 R3

(c)

R2 R4 = = 5 ; Set R 2 = R 4 = 250 k Ω , R1 = R3 = 50 k Ω R1 R3

R2 R4 = = 0.5 ; Set R 2 = R 4 = 125 k Ω , R1 = R3 = 250 k Ω R1 R3 ______________________________________________________________________________________

(d)

9.62 We have ⎞ ⎛ R ⎞⎛ R / R ⎞ ⎛R ⎞ ⎛ R ⎞⎛ ⎛ R2 ⎞ 1 vO = ⎜1 + 2 ⎟ ⎜ 4 3 ⎟ vI 2 − ⎜ 2 ⎟ vI 1 or vO = ⎜1 + 2 ⎟ ⎜ ⎟ vI 2 − ⎜ ⎟ vI 1 R1 ⎠ ⎝ 1 + R4 / R3 ⎠ R1 ⎠ ⎝ 1 + R3 / R4 ⎠ ⎝ ⎝ R1 ⎠ ⎝ ⎝ R1 ⎠ Set R2 = 50 (1 + x ) , R1 = 50 (1 − x ) R3 = 50 (1 − x ) , R4 = 50 (1 + x ) ⎡ ⎤ ⎥ ⎡ ⎛ 1 + x ⎞⎤ ⎢ 1 1+ x ⎞ ⎢ ⎥ vI 2 − ⎛⎜ vO = ⎢1 + ⎜ ⎟⎥ ⎟ vI 1 ⎝ 1− x ⎠ ⎣ ⎝ 1 − x ⎠ ⎦ ⎢1 + ⎛ 1 − x ⎞ ⎥ ⎢ ⎜⎝ 1 + x ⎟⎠ ⎥ ⎣ ⎦ ⎤ ⎡1 − x + (1 + x ) ⎤ ⎡ 1+ x ⎛ 1+ x ⎞ vO = ⎢ ⎥ vI 2 − ⎜ ⎥⋅⎢ ⎟ vI 1 1 − x 1 + x + 1 − x ( ) ⎦⎥ ⎝ 1− x ⎠ ⎣ ⎦ ⎣⎢ ⎛1+ x ⎞ ⎛ 1+ x ⎞ =⎜ ⎟ vI 2 − ⎜ ⎟ vI 1 1 − x ⎝ ⎠ ⎝ 1− x ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For vI 1 = vI 2 ⇒ vO = 0 Set

R2 = 50 (1 + x ) R1 = 50 (1 − x ) R3 = 50 (1 + x ) R4 = 50 (1 − x )

⎛ ⎞ ⎛ 1+ x ⎞⎜ 1 ⎟ ⎛ 1+ x ⎞ vO = ⎜1 + ⎟ ⎜ 1 + x ⎟ vI 2 − ⎜ ⎟ vI 1 1 x − ⎝ ⎠⎜1+ ⎝ 1− x ⎠ ⎟⎟ ⎜ ⎝ 1− x ⎠ ⎛1+ x ⎞ = vI 2 − ⎜ ⎟ vI 1 ⎝1− x ⎠ vI 1 = vI 2 = vcm

vO 1 + x 1 − x − (1 + x ) −2 x = 1− = = vcm 1− x 1− x 1− x

Set

R2 = 50 (1 − x )

R1 = 50 (1 + x )

R3 = 50 (1 − x ) R4 = 50 (1 + x )

⎛ ⎞ ⎛ 1− x ⎞⎜ 1 ⎟ ⎛ 1− x ⎞ vO = ⎜1 + ⎟ ⎜ 1 − x ⎟ vI 2 − ⎜ ⎟ vI 1 ⎝ 1+ x ⎠⎜1+ ⎝ 1+ x ⎠ ⎟⎟ ⎜ ⎝ 1+ x ⎠ 1 x − ⎛ ⎞ = ⎜1 − ⎟ vcm ⎝ 1+ x ⎠ 1 + x − (1 − x ) 2x Acm = = 1+ x 1+ x Worst common-mode gain

Acm = (b)

−2 x 1− x

−2 x −2 ( 0.01) = = −0.0202 1 − x 1 − 0.01 −2 ( 0.02 ) For x = 0.02, Acm = = −0.04082 1 − 0.02 −2 ( 0.05 ) For x = 0.05, Acm = = −0.1053 1 − 0.05 1 1 For this condition, set vI 2 = + , vI 1 = − ⇒ vd = 1 V 2 2 1 1 ⎡ ⎛ 1 + x ⎞ ⎤ 1 ⎡1 − x + (1 + x ) ⎤ 1 2 = Ad = ⎢1 + ⎜ ⎥= ⋅ ⎟ = ⎢ 2 1 1 − x − x 2 ⎣ ⎝ 1 − x ⎠ ⎥⎦ 2 ⎣ 1− x ⎦ 1.010 C M R RdB = 20 log10 = 33.98 dB For x = 0.01 Ad = 1.010 0.0202 For x = 0.01,

Acm =

For x = 0.02, Ad =

1 = 1.020 0.98

C M R RdB = 20 log10

1.020 = 27.96 dB 0.04082

1 1.0526 = 1.0526 C M R RdB = 20 log10 ≅ 20 dB 0.95 0.1053 ______________________________________________________________________________________ For x = 0.05 Ad =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.63 (a) υ O = 10(υ 2 − υ1 ) = 10(1.4 − 1.8) = −4 V

i3 = i 4 =

υ2 R + 10 R

=

1.4

(11)(10)

= 0.0127 mA

⎛ 10 ⎞ ⎛ 10 ⎞ ⎟ ⋅υ 2 = ⎜ ⎟(1.4 ) = 1.273 V ⎝ 11 ⎠ ⎝ 11 ⎠ υ − υ X 1.8 − 1.273 = = 0.0527 mA i1 = i 2 = 1 10 R (b) υ O = 10(υ 2 − υ1 ) = 10(3.6 − 3.2 ) = 4 V

υ X = υY = ⎜

⎛ 10 ⎞ ⎟(3.6 ) = 3.273 V ⎝ 11 ⎠ υ2 3.6 i3 = i 4 = = = 0.0327 mA (11)(10) 110 3.2 − 3.273 i1 = i 2 = = −0.00727 mA 10 (c) υ O = 10(− 1.35 − (− 1.20 )) = −1.5 V

υ X = υY = ⎜

i3 = i 4 =

−1.35 = −0.0123 mA (11)(10)

⎛ 10 ⎞ ⎟(− 1.35) = −1.227 V ⎝ 11 ⎠ − 1.2 − (− 1.227 ) i1 = i 2 = = 0.00273 mA 10 ______________________________________________________________________________________

υ X = υY = ⎜

9.64 (a) I E = (1 + β ) ⋅ I B = (76)(1.2) = 91.2 mA 10 ⇒ R = 109.6 Ω 91.2 (b) I E = (101)(0.2 ) = 20.2 mA R=

10 = 0.495 k Ω 20.2 6 (c) I E = = 54.74 mA 0.1096 54.74 IO = = 0.72 mA 76 4 (d) I E = = 8.08 mA 0.495 8.08 IO = = 0.080 mA 101 ______________________________________________________________________________________ R=

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.65

(a) υ O1 =

− R2 ⋅υ CM R1 ⎛

R

4 υ O 2 = ⎜⎜ + R ⎝ 3 R4

υ O = υ O1 + υ O 2

ACM =

(b) ACM

(c) ACM

υO = υ CM

⎞ ⎛ R ⎞ ⎟ ⋅υ CM ⋅ ⎜⎜1 + 2 ⎟⎟ ⎟ R1 ⎠ ⎝ ⎠ ⎡ R4 ⎢ R ⎛ R 3 =⎢ ⋅ ⎜⎜1 + 2 ⎢ R4 ⎝ R1 ⎢1 + R 3 ⎣

⎛ R4 ⎜ ⎜R ⎝ 3

⎤ ⎞ R 2 ⎥⎥ ⎟⎟ − ⋅υ CM ⎠ R1 ⎥ ⎥ ⎦

⎞⎛ R 2 ⎞ ⎛ R 2 ⎞⎛ R 4 ⎞ R 4 R 2 ⎟⎜1 + ⎟ ⎟−⎜ ⎟⎜1 + − ⎟⎜ R1 ⎟⎠ ⎜⎝ R1 ⎟⎠⎜⎝ R3 ⎟⎠ R3 R1 ⎠⎝ = R ⎛ R4 ⎞ 1+ 4 ⎜1 + ⎟ ⎜ R ⎟ R3 3 ⎠ ⎝

86.4 62.4 − 9−6 = 9.6 10.4 = = 0.3 86.4 10 1+ 9.6 80.8 79.2 − = 19.8 20.2 = 0.03149 80.8 1+ 19.8

or 79.2 80.8 − ACM = 20.2 19.8 = −0.0325 79.2 1+ 20.2 ⇒ ACM max = 0.0325 ______________________________________________________________________________________ 9.66

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ vI 1 − v A v A − vB v A − v0 = + R1 + R2 Rv R2

(1)

vI 2 − vB vB − v A vB = + R1 + R2 Rv R2

(2)

⎛ R1 ⎞ ⎛ R2 ⎞ v− = ⎜ ⎟ vA + ⎜ ⎟ vI 1 ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠

(3)

⎛ R1 ⎞ ⎛ R2 ⎞ v+ = ⎜ ⎟ vB + ⎜ ⎟ vI 2 ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ Now v− = v+ ⇒ R1v A + R2 vI 1 = R1vB + R2 vI 2 So that v A = vB +

(4)

R2 ( vI 2 − vI 1 ) R1

⎛ 1 v vI 1 1 1 ⎞ v = vA ⎜ + + ⎟− B − 0 R1 + R2 ⎝ R1 + R2 RV R2 ⎠ RV R2

(1)

⎛ 1 vI 2 1 1 ⎞ v = vB ⎜ + + ⎟− A R1 + R2 ⎝ R1 + R2 RV R2 ⎠ RV Then

( 2)

⎛ 1 v ⎛ R ⎞⎛ 1 vI 1 1 ⎞ 1 1 ⎞ v 1 = vB ⎜ + + ⎟ − B − 0 + ⎜ 2 ⎟⎜ + + ⎟ ( v I 2 − vI 1 ) R1 + R2 ⎝ R1 + R2 RV R2 ⎠ RV R2 ⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠ ⎛ 1 ⎤ vI 2 R2 1 1 ⎞ 1 ⎡ = vB ⎜ + + ⎟− ⎢ vB + ( vI 2 − vI 1 ) ⎥ R1 + R2 R1 ⎦ ⎝ R1 + R2 RV R2 ⎠ RV ⎣ Subtract (2) from (1) ⎛ R ⎞⎛ 1 v 1 1 1 ⎞ 1 R2 + + ⎟ ( vI 2 − vI 1 ) − 0 + ⋅ ( vI 2 − vI 1 ) ( vI 1 − vI 2 ) = ⎜ 2 ⎟ ⎜ R1 + R2 R2 RV R1 ⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠

(1) (2)

⎧⎪⎛ R ⎞ ⎛ 1 v0 1 1 ⎞ 1 1 R2 ⎫⎪ = ( vI 2 − vI 1 ) ⎨⎜ 2 ⎟ ⎜ + + ⎟+ + ⋅ ⎬ R2 ⎩⎪⎝ R1 ⎠⎝ R1 + R2 RV R2 ⎠ R1 + R2 RV R1 ⎪⎭ ⎛ R ⎞ ⎧ R2 R R1 R ⎫ v0 = ( vI 2 − vI 1 ) ⎜ 2 ⎟ ⎨ + 2 +1+ + 2⎬ + + R R R R R R R 2 1 2 V V ⎭ ⎝ 1 ⎠⎩ 1 R ⎞ 2R ⎛ v0 = 2 ⎜ 1 + 2 ⎟ ( vI 2 − vI 1 ) R1 ⎝ RV ⎠

______________________________________________________________________________________ 9.67

(a) i1 =

υ I1 − υ I 2 R1

=

(1.2 − 0.08 sin ω t ) − (1.2 + 0.08 sin ω t ) ⇒ i 10

1

= −16 sin ω t ( μ A)

υ O1 = (1.2 − 0.08 sin ω t ) − (0.016 sin ω t )(40 ) = 1.2 − 0.72 sin ω t (V) υ O 2 = (1.2 + 0.08 sin ω t ) − (− 0.016 sin ω t )(40 ) = 1.2 + 0.72 sin ω t (V) υO =

R4 (υ O 2 − υ O1 ) = ⎛⎜ 120 ⎞⎟(2)(0.72 sin ω t ) = 4.32 sin ω t (V) R3 ⎝ 40 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(− 0.65 + 0.05 sin ω t ) − (− 0.60 − 0.05 sin ω t )

(b) ii =

10 i1 = −5 + 10 sin ω t ( μ A)

υ O1 = −0.65 + 0.05 sin ω t + (40)(− 0.005 + 0.010 sin ω t ) = −0.85 + 0.45 sin ω t (V) υ O 2 = −0.60 − 0.05 sin ω t − (40 )(− 0.005 + 0.01 sin ω t ) = −0.40 − 0.45 sin ω t (V) ⎛ 120 ⎞ ⎟[(− 0.40 − 0.45 sin ω t ) − (− 0.85 + 0.45 sin ω t )] = 1.35 − 2.7 sin ω t (V) ⎝ 40 ⎠ ______________________________________________________________________________________

υO = ⎜

9.68

(a) (b) (c)

⎛ 40 ⎞ vOB = ⎜1 + ⎟ vI = 2.1667sin ω t ⎝ 12 ⎠ vOC = −

30 vI = −1.25sin ω t 12

vO = vOB − vOC = 2.1667 sin ω t − ( −1.25sin ω t ) vO = 3.417 sin ω t

vO 3.417 = = 6.83 vI 0.5

(d) ______________________________________________________________________________________ 9.69

(a) iO =

υ I1 − υ I 2

R 0.25 − (− 0.25) (b) R = ⇒ R = 100 Ω 5 (c) υ O1 = υ I 1 + i O R L = 0.25 + (5)(1) = 5.25 V

υ O 2 = υ I 2 = −0.25 V υ I 1 − υ I 2 1.25 − 1.75

(d) iO =

= = −1 mA R 0. 5 = υ I 1 + i O R L = 1.25 − (1)(3) = −1.75 V

υ O1 υ O 2 = υ I 2 = 1.75 V ______________________________________________________________________________________ 9.70

vO R ⎛ 2R ⎞ = 4 ⎜1 + 2 ⎟ vI 2 − vI 1 R3 ⎝ R1 ⎠ 200 ⎛ 2 (115 ) ⎞ vO = ⎜1 + ⎟ ( 0.06sin ω t ) 50 ⎝ R1 ⎠ 230 For vO = 0.5 = 1.0833 ⇒ R1 = 212.3 K R1 Ad =

230 = 32.33 ⇒ R1 = 7.11 K ⇒ R1 f = 7.11 K, R1 (potentiometer) = 205.2 K R1 ______________________________________________________________________________________ vO = 8 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.71

For υ O = 10 V, Ad = 200 ⇒ υ I 2 − υ I 1 =

10 = 0.05 V 200

R1 ( fixed ) =

0.05 ⇒ R1 ( fixed ) = 1 k Ω 50 × 10 − 6 R ⎛ 2R ⎞ Ad = 4 ⎜⎜1 + 2 ⎟⎟ R3 ⎝ R1 ⎠

⎛ 2R 200 = (2.5)⎜⎜1 + 2 1 ⎝ For Ad = 5

⎞ ⎟⎟ ⇒ R 2 = 39.5 k Ω ⎠

⎡ 2(39.5) ⎤ 5 = (2.5)⎢1 + ⎥ ⇒ R1 = 79 = R1 (var ) + R1 ( fixed ) = R1 (var ) + 1 R1 ⎦ ⎣ R1 (var ) = 78 k Ω ______________________________________________________________________________________

9.72

υ O (υ O1 ) = −

R4 ⋅υ O1 R3′

⎛

⎞⎛ R 4 ⎞ ⎟⎜1 + ⎟ ⎟⎜ R ′ ⎟ ⋅υ O 2 3 ⎠ ⎠⎝ υ O = υ O (υ O1 ) + υ O (υ O 2 ) and υ O1 = υ O 2 ≡ υ CM Then ⎛ R 4 ⎞⎛ R 4 ⎞ ⎛ R 4 ⎞ υ ⎟⎜1 + ⎟−⎜ ⎟ ACM = O = ⎜⎜ υ CM ⎝ R3 + R 4 ⎟⎠⎜⎝ R3′ ⎟⎠ ⎜⎝ R3′ ⎟⎠ R 4 = 2 R3 = 60 k Ω , R3 = 30 k Ω , R3′ = 30 k Ω ±5% For R3′ = 30 k Ω −5% = 28.5 k Ω R4 ⎝ R3 + R 4

υ O (υ O 2 ) = ⎜⎜

60 ⎞ ⎛ 60 ⎞ ⎛ 60 ⎞⎛ ACM = ⎜ ⎟⎜1 + ⎟−⎜ ⎟ = −0.03509 ⎝ 60 + 30 ⎠⎝ 28.5 ⎠ ⎝ 28.5 ⎠ For R3′ = 30 k Ω +5% = 31.5 k Ω 60 ⎞ ⎛ 60 ⎞ ⎛ 60 ⎞⎛ ACM = ⎜ ⎟⎜1 + ⎟−⎜ ⎟ = +0.03175 31 .5 ⎠ ⎝ 31.5 ⎠ 60 30 + ⎝ ⎠⎝ Then −0.03509 ≤ ACM ≤ +0.03175 ______________________________________________________________________________________

9.73 (a) R1C 2 = 20 ×10 3 0.02 ×10 −6 = 4 × 10 −4 s

(

υO =

−1 4 × 10 − 4

)(

)

−0.25

∫ (0.25) cos ω tdt = (4 ×10 )ω ⋅ sin ω t −4

(

)

For υ O = 0.25 ⇒ 4 × 10 −4 (2π f ) = 1 ⇒ f = 398 Hz Phase = 90°

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) (i) υ O = 1.5 =

0.25 ⇒ f = 66.3 Hz 2π f 4 × 10 − 4

(

)

0.25 ⇒ f = 663 Hz 2π f 4 × 10 − 4 ______________________________________________________________________________________

(ii) υ O = 0.15 =

(

)

9.74

−1 R1C 2

(a) υ O =

t

∫υ

I

(t ′)dt ′ =

0

1.2 −1 (0.25)t ′ R1C 2 0

− (0.25)(1.2) −5 = ⇒ R1C 2 = 60 ms R1C 2

(0.10) ⋅ t ′′ ⇒ t ′′ = 3 s,

t = 4.2 s 0.06 (0.10) ⋅ t ′′ ⇒ t ′′ = 6 s, t = 7.2 s (ii) 5 = −5 + 0.06 ______________________________________________________________________________________

(b) (i) 0 = −5 +

9.75 −Z 2 , where Z 2 = R 2 R1

(a) Aυ =

− R2 1 ⋅ R1 1 + jωR 2 C 2

Aυ =

− R2 R1

(b) At ω = 0 , Aυ (0 ) = (c)

⎛ 1 ⎞ ⎟⎟ R 2 ⎜⎜ R2 1 ⎝ jωC 2 ⎠ = = 1 jωC 2 1 + j ωR 2 C 2 R2 + j ωC 2

Aυ =

R2 1 ⋅ R1 1 + (ωR C )2 2 2

Set 1 + (ωR 2 C 2 ) = 2 ⇒ ω =

1 1 ⇒ f = R2 C 2 2πR 2 C 2 ______________________________________________________________________________________ 2

9.76 (a) R1 = 20 k Ω R2 = 15 ⇒ R 2 = 300 k Ω R1

ω=

1 = 2π f R2 C 2

C2 =

1 1 = ⇒ C 2 = 106 pF 3 2π fR 2 2π 5 × 10 300 × 10 3

(

)(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) R1 = 15 k Ω R2 = 25 ⇒ R 2 = 375 k Ω R1 C2 =

1

⇒ C 2 = 28.3 pF 2π 15 × 10 375 × 10 3 ______________________________________________________________________________________

(

9.77

)(

− jωR 2 C1 − R2 jωR1C1 = ⋅ 1 + jωR1C1 R1 1 + jωR1C1

(b) As ω ⇒ ∞ , Aυ = (c)

Aυ = Set

)

− R2 1 + jωR1C1 1 , where Z 1 = R1 + = Z1 j ωC 1 j ωC 1

(a) Aυ = Aυ =

3

− R2 R1

R2 ωR1C1 ⋅ R1 1 + (ωR C )2 1 1

ωR1C1 1 + (ωR1C1 )

= 2

1

⇒ω =

2

1 1 ⇒ f = R1C1 2π R1C1

______________________________________________________________________________________ 9.78 (a) Set R 2 = 350 k Ω R2 = 15 ⇒ R1 = 23.33 k Ω R1 1 1 1 ⇒ C1 = = ⇒ C1 = 341 pF R1C1 2π R1 f 2π 23.33 × 10 3 20 × 10 3 (b) Set R1 = 20 k Ω 2π f =

(

)(

)

R2 = 25 ⇒ R 2 = 500 k Ω R1 1 ⇒ C1 = 227 pF 2π 20 × 10 3 35 × 10 3 ______________________________________________________________________________________ C1 =

(

)(

)

9.79 Assuming the Zener diode is in breakdown, R 1 vO = − 2 ⋅ Vz = − ( 6.8 ) ⇒ vO = −6.8 V R1 1 i2 =

0 − vO 0 − ( −6.8 ) = ⇒ i2 = 6.8 mA R2 1

10 − Vz 10 − 6.8 − i2 = − 6.8 ⇒ iz = −6.2 mA!!! Rs 5.6 Circuit is not in breakdown. Now iz =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10 − 0 10 = i2 = ⇒ i2 = 1.52 mA Rs + R1 5.6 + 1

vO = −i2 R2 = − (1.52 )(1) ⇒ vO = −1.52 V iz = 0

______________________________________________________________________________________ 9.80 ⎡ ⎤ ⎛ v ⎞ v ⎛ v ⎞ vO = −VT ln ⎜ I ⎟ = − ( 0.026 ) ln ⎢ −14 I 4 ⎥ ⇒ vO = −0.026 ln ⎜ −I 10 ⎟ I R ⎝ 10 ⎠ ⎝ s 1⎠ ⎣⎢ (10 )(10 ) ⎥⎦ For vI = 20 mV , vO = 0.497 V For vI = 2 V , vO = 0.617 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.81 ⎛ 333 ⎞ v0 = ⎜ ⎟ ( v01 − v02 ) = 16.65 ( v01 − v02 ) ⎝ 20 ⎠ ⎛i ⎞ v01 = −vBE1 = −VT ln ⎜ C1 ⎟ ⎝ IS ⎠ ⎛i ⎞ v02 = −vBE 2 = −VT ln ⎜ C 2 ⎟ ⎝ IS ⎠ ⎛i ⎞ ⎛i ⎞ v01 − v02 = −VT ln ⎜ C1 ⎟ = VT ln ⎜ C 2 ⎟ i ⎝ C2 ⎠ ⎝ iC1 ⎠ v v iC 2 = 2 , iC1 = 1 R2 R1 ⎛v R ⎞ So v01 − v02 = VT ln ⎜ 2 ⋅ 1 ⎟ ⎝ R2 v1 ⎠ Then ⎛v R ⎞ v0 = (16.65 )( 0.026 ) ln ⎜ 2 ⋅ 1 ⎟ ⎝ v1 R2 ⎠ ⎛v R ⎞ v0 = 0.4329 ln ⎜ 2 ⋅ 1 ⎟ ⎝ v1 R2 ⎠ ln ( x ) = log e ( x ) = ⎡⎣ log10 ( x ) ⎤⎦ ⋅ ⎡⎣log e (10 ) ⎤⎦ = 2.3026 log10 ( x )

⎛v R ⎞ Then v0 ≅ (1.0 ) log10 ⎜ 2 ⋅ 1 ⎟ ⎝ v1 R2 ⎠ ______________________________________________________________________________________

9.82

(

)

vO = − I s R evI / VT = − (10−14 )(104 ) evI / VT vO = (10−10 ) evI / 0.026 For vI = 0.30 V ,

vo = 1.03 × 10−5 V

For vI = 0.60 V ,

vo = 1.05 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.83

From Figure 9.40 ⎛ R ⎞⎡ R ⎤ R R R υ O = − F ⋅υ I 2 − F ⋅υ I 4 + ⎜⎜1 + F ⎟⎟ ⎢ P ⋅υ I 1 + P ⋅υ I 3 ⎥ R1 R2 RB ⎦ ⎝ R N ⎠⎣ R A = −10υ I 2 − υ I 4 + 2υ I 1 + 3υ I 3 ⎛ R F ⎞⎛ R P ⎞ ⎜1 + ⎟⎜ ⎟ ⎜ R ⎟⎜ R ⎟ = 2 , N ⎠⎝ A ⎠ ⎝ Set R F = 500 k Ω , R1 = 50 k Ω , R 2 = 500 k Ω

Then

RF R = 10 , F = 1 , R1 R2

⎛ R F ⎞⎛ R P ⎞ ⎜1 + ⎟⎜ ⎟ ⎜ R ⎟⎜ R ⎟ = 3 N ⎠⎝ B ⎠ ⎝

Now R N = R1 R 2 = 50 500 = 45.45 k Ω 500 ⎞⎛ R P ⎛ Then ⎜1 + ⎟⎜⎜ ⎝ 45.45 ⎠⎝ R A

⎞ ⎛R ⎞ 500 ⎞⎛ R P ⎛ ⎟⎟ = 12⎜⎜ P ⎟⎟ = 2 , Also ⎜1 + ⎟⎜⎜ ⎝ 45.45 ⎠⎝ R B ⎠ ⎝ RA ⎠ 2 Let R A = 500 k Ω , then R B = R A = 333.3 k Ω 3 Then R P = 83.33 k Ω = R A R B RC

⎞ ⎛R ⎟⎟ = 12⎜⎜ P ⎠ ⎝ RB

⎞ ⎟⎟ = 3 ⎠

We find R A R B = 500 333.3 = 200 k Ω So 200 RC = 83.33 ⇒ RC = 142.8 k Ω ______________________________________________________________________________________ 9.84 ⎛

R ⎞⎡ R

R

R

⎤

R

R

υ O = ⎜⎜1 + F ⎟⎟ ⎢ P ⋅υ I 1 + P ⋅υ I 2 + P ⋅υ I 3 ⎥ − R ⋅υ I 4 − F ⋅υ I 5 RB RC R2 ⎝ R N ⎠⎣ R A ⎦ R1 = 3υ I 1 + 1.5υ I 2 + 2υ I 3 − 4υ I 4 − 6υ I 5 We have

RF R = 4 , F = 6 ; Set R F = 250 k Ω , R1 = 62.5 k Ω , R 2 = 41.67 k Ω R1 R2

Now R N = R1 R 2 = 62.5 41.67 = 25 k Ω ⎛ R ⎞ ⎛ 250 ⎞ Also ⎜⎜1 + F ⎟⎟ = ⎜1 + ⎟ = 11 25 ⎠ ⎝ RN ⎠ ⎝ (11)R P (11)R P (11)R P R 1 RA 2 Now = 3, = 1.5 , =2 ⇒ A = , = RA RB RC R B 2 RC 3

Set R B = 250 k Ω , R A = 125 k Ω , RC = 187.5 k Ω This yields R P = 34.09 k Ω , We have R p = R A R B RC R D We find R A R B RC = 125 250 187.5 = 57.69 k Ω Then 57.69 R D = 34.09 ⇒ R D = 83.3 k Ω ______________________________________________________________________________________ 9.85 VO ⎛ R 2 ⎞ 12 R ⎟⎟ = = ⎜⎜1 + ⇒ 2 = 1.143 VZ ⎝ R1 ⎠ 5.6 R1 V − VZ IF = O ; Set I F = I Z (min ) = 1.2 mA RF

Then R F =

12 − 5.6 = 5.33 k Ω 1. 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Set I D1 = 0.15 mA V Z′ = V Z + Vγ = 5.6 + 0.7 = 6.3 V 6. 3 = 31.5 k Ω 0.2 Then I 3 = 0.2 + 0.15 = 0.35 mA V − V Z′ 10 − 6.3 So R3 = S = = 10.6 k Ω I3 0.35 ______________________________________________________________________________________

Let I 4 = 0.2 mA, ⇒ R 4 =

9.86 R1 =

VO − V Z 12 − 5.6 = = 3.2 k Ω IZ 2

VO ⎛ R 2 ⎞ 12 R ⎟= = ⎜⎜1 + ⇒ 2 = 1.143 ⎟ V Z ⎝ R3 ⎠ 5.6 R3 V 12 Let I R = 2 mA, ⇒ R 2 + R3 = O = = 6kΩ IR 2

Then 1.143R3 + R3 = 6 , ⇒ R3 = 2.8 k Ω and R 2 = 3.2 k Ω V − VO 15 − 12 Let I R 4 = 4 mA, R 4 = IN = = 0.75 k Ω I R4 4 ______________________________________________________________________________________ 9.87

Let R1 = R 2 = R3 = 20 k Ω Let RT = 20(1 + δ ) k Ω

⎛ R3 ⎞ + 1 ⎟ ⋅ V = (10 ) = 5 V Now υ O1 = υ A = ⎜⎜ ⎟ 2 ⎝ R3 + R1 ⎠ ⎛ RT ⎞ + ⎡ 20(1 + δ ) ⎤ 10(1 + δ ) ⎟⎟ ⋅ V = ⎢ υ O 2 = υ B = ⎜⎜ ⎥ (10 ) = ( ) + δ + 2+δ R R + 20 1 20 ⎣ ⎦ 2 ⎠ ⎝ T

So υ OA = υ A − υ B = 5 −

10(1 + δ ) 5(2 + δ ) − 10(1 + δ ) 5δ ≅− = −2.5δ = 2+δ 2+δ 2

⎛ T − 300 ⎞ We have δ = y⎜ ⎟ ; At T = 350 , RT = 21 k Ω , ⇒ 21 = 20(1 + δ ) ⇒ δ = 0.05 ⎝ 300 ⎠ ⎛ 350 − 300 ⎞ Then 0.05 = y⎜ ⎟ ⇒ y = 0.30 ⎝ 300 ⎠ For δ = 0.05 , υ OA = (2.5)(0.05) = 0.125 V

For the instrumentation amplifier, υ O = 5 =

R4 R3

⎛ 2 R2 ⎜⎜1 + R1 ⎝

⎞ ⎟⎟(0.125) ⎠

R4 R = 4 and 2 = 4.5 R3 R1 ______________________________________________________________________________________

For example, set

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 9.88 R − ΔR ⎛ ⎞ + ⎛ R − ΔR ⎞ + (a) υ A = ⎜ ⎟ ⋅V = ⎜ ⎟ ⋅V ⎝ R − ΔR + R + ΔR ⎠ ⎝ 2R ⎠ R + ΔR ⎞ + ⎛ R + ΔR ⎞ + ⎟ ⋅V = ⎜ ⎟ ⋅V R + Δ R + R − Δ R ⎝ ⎠ ⎝ 2R ⎠ ⎛

υB = ⎜

⎡ R − ΔR

υ O1 = υ A − υ B = ⎢ ⎣ 2R

(

)

−

9 ΔR + R + ΔR ⎤ + ⎛ ⎞ ⋅V = − ⋅ V = − ΔR⎜ ⎟ 3 2 R ⎥⎦ R ⎝ 20 × 10 ⎠

or υ O1 = − 4.5 ×10 −4 (ΔR ) (b) For an instrumentation amplifier, R ⎛ 2R ⎞ υ O = 4 ⎜⎜1 + 2 ⎟⎟ ⋅υ O1 R3 ⎝ R1 ⎠ For ΔR = 200 Ω , υ O = −5 V −5 =

R4 R3

⎛ 2R2 ⎜⎜1 + R1 ⎝

⎞ ⎟⎟ − 4.5 × 10 − 4 (200 ) ⎠

(

)

⎛ 2R2 ⎞ ⎜⎜1 + ⎟ R1 ⎟⎠ ⎝ R R For example, set 4 = 6 and 2 = 4.13 R3 R1 ______________________________________________________________________________________

or 55.55 =

R4 R3

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 10 10.1 I1 = I 2 =

0 − 2Vγ − V − R1 + R2

2Vγ + I 2 R2 = VBE + I C R3 2Vγ + IC =

a.

R2 ( −2Vγ − V − ) = VBE + IC R3 R1 + R2

1 R3

⎧⎪ ⎫⎪ R2 ⎞ − ⎛ ⎨2Vγ − ( 2Vγ + V ) ⎜ ⎟ − VBE ⎬ ⎪⎩ ⎪⎭ ⎝ R1 + R2 ⎠

Vγ = VBE and R1 = R2 IC =

1 ⎧ 1 ⎫ − ⎨2Vγ − ( 2Vγ + V ) − VBE ⎬ R3 ⎩ 2 ⎭

or I C =

b.

−V − 2 R3

I C = 2 mA =

− ( −10 ) 2 R3

I1 = I 2 = 2 mA =

⇒ R3 = 2.5 kΩ

−2 ( 0.7 ) − ( −10 )

⇒ R1 + R2 = 4.3 kΩ ⇒ R1 = R2 = 2.15 kΩ R1 + R2 c. ______________________________________________________________________________________ 10.2 ⎛ 50 × 10 −6 ⎞ ⎟ = 0.7004 V (a) (i) I O = 50 μ A, V BE1 = (0.026) ln⎜⎜ −16 ⎟ ⎝ 10 ⎠ −6 ⎛ 150 × 10 ⎞ ⎟ = 0.7289 V (ii) I O = 150 μ A, V BE1 = (0.026 ) ln⎜⎜ −16 ⎟ ⎝ 10 ⎠ −3 ⎛ 1.5 × 10 ⎞ ⎟ = 0.7888 V (iii) I O = 1.5 mA, V BE1 = (0.026) ln⎜⎜ −16 ⎟ ⎝ 10 ⎠ ⎛ 48.0769 × 10 −6 ⎞ 50 ⎟ = 0.6994 V = 48.08 μ A, V BE1 = (0.026) ln⎜⎜ (b) (i) I O = ⎟ 2 10 −16 ⎝ ⎠ 1+ 50 ⎛ 144.23 × 10 −6 ⎞ 150 ⎟ = 0.7279 V = 144.23 μ A, V BE1 = (0.026) ln⎜⎜ (ii) I O = −16 ⎟ 2 10 ⎝ ⎠ 1+ 50 ⎛ 1.4423 × 10 −3 ⎞ 1.5 ⎟ = 0.7878 V = 1.4423 mA, V BE1 = (0.026) ln⎜⎜ (iii) I O = −16 ⎟ 2 10 ⎝ ⎠ 1+ 50 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.3 I C1 ≅

I REF 200 = = 195.12 μ A 2 2 1+ 1+ β 80

⎛ 195.12 × 10 −6 V BE1 = V BE 2 = (0.026) ln⎜⎜ −15 ⎝ 5 × 10

⎞ ⎟ = 0.6341 V ⎟ ⎠

⎛V ⎞ ⎛ 0.6341 ⎞ I O = I S 2 exp⎜⎜ BE 2 ⎟⎟ = 2 × 10 −15 exp⎜ ⎟ ⇒ I O = 78.05 μ A ⎝ 0.026 ⎠ ⎝ VT ⎠ ______________________________________________________________________________________

(

)

10.4 I C1 =

I REF 150 = = 147.54 μ A 2 2 1+ 1+ β 120

⎛ 147.54 × 10 −6 V BE1 = V BE 2 = (0.026) ln⎜⎜ 10 −16 ⎝

⎞ ⎟ = 0.7285 V ⎟ ⎠

⎛V ⎞ ⎛ 0.7285 ⎞ I O = I S 2 exp⎜⎜ BE 2 ⎟⎟ = 3 × 10 −16 exp⎜ ⎟ ⇒ I O = 0.4426 mA ⎝ 0.026 ⎠ ⎝ VT ⎠ ______________________________________________________________________________________

(

)

10.5

Approximation: I C1 ≅

I REF 200 = = 190.48 μ A 2 2 1+ 1+ β 40

⎛ 190.48 × 10 −6 ⎞ ⎟ = 0.63345 V V BE1 = (0.026) ln⎜⎜ −15 ⎟ ⎝ 5 × 10 ⎠ I C1 190.48 I B2 ≅ = = 4.762 μ A β 40 V BE 2 = V BE1 − I B 2 R = 0.63345 − (0.004762 )(2) = 0.62393 V

(

)

⎛ 0.62393 ⎞ I O = 5 × 10 −15 exp⎜ ⎟ ⇒ I O = 132.07 μ A ⎝ 0.026 ⎠ ______________________________________________________________________________________

10.6 I REF =

V + − VBE ( on ) − V − R1

⇒ 0.250 =

3 − 0.7 − ( −3) R1

R1 = 21.2 K I REF 0.250 = ⇒ I C1 = I C 2 = 0.2419 mA 2 2 1+ 1+ 60 β I B1 = I B 2 = 4.03 μ A ______________________________________________________________________________________ I C1 = I C 2 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.7 I REF =

V + − VBE ( on ) − V −

=

R1

5 − 0.7 − ( −5 ) 18.3

I REF = 0.5082 mA I 0.5082 I C1 = I C 2 = REF = ⇒ I C1 = I C 2 = 0.4958 mA 2 2 1+ 1+ 80 β I B1 = I B 2 = ( 6.198 μ A )

______________________________________________________________________________________ 10.8 (a) P = (I O + I REF ) V + − V − 1.8 = (0.25 + I REF )(5) ⇒ I REF = 0.11 mA

(

)

⎛ 0.25 × 10 −3 ⎞ ⎟ = 0.68236 V V BE1 = V BE 2 = (0.026) ln⎜⎜ −15 ⎟ ⎝ 10 ⎠ 0 . 68236 ⎛ ⎞ −16 (b) 0.11 × 10 −3 = I S 1 exp⎜ ⎟ ⇒ I S 1 = 4.4 × 10 A ⎝ 0.026 ⎠ 5 − 0.68236 = 39.25 k Ω 0.11 ______________________________________________________________________________________

(c) R1 =

10.9 ⎛ 2⎞ 2 ⎞ ⎛ (a) I REF = ⎜⎜1 + ⎟⎟ ⋅ I O = ⎜1 + ⎟(0.5) = 0.5083 mA β 120 ⎝ ⎠ ⎝ ⎠ 5 − 0.7 − (− 5) R1 = = 18.3 k Ω 0.5083 V 100 = 200 k Ω (b) Ro = A = 0.5 IO

ΔI O =

ΔVCE 2 7 − 0.7 = = 0.0315 mA 200 Ro

ΔI O 0.0315 = ⇒ 6.3% 0.5 IO ______________________________________________________________________________________

10.10

I 0 = nI C1 I REF = I C1 + I B1 + I B 2 = I C1 +

I C1

β

+

I0

β

⎛ ⎛ 1+ n ⎞ 1 n⎞ I REF = I C1 ⎜ 1 + + ⎟ = I C1 ⎜ 1 + ⎟ β β β ⎠ ⎝ ⎠ ⎝ I ⎛ 1+ n ⎞ nI REF = 0 ⎜1 + ⎟ or I 0 = β ⎠ n⎝ ⎛ 1+ n ⎞ ⎜1 + β ⎟⎠ ⎝ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.11

IO =

I REF 2 ⎞ ⎛ ⇒ I REF = ( 0.20 ) ⎜1 + ⎟ = 0.210 mA 2 40 ⎝ ⎠ 1+

β

5 − 0.7 4.3 = ⇒ R1 = 20.5 K R1 = 0.21 I REF ______________________________________________________________________________________ 10.12 ⎛ 2⎞ 2 ⎞ ⎛ (a) I REF = ⎜⎜1 + ⎟⎟ ⋅ I O = ⎜1 + ⎟(0.12 ) = 0.123 mA β 80 ⎝ ⎠ ⎝ ⎠ 5 − 0. 7 R1 = = 34.96 k Ω 0.123 V 80 = 666.7 k Ω (b) ro = A = I O 0.12

(i) ΔI O =

ΔV EC 2 − 0.7 = ⇒ ΔI O = 1.95 μ A ro 666.7

4 − 0. 7 ⇒ ΔI O = 4.95 μ A 666.7 ______________________________________________________________________________________

(ii ) ΔI O =

10.13 I REF = 1 =

a. b.

5 − 0.7 − ( −5 ) R1

⇒ R1 = 9.3 kΩ

I 0 = 2 I REF ⇒ I 0 = 2 mA

5 − 0.7 ⇒ RC 2 = 2.15 kΩ 2 c. ______________________________________________________________________________________ For VEC 2 ( min ) = 0.7 ⇒ RC 2 =

10.14

I O = 0.50 mA ⇒ I OA = I OB = 0.25 mA ⎛ 3⎞ 3 ⎞ ⎛ I REF = I OA ⎜ 1 + ⎟ = 0.25 ⎜ 1 + ⎟ β 60 ⎝ ⎠ ⎝ ⎠ I REF = 0.2625 mA 2.5 − 0.7 ⇒ R1 = 6.86 K 0.2625 ______________________________________________________________________________________ R1 =

10.15

Similar to Figure P10.14 biased at V + and V − . 2.5 − 0.7 − (− 2.5) R1 = = 21.5 k Ω 0. 2 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.16

I 2 = 2 I1 and I3 = 3I1 I (a) 2 = 1.0 mA, I 3 = 1.5 mA (b) I1 = 0.25 mA, I 3 = 0.75 mA

(c) I1 = 0.167 mA, I 2 = 0.333 mA ______________________________________________________________________________________ 10.17 a. I 0 = I C1 and I REF = I C1 + I B 3 = I C1 + I E 3 = I B1 + I B 2 + I REF = I C1 + I REF −

VBE 2 I C1 VBE = + R2 R2 β

2 I C1

β (1 + β )

+

VBE

(1 + β ) R2

⎛ ⎞ VBE 2 = I 0 ⎜⎜ 1 + ⎟⎟ (1 + β ) R2 ⎝ β (1 + β ) ⎠

I REF − I0 =

IE3 1+ β

VBE

(1 + β ) R2

⎛ ⎞ 2 ⎜⎜ 1 + ⎟⎟ ⎝ β (1 + β ) ⎠

⎛ ⎞ 2 0.7 I REF = ( 0.70 ) ⎜⎜ 1 + ⎟⎟ + ⎝ ( 80 )( 81) ⎠ ( 81)(10 ) I REF = 0.700216 + 0.000864 I REF = 0.7011 mA =

10 − 2 ( 0.7 )

⇒ R1 = 12.27 kΩ R1 b. ______________________________________________________________________________________

10.18 a. I ES 1+ β = (1 + N ) I BR

I 0i = I CR and I REF = I CR + I BS = I CR + I ES = I BR + I B1 + I B 2 + ... + I BN =

(1 + N ) ICR β

Then I REF = I CR + or I 0i =

(1 + N ) I CR β (1 + β )

I REF ⎛ (1 + N ) ⎞ ⎜⎜ 1 + β (1 + β ) ⎟⎟ ⎝ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎡ ⎤ 6 I REF = ( 0.5 ) ⎢1 + ⎥ = 0.5012 mA ⎣⎢ ( 50 )( 51) ⎦⎥ R1 =

5 − 2 ( 0.7 ) − ( −5 )

⇒ R1 = 17.16 kΩ 0.5012 b. ______________________________________________________________________________________

10.19 ⎡ ⎡ 2 ⎤ 2 ⎤ I REF = I O ⎢1 + ⎥ = (0.15)⎢1 + ⎥ = 0.15018 mA ⎣ β (1 + β ) ⎦ ⎣ (40 )(41) ⎦ 3 − 0.7 − 0.7 − (− 3) R1 = = 30.63 k Ω 0.15018 ______________________________________________________________________________________

10.20

I 0 = I REF ⋅

1

⎛ ⎞ 2 ⎜⎜1 + ⎟⎟ ⎝ β (2 + β ) ⎠ For I 0 = 0.8 mA ⎛ 2 ⎞ I REF = ( 0.8 ) ⎜⎜1 + ⎟⎟ ⇒ I REF = 0.8024 mA ⎝ 25 ( 27 ) ⎠ 18 − 2 ( 0.7 ) R1 = ⇒ R1 = 20.69 kΩ 0.8024 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.21 The analysis is exactly the same as in the text. We have 1 I 0 = I REF ⋅ ⎛ ⎞ 2 ⎜⎜1 + ⎟⎟ ⎝ β (2 + β ) ⎠ ______________________________________________________________________________________ 10.22

(a) I O = 0.5 mA, I B 2 =

0. 5 = 0.005556 mA 90

I B2 = 0.002778 mA 2 = I B1 + I B 2 = 0.008334 mA

I B1 = I E3

0.008334 = 0.0001366 mA 61 I I C1 = O = 0.25 mA 2 I REF = I C1 + I B 3 = 0.2501366 mA I B3 =

5 − 0.6 − 0.7 − (− 5) = 34.78 k Ω 0.2501366 = 0.002778 mA

R1 =

(b) I B1 I B 2 = 0.005556 mA I E 3 = 0.008334 mA I B 3 = 0.0001366 mA ______________________________________________________________________________________ 10.23 (a)

Assuming RO ≈

β ro3

2 VA VA 100 = = = 400 K rO 3 = I O I REF 0.25 RO =

(100 )( 400 ) 2

⇒ RO = 20 MΩ

(b) RO =

ΔV ΔV 5 ⇒ ΔI O = = ΔI O 20 MΩ 20 MΩ

ΔI O = 0.25 μ A ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.24 I REF =

V + − VBE1 − V − 5 − 0.7 = 9.3 R1

I REF = 0.4624 mA VT ⎛ I REF ⎞ 0.026 ⎛ 0.4624 ⎞ ln ⎜ ln ⎜ ⎟ ⎟= 1.5 RE ⎝ I O ⎠ ⎝ IO ⎠ ⎛ 0.4624 ⎞ I O = 0.01733ln ⎜ ⎟ ⎝ IO ⎠ IO =

By trial and error I O = 41.7 μ A V BE 2 = 0.7 − I O R E = 0.7 − (0.0417 )(1.5) V BE 2 = 0.6375 V ______________________________________________________________________________________ 10.25 ⎛ 200 × 10 −6 (a) V BE1 = (0.026) ln⎜⎜ −15 ⎝ 5 × 10 ⎛I ⎞ I O R E = VT ln⎜⎜ REF ⎟⎟ ⎝ IO ⎠

⎞ ⎟ = 0.6347 V ⎟ ⎠

⎛ 0.2 ⎞ ⎟ I O (0.5) = (0.026 ) ln⎜⎜ ⎟ ⎝ IO ⎠ I O ≅ 61.4 μ A ⎛ 61.4 × 10 −6 V BE 2 = (0.026) ln⎜⎜ −15 ⎝ 5 × 10 ⎛ 200 × 10 −6 (b) V BE1 = (0.026) ln⎜⎜ −15 ⎝ 5 × 10 V BE1 − V BE 2 = I O R E

⎡ ⎛I VT ⎢ln⎜⎜ REF ⎣⎢ ⎝ I S1

⎞ ⎛I ⎟ − ln⎜ O ⎟ ⎜I ⎠ ⎝ S2

⎛I I VT ln⎜⎜ REF ⋅ S 2 ⎝ I O I S1

⎡⎛

⎞

⎞ ⎟ = 0.6040 V ⎟ ⎠

⎞ ⎟ = 0.6347 V ⎟ ⎠

⎞⎤ ⎟⎥ = I O R E ⎟ ⎠⎦⎥

⎞ ⎟ = I O RE ⎟ ⎠

⎤

(0.026) ln ⎢⎜⎜ 0.2 ⎟⎟⎛⎜ 7 ⎞⎟⎥ = I O (0.5) ⎣⎢⎝ I O ⎠⎝ 5 ⎠⎦⎥ I O ≅ 71.2 μ A

⎛ 71.2 × 10 −6 ⎞ ⎟ = 0.5991 V V BE 2 = (0.026) ln⎜⎜ −15 ⎟ ⎠ ⎝ 7 × 10 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.26 ⎛ 100 × 10 −6 ⎞ ⎟ = 0.61669 V (a) V BE1 = (0.026) ln⎜⎜ −15 ⎟ ⎠ ⎝ 5 × 10 V BE 2 = V BE1 + I REF R E = 0.61669 + (0.1)(0.7 ) = 0.68669 V

(

)

⎛ 0.68669 ⎞ I O = 5 × 10 −15 exp⎜ ⎟ ⇒ I O = 1.477 mA ⎝ 0.026 ⎠ (b) V BE1 = 0.61669 V

V BE 2 = 0.68669 V

(

)

⎛ 0.68669 ⎞ I O = 2 × 10 −15 exp⎜ ⎟ ⇒ I O = 0.5906 mA ⎝ 0.026 ⎠ ______________________________________________________________________________________

10.27

(a) I REF =

5 − 0.7 − (− 5) = 0.186 mA 50

⎛I I O R E = VT ln⎜⎜ REF ⎝ IO

⎞ ⎟ ⎟ ⎠

⎛ 0.186 ⎞ ⎟ I O (3) = (0.026) ln⎜⎜ ⎟ ⎝ IO ⎠ I O ≅ 19.53 μ A V BE 2 = V BE1 − I O R E = 0.7 − (0.01953)(3) = 0.6414 V

(b) ro 2 =

VA 80 = ⇒ ro 2 = 4.096 M Ω I O 0.01953

0.01953 = 0.7512 mA/V 0.026 (120 )(0.026 ) = 159.8 k Ω rπ 2 = 0.01953 R E rπ 2 = 3 159.8 = 2.945 k Ω g m2 =

Ro = (4.096 )[1 + (0.7512 )(2.945)] = 13.16 M Ω ______________________________________________________________________________________ 10.28 Ro = ro 2 1 + g m 2 (R E rπ 2 )

[

]

From 10.27, I O = 19.53 μ A I 0.01953 = 0.7512 mA/V g m2 = O = 0.026 VT rπ 2 = ro 2

β VT IO

=

(80)(0.026) = 106.5 k Ω 0.01953

V 80 = A = ⇒ 4.096 M Ω I O 0.01953

[

]

Then Ro = (4.096) 1 + (0.7512)(3 106.5) = 13.07 M Ω ΔVO 5 = = 0.382 μ A 13.07 Ro ______________________________________________________________________________________ ΔI O =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.29

I REF =

5 − 0.7 − ( −5 ) R1

= 0.50

R1 = 18.6 K

(a)

⎛I ⎞ I O RE = VT ln ⎜ REF ⎟ ⎝ IO ⎠ 0.026 ⎛ 0.50 ⎞ ln ⎜ RE = ⎟ 0.050 ⎝ 0.050 ⎠ RE = 1.20 K RO = rc 2 [1 + RE′ g m 2 ] RE′ = RE rπ 2 rπ 2 =

( 75)( 0.026 )

= 39 K 0.050 V 100 ro 2 = A = ⇒ 2 MΩ I O 0.05

(b)

gm2 =

0.050 = 1.923 mA/V 0.026

RE′ = 1.20 39 = 1.164 K

RO = 2 ⎡⎣1 + (1.164 )(1.923) ⎤⎦ ⇒ RO = ( 6.477 ) MΩ ΔI O =

ΔV 5 = = 0.772 μ A RO 6.477

ΔI O 0.772 × 100% = × 100 = 1.54% IO 50

(c) ______________________________________________________________________________________ 10.30

Let R1 = 10 k Ω

Then I REF =

3 − 0.7 − (− 3) = 0.53 mA 10

VT ⎛ I REF ⎞ (0.026 ) ⎛ 0.53 ⎞ ⎟= ln⎜ ln⎜ ⎟ ⇒ R E = 1.228 k Ω I O ⎜⎝ I O ⎟⎠ (0.05) ⎝ 0.05 ⎠ ______________________________________________________________________________________ RE =

10.31 ⎛I ⎞ VBE = VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛ 10−3 ⎞ −15 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 × 10 A ⎝ IS ⎠ ⎛ 2 × 10−3 ⎞ At 2 mA, VBE = ( 0.026 ) ln ⎜ −15 ⎟ ⎝ 2.03 × 10 ⎠ = 0.718 V 15 − 0.718 ⇒ R1 = 7.14 kΩ 2 ⎛ I ⎞ 0.026 ⎛ 2 ⎞ V ⋅ ln ⎜ RE = T ln ⎜ REF ⎟ = ⎟ ⇒ RE = 1.92 kΩ I 0 ⎝ I 0 ⎠ 0.050 ⎝ 0.050 ⎠ ______________________________________________________________________________________ R1 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.32

(a) I REF =

3 − 0.7 − (− 3) = 0.265 mA 20

VT ⎛ I REF ⎞ (0.026 ) ⎛ 0.265 ⎞ ⎟= ln⎜ ln⎜ ⎟ ⇒ R E = 253 Ω (0.1) ⎝ 0.1 ⎠ I O ⎜⎝ I O ⎟⎠ ______________________________________________________________________________________

(b) R E =

10.33 I REF ≈

10 − 0.7 − ( −10 ) 40

= 0.4825 mA

⎛I ⎞ VBE ≅ VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛ 10−3 ⎞ −15 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 × 10 A I ⎝ S ⎠

Now ⎛ 0.4825 × 10−3 ⎞ = 0.681 V VBE = ( 0.026 ) ln ⎜ −15 ⎟ ⎝ 2.03 × 10 ⎠ VBE1 = 0.681 V So I REF ≅

10 − 0.681 − ( −10 )

⇒ I REF = 0.483 mA 40 ⎛I ⎞ I 0 RE = VT ln ⎜ REF ⎟ ⎝ I0 ⎠ ⎛ 0.483 ⎞ I 0 (12 ) = ( 0.026 ) ln ⎜ ⎟ ⎝ I0 ⎠

By trial and error. ⇒ I 0 ≅ 8.7 μ A VBE 2 = VBE1 − I 0 RE = 0.681 − ( 0.0087 )(12 ) ⇒ VBE 2 = 0.5766 V ______________________________________________________________________________________ 10.34 VBE1 + I REF RE1 = VBE 2 + I 0 RE 2 VBE1 − VBE 2 = I 0 RE 2 − I REF RE1 For matched transistors ⎛I ⎞ VBE1 = VT ln ⎜ REF ⎟ ⎝ IS ⎠

⎛I ⎞ VBE 2 = VT ln ⎜ 0 ⎟ ⎝ IS ⎠ ⎛I ⎞ Then VT ln ⎜ REF ⎟ = I 0 RE 2 − I REF RE1 ⎝ I0 ⎠ Output resistance looking into the collector of Q2 is increased. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.35 ⎛I (a) V BE1 = VT ln⎜⎜ REF ⎝ I S1

⎞ ⎛ 0.5 × 10 −3 ⎟ = (0.026 ) ln⎜ ⎟ ⎜ 10 −15 ⎠ ⎝

⎞ ⎟ = 0.70038 V ⎟ ⎠

V + = I REF R1 + V BE1 + I REF R E1 + V − Then R1 =

3 − 0.70038 − (0.5)(0.5) − (− 3) = 10.1 k Ω 0.5

⎛ 0.2 × 10 −3 ⎞ ⎟ = 0.67656 V V BE 2 = (0.026) ln⎜⎜ −15 ⎟ ⎝ 10 ⎠ V BE1 + I REF R E1 = V BE 2 + I O R E 2

0.70038 + (0.5)(0.5) − 0.67656 = 1.37 k Ω 0. 2 (b) R1 = 10.1 k Ω

Then R E 2 =

⎛ 0.2 × 10 −3 ⎞ ⎟ = 0.65854 V V BE 2 = (0.026) ln⎜⎜ −15 ⎟ ⎝ 2 × 10 ⎠ 0.70038 + (0.5)(0.5) − 0.65854 RE 2 = = 1.46 k Ω 0. 2 ______________________________________________________________________________________

10.36 Assume all transistors are matched. a. 2VBE1 = VBE 3 + I 0 RE

⎛I ⎞ VBE1 = VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛I ⎞ VBE 3 = VT ln ⎜ 0 ⎟ ⎝ IS ⎠ ⎛I ⎞ ⎛I ⎞ 2VT ln ⎜ REF ⎟ − VT ln ⎜ 0 ⎟ = I 0 RE ⎝ IS ⎠ ⎝ IS ⎠ 2 ⎡ ⎛I ⎞ ⎛ I ⎞⎤ VT ⎢ln ⎜ REF ⎟ − ln ⎜ 0 ⎟ ⎥ = I 0 RE ⎢⎣ ⎝ I S ⎠ ⎝ I S ⎠ ⎥⎦ ⎛ I 2 REF ⎞ VT ln ⎜ ⎟ = I 0 RE ⎝ I0 I S ⎠ b. ⎛ 0.7 ⎞ −15 VBE = 0.7 V at 1 mA ⇒ 10−3 = I S exp ⎜ ⎟ or I S = 2.03 × 10 A 0.026 ⎝ ⎠ ⎛ 0.1× 10−3 ⎞ VBE at 0.1 mA ⇒ VBE = ( 0.026 ) ln ⎜ = 0.640 V −15 ⎟ ⎝ 2.03 × 10 ⎠ 0.640 Since I 0 = I REF , then VBE = I 0 RE ⇒ RE = or RE = 6.4 kΩ 0.1 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.37

(a) I REF =

5 − 0.7 − (− 5) = 0.93 mA 10

⎛I I O 2 R E 2 = VT ln⎜⎜ REF ⎝ I O2

⎞ ⎟ ⎟ ⎠

⎛ 0.93 ⎞ ⎟ I O 2 (1) = (0.026 ) ln⎜⎜ ⎟ ⎝ I O2 ⎠ I O 2 ≅ 68 μ A ⎛ 0.93 ⎞ ⎟ I O 3 (2 ) = (0.026 ) ln⎜⎜ ⎟ ⎝ I O3 ⎠ I O 3 ≅ 40.7 μ A VT ⎛ I REF ⎞ (0.026 ) ⎛ 0.93 ⎞ ⎟= ln⎜ ln⎜ ⎟ ⇒ R E 2 = 4.99 k Ω I O 2 ⎜⎝ I O 2 ⎟⎠ (0.02 ) ⎝ 0.02 ⎠ (0.026 ) ln⎛ 0.93 ⎞ ⇒ R = 0.797 k Ω RE3 = ⎜ ⎟ E3 (0.08) ⎝ 0.08 ⎠ ______________________________________________________________________________________

(b) R E 2 =

10.38 (a)

VBE1 = VBE 2 I REF =

V + − 2VBE1 − V − R1 + R2

Now 2VBE1 + I REF R2 = VBE 3 + I O RE or I O RE = 2VBE1 − VBE 3 + I REF R2 We have ⎛I ⎞ ⎛I ⎞ VBE1 = VT ln ⎜ REF ⎟ and VBE 3 = VT ln ⎜ O ⎟ I ⎝ S ⎠ ⎝ IS ⎠ (b) Let R1 = R2 and I O = I REF ⇒ VBE1 = VBE 3 ≡ VBE Then VBE = I O RE − I REF R2 = I O ( RE − R2 ) so I REF = I O = =

V + − V − − 2 I O ( RE − R2 )

Then IO =

2 R2

⎛R V −V − IO ⎜ E 2 R2 ⎝ R2 +

V + −V − 2 Rε

−

⎞ ⎟ + IO ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) Want I O = 0.5 mA So RE = 2 R2 =

5 − ( −5 ) 2 ( 0.5 )

⇒ RE = 10 k Ω

5 − 2 ( 0.7 ) − ( −5 )

0.5 Then R1 = R2 = 8.6 k Ω

= 17.2 k Ω

______________________________________________________________________________________ 10.39 a. 20 − 0.7 − 0.7 = 1.55 mA 12 I 01 = 2 I REF = 3.1 mA I 02 = I REF = 1.55 mA I REF =

I 03 = 3I REF = 4.65 mA

b.

VCE1 = − I 01 RC1 − ( −10 ) = − ( 3.1)( 2 ) + 10 ⇒ VCE1 = 3.8 V VEC 2 = 10 − I 02 RC 2 = 10 − (1.55 )( 3 ) ⇒ VEC 2 = 5.35 V VEC 3 = 10 − I 03 RC 3 = 10 − ( 4.65 )(1) ⇒ VEC 3 = 5.35 V

______________________________________________________________________________________ 10.40

10 − 0.7 − 0.7 − (− 10 ) = 0.775 mA 24 = 2 I REF = 1.55 mA

(a) I REF = I O1

I O 2 = I REF = 0.775 mA I O 3 = 3I REF = 2.325 mA

0 − 0.7 − (− 10 ) = 6kΩ 1.55 10 − 0.7 RC 2 = = 12 k Ω 0.775 10 − 0.7 RC 3 = = 4 kΩ 2.325 ______________________________________________________________________________________

(b) RC1 =

10.41 I C1 = I C 2 =

10 − 0.7 − 0.7 − ( −10 ) 10

I C 3 = I C 4 = 1.86 mA ⎛ 1.86 ⎞ I C 5 ( 0.5 ) = 0.026 ln ⎜ ⎟ ⎝ IC 5 ⎠ By Trial and error. ⇒ I C 5 = 0.136 mA = I C 6 = I C 7

= 1.86 mA

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2 I C 3 ( 0.8 ) + VCE 3 = 10 ⇒ VCE 3 = 10 − 2 (1.86 )( 0.8 ) VCE 3 = 7.02 V 5 = VEB 6 + VCE 5 + I C 5 ( 0.5 ) − 10

VCE 5 = 5 + 10 − 0.7 − ( 0.136 )( 0.5 ) VCE 5 = 14.2 V 5 = VEC 7 + I C 7 ( 0.8 )

VEC 7 = 5 − ( 0.136 )( 0.8 ) VEC 7 = 4.89 V

______________________________________________________________________________________ 10.42 I C1 = I C 2 =

10 − 0.7 − 0.7 − ( −10 ) 10

⇒ I C1 = I C 2 = 1.86 mA

I C 4 = I C 5 = 1.86 mA ⎛I ⎞ ⎛ 1.86 ⎞ I C 3 RE1 = VT ln ⎜ C1 ⎟ ⇒ I C 3 ( 0.3) = 0.026 ln ⎜ ⎟ I ⎝ C3 ⎠ ⎝ IC 3 ⎠ By trial and error I C 3 = 0.195 mA ⎛I ⎞ ⎛ 1.86 ⎞ I C 6 RE 2 = VT ln ⎜ C 5 ⎟ ⇒ I C 6 ( 0.5 ) = 0.026 ln ⎜ ⎟ ⎝ IC 6 ⎠ ⎝ IC 6 ⎠ By trial and error I C 6 = 0.136 mA

______________________________________________________________________________________ 10.43

10 − 0.7 = 1 mA 6.3 + 3 VBE ( QR ) = 0.7 V as assumed VRER = I REF ⋅ RER = (1)( 3) = 3 V I REF =

VRE1 = 3 V ⇒ RE1 =

VRE1 3 = ⇒ RE1 = 3 kΩ I 01 1

VRE 2 = 3 V ⇒ RE 2 =

VRE 2 3 = ⇒ RE 2 = 1.5 kΩ 2 I 02

VRE 3 = 3 V ⇒ RE 3 =

VRE 3 3 = ⇒ RE 3 = 0.75 kΩ 4 I 03

I 01 = 1 mA I 02 = 2 mA I 03 = 4 mA ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.44 2.5 − VGS ⎛ 0.08 ⎞ 2 I REF = =⎜ ⎟ ( 6 )(VGS − 0.5 ) 15 ⎝ 2 ⎠ 2.5 − VGS = 3.6 (VGS2 − VGS + 0.25 )

3.6VGS2 − 2.6VGS − 1.6 = 0

VGS =

2.6 ± 6.76 + 23.04 2 ( 3.6 )

VGS = 1.12 V (1.1193) 2.5 − 1.1193 ⇒ I REF = 92.0 μ A ( 92.05 ) 15 I o = 92.0 μ A I REF =

VDS 2 ( sat ) = VGS − VTN = 1.1193 − 0.5

VDS 2 ( sat ) = 0.619 V ______________________________________________________________________________________

10.45 (a) 2 ⎛ 80 ⎞⎛ W ⎞ I REF = 50 = ⎜ ⎟⎜ ⎟ (VGS − 0.5 ) 2 L ⎝ ⎠⎝ ⎠1 2.0 − VGS I REF = 0.050 = R

Design such that VGS = 0.75 V So

0.050 =

VDS 2 ( sat ) = 0.25 = VGS − 0.5

2 − 0.75 ⇒ R = 25 K R

2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 50 = ⎜ ⎟ ⎜ ⎟ ( 0.75 − 0.5 ) ⇒ ⎜ ⎟ = 20 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1

(b)

⎛W ⎞ ⎜ ⎟ 20 50 ⎛W ⎞ ⎝ L ⎠1 I REF = ⇒ = ⇒ ⎜ ⎟ = 40 W W I 100 ⎛ ⎞ ⎛ ⎞ ⎝ L ⎠2 O ⎜ ⎟ ⎜ ⎟ ⎝ L ⎠2 ⎝ L ⎠2 1 1 RO = = ⇒ RO = 667 K λ I O ( 0.015 )( 0.1) ΔI O =

ΔV 1 = ⇒ 1.5 μ A RO 666

ΔI O ⎛ 1.5 ⎞ × 100% = ⎜ ⎟ × 100% ⇒ 1.5% IO ⎝ 100 ⎠

(c) ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.46 2 ⎛ 80 ⎞ I REF = 250 = ⎜ ⎟ ( 3)(VGS − 1) ⎝ 2⎠ VGS = 2.44 V I O = 250 μ A at VDS 2 = VGS = 2.44 V 1 1 = = 200 K RO = λ I O ( 0.02 )( 0.25 ) (a) ΔV 3 − 2.44 ΔI O = = ⇒ 2.8 μ A RO 200 (i)

I O = 252.8 μ A ΔI O =

(ii)

I O = 260.3 μ A ΔI O =

(iii)

ΔV 4.5 − 2.44 = ⇒ 10.3 μ A RO 200 ΔV 6 − 2.44 = ⇒ 17.8 μ A RO 200

I O = 267.8 μ A 4.5 ( 250 ) = 375 μ A at VDS = 2.44 V 3 1 1 RO = = = 133.3 K λ I O ( 0.02 )( 0.375 ) IO =

(b)

ΔI O = (i)

I O = 379.2 μ A ΔI O =

(ii)

ΔV 3 − 2.44 = ⇒ 4.20 μ A RO 133.3 ΔV 4.5 − 2.44 = ⇒ 15.5 μ A RO 133.3

I O = 390.5 μ A ΔI O =

ΔV 6 − 2.44 = ⇒ 26.7 μ A RO 133.3

I = 401.7 μ A (iii) O ______________________________________________________________________________________ 10.47 2 (a) I REF = K n1 (VGS1 − VTN 1 )

0.2 = 0.2(VGS1 − 0.4) ⇒ VGS1 = VGS 2 = 1.4 V 2

I O = K n 2 (VGS 2 − VTN 2 )

2

Now I O = (0.2 − 0.01)(1.4 − 0.4) = 0.19 mA 2

I O = (0.2 + 0.01)(1.4 − 0.4) = 0.21 mA So 0.19 ≤ I O ≤ 0.21 mA 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) I O = (0.2)[1.4 − (0.4 − 0.02)] = 0.2081 mA 2

I O = (0.2)[1.4 − (0.4 + 0.02)] = 0.1921 mA So 0.1921 ≤ I O ≤ 0.2081 mA ______________________________________________________________________________________ 2

10.48 2 (a) I REF = K n1 (VGS1 − VTN )

0.2 = 0.2(VGS1 − 0.5) ⇒ VGS1 = 1.5 V 2

VGS1 = VGS 2 + I O R S = VGS 2 + K n 2 R S (VGS 2 − VTN )

(

2

)

2 1.5 = VGS 2 + (0.2)(10) VGS 2 − VGS 2 + 0.25

2 or 2VGS 2 − VGS 2 − 1 = 0 ⇒ VGS 2 = 1.0 V V − VGS 2 1.5 − 1.0 I O = GS1 = ⇒ I O = 50 μ A RS 10

(b) I O = 0.5 I REF = 0.1 mA

0.1 = 0.2(VGS 2 − 0.5) ⇒ VGS 2 = 1.207 V V − VGS 2 1.5 − 1.207 = = 2.93 k Ω R S = GS1 0.1 IO ______________________________________________________________________________________ 2

10.49

Ix =

(1)

Vx − VA + g mVgs 2 ro

Ix =

(2) So

Vgs1 = Vx , Vgs 2 = −VA

Ix =

⎛1 ⎞ Vx − VA ⎜ + g m ⎟ ro ⎝ ro ⎠

Ix =

VA + g mVx ⇒ VA = ro [ I x − g mVx ] ro

(1) (2)

VA + g mVgs1 ro

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then Ix =

⎛1 ⎞ Vx − ro ( I x − g mVx ) ⎜ + g m ⎟ ro ⎝ ro ⎠

Ix =

⎡I ⎤ Vx g − ro ⎢ x + g m I x − m ⋅ Vx − g m2 Vx ⎥ ro ro ⎣ ro ⎦

Ix =

Vx − I x − g m ro I x + g mVx + g m2 roVx ro

⎡1 ⎤ I x [ 2 + g m ro ] = Vx ⎢ + g m + g m2 ro ⎥ r ⎣ o ⎦ 1 g m >> ro Since I x [ 2 + g m ro ] ≅ Vx ( g m )(1 + g m ro ) Then

Vx 2 + g m ro = Ro = Ix g m (1 + g m ro )

Ro ≅

1 gm

Usually, g m ro >> 2, so that ______________________________________________________________________________________ 10.50

VGS 2 = V DS 2 (sat ) + VTN = 1 + 0.5 = 1.5 V ⎛ k ′ ⎞⎛ W ⎞ 2 I O = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 2 − VTN ) ⎝ 2 ⎠⎝ L ⎠ 2 ⎛ 0.08 ⎞⎛ W ⎞ ⎛W ⎞ 2 0.15 = ⎜ ⎟⎜ ⎟ (1.5 − 0.5) ⇒ ⎜ ⎟ = 3.75 L 2 ⎝ ⎠⎝ ⎠ 2 ⎝ L ⎠2 ⎛ 0.08 ⎞⎛ W ⎞ ⎛W ⎞ 2 I REF = 0.5 = ⎜ ⎟⎜ ⎟ (1.5 − 0.5) ⇒ ⎜ ⎟ = 12.5 ⎝ 2 ⎠⎝ L ⎠ 1 ⎝ L ⎠1

(

)

VGS 3 = V + − V − − VGS1 = 1.8 − (− 1.8) − 1.5 = 2.1 V

⎛ 0.08 ⎞⎛ W ⎞ ⎛W ⎞ 2 I REF = 0.5 = ⎜ ⎟⎜⎜ ⎟⎟(2.1 − 0.5) ⇒ ⎜ ⎟ = 4.88 L 2 ⎝ ⎠⎝ 3 ⎠ ⎝ L ⎠3 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.51 (a) 2 2 ⎛ 60 ⎞ ⎛ 60 ⎞ I REF = ⎜ ⎟ ( 20 )(VGS 1 − 0.7 ) = ⎜ ⎟ ( 3)(VGS 3 − 0.7 ) ⎝ 2 ⎠ ⎝ 2 ⎠ VGS1 + VGS 3 = 5 20 (VGS1 − 0.7 ) = 5 − VGS1 − 0.7 3 3.582VGS 1 = 6.107 ⇒ VGS1 = VGS 2 = 1.705 V 2 ⎛ 60 ⎞ I O = ⎜ ⎟ (12 )(1.705 − 0.7 ) = 363.6 μ A at VDS 2 = 1.705 V ⎝ 2 ⎠ 2 ⎛ 60 ⎞ I REF = ⎜ ⎟ ( 20 )(1.705 − 0.7 ) = 606 μ A 2 ⎝ ⎠

RO = ΔI O =

(b)

1 1 = = 183.4 K λ I O ( 0.015 )( 0.3636 ) ΔV 1.5 − 1.705 = ⇒ −1.12 μ A RO 183.4

I O = 362.5 μ A ΔI O =

ΔV 3 − 1.705 = ⇒ 7.06 μ A RO 183.4

I O = 370.7 μ A (c) ______________________________________________________________________________________

10.52 2 2 ⎛ 50 ⎞ ⎛ 50 ⎞ I REF = ⎜ ⎟ (15 )(VSG1 − 0.5 ) = ⎜ ⎟ ( 3)(VSG 3 − 0.5 ) ⎝ 2⎠ ⎝ 2⎠ VSG1 + VSG 3 = 10 ⇒ VSG 3 = 10 − VSG1

15 (VSG1 − 0.5) = 10 − VSG1 − 0.5 3 3.236VSG1 = 10.618 ⇒ VSG1 = 3.28 V 2 ⎛ 50 ⎞ I REF = ⎜ ⎟ (15 )( 3.28 − 0.5 ) ⇒ I REF = 2.90 mA ⎝ 2⎠ I O = I REF = 2.90 mA

VSD 2 (sat) = VSG 2 + VTP = 3.28 − 0.5 ⇒ VSD 2 (sat) = 2.78 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.53 VSD 2 (sat) = 1.2 = VSG 2 − 0.35 ⇒ VSG 2 = 1.55 V 2 ⎛ 50 ⎞ ⎛ W ⎞ ⎛W ⎞ I O = 100 = ⎜ ⎟ ⎜ ⎟ (1.55 − 0.35 ) ⇒ ⎜ ⎟ = 2.78 2 L ⎝ ⎠ ⎝ ⎠2 ⎝ L ⎠2 W W I REF 200 ⎛W ⎞ L1 L1 = ⇒ = ⇒ ⎜ ⎟ = 5.56 W IO 100 2.78 ⎝ L ⎠1 L 2 VSG1 + VSG 3 = 4 ⇒ VSG 3 = 2.45 V

( ) ( )

( )

2 ⎛ 50 ⎞ ⎛ W ⎞ ⎛W ⎞ I REF = 200 = ⎜ ⎟ ⎜ ⎟ ( 2.45 − 0.35 ) ⇒ ⎜ ⎟ = 1.81 2 L ⎝ ⎠ ⎝ ⎠3 ⎝ L ⎠3 ______________________________________________________________________________________

10.54 2 2 ⎛ 80 ⎞ ⎛ 80 ⎞ I REF = ⎜ ⎟ ( 25 )(VSG1 − 1.2 ) = ⎜ ⎟ ( 4 )(VSG 3 − 1.2 ) 2 2 ⎝ ⎠ ⎝ ⎠ 10 − VSG1 VSG1 + 2VSG 3 = 10 ⇒ VSG 3 = 2 10 − VSG1 25 Then − 1.2 (VSG1 − 1.2 ) = 4 2 3VSG1 = 6.8 ⇒ VSG1 = 2.27 V 2 ⎛ 80 ⎞ I REF = ⎜ ⎟ ( 25 )( 2.267 − 1.2 ) ⇒ I REF = I O = 1.14 mA ⎝ 2⎠ VSD 2 (sat) = VSG 2 + VTP = 2.27 − 1.2 ⇒ VSD 2 ( sat ) = 1.07 V

______________________________________________________________________________________ 10.55

V SG 2 = V SD 2 (sat ) − VTP = 1 + 0.5 = 1.5 V ⎛ k ′p ⎞⎛ W ⎞ 2 I O = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 2 + VTP ) ⎝ 2 ⎠⎝ L ⎠ 2 ⎛ 60 ⎞⎛ W ⎞ ⎛W ⎞ 2 80 = ⎜ ⎟⎜ ⎟ (1.5 − 0.5) ⇒ ⎜ ⎟ = 2.67 ⎝ 2 ⎠⎝ L ⎠ 2 ⎝ L ⎠2 ⎛ 60 ⎞⎛ W ⎞ ⎛W ⎞ 2 I REF = 250 = ⎜ ⎟⎜ ⎟ (1.5 − 0.5) ⇒ ⎜ ⎟ = 8.33 L 2 ⎝ ⎠⎝ ⎠ 1 ⎝ L ⎠1 V SG 3 = V SG 4

Then 2V SG 3 = V + − V − − V SG1 = 3 − (− 3) − 1.5 = 4.5 V So V SG 3 = 2.25 V

⎛ 60 ⎞⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ 2 250 = ⎜ ⎟⎜ ⎟ (2.25 − 0.5) ⇒ ⎜ ⎟ = ⎜ ⎟ = 2.72 ⎝ 2 ⎠⎝ L ⎠ 3, 4 ⎝ L ⎠3 ⎝ L ⎠4 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.56 ⎛ k ′p ⎞⎛ W ⎞ ⎛ k ′p ⎞⎛ W ⎞ 2 2 (a) I REF = ⎜⎜ ⎟⎟⎜ ⎟ (V SG1 + VTP ) = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 3 + VTP ) 2 2 L L ⎝ ⎠ ⎝ ⎠ 1 3 ⎝ ⎠ ⎝ ⎠ V SG 3 = 3 − V SG1 25 (V SG1 − 0.4 ) = 5 (3 − V SG1 − 0.4 ) 3.236V SG1 = 3.4944 ⇒ V SG1 = 1.08 V and V SG 3 = 1.92 V ⎛ 60 ⎞ 2 I REF = ⎜ ⎟(25)(1.08 − 0.4 ) ⇒ I REF = 0.347 mA ⎝ 2 ⎠ ⎛ 60 ⎞ 2 I O = ⎜ ⎟(15)(1.08 − 0.4 ) ⇒ I O = 0.208 mA ⎝ 2 ⎠ ( (b) V SD 2 sat ) = V SG 2 + VTP = 1.08 − 0.4 = 0.68 V 3 − 0.68 = 11.15 k Ω 0.208 ______________________________________________________________________________________ R=

10.57

V SD 2 (sat ) = 0.35 = V SG 2 + VTP = V SG 2 − 0.4 ⇒ V SG 2 = 0.75 V ⎛ 60 ⎞⎛ W ⎞ ⎛W ⎞ 2 I REF = 220 = ⎜ ⎟⎜ ⎟ (0.75 − 0.4 ) ⇒ ⎜ ⎟ = 59.9 L 2 ⎝ ⎠⎝ ⎠ 1 ⎝ L ⎠1 ⎛ 60 ⎞⎛ W ⎞ ⎛W ⎞ 2 I O = 80 = ⎜ ⎟⎜ ⎟ (0.75 − 0.4 ) ⇒ ⎜ ⎟ = 21.8 L 2 ⎝ ⎠⎝ ⎠ 2 ⎝ L ⎠2 V SG 3 = 3 − 0.75 = 2.25 V

⎛ 60 ⎞⎛ W ⎞ ⎛W ⎞ 2 220 = ⎜ ⎟⎜ ⎟ (2.25 − 0.4 ) ⇒ ⎜ ⎟ = 2.14 ⎝ 2 ⎠⎝ L ⎠ 3 ⎝ L ⎠3 ______________________________________________________________________________________

10.58 2 (a) I REF = 100 = 100(VGS1 − 0.5) ⇒ VGS1 = 1.5 V For V D 4 = −2 V, V DS 4 + V DS 2 = 3 V = V DS 3 + V DS1 Then I O = 100 μ A (b) Ro = ro 4 + ro 2 (1 + g m ro 4 ) g m = 2 K n I O = 2 (0.1)(0.1) = 0.2 mA/V

ro 2 = ro 4 =

1 1 = = 500 k Ω λ I O (0.02)(0.1)

R o = 500 + 500[1 + (0.2 )(500 )] ⇒ R o = 51 M Ω ΔI O =

ΔV D 4 4 = = 0.07843 μ A Ro 51

ΔI O ⎛ 0.07843 ⎞ × 100% = ⎜ ⎟ × 100% = 0.07843% IO ⎝ 100 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.59

Vgs 4 = − I X r02

VS 6 = ( I X − g mVgs 4 ) r04 + I X r02

= ( I X + g m I X r02 ) r04 + I X r02 VS 6 = I X ⎡⎣ r02 + (1 + g m r02 ) r04 ⎤⎦ = −Vgs 6 I X = g mVgs 6 +

⎛ VX − VS 6 VX 1 ⎞ = − VS 6 ⎜ g m + ⎟ r06 r06 r06 ⎠ ⎝

⎛ 1 ⎞ ⎜ g m + ⎟ ⎡⎣ r02 + (1 + g m r02 ) r04 ⎤⎦ r06 ⎠ ⎝ ⎧⎪ ⎛ ⎫⎪ V 1 ⎞ I X ⎨1 + ⎜ g m + ⎟ ⎡⎣ r02 + (1 + g m r02 ) r04 ⎤⎦ ⎬ = X r06 ⎠ ⎩⎪ ⎝ ⎭⎪ r06 VX = R0 = r06 + (1 + g m r06 ) ⎡⎣ r02 + (1 + g m r02 ) r04 ⎤⎦ IX IX =

VX − IX r06

I 0 ≈ I REF = 0.2 mA = 0.2 (VGS − 1)

2

VGS = 2 V

g m = 2 K n (VGS − VTN ) = 2 ( 0.2 )( 2 − 1) = 0.4 mA / V

r02 = r04 = r06 =

1

λ I0

=

1 = 250 kΩ 0.02 ( )( 0.2 )

{

}

R0 = 250 + ⎡⎣1 + ( 0.4 )( 250 ) ⎤⎦ × 250 + ⎡⎣1 + ( 0.4 )( 250 ) ⎤⎦ ( 250 ) R0 = 2575750 kΩ ⇒ R0 = 2.58 × 109 Ω

______________________________________________________________________________________ 10.60 ⎛ k ′p ⎞⎛ W ⎞ ⎛ k ′ ⎞⎛ W ⎞ 2 2 I REF = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 3 − VTN ) = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 4 + VTP ) 2 2 L L ⎝ ⎠⎝ ⎠ 3 ⎝ ⎠⎝ ⎠ 4 ⎛ 60 ⎞ ⎛ 100 ⎞ 2 2 ⎟(5)(VGS 3 − 0.4 ) = ⎜ ⎟(10 )(V SG 4 − 0.4 ) ⎜ 2 2 ⎠ ⎝ ⎠ ⎝

We find V SG 4 = 0.91287VGS 3 + 0.03485 ⎛ k′ I REF = ⎜⎜ n ⎝ 2

⎛ k′ ⎞⎛ W ⎞ 2 ⎟⎟⎜ ⎟ (VGS 1 − VTN ) = ⎜⎜ n ⎝ 2 ⎠⎝ L ⎠ 1

⎞⎛ W ⎞ 2 ⎟⎟⎜ ⎟ (VGS 3 − VTN ) ⎠⎝ L ⎠ 3

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 20 (VGS1 − 0.4 ) = 5 (VGS 3 − 0.4 )

Now V SG 4 + VGS 3 + VGS1 = 6 0.91287VGS 3 + 0.03485 + VGS 3 + VGS 1 = 6 Then VGS 3 = 3.1184 − 0.52277VGS1 And 2(VGS1 − 0.4) = 3.1184 − 0.52277VGS 1 − 0.4 So VGS 1 = 1.395 V VGS 3 = 2.389 V V SG 4 = 2.216 V ⎛ 0. 1 ⎞ 2 I REF = ⎜ ⎟(20 )(1.395 − 0.4 ) = 0.99 mA ⎝ 2 ⎠ ⎛ 0.1 ⎞ 2 IO = ⎜ ⎟(20 )(1.395 − 0.4 ) = 0.99 mA 2 ⎠ ⎝ V DS 2 (sat ) = 1.395 − 0.4 = 0.995 V ______________________________________________________________________________________

10.61

V DS 2 (sat ) = 0.5 V = VGS 2 − VTN = VGS 2 − 0.4 ⇒ VGS 2 = 0.9 V ⎛ k ′ ⎞⎛ W ⎞ 2 I O = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 2 − VTN ) 2 L ⎝ ⎠⎝ ⎠ 2 ⎛W ⎞ ⎛ 100 ⎞⎛ W ⎞ 2 50 = ⎜ ⎟⎜ ⎟ (0.9 − 0.4 ) ⇒ ⎜ ⎟ = 4 ⎝ L ⎠2 ⎝ 2 ⎠⎝ L ⎠ 2 ⎛W ⎞ ⎛ 100 ⎞⎛ W ⎞ 2 I REF = 500 = ⎜ ⎟⎜ ⎟ (0.9 − 0.4 ) ⇒ ⎜ ⎟ = 40 ⎝ L ⎠1 ⎝ 2 ⎠⎝ L ⎠ 1 VGS 3 = V SG 4

6 = VGS1 + VGS 3 + V SG 4 = 0.9 + 2VGS 3 ⇒ VGS 3 = V SG 4 = 2.55 V ⎛W ⎞ ⎛ 100 ⎞⎛ W ⎞ 2 500 = ⎜ ⎟⎜ ⎟ (2.55 − 0.4 ) ⇒ ⎜ ⎟ = 2.16 ⎝ L ⎠3 ⎝ 2 ⎠⎝ L ⎠ 3 ⎛W ⎞ ⎛ 60 ⎞⎛ W ⎞ 2 500 = ⎜ ⎟⎜ ⎟ (2.55 − 0.4 ) ⇒ ⎜ ⎟ = 3.61 2 L ⎝ L ⎠4 ⎝ ⎠⎝ ⎠ 4 ______________________________________________________________________________________

10.62 a.

As a first approximation 2 I REF = 80 = 80 (VGS 1 − 1) ⇒ VGS 1 = 2 V

Then VDS 1 ≅ 2 ( 2 ) = 4 V The second approximation

80 = 80 (VGS 1 − 1) ⎡⎣1 + ( 0.02 )( 4 ) ⎦⎤ 80 2 = (VGS 1 − 1) ⇒ VGS1 = 1.962 Or 86.4 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then I O = K n (VGS 1 − VTN ) (1 + λnVGS 1 ) 2

= 80 (1.962 − 1) ⎡⎣1 + ( 0.02 )(1.962 ) ⎤⎦ Or I 0 = 76.94 μ A 2

From a PSpice analysis, I 0 = 77.09 μ A for VD 3 = −1 V and I 0 = 77.14 μ A for VD 3 = 3 V. The change is ΔI 0 ≈ 0.05 μ A or 0.065%.

b.

______________________________________________________________________________________ 10.63 a.

For a first approximation, I REF = 80 = 80 (VGS 4 − 1) ⇒ VGS 4 = 2 V 2

As a second approximation I REF = 80 = 80 (VGS 4 − 1) ⎡⎣1 + ( 0.02 )( 2 ) ⎤⎦ Or VGS 4 = 1.98 V = VGS 1 2

I O = K n (VGS 2 − VTN ) (1 + λVGS 2 ) 2

To a very good approximation I 0 = 80 μ A From a PSpice analysis, I 0 = 80.00 μ A for VD 3 = −1 V and the output resistance is b. R0 = 76.9 MΩ. Then For VD = +3 V 1 4 ΔI 0 = ⋅ VD 3 = = 0.052 μ A 76.9 R0 I 0 = 80.05 μ A ______________________________________________________________________________________

10.64

VDS 3 ( sat ) = VGS 3 − VTN or VGS 3 = VDS 3 ( sat ) + VTN = 0.2 + 0.8 = 1.0 k n′ ⎛ W ⎞ 2 ⎜ ⎟ (VGS 3 − VTN ) 2⎝L⎠ 2 ⎛W ⎞ ⎛W ⎞ 50 = 48 ⎜ ⎟ ( 0.2 ) ⇒ ⎜ ⎟ = 26 L ⎝ ⎠ ⎝ L ⎠3 ID =

(a) (b)

VGS 5 − VTN = 2 (VGS 3 − VTN )

VGS 5 = 0.8 + 2 ( 0.2 ) ⇒ VGS 5 = 1.2 V

VD1 ( min ) = 2VDS ( sat ) = 2 ( 0.2 ) ⇒ VD1 ( min ) = 0.4 V (c) ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.65 (a) k′ ⎛ W ⎞ K n1 = n ⎜ ⎟ = 50 ( 5 ) = 250 μ A / V 2 2 ⎝ L ⎠1 R= =

1 K n1 I D1

⎛ ⎜1 − ⎜ ⎝

(W / L )1 ⎞⎟ (W / L )2 ⎟⎠

⎛ 5 ⎞ ⎜1 − ⎟ = ( 8.944 )( 0.6838 ) ⎜ 50 ⎟⎠ ( 0.25)( 0.05) ⎝ 1

R = 6.12 k Ω

(b)

V + − V − = VSD 3 ( sat ) + VGS 1 VSD 3 ( sat ) = VSG 3 + VTP I D1 = 50 = 20 ( 5 )(VSG 3 − 0.5 ) ⇒ VSG 3 = 1.207 V Then VSD 3 ( sat ) = 1.21 − 0.5 = 0.707 V 2

Also I D1 = 50 = 50 ( 5 )(VGS1 − 0.5 ) ⇒ VGS 1 = 0.9472 V 2

Then (V + − V − )

min

= 0.71 + 0.947 = 1.66 V

(c) 2 ⎛W ⎞ ⎛W ⎞ I O1 = 25 = 50 ⎜ ⎟ ( 0.947 − 0.5 ) ⇒ ⎜ ⎟ = 2.5 ⎝ L ⎠5 ⎝ L ⎠5 2 ⎛W ⎞ ⎛W ⎞ I O 2 = 75 = 20 ⎜ ⎟ (1.207 − 0.5) ⇒ ⎜ ⎟ = 7.5 ⎝ L ⎠6 ⎝ L ⎠6

______________________________________________________________________________________ 10.66 V GS 3 = VGS 4 = VGS 5 ⇒ VGS 3 =

5 V 3 2

⎛ 0.08 ⎞⎛ W ⎞ ⎛ 5 ⎞ ⎛W ⎞ I REF = 0.1 = ⎜ = 2.68 ⎟⎜ ⎟ ⎜ − 0.7 ⎟ ⇒ ⎜ ⎟ 2 L 3 ⎠⎝ ⎠ 3 ⎝ ⎠ ⎝ L ⎠ 3, 4 , 5 ⎝ 2

⎛ 0.08 ⎞⎛ W ⎞ ⎛ 5 ⎞ ⎛W ⎞ I O1 = 0.2 = ⎜ ⎟⎜ ⎟ ⎜ − 0.7 ⎟ ⇒ ⎜ ⎟ = 5.35 ⎝ 2 ⎠⎝ L ⎠ 1 ⎝ 3 ⎠ ⎝ L ⎠1 2

⎞ ⎛W ⎞ ⎛ 0.08 ⎞⎛ W ⎞ ⎛ 5 I O 2 = 0.3 = ⎜ ⎟⎜ ⎟ ⎜ − 0.7 ⎟ ⇒ ⎜ ⎟ = 8.03 ⎝ 2 ⎠⎝ L ⎠ 2 ⎝ 3 ⎠ ⎝ L ⎠2 ______________________________________________________________________________________

10.67

(

)

P = (I REF + I O1 + I O 2 + I O 3 ) V + − V − 5 = (I REF + 0.1 + 0.2 + 0.4 )[1.8 − (− 1.8)] ⇒ I REF = 0.689 mA V DS 2 (sat ) = 0.4 = VGS 2 − 0.4 ⇒ VGS 2 = 0.8 V = VGS ⎛ k ′ ⎞⎛ W ⎞ 2 I REF = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 1 − VTN ) L 2 ⎝ ⎠⎝ ⎠ 1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 2 0.689 = ⎜ ⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 86.1 ⎝ 2 ⎠⎝ L ⎠ 1 ⎝ L ⎠1 ⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 2 I O1 = 0.1 = ⎜ ⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 12.5 ⎝ 2 ⎠⎝ L ⎠ 2 ⎝ L ⎠2 ⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 2 I O 2 = 0.2 = ⎜ ⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 25 2 L ⎝ ⎠⎝ ⎠ 3 ⎝ L ⎠3 ⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 2 I O 3 = 0.4 = ⎜ ⎟⎜ ⎟ (0.8 − 0.4 ) ⇒ ⎜ ⎟ = 50 ⎝ 2 ⎠⎝ L ⎠ 4 ⎝ L ⎠4 ______________________________________________________________________________________

10.68 24 = I REF R + V SG + VGS V SG =

I REF − VTP = 5.7735 I REF + 0.8 ⎛ k ′p ⎞ ⎜ ⎟(1) ⎜ 2 ⎟ ⎝ ⎠

VGS =

I REF + VTN = 4.472 I REF + 0.8 ⎛ k n′ ⎞ ⎜⎜ ⎟⎟(1) ⎝ 2 ⎠

So 24 = I REF (100 ) + 5.7735 I REF + 0.8 + 4.4721 I REF + 0.8 I REF = x

Let

Then 100 x 2 + 10.2456x − 22.4 = 0 ⇒ x = 0.4248 ⇒ x 2 = I REF = 0.1805 mA I 1 = (0.2)I REF = 0.0361 mA I 2 = (1.25)I REF = 0.2256 mA I 3 = (0.8)I REF = 0.1444 mA I 4 = (4 )I REF = 0.722 mA ______________________________________________________________________________________ 10.69 10 = I REF R + V SG + VGS V SG =

I REF + 0.8 = 5.7735 I REF + 0.8 ⎛ 0.06 ⎞ 1 ( ) ⎜ ⎟ ⎝ 2 ⎠

VGS =

I REF + 0.8 = 4.472 I REF + 0.8 ⎛ 0.1 ⎞ 1 ( ) ⎜ ⎟ ⎝ 2 ⎠

8.4 = 100 I REF + 10.2456 I REF

Let

I REF = x

Then 100 x 2 + 10.2456x − 8.4 = 0 ⇒ x = 0.2431 ⇒ x 2 = I REF = 59.09 μ A I 1 = (0.2)I REF = 11.82 μ A, I 2 = (1.25)I REF = 73.87 μ A I 3 = (0.8)I REF = 47.27 μ A, I 4 = (4)I REF = 236.4 μ A ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.70 W 9 L 2 ⋅ I REF = ( 200 ) ⇒ I D 2 = 120 μ A ID2 = W 15 L1 W ⎛ 20 ⎞ L 4 ⋅ I D 2 = ⎜ ⎟ (120 ) ⇒ I O = 267 μ A IO = W ⎝ 9 ⎠ L 3 2 ⎛ 40 ⎞ I O = 266.7 = ⎜ ⎟ ( 20 )(VSG 4 − 0.6 ) 2 ⎝ ⎠ VSG 4 = 1.416 V VSD 4 (sat) = 1.416 − 0.6 ⇒ VSD 4 ( sat ) = 0.816 V

( ( ( (

) ) ) )

______________________________________________________________________________________ 10.71 ⎛ k ′ ⎞⎛ W ⎞ 2 For M 1 : I REF = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 1 − VTN ) ⎝ 2 ⎠⎝ L ⎠ 1 ⎛ 100 ⎞ 2 100 = ⎜ ⎟(4 )(VGS 1 − 0.4) ⇒ VGS 1 = VGS 2 = 1.107 V 2 ⎝ ⎠ ⎛ k ′ ⎞⎛ W ⎞ ⎛ 100 ⎞ 2 2 I D 2 = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 2 − VTN ) = ⎜ ⎟(2.5)(1.107 − 0.4) ⇒ I D 2 = 62.5 μ A 2 L 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 ⎛ k ′p ⎞⎛ W ⎞ 2 Also I D 2 = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 3 + VTP ) 2 L ⎝ ⎠ 3 ⎝ ⎠ ⎛ 60 ⎞ 2 62.5 = ⎜ ⎟(6 )(V SG 3 − 0.4 ) ⇒ V SG 3 = V SG 4 = 0.9893 V ⎝ 2 ⎠ ⎛ 60 ⎞ 2 I O = ⎜ ⎟(4 )(0.9893 − 0.4 ) = 41.67 μ A ⎝ 2 ⎠ ______________________________________________________________________________________

10.72 2 ⎛ 40 ⎞ ⎛ W ⎞ I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VSG1 − 0.6 ) ⎝ 2 ⎠ ⎝ L ⎠1 1.75 − VSG1 I REF = = 50 R VSD 2 (sat) = 0.35 = VSG 2 − 0.6 ⇒ VSG 2 = 0.95 V

1.75 − 0.95 ⇒ R = 16 K 0.05 2 ⎛ 40 ⎞⎛ W ⎞ ⎛W ⎞ 50 = ⎜ ⎟⎜ ⎟ ( 0.95 − 0.6 ) ⇒ ⎜ ⎟ = 20.4 ⎝ 2 ⎠⎝ L ⎠1 ⎝ L ⎠1 R=

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ W I O1 120 ⎛W ⎞ L 2 = = ⇒ ⎜ ⎟ = 49 I REF 50 20.4 ( ) ⎝ L ⎠2

( )

( )

W I D3 25 ⎛W ⎞ L 3 = = ⇒ ⎜ ⎟ = 10.2 I REF 50 ( 20.4 ) ⎝ L ⎠3 VDS 5 (sat) = 0.35 = VGS 5 − 0.4 ⇒ VGS 5 = 0.75 V 2 ⎛ 100 ⎞⎛ W ⎞ ⎛W ⎞ IO 2 = ⎜ ⎟⎜ ⎟ ( 0.75 − 0.4 ) = 150 ⇒ ⎜ ⎟ = 24.5 ⎝ 2 ⎠⎝ L ⎠5 ⎝ L ⎠5 W I D4 I D3 25 ⎛W ⎞ L 4 = = = ⇒ ⎜ ⎟ = 4.08 I O 2 I O 2 150 24.5 ⎝ L ⎠4 ______________________________________________________________________________________

( )

10.73

For a.

vGS = 0, iD = I DSS (1 + λ vDS ) VD = −5 V, vDS = 5 iD = ( 2 ) ⎡⎣1 + ( 0.05 )( 5) ⎤⎦ ⇒ iD = 2.5 mA VD = 0, vDS = 10

b.

iD = ( 2 ) ⎣⎡1 + ( 0.05 )(10 ) ⎦⎤ ⇒ iD = 3 mA

VD = 5 V, vDS = 15 V

iD = ( 2 ) ⎡⎣1 + ( 0.05 )(15 ) ⎤⎦ ⇒ iD = 3.5 mA c. ______________________________________________________________________________________ 10.74 ⎛ V ⎞ I 0 = I DSS ⎜1 − GS ⎟ VP ⎠ ⎝ ⎛ V ⎞ 2 = 4 ⎜ 1 − GS ⎟ ⎝ VP ⎠

2

2

VGS 2 = 1− = 0.293 VP 4

So VGS = ( 0.293)( −4 ) = −1.17 V V Then I 0 = S and VS = −VGS R ( −1.17 ) −V ⇒ R = 0.586 kΩ R = GS = − I0 2

Need vDS ≥ vDS ( sat ) = vGS − VP = −1.17 − ( −4 ) vDS ≥ 2.83 V So VD ≥ vDS ( sat ) + VS = 2.83 + 1.17 ⇒ VD ≥ 4 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.75 1 1 − − VT − 38.46 = 0.026 = (a) Aυ = ⇒ Aυ = −1846 1 1 1 1 0.00833 + 0.0125 + + V AN V AP 120 80

⎡ ⎛V (b) I CQ = I S1 ⎢exp⎜⎜ I ⎢⎣ ⎝ VT

⎞⎤⎛ VCE1 ⎞ ⎟ ⎟⎟⎥⎜1 + ⎟ ⎜ ⎠⎥⎦⎝ V AN ⎠

⎡ ⎛V 200 × 10 − 6 = 5 × 10 −16 ⎢exp⎜⎜ I ⎣⎢ ⎝ VT

(

)

⎞⎤⎛ 1.25 ⎞ ⎟⎟⎥⎜1 + ⎟ ⎠⎦⎥⎝ 120 ⎠

⎡ ⎤ 200 × 10 − 6 or V I = (0.026 ) ln ⎢ ⎥ = 0.6943 V −16 ⎣ 5 × 10 (1 + 1.25 120) ⎦ ⎡ ⎛ V ⎞⎤⎛ 1.25 ⎞ (c) 200 × 10 − 6 = 10 −15 ⎢exp⎜⎜ EB ⎟⎟⎥⎜1 + ⎟ 80 ⎠ ⎣⎢ ⎝ VT ⎠⎦⎥⎝

(

)

(

)

⎡ ⎤ 200 × 10 − 6 V EB = (0.026 ) ln ⎢ −15 ⎥ = 0.6762 V ⎣ 10 (1 + 1.25 80 ) ⎦

(

)

+

V EB = 0.6762 = V − V B ⇒ V B = 1.824 V ______________________________________________________________________________________ 10.76 (a) Aυ = − g m1 ro1 ro 2

(

ro1 =

)

1 1 1 1 = = 250 k Ω , ro 2 = = = 166.7 k Ω λ1 I DQ (0.02 )(0.2 ) λ 2 I DQ (0.03)(0.2)

− 100 = − g m1 (250 166.7 ) ⇒ g m1 = 1 mA/V

⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ g m1 = 1 = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜ ⎟⎜ ⎟ (0.2) ⇒ ⎜ ⎟ = 25 ⎝ 2 ⎠⎝ L ⎠ 1 ⎝ L ⎠1 ⎝ 2 ⎠⎝ L ⎠ 1 ⎛ k ′ ⎞⎛ W ⎞ (b) I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (V I − VTN )2 (1 + λ1V DS1 ) ⎝ 2 ⎠⎝ L ⎠ 1 ⎛ 0.1 ⎞ 2 0.2 = ⎜ ⎟(25)(V I − 0.5) [1 + (0.02 )(1.25)] ⇒ V I = 0.895 V ⎝ 2 ⎠ ⎞⎛ W ⎞ ⎛ k ′p ⎞⎛ W ⎞ ⎟⎟⎜ ⎟ = ⎜⎜ ⎟⎟⎜ ⎟ ⎠⎝ L ⎠ 1 ⎝ 2 ⎠⎝ L ⎠ 2 ⎛W ⎞ ⎛ 100 ⎞ Then ⎜ ⎟ = ⎜ ⎟(25) = 41.67 ⎝ L ⎠ 2 ⎝ 60 ⎠ ⎛ k ′p ⎞⎛ W ⎞ 2 I DQ = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 2 + VTP ) (1 + λ 2V SD 2 ) ⎝ 2 ⎠⎝ L ⎠ 2 ⎛ 0.06 ⎞ 2 0.2 = ⎜ ⎟(41.67 )(V SG 2 − 0.5) [1 + (0.03)(1.25)] ⇒ V SG 2 = 0.8927 V ⎝ 2 ⎠

⎛ k′ (c) K n1 = K p 2 ⇒ ⎜⎜ n ⎝ 2

V SG 2 = 0.8927 = V + − VG ⇒ VG = 1.607 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.77 ⎛V ⎞ I REF = I S 1 exp ⎜ EB1 ⎟ ⎝ VT ⎠ a. b. c.

⎛I or VEB1 = VT ln ⎜ REF ⎝ I S1 R1 =

⎞ ⎛ 1× 10−3 ⎞ ⇒ VEB1 = 0.5568 ⎟ = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 5 × 10 ⎠ ⎠

5 − 0.5568 ⇒ R1 = 4.44 kΩ 1

From Equations (10.79) and (10.80) and letting VCE 0 = VEC 2 = 2.5 V 2.5 ⎞ ⎛ 1+ ⎜ ⎛ ⎞ V 2.5 ⎡ ⎤ 80 ⎟ = 10−3 ⎜ 10−12 exp ⎜ I ⎟ ⎢1 + ⎟ ⎥ ⎝ VT ⎠ ⎣ 120 ⎦ ⎜⎜ 1 + 0.5568 ⎟⎟ 80 ⎠ ⎝ ⎛V ⎞ ⎛ 1.03125 ⎞ 1.0208333 ×10−12 exp ⎜ I ⎟ = (10−3 ) ⎜ ⎟ ⎝ 1.00696 ⎠ ⎝ VT ⎠ Then VI = 0.026 ln (1.003222 × 109 )

So VI = 0.5389 V Av =

− (1/ VT )

(1/ VAN ) + (1/ VAP )

1 −38.46 Av = 0.026 = 1 1 0.00833 + 0.0125 + 120 80 Av = −1846 −

d. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.78 ⎛I ⎞ ⎛ 0.5 × 10−3 ⎞ VBE = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ ⎟ ⇒ VBE = 0.5208 I S1 ⎠ 10−12 ⎠ ⎝ ⎝ a. 5 − 0.5208 R1 = ⇒ R1 = 8.96 kΩ 0.5 b. c. Modify Eqs. 10.79 and 10.80 to apply to pnp and npn, and set the two equation equal to each other. ⎛V ⎞⎛ V ⎞ ⎛V ⎞⎛ V ⎞ I CO = I SO exp ⎜ EBO ⎟ ⎜ 1 + ECO ⎟ = I C 2 = I S 2 exp ⎜ BE ⎟ ⎜1 + CE 2 ⎟ VAP ⎠ ⎝ VT ⎠ ⎝ ⎝ VT ⎠ ⎝ VAN ⎠ ⎛V 5 × 10−13 exp ⎜ EBO ⎝ VT

⎞ ⎛ 2.5 ⎞ ⎛ VBE ⎞ ⎛ 2.5 ⎞ −12 ⎟ ⎜1 + ⎟ ⎜1 + ⎟ = 10 exp ⎜ ⎟ V 80 120 ⎝ ⎠ ⎝ ⎠ ⎠ ⎝ T ⎠ ⎛V ⎞ ⎛V ⎞ 5.15625 × 10−13 exp ⎜ EBO ⎟ = 1.020833 × 10−12 exp ⎜ BE ⎟ V ⎝ T ⎠ ⎝ VT ⎠ ⎛V ⎞ exp ⎜ EBO ⎟ ⎝ VT ⎠ = 1.9798 = exp ⎛ VEBO − VBE ⎞ ⎜ ⎟ VT ⎛V ⎞ ⎝ ⎠ exp ⎜ BE ⎟ ⎝ VT ⎠

VEBO = VBE + VT ln (1.9798 ) = 0.5208 + ( 0.026 ) ln (1.9798 ) VEBO = 0.5386 ⇒ VI = 5 − 0.5386 ⇒ VI = 4.461 V Av =

− (1/ VT )

(1/ VAN ) + (1/ VAP )

1 −38.46 0.026 = Av = 1 1 0.00833 + 0.0125 + 120 80 Av = −1846 −

d. ______________________________________________________________________________________ 10.79 ⎛ k ′ ⎞⎛ W ⎞ (a) For M O : I DQ = ⎜⎜ n ⎟⎟⎜ ⎟ (V I − VTN )2 (1 + λ nV DSO ) ⎝ 2 ⎠⎝ L ⎠ O ⎛ 100 ⎞⎛ W ⎞ ⎛W ⎞ 2 100 = ⎜ ⎟⎜ ⎟ (1.2 − 0.5) [1 + (0.02 )(1.5)] ⇒ ⎜ ⎟ = 3.96 2 L ⎝ L ⎠O ⎝ ⎠⎝ ⎠ O For M 1 , M 2 : For I REF = I O ⇒ V SD 2 = V SD1 = V SG = 1.5 V ⎛ 60 ⎞⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ 2 100 = ⎜ ⎟⎜ ⎟ (1.5 − 0.5) [1 + (0.02 )(1.5)] ⇒ ⎜ ⎟ = ⎜ ⎟ = 3.24 ⎝ 2 ⎠⎝ L ⎠ 2 ⎝ L ⎠ 2 ⎝ L ⎠1 ⎛W ⎞ For M 3 : V SG 3 = 3 − 1.5 = 1.5 V = V SD 3 ⇒ ⎜ ⎟ = 3.24 ⎝ L ⎠3

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) Aυ = − g mO roO ro 2

(

)

⎛ k′ g mO = 2 ⎜⎜ n ⎝ 2

⎞⎛ W ⎞ ⎛ 0.1 ⎞ ⎟⎟⎜ ⎟ I DQ = 2 ⎜ ⎟(3.96)(0.1) = 0.2814 mA/V ⎝ 2 ⎠ ⎠⎝ L ⎠ O 1 1 = = = 500 k Ω λ I DQ (0.02 )(0.1)

roO = ro 2

Aυ = −(0.2814)(500 500) = −70.35

______________________________________________________________________________________ 10.80 2 (a) I REF = K p (V SG + VTP ) 1 + λ pV SD1

(

)

0.080 = 0.1(V SG − 0.5) [1 + (0.02)V SG ] Approximation, for λ = 0 ⇒ V SG = 1.394 V (b) I O = I 2 2

(

K n (V I − VTN ) (1 + λ nV DSO ) = K p (V SG + VTP ) 1 + λ pV SD 2 2

2

)

K n = K p , V DSO = V SD 2 , λ n = λ p

So V I = 1.394 V (c) I O ≅ I REF = 80 μ A

Aυ = − g mo (roo ro 2 )

g mo = 2 K n I O = 2 (0.1)(0.08) = 0.1789 mA/V

roo = ro 2 =

1 1 = = 625 k Ω λ I O (0.02)(0.08)

Aυ = −(0.1789)(625 625) = −55.9

______________________________________________________________________________________ 10.81 3 − 0.6 ⇒ I CO = 51.06 μ A 47 V 120 roo = AN = ⇒ roo = 2.35 M Ω I CO 0.05106

(a) I REF = I CO =

ro 2 =

V AP 90 = ⇒ ro 2 = 1.763 M Ω I CO 0.05106

gm =

I CO 0.05106 = = 1.964 mA/V 0.026 VT

Aυ = − g m (roo ro 2 ) = −(1.964)(2350 1763) = −1978

(

)

(

)

(b) Aυ = − g m roo ro 2 R L = −(1.964 ) 2350 1763 300 = −454

(

)

(c) Aυ = −(1.964 ) 2350 1763 150 = −256 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.82 5 − 0.6 I REF = = 0.1257 mA 35 Then I CO = 2 I REF = 0.2514 mA From Eq. 10.96 −0.2514 −9.6692 0.026 Av = = 0.2514 0.2514 1 1 0.002095 + 0.0031425 + + + 120 80 RL RL Av =

(a)

−9.6692 0.0052375 +

1 RL

RL = ∞ Av = −1846

(b) RL = 250 K, Av = −1047 ______________________________________________________________________________________ 10.83 (a)

To a good approximation, output resistance is the same as the widlar current source. R0 = r02 ⎡⎣1 + g m 2 ( rπ 2 || RE ) ⎤⎦

Av = − g m 0 ( r0 || RL || R0 ) (b) ______________________________________________________________________________________ 10.84 Output resistance of Wilson source βr R0 ≅ 03 2 Then Av = − g m ( r0 || R0 ) V 80 = 400 kΩ r03 = AP = I REF 0.2 r0 =

VAN 120 = = 600 kΩ I REF 0.2

gm =

I REF 0.2 = = 7.692 mA/V 0.026 VT

⎡ (80 )( 400 ) ⎤ Av = −7.69 ⎢600 ⎥ = −7.69 [ 600 || 16, 000] ⇒ Av = −4448 2 ⎢⎣ ⎥⎦ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.85 (a) I D 2 = I D 0 = I REF = 200 μ A For M 2 ; ro 2 =

1 1 = = 250 K λP I D 2 ( 0.02 )( 0.2 )

⎛ 0.04 ⎞ gm2 = 2 K P I D 2 = 2 ⎜ ⎟ ( 35 )( 0.2 ) ⎝ 2 ⎠ g m 2 = 0.748 mA/V For M 0 ; r∞ =

(b)

1

λn I Do

=

1

( 0.015)( 0.2 )

= 333 K

⎛ 0.08 ⎞ g mo = 2 ⎜ ⎟ ( 20 )( 0.2 ) ⇒ g mo = 0.80 mA/V ⎝ 2 ⎠ Av = − g mo ( ro 2 || roo ) = − ( 0.80 )( 250 || 333) Av = −114.3 Want Av = −57.15 = −0.80 (142.8 || RL ) 142.8 || RL = 71.375 =

142.8RL ⇒ RL = 143 K 142.8 + RL

(c) ______________________________________________________________________________________ 10.86 Assume M1, M2 matched I REF = I D 2 = I Do = 200 μ A 1 1 ro 2 = = = 250 K λ p I D 2 ( 0.02 )( 0.2 ) roo =

1 1 = = 333 K λn I D 0 ( 0.015 )( 0.2 )

Av = − g mo ( ro 2 roo )

−100 = − g mo ( 250 333) ⇒ g mo = 0.70 mA/V ⎛ 0.08 ⎞⎛ W ⎞ g mo = 2 ⎜ ⎟⎜ ⎟ ( 0.2 ) = 0.70 ⎝ 2 ⎠⎝ L ⎠0 ⎛W ⎞ ⎜ ⎟ = 15.3 ⎝ L ⎠0 ⎛ k ′ ⎞⎛ W ⎞ ⎛ k ′p ⎞ ⎛ W ⎞ Now ⎜ n ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ L ⎠0 ⎝ 2 ⎠ ⎝ L ⎠ 2 ⎛ 80 ⎞ ⎛ 40 ⎞⎛ W ⎞ ⎜ ⎟ (15.3) = ⎜ ⎟⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠⎝ L ⎠ 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 30.6 ⎝ L ⎠ 2 ⎝ L ⎠1 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.87

⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞ g m1 = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I D1 = 2 ⎜ ⎟(20)(0.1) = 0.6325 mA/V ⎝ 2 ⎠ ⎝ 2 ⎠⎝ L ⎠1 1 1 ro1 = = = 500 k Ω λ n I D1 (0.02 )(0.1) R o 2 = ro 2 + ro3 (1 + g m 2 ro 2 ) ⎛ 0.06 ⎞ g m2 = 2 ⎜ ⎟(80 )(0.1) = 0.9798 mA/V ⎝ 2 ⎠

1

ro 2 = ro3 =

= 500 k Ω

(0.02)(0.1) R o 2 = 500 + 500[1 + (0.9798)(500 )] = 245,949 k Ω Aυ = − g m1 (ro1 Ro 2 ) = −(0.6325)(500 245,949) = −316 ______________________________________________________________________________________ 10.88 − g m2

Aυ =

1 1 + ro3 ro 4 ro1 ro 2

⎛ 0.1 ⎞ gm = 2 ⎜ ⎟(25)(0.08) = 0.6325 mA/V ⎝ 2 ⎠

ro =

1 1 = = 312.5 k Ω λ I D (0.04)(0.08)

− (0.6325) − 0.40 = = −19,531 1 1 2(0.00001024 ) + (312.5)2 (312.5)2 ______________________________________________________________________________________ 2

Aυ =

10.89

g m1Vi =

(1) (2)

V − ( −Vπ 2 ) Vπ 2 Vπ 2 + + g m 2Vπ 2 + O rπ 2 ro1 ro 2

VO VO − ( −Vπ 2 ) + + g m 2Vπ 2 = 0 RO 3 ro 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(1)

⎛ 1 1 1 ⎞ V + + gm2 + ⎟ + O g m1Vi = Vπ 2 ⎜ ro 2 ⎠ ro 2 ⎝ rπ 2 ro1

(2)

⎛ 1 ⎛ 1 ⎞ 1 ⎞ VO ⎜ + ⎟ + Vπ 2 ⎜ + gm2 ⎟ = 0 ⎝ RO 3 ro 2 ⎠ ⎝ ro 2 ⎠ g m >>

1 ro

(1)

⎛1+ β g m1Vi = Vπ 2 ⎜ ⎝ rπ 2

(2)

⎛ 1 1 ⎞ VO ⎜ + ⎟ + Vπ 2 ⋅ g m 2 = 0 ⎝ RO 3 ro 2 ⎠ Vπ 2 = −

VO gm2

⎞ VO ⎟+ ⎠ ro 2

⎛ 1 1 ⎞ + ⎟ ⎜ ⎝ RO 3 ro 2 ⎠

Then (1) g m1Vi = −

VO gm2

⎛ 1 1 ⎞⎛ 1+ β + ⎟⎜ ⎜ ⎝ RO 3 ro 2 ⎠ ⎝ rπ 2

⎛ 1 1 ⎞⎛1+ β = −VO ⎜ + ⎟⎜ ⎝ RO 3 ro 2 ⎠ ⎝ β ≈−

⎞ VO ⎟+ ⎠ ro 2

⎞ VO ⎟+ ⎠ ro 2

VO ⎛ 1 + β ⎞ ⎜ ⎟ RO 3 ⎝ β ⎠

⎛ β ⎞ VO = − g m1 RO 3 ⎜ ⎟ Vi ⎝1+ β ⎠ From Equation (10.20) RO 3 ≈ β rO 3

So Av =

Av =

VO − g m1ro3 β 2 = 1+ β Vi

gm =

0.25 = 9.615 mA/V 0.026

ro3 =

80 = 320 K 0.25

− ( 9.615 )( 320 )(120 )

2

= −366,165 121 ______________________________________________________________________________________ 10.90 Design Problem ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.91 Let (W L )1, 2 = 50 and (W L ) = 25 for all other transistors Let I REF = 80 μ A ro =

1

λI D

=

1

(0.04)(0.08)

= 312.5 k Ω

⎛ 0.06 ⎞ gm = 2 ⎜ ⎟(50 )(0.08) = 0.6928 mA/V ⎝ 2 ⎠ Aυ =

− g m2 1 1 + ro 3 ro 4 ro1 ro 2

− (0.6928) − 0.480 = 1 1 2(0.00001024 ) + (312.5)2 (312.5)2 2

=

Aυ = −23,438 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 11 11.1 (a) CMRR dB = ∞ ⇒ υ o = Ad υ d = (250)(1.5 sin ω t ) (mV) υ o = 0.375 sin ω t (V) (b) CMRR dB = 80 dB ⇒ CMRR = 10 4 =

250 ⇒ Acm = 0.025 Acm

υ o = (250 )(0.0015 sin ω t ) + (0.025)(3 sin ω t ) υ o = 0.45 sin ω t (V) (c) CMRR dB = 50 dB ⇒ CMRR = 316.2 =

250 ⇒ Acm = 0.791 Acm

υ o = (250)(0.0015 sin ω t ) + (0.791)(3 sin ω t ) υ o = 2.75 sin ω t (V)

______________________________________________________________________________________ 11.2 (a) (i) υ o1 = − g m Rυ1 = −(1)(5)(0.7 + 0.1 sin ω t ) υ o1 = −3.5 − 0.5 sin ω t (V) (ii) υ o 2 = − g m Rυ 2 = −(1)(5)(0.7 − 0.1 sin ω t ) υ o 2 = −3.5 + 0.5 sin ω t (V) (iii) υ o1 − υ o 2 = −1.0 sin ω t (V) (b) Δ(υ1 − υ 2 ) = (0.7 + 0.1 sin ω t ) − (0.7 − 0.1 sin ω t ) = 0.2 sin ω t −0.5 = −2.5 0.2 +0.5 (ii) Ad 2 = = +2.5 0.2 −1 (iii) Ad 3 = = −5 0.2 ______________________________________________________________________________________

(c) (i) Ad 1 =

11.3 (a) Neglect dc base currents I E = I C1 + I C 2 = 0.2 mA =

− 0.7 − (− 3) ⇒ R E = 11.5 k Ω RE

3 − 1.2 = 18 k Ω 0.1 (c) For VCB = 0 ⇒ VCE = 0.7 V 3 = I C (18) + 0.7 + 2 I C (11.5) − 3 So I C = 0.1293 mA υ o1 = υ cm + = 3 − (0.1293)(18) = 0.673 V υ cm − = −2.3 V So −2.3 ≤ υ cm ≤ 0.673 V ______________________________________________________________________________________

υ o1 = 1.2 V, ⇒ RC =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.4 a. I1 =

10 − 2 ( 0.7 )

IC 2 =

⇒ I1 = 1.01 mA

8.5 I1

2 1+ β (1 + β )

=

1.01 ⇒ I C 2 ≅ 1.01 mA 2 1+ (100 )(101)

⎛ 100 ⎞⎛ 1.01 ⎞ IC 4 = ⎜ ⎟⎜ ⎟ ⇒ I C 4 ≅ 0.50 mA ⎝ 101 ⎠⎝ 2 ⎠ VCE 2 = ( 0 − 0.7 ) − ( −5 ) ⇒ VCE 2 = 4.3 V VCE 4 = ⎡⎣5 − ( 0.5 )( 2 ) ⎤⎦ − ( −0.7 ) ⇒ VCE 4 = 4.7 V

b.

For VCE 4 = 2.5 V ⇒ VC 4 = −0.7 + 2.5 = 1.8 V 5 − 1.8 ⇒ I C 4 = 1.6 mA 2 ⎛ 1+ β ⎞ ⎛ 101 ⎞ IC 2 + ⎜ ⎟ ( 2 IC 4 ) = ⎜ ⎟ ( 2 )(1.6 ) ⇒ I C 2 = 3.23 mA ⎝ 100 ⎠ ⎝ β ⎠ I1 ≈ I C 2 = 3.23 mA IC 4 =

R1 =

10 − 2 ( 0.7 ) 3.23

⇒ R1 = 2.66 kΩ

______________________________________________________________________________________ 11.5 a.

Neglecting base currents 30 − 0.7 ⇒ R1 = 73.25 kΩ 0.4 VCE1 = 10 V ⇒ VC1 = 9.3 V 15 − 9.3 RC = ⇒ RC = 28.5 kΩ 0.2 I1 = I 3 = 400 μ A ⇒ R1 =

b. rπ =

(100 )( 0.026 )

= 13 kΩ 0.2 50 r0 ( Q3 ) = = 125 kΩ 0.4 We have Ad =

(100 )( 28.5 ) ⇒ Ad 2 ( rπ + RB ) 2 (13 + 10 ) β RC

=

= 62

⎧ ⎫ ⎪ ⎪ 1 ⎪ Acm = − ⎨ ⎬ rπ + RB ⎪ 2r0 (1 + β ) ⎪ 1+ rπ + RB ⎭⎪ ⎩⎪ ⎧ ⎫ (100 )( 28.5) ⎪⎪ 1 ⎪⎪ =− ⇒ Acm = −0.113 ⎨ 2 125 101 13 + 10 ⎪ ( )( ) ⎬⎪ 1+ 13 + 10 ⎭⎪ ⎩⎪ ⎛ 62 ⎞ C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 54.8 dB ⎝ 0.113 ⎠

β RC ⎪

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ c.

Rid = 2 ( rπ + RB ) = 2 (13 + 10 ) ⇒ Rid = 46 kΩ 1 ⎡ rπ + RB + 2 (1 + β ) r0 ⎤⎦ 2⎣ 1 = ⎡⎣13 + 10 + 2 (101)(125 ) ⎤⎦ ⇒ Ricm = 12.6 MΩ 2

Ricm =

______________________________________________________________________________________ 11.6

(a)

vCM ( max ) ⇒ VCB = 0 vCM ( max ) = 3 V

so that

vCM ( max ) = 5 −

IQ 2

( RC ) = 5 −

( 0.5) 2

(8)

(b) Vd ⎛ I CQ ⎞ Vd ⎛ 0.25 ⎞ ⎛ 0.018 ⎞ =⎜ =⎜ ⎟⋅ ⎟⎜ ⎟ = 0.08654 mA 2 ⎝ VT ⎠ 2 ⎝ 0.026 ⎠ ⎝ 2 ⎠ = ΔI ⋅ RC = ( 0.08654 )( 8 ) = 0.692 V

ΔI = g m ⋅ ΔVC 2

(c) ⎛ 0.25 ⎞ ⎛ 0.010 ⎞ ΔI = ⎜ ⎟⎜ ⎟ = 0.04808 mA ⎝ 0.026 ⎠ ⎝ 2 ⎠ ΔVC 2 = ( 0.04808 )( 8 ) = 0.385 V ______________________________________________________________________________________

11.7

P = ( I1 + I C 4 ) (V + − V − ) I1 ≅ I C 4 so 1.2 = 2 I1 ( 6 ) ⇒ I1 = I C 4 = 0.1 mA R1 =

3 − 0.7 − ( −3) 0.1

⇒ R1 = 53 k Ω

For vCM = +1V ⇒ VC1 = VC 2 = 1 V ⇒ RC =

3 −1 ⇒ RC = 40 k Ω 0.05

One-sided output Ad =

1 0.05 g m RC where g m = = 1.923 mA / V 2 0.026

Then Ad =

1 (1.923)( 40 ) ⇒ Ad = 38.5 2

______________________________________________________________________________________ 11.8 a. IE ( 2 ) + I E (85 ) − 5 2 5 − 0.7 IE = ⇒ I E = 0.050 mA 85 + 1 ⎛ β ⎞ ⎛ I E ⎞ ⎛ 100 ⎞⎛ 0.050 ⎞ I C1 = I C 2 = ⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ ⎝ 1 + β ⎠ ⎝ 2 ⎠ ⎝ 101 ⎠⎝ 2 ⎠

0 = 0.7 +

Or I C1 = I C 2 = 0.0248 mA VCE1 = VCE 2 = ⎡⎣5 − I C1 (100 ) ⎤⎦ − ( −0.7 ) So VCE1 = VCE 2 = 3.22 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ vcm ( max ) for VCB = 0 and VC = 5 − I C1 (100 ) = 2.52 V

b.

So vcm ( max ) = 2.52 V vcm ( min ) for Q1 and Q2

at the edge of cutoff (c) Differential-mode half circuits −

⇒ vcm ( min ) = −4.3 V

⎛V ⎞ vd = Vπ + ⎜ π + g mVπ ⎟ .RE′ 2 ⎝ rπ ⎠

⎡ (1 + β ) ⎤ = Vπ ⎢1 + RE′ ⎥ rπ ⎣ ⎦ Then Vπ =

− ( vd / 2 )

⎡ (1 + β ) ⎤ RE′ ⎥ ⎢1 + rπ ⎣ ⎦

vo = − g mVπ RC ⇒ Ad = rπ =

β VT I CQ

=

β RC 1 ⋅ 2 rπ + (1 + β ) RE′

(100 )( 0.026 )

= 105 k Ω RE′ = 2 k Ω

0.0248

Then Ad =

1 (100 )(100 ) ⋅ ⇒ Ad = 16.3 2 105 + (101)( 2 )

______________________________________________________________________________________ 11.9

a.

For v1 = v2 = 0 and neglecting base currents RE =

b. Ad = rπ = Ad =

−0.7 − ( −10 ) 0.15

⇒ RE = 62 kΩ

v02 β RC = vd 2 ( rπ + RB )

β VT I CQ

=

(100 )( 0.026 ) 0.075

(100 )( 50 ) ⇒ Ad 2 ( 34.7 + 0.5 ) ⎡ ⎢

= 34.7 kΩ

= 71.0

⎤ ⎥ ⎥ rπ + RB ⎢ 2 RE (1 + β ) ⎥ ⎢1 + ⎥ rπ + RB ⎦⎥ ⎣⎢

β RC ⎢ Acm = −

1

⎡

⎤ ⎥ 1 =− ⎢ ⎥ ⇒ Acm = −0.398 34.7 + 0.5 ⎢ 2 ( 62 )(101) ⎥ ⎢⎣1 + 34.7 + 0.5 ⎥⎦ 71.0 ⇒ C M RRdB = 45.0 dB C M RRdB = 20 log10 0.398

(100 )( 50 ) ⎢

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ c.

Rid = 2 ( rπ + RB ) Rid = 2 ( 34.7 + 0.5 ) ⇒ Rid = 70.4 kΩ

Common-mode input resistance 1 ⎡ rπ + RB + 2 (1 + β ) RE ⎤⎦ 2⎣ 1 = ⎡⎣34.7 + 0.5 + 2 (101)( 62 ) ⎤⎦ ⇒ Ricm = 6.28 MΩ 2

Ricm =

______________________________________________________________________________________ 11.10 ⎛ 81 ⎞ (a) I E1 = I E 2 = ⎜ ⎟(50 ) = 50.625 ⇒ I E = 101.25 μ A ⎝ 80 ⎠ 3 − 0.6 RE = = 23.7 k Ω 0.10125 − 1.5 − (− 3) RC = = 30 k Ω 0.05 3 − (1 + 0.6 ) (b) (i) I E = ⇒ I E = 59.07 μ A 23.7 1 ⎛ 80 ⎞ I C1 = I C 2 = ⎜ ⎟(59.07 ) = 29.17 μ A 2 ⎝ 81 ⎠

υ C1 = υ C 2 = I C RC − 3 = (0.02917 )(30 ) − 3 = −2.125 V

(ii) g m =

0.02917 = 1.122 mA/V 0.026

υ d = 12 mV,

υd

= 6 mV 2 υ C 2 = −2.125 − (1.122 )(30 )(0.006 ) = −2.327 V υ C1 = −2.125 + (1.122)(30 )(0.006 ) = −1.923 V ______________________________________________________________________________________

11.11 (a) v1 = v2 = 0 I E1 = I E 2 ≅ 6 μ A β = 60 I C1 = I C 2 = 5.90 μ A

vC1 = vC 2 = ( 5.90 )( 0.360 ) − 3 = −0.875 V VEC1 = VEC 2 = +0.6 − ( −0.875 ) = 1.475 V

(b) (i) 5.90 ⇒ 0.227 mA/V 0.026 Ad = g m RC = ( 0.227 )( 360 ) = 81.7 Acm = 0

gm =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (ii) Ad =

Acm =

( 60 )( 0.026 ) g m RC = 40.8 rπ = 2 0.0059 = 264 K − ( 0.227 )( 360 ) = −0.0442 2 ( 61)( 4000 ) 1+ 264

______________________________________________________________________________________ 11.12

I Q = 0.2 mA, ⇒ I C1 = I C 2 =

1 ⎛ 80 ⎞ ⎜ ⎟(0.2) = 0.09877 mA 2 ⎝ 81 ⎠

(a) For VCB = 0 , υ C1 = υ C 2 = υ CM = −2.5 = I C RC − 5 5 − 2.5 = 25.3 k Ω 0.09877 0.09877 (b) Ad = g m RC , g m = = 3.799 mA/V 0.026 Ad = (3.799 )(25.3) = 96.1

So RC =

(c) υ d = 14 mV,

υd 2

= 7 mV

υd

= −2.5 − (3.799 )(25.3)(0.007 ) = −3.173 V 2 υ C 2 = −2.5 + (3.799)(25.3)(0.007 ) = −1.827 V (d) CMRR dB = 60 dB ⇒ CMRR = 1000

υ C1 = −2.5 − g m RC ⋅

1 ⎡ (81)(0.2 )R o ⎤ ⎢1 + ⎥ ⇒ R o = 257 k Ω 2 ⎣ (0.026)(80 ) ⎦ ______________________________________________________________________________________ 1000 =

11.13 (a) Neglect dc base currents I E = I C1 + I C 2 = 240 μ A

− 0.7 − (− 5) = 17.9 k Ω 0.24 5−3 RC = = 16.7 k Ω 0.12 0.12 (b) g m = = 4.615 mA/V 0.026 Ad = g m RC = (4.615)(16.667 ) = 76.9 ΔRC 0.5 (c) Acm ≅ = = 0.0279 17.9 RE RE =

⎛ 76.9 ⎞ CMRRdB = 20 log 10 ⎜ ⎟ = 68.8 dB ⎝ 0.0279 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.14 v1 = v2 = 0 IE =

−0.7 − ( −10 )

75 = 0.124 mA

I C1 = I C 2 = 0.0615 mA gm =

0.0615 = 2.365 mA/V 0.026

Δg m = 0.01 gm g m1 = 2.377 mA/V g m 2 = 2.353 mA/V rπ =

(120 )( 0.026 )

0.0615 V ΔI = g m d 2 V ΔvC1 = − g m1 d Rc 2 Vd ΔvC 2 = + g m 2 Rc 2

= 50.7 K

vo = ΔvC1 − ΔvC 2 = − g m1

Vd V RC − g m 2 d RC 2 2

Vd RC ( g m1 + g m 2 ) 2 R −50 Ad = − C ( g m1 + g m 2 ) = ( 2.377 + 2.353) ⇒ Ad = −118.25 2 2 Common-Mode − g m1 RC vcm − g m 2 RC vcm ΔvC1 = ΔvC 2 = ⎛ 1+ β ⎞ ⎛ 1+ β ⎞ 1+ ⎜ 1+ ⎜ ⎟ ( 2 RE ) ⎟ ( 2 RE ) ⎝ rπ ⎠ ⎝ rπ ⎠ =−

− ( g m1 − g m 2 ) RC − ( 2.377 − 2.353) ( 50 ) vo = Acm = = vcm ⎛ 1+ β ⎞ ⎛ 121 ⎞ 1+ ⎜ 1+ ⎜ ⎟ ( 2 )( 75 ) ⎟ ( 2 RE ) ⎝ 50.7 ⎠ ⎝ rπ ⎠ −1.2 = ⇒ Acm = −0.003343 358.99 C M R R ∫ = 91 dB dB

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.15 (a) v1 = v2 = 0 vE = +0.7 V 5 − 0.7 IE = = 4.3 mA 1 I C1 = I C 2 = 2.132 mA

vC1 = vC 2 = ( 2.132 )(1) − 5 = −2.87 V v1 = 0.5, v2 = 0 Q2 on Q1 off

(b)

⎛ 120 ⎞ I C1 = 0, I C 2 = 4.3 ⎜ ⎟ mA = 4.264 mA ⎝ 121 ⎠ vC1 = −5 V vC 2 = ( 4.264 ) (1) − 5 vC 2 = −0.736 V

(c)

vE ≈ 0.7 V ΔI = g m

vd 2

gm =

2.132 = 82.0 mA/V 0.026

ΔvC = ΔI ⋅ RC = g m

Vd = 0.015 ⇒ Δvc = 0.615 V

( 82.0 ) Vd ⋅ RC = ⋅ Vd (1) = 41.0Vd 2 2

vC 2 ↓ vC1 ↑

vC1 = −2.87 + 0.615 = −2.255 V vC 2 = −2.87 − 0.615 = −3.485 V ______________________________________________________________________________________ 11.16 (a) gm =

IC 1 = = 38.46 mA/V VT 0.026

Ad =

vo 1 = = 100 vd 0.01

Ad = g m RC 100 = 38.46 RC Rc = 2.6 K

(b)

With v1 = v2 = 0

vC1 = vC 2 = 10 − (1)( 2.6 ) = 7.4 V ⇒ vcm ( max ) = 7.4 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.17 (a) I O = I E1 + I E 2 ⇒ I E1 = I E 2 = 0.4 mA

[

] [

]

(i) υ O1 − υ O 2 = V + − I E1 RC1 − V + − I E 2 RC 2 = I E 2 RC 2 − I E1 RC1 υ O1 − υ O 2 = 0 (ii) υ O1 − υ O 2 = (0.4)(7.6 − 7.4) = 0.08 V (b) V BE1 = V BE 2 ⎛V 0.8 = 2.9 × 10 −15 + 3.1× 10 −15 ⋅ exp⎜⎜ BE ⎝ VT

(

)

⎞ ⎛V ⎟⎟ ⇒ exp⎜⎜ BE ⎠ ⎝ VT

( )( ) = (3.1× 10 )(1.333 × 10 ) = 0.4133 mA

⎞ ⎟⎟ = 1.333 × 1014 ⎠

Then I E1 = 2.9 × 10 −15 1.333 × 1014 = 0.3867 mA I E2

−15

14

(i) υ O1 − υ O 2 = (0.4133 − 0.3867 )(7.5) = 0.1995 V (ii) υ O1 − υ O 2 = (0.4133)(7.6) − (0.3867 )(7.4) = 0.2795 V ______________________________________________________________________________________ 11.18

(a)

⎛ −υ d ⎞ 1 ⎟=4 ⇒ exp⎜⎜ VT ⎟⎠ ⎛ −υ d ⎞ ⎝ ⎟⎟ 1 + exp⎜⎜ ⎝ VT ⎠ − υ d = (0.026 ) ln (4 ) ⇒ υ d = −36.0 mV

i C1 = 0.20 = IQ

so that

⎛ +υd ⎞ 1 ⎟ = 0.1111 ⇒ exp⎜⎜ VT ⎟⎠ ⎛ +υd ⎞ ⎝ ⎟⎟ 1 + exp⎜⎜ ⎝ VT ⎠ so that υ d = (0.026 ) ln (0.1111) = −57.1 mV ______________________________________________________________________________________

(b)

iC 2 = 0.90 = IQ

11.19 ⎡ ⎤ IQ ⎛ IQ ⎞ ⎟ ⋅υ d (max )⎥ − ⎢0.5 I Q + ⎜⎜ ⎟ ⎝ 4VT ⎠ ⎣⎢ ⎦⎥ 1 + exp(− υ d (max ) VT ) = 0.005 (a) ⎡ ⎤ ⎛ IQ ⎞ ⎟ ⋅υ d (max )⎥ ⎢0.5 I Q + ⎜⎜ ⎟ ⎢⎣ ⎥⎦ ⎝ 4VT ⎠ ⎡

⎛ 1 ⎝ 4VT

⎤ ⎞ 1 ⎟⎟ ⋅υ d (max )⎥ = ⎢⎣ ⎥⎦ 1 + exp(− υ d (max ) VT ) ⎠ By trial and error, υ d (max ) ≅ 14 mV ⎡

⎛ 1 ⎝ 4VT

(0.995)⎢0.5 + ⎜⎜

⎤ ⎞ 1 ⎟⎟ ⋅υ d (max )⎥ = ( 1 + exp − υ d (max ) VT ) ⎠ ⎣⎢ ⎦⎥ By trial and error, υ d (max ) ≅ 21.2 mV ______________________________________________________________________________________

(b)

(0.985)⎢0.5 + ⎜⎜

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.20

For vCM = 3.5 V and a maximum peak-to-peak swing in the output voltage of 2 V, we need the quiescent collector voltage to be VC = 3.5 + 1 = 4.5 V I = 0.5 mA. Assume the bias is ±10 V , and Q

Then I C = 0.25 mA Now

10 − 4.5 ⇒ RC = 22 k Ω 0.25

RC =

In this case, Then Ad =

rπ =

(100 )( 0.026 ) 0.25

(100 )( 22 )

2 (10.4 + 0.5 )

= 10.4 k Ω

= 101

So gain specification is met.

For CMRRdB = 80 dB ⇒ CMRR = 104 =

1 ⎡ (1 + β ) I Q Ro ⎤ 1 ⎡ (101)( 0.5 ) Ro ⎤ ⎥ ⇒ Ro = 1.03 M Ω ⎢1 + ⎥ = ⎢1 + 2⎣ VT β ⎦ 2 ⎢⎣ ( 0.026 )(100 ) ⎥⎦

Need to use a Modified Widlar current source. R o = ro 1+ g m (R E1 rπ )

[

]

If V A = 100 V, then ro = rπ =

100 = 200 k Ω 0.5

(100)(0.026) = 5.2 k Ω

0.5 0.5 gm = = 19.23 mA/V 0.026 Then 1030 = 200 1 + (19.23)(R E1 rπ ) ⇒ R E1 rπ = 0.216 k Ω ⇒ R E1 5.2 = 0.216

[

]

So, R E1 = 225 Ω ; also I REF ≅ 0.5 mA ______________________________________________________________________________________ 11.21

(a)

RE =

−0.7 − ( −10 ) 0.25

⇒ RE = 37.2 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) ⎛1+ β ⎞ Vπ 1 V V Ve + g mVπ 1 + π 2 + g mVπ 2 = e or (1) ⎜ ⎟ (Vπ 1 + Vπ 2 ) = rπ rπ RE r R E ⎝ π ⎠ ⎛ r ⎞ Vπ 1 V1 − Ve = ⇒ Vπ 1 = ⎜ π ⎟ (V1 − Ve ) rπ RB + rπ ⎝ rπ + RB ⎠ Vπ 2 = V2 − Ve Then ⎛ 1+ β ⎝ rπ

⎞ ⎡ rπ ⎤ V (V1 − Ve ) + (V2 − Ve )⎥ = e ⎟⎢ ⎠ ⎣ rπ + RB ⎦ RE From this, we find

(1) ⎜

V1 + Ve =

rπ + RB ⋅ V2 rπ

⎡ rπ + RB r + RB ⎤ +1+ π ⎢ ⎥ rπ ⎦ ⎣ RE (1 + β )

Now Vo = − g mVπ 2 RC = − g m RC (V2 − Ve ) We have rπ =

(120 )( 0.026 ) 0.125

≅ 25 k Ω,

gm =

0.125 = 4.81 mA / V 0.026

(i) Set V1 =

Vd V and V2 = − d 2 2

Then ⎛ ⎛ 25 + 0.5 ⎞ ⎞ Vd ⎜1 − ⎜ 25 ⎟ ⎟ ( −0.02 ) ⎝ ⎠⎠ ⎝ = 2 Ve = 2.026 ⎡ 25 + 0.5 25 + 0.5 ⎤ +1+ ⎢ ⎥ 25 ⎦ ⎣ ( 37.2 )(121) Vd 2

So Ve = −0.00494Vd Now V ⎛ V ⎞ Vo = − ( 4.81)( 50 ) ⎜ − d − ( −0.00494 ) Vd ⎟ ⇒ Ad = o = 119 V 2 ⎝ ⎠ d

(ii) Set V1 = V2 = Vcm Then ⎛ 25 + 0.5 ⎞ Vcm ⎜ 1 + ⎟ V ( −2.02 ) 25 ⎠ ⎝ = cm Ve = 2.02567 ⎡ 25 + 0.5 ⎤ 25 + 0.5 +1+ ⎢ ⎥ 25 ⎦ ⎣ ( 37.2 )(121) Ve = Vcm ( 0.9972 ) Then

Vo = − ( 4.81)( 50 ) ⎡⎣Vcm − Vcm ( 0.9972 ) ⎤⎦ or Acm =

Vo = −0.673 Vcm

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.22 (a) Ad = g m RC 1.2 = 75 0.016 0.125 gm = = 4.808 mA/V 0.026 75 Then RC = = 15.6 k Ω 4.808 (b) For VCB = 0 ⇒ υ C1 = υ C 2 = υ CM = 3 − (0.125)(15.6 ) = 1.05 V Ad =

1 ⎡ (0.25)(4000 ) ⎤ = 19,231 ⇒ CMRR dB = 85.7 dB ⎢1 + (0.026) ⎥⎦ 2⎣ ______________________________________________________________________________________

(c) CMRR =

11.23 The small-signal equivalent circuit is

A KVL equation: v1 = Vπ 1 − Vπ 2 + v2 v1 − v2 = Vπ 1 − Vπ 2

A KCL equation Vπ 1 V + g mVπ 1 + π 2 + g mVπ 2 = 0 rπ rπ ⎛1 ⎞ + g m ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 r ⎝ π ⎠

(Vπ 1 + Vπ 2 ) ⎜

Then v1 − v2 = 2Vπ 1 ⇒ Vπ 1 =

1 1 ( v1 − v2 ) and Vπ 2 = − ( v1 − v2 ) 2 2

At the v01 node: v01 v01 − v02 + + g mVπ 1 = 0 RC RL

⎛ 1 ⎛ 1 ⎞ 1 1 ⎞ v01 ⎜ + ⎟ − v02 ⎜ ⎟ = g m ( v2 − v1 ) R R L ⎠ ⎝ RL ⎠ 2 ⎝ C (1) v02

At the

node:

v02 v02 − v01 + + g mVπ 2 = 0 RC RL

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ 1 ⎛ 1 ⎞ 1 1 ⎞ v02 ⎜ + ⎟ − v01 ⎜ ⎟ = g m ( v1 − v2 ) R R L ⎠ ⎝ RL ⎠ 2 ⎝ C (2)

From (1): ⎛ R ⎞ 1 v02 = v01 ⎜1 + L ⎟ − g m RL ( v2 − v1 ) ⎝ RC ⎠ 2

Substituting into (2) ⎛ R ⎞⎛ 1 ⎛ 1 ⎛ 1 ⎞ 1 1 ⎞ 1 1 ⎞ v01 ⎜1 + L ⎟ ⎜ + + ⎟ − g m RL ( v2 − v1/ ) ⎜ ⎟ − v01 ⎜ ⎟ = g m ( v1 − v2 ) ⎝ RL ⎠ 2 ⎝ RC ⎠ ⎝ RC RL ⎠ 2 ⎝ RC RL ⎠ ⎡ ⎛ RL ⎛ 1 RL ⎞⎤ 1 ⎞ 1 v01 ⎜ + 2+ + 1⎟ ⎥ ⎟ = g m ( v1 − v2 ) ⎢1 − ⎜ R R R 2 R C ⎠ C ⎠⎦ ⎝ C ⎣ ⎝ C v01 ⎛ RL ⎞ 1 ⎛ RL ⎞ ⎜2+ ⎟ = − gm ⎜ ⎟ ( v1 − v2 ) 2 ⎝ RC ⎠ RC ⎝ RC ⎠ For v1 − v2 = vd 1 − g m RL v01 Av1 = = 2 vd ⎛ RL ⎞ ⎜2+ ⎟ R C ⎠ ⎝

From symmetry: Av =

1 g m RL v02 Av 2 = = 2 vd ⎛ RL ⎞ ⎜2+ ⎟ R C ⎠ ⎝

v02 − v01 g m RL = vd ⎛ RL ⎞ ⎜2+ ⎟ RC ⎠ ⎝

Then ______________________________________________________________________________________ 11.24 The small-signal equivalent circuit is

KVL equation: v1 = Vπ 1 − Vπ 2 + v2 or v1 − v2 = Vπ 1 − Vπ 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ KCL equation: V Vπ 1 + g m Vπ 1 + g m Vπ 2 + π 2 = 0 rπ rπ ⎞ ⎛ 1 + g m ⎟⎟ = 0 ⇒ Vπ 1 = −Vπ 2 r ⎠ ⎝ π

(Vπ 1 + Vπ 2 )⎜⎜

Then υ1 − υ 2 = −2Vπ 2 , or Vπ 2 = − Now υ o = − g mVπ 2 (RC R L )

1 (υ1 − υ 2 ) 2

1 g m (RC RL )(υ1 − υ 2 ) 2 υ 1 For υ1 − υ 2 ≡ υ d ⇒ Ad = o = g m (RC R L ) υd 2 ______________________________________________________________________________________ =

11.25

We have VC 2 = − g mVπ 2 RC = − g m (Vb 2 − Ve ) RC

and VC1 = − g mVπ 1 RC = − g m (Vb1 − Ve ) RC Then V0 = VC 2 − VC1 = − g m (Vb 2 − Ve ) RC − ⎣⎡ − g m (Vb1 − Ve ) RC ⎦⎤ = g m RC (Vb1 − Vb 2 )

Ad =

Differential gain

V0 = g m RC Vb1 − Vb 2

A =0

Common-mode gain cm ______________________________________________________________________________________ 11.26 (a) vcm = 3 V ⇒ VC1 = VC 2 = 3 V 10 − 3 Then RC = ⇒ RC = 70 k Ω 0.1

(b)

CMRRdB = 75 dB ⇒ CMRR = 5623 Now CMRR = 5623 =

1 ⎡ (1 + β ) I Q Ro ⎤ ⎢1 + ⎥ β VT 2⎣ ⎦

1 ⎡ (151)( 0.2 ) Ro ⎤ ⎢1 + ⎥ ⇒ Ro = 1.45 M Ω 2 ⎢⎣ (150 )( 0.026 ) ⎥⎦

Use a Widlar current source. Ro = ro [1 + g m RE′ ]

Let VA of current source transistor be 100 V.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 100 0.2 = 500 k Ω , g m = = 7.69 mA/V 0.2 0.026 (150)(0.026) = 19.5 k Ω rπ = 0.2 So 1450 = 500[1 + (7.69)R E′ ] ⇒ R E′ = 0.247 k Ω Now ro =

Now R E′ = R E rπ ⇒ 0.247 = R E 19.5 ⇒ R E = 250 Ω ⎛I ⎞ We have I Q R E = VT ln⎜ REF ⎟ ⎜ IQ ⎟ ⎝ ⎠ I (0.2)(0.250) = (0.026) ln ⎡⎢ REF ⎤⎥ ⇒ I REF = 1.37 mA ⎣ (0.2) ⎦

10 − 0.7 − (− 10 ) ⇒ R1 = 14.1 k Ω 1.37 ______________________________________________________________________________________

Then R1 =

11.27

⎛ R A ⎞ + ⎡ R(1 + δ ) ⎤ + ⎛ 1 + δ ⎞ + ⎟⎟ ⋅ V = ⎢ υ A = ⎜⎜ ⎟ ⋅V ⎥ ⋅V = ⎜ ⎝ 2+δ ⎠ ⎣ R(1 + δ ) + R ⎦ ⎝ RA + R ⎠ 1 2

υ B = V + = 2.5 V ⎛ 1.01 ⎞ For δ = +0.01 , υ A = ⎜ ⎟(5) = 2.5124 V ⎝ 2.01 ⎠

⎛ 0.99 ⎞ For δ = −0.01 , υ A = ⎜ ⎟(5) = 2.4874 V ⎝ 1.99 ⎠ υ d ≅ 12.5 mV R B ≅ R R = 20 k Ω

rπ =

β VT I CQ

(120)(0.026) = 31.2 k Ω 0.1

(120)(15) = 17.58 2(rπ + R B ) 2(31.2 + 20 ) = Ad ⋅υ d = (17.58)(0.0125) = 0.22 V

Ad =

υo

=

β RC

=

−0.22 ≤ υ O 2 ≤ +0.22 V ______________________________________________________________________________________

11.28

(a) Rid = 2rπ =

2 β VT 2(120 )(0.026 ) = = 49.9 k Ω I CQ 0.125

1 [rπ + (1 + β )(2 Ro )] = 1 ⎡⎢ (120)(0.026) + (121)(2)(8000)⎤⎥ 2 ⎣ 0.125 2 ⎦ Ricm = 968 M Ω ______________________________________________________________________________________

(b) Ricm =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.29 (a) I1 = R2 =

(b)

10 − 0.7 − ( −10 ) R1

= 0.5 ⇒ R1 = 38.6 K

0.026 ⎛ 0.5 ⎞ ln ⎜ ⎟ ⇒ R2 = 236 Ω 0.14 ⎝ 0.14 ⎠

Ricm ≈ (1 + β ) Ro 0.14 = 5.385 mA/V 0.026 (180 )( 0.026 ) = 33.4 K rπ 4 = 0.14 RE′ = 33.4 0.236 = 0.234 K

Ro = ro 4 (1 + g m 4 RE′ ) g m 4 =

100 = 714 K 0.14 Ro = 714 ⎡⎣1 + ( 5.385 )( 0.234 ) ⎤⎦ ro 4 =

Ricm

= 1614 K = (181)(1614 ) ≈ 292 MΩ

(c) Acm =

− g m1 RC 2 (1 + β ) Ro 1+ rπ 1

g m1 =

rπ 1 =

0.07 = 2.692 mA/V 0.026

(180 )( 0.026 ) 0.07

− ( 2.692 )( 40 ) 2 (181)(1614 ) 1+ 66.86 Acm = −0.0123

= 66.86 K

Acm =

______________________________________________________________________________________ 11.30

Ad 1 = g m1 ( R1 rπ 3 ) g m1 = rπ 3 = Ad 2 =

I Q1 / 2 VT

β VT IQ 2 / 2

= 19.23I Q1 =

2 (100 )( 0.026 ) IQ 2

=

5.2 IQ 2

IQ2 / 2 g m 3 R2 , g m3 = = 19.23I Q 2 2 VT

Then 30 =

Maximum

(19.23) IQ 2

⋅ R2 ⇒ I Q 2 R2 = 3.12 V 2 vo 2 − vo1 = ±18 mV

for linearity

vo 3 ( max ) = ( ±18 )( 30 ) mV ⇒ ±0.54 V

so I Q 2 R2 = 3.12 V is OK.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ From Ad 1 : ⎡ ⎛ 5.2 ⎞ ⎤ ⎟ ⎥ ⎢ R1 ⎜ ⎜ I Q2 ⎟ ⎥ ⎢ ⎠ ⎝ 20 = 19.23I Q1 (R1 rπ 3 ) = 19.23I Q1 ⎢ ⎥ ⎞⎥ ⎛ 5 . 2 ⎢ ⎟ ⎜ ⎢ R1 + ⎜ I ⎟ ⎥ ⎝ Q2 ⎠ ⎦ ⎣ 19.23I Q1 R1 (5.2 ) 20 = I Q 2 R1 + 5.2

Let

I Q1 2

⋅ R1 = 5 V, ⇒ I Q1 R1 = 10 V

Then 20 =

19.23(10)(5.2) ⇒ I Q 2 R1 = 44.8 V I Q 2 R1 + 5.2

Now I Q1 R1 = 10 ⇒ R1 =

10 I Q1

⎛ 10 ⎞ I ⎟ = 44.8 ⇒ Q 2 = 4.48 So I Q 2 ⎜ ⎜ I Q1 ⎟ I Q1 ⎝ ⎠ Let I Q1 = 100 μ A, then I Q 2 = 448 μ A

Then I Q 2 R 2 = 3.12 ⇒ R 2 = 6.96 k Ω I Q1 R1 = 10 ⇒ R1 = 100 k Ω

______________________________________________________________________________________ 11.31 a. 20 − VGS 3 2 = 0.25 (VGS 3 − 2 ) 50 20 − VGS 3 = 12.5 (VGS2 3 − 4VGS 3 + 4 ) I1 =

12.5VGS2 3 − 49VGS 3 + 30 = 0 VGS 3 =

49 ±

( 49 )

2

− 4 (12.5 )( 30 )

2 (12.5 )

⇒ VGS 3 = 3.16 V

20 − 3.16 ⇒ I1 = I Q = 0.337 mA 50 IQ = ⇒ I D1 = 0.168 mA 2

I1 = I D1

0.168 = 0.25 (VGS 1 − 2 ) ⇒ VGS 1 = 2.82 V VDS 4 = −2.82 − ( −10 ) ⇒ VDS 4 = 7.18 V 2

VD1 = 10 − ( 0.168 )( 24 ) = 5.97 V VDS1 = 5.97 − ( −2.82 ) ⇒ VDS 1 = 8.79 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

(c)

Max vCM ⇒ VDS 1 = VDS 2 = VDS ( sat ) = VGS 1 − VTN 2.82 − 2 = 0.82 V Now VD1 = 10 − ( 0.168 )( 24 ) = 5.97 V

VS ( max ) = 5.97 − VDS1 ( sat ) = 5.97 − 0.82 VS ( max ) = 5.15 V

vCM ( max ) = VS ( max ) + VGS 1 = 5.15 + 2.82 vCM ( max ) = 7.97 V vCM ( min ) = V − + VDS 4 ( sat ) + VGS 1

VDS 4 ( sat ) = VGS 4 − VTN = 3.16 − 2 = 1.16 V Then vCM ( min ) = −10 + 1.16 + 2.82 ⇒ vCM ( min ) = −6.02 V

______________________________________________________________________________________ 11.32 2 (a) I D1 = K n1 (VGS1 − VTN )

60 = 100(VGS1 − 0.3) ⇒ VGS1 = 1.075 V 2

V D1 = V DS1 − VGS1 + V1 = 4 − 1.075 − 1.15 = 1.775 V 3 − 1.775 = 20.4 k Ω 0.06 (i) I Q = I 1 = I D1 + I D 2 = 120 μ A RD =

I 1 = K 3 (VGS 3 − VTN )

2

120 = 200(VGS 3 − 0.3) ⇒ VGS 3 = 1.075 V 2

3 − 1.075 − (− 3) = 41 k Ω 0.12 (iii) VGS1 = VGS 4 = 1.075 V

(ii) R1 =

(b) ro =

1 1 = = 833.3 k Ω λ 4 I Q (0.01)(0.12)

ΔV DS 4 2.3 = ⇒ ΔI Q = 2.76 μ A ro 833.3 ______________________________________________________________________________________ ΔI Q =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.33 (a) I Q = 160 μ A k′ ⎛W ⎞ 2 I D = n ⎜ ⎟ (VGS − VTN ) 2⎝L⎠ 80 2 80 = ( 4 )(VOS − 0.5 ) 2 80 = 160 (Vo5 − 0.5 )

2

80 + 0.5 = 1.207 V 160 5−2 RD = = 37.5 K VDS = 2 − ( −1.207 ) = 3.21 V 0.08

VGS =

(c)

VDS ( sat ) = VGS − VTN = 1.207 − 0.5 = 0.707 V

Then VS = VO 2 − VDS ( sat ) = 2 − 0.707 = +1.29 V And v1 = v2 = vcm = VGS + VS = 1.207 + 1.29 vcm = 2.50 V

(b)

______________________________________________________________________________________ 11.34 (a) υ CM (max ) = υ D 2 − υ DS 2 (sat ) + υ GS 2 ⎛ k ′ ⎞⎛ W ⎞ 2 i D 2 = ⎜⎜ n ⎟⎟⎜ ⎟(υ GS 2 − VTN ) ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.1 ⎞ 2 0.09 = ⎜ ⎟(4 )(υ GS 2 − 0.35) ⇒ υ GS 2 = 1.021 V 2 ⎠ ⎝

υ DS 2 (sat ) = 1.0208 − 0.35 = 0.6708 V Then 2.25 = υ D 2 − 0.6708 + 1.021 ⇒ υ D 2 = 1.90 V 3 − 1.90 = 12.2 k Ω 0.09 (b) (i) υ D 2 = 1.90 V RD =

⎛ k′ (ii) g m = 2 ⎜⎜ n ⎝ 2

⎞⎛ W ⎞ ⎛ 0.1 ⎞ ⎟⎟⎜ ⎟ I DQ = 2 ⎜ ⎟(4 )(0.09 ) = 0.2683 mA/V L ⎝ 2 ⎠ ⎠⎝ ⎠

υ D 2 = 1.90 + g m ⋅

υd 2

⎛ 0.12 ⎞ ⋅ R D = 1.90 + (0.2683)⎜ ⎟(12.2) = 2.096 V ⎝ 2 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

υd

⎛ 0.05 ⎞ ⋅ R D = 1.90 − (0.2683)⎜ ⎟(12.2) = 1.818 V 2 ⎝ 2 ⎠ ______________________________________________________________________________________ (iii) υ D 2 = 1.90 − g m ⋅

11.35 2 (a) I Q = I 1 = K n3 (VGS 3 − VTN ) 0.25 = 0.20(VGS 3 − 0.4) ⇒ VGS 3 = 1.518 V 2

5 − 1.518 − (− 5) = 33.9 k Ω 0.25 0.25 (b) I D1 = I D 2 = = 0.125 mA 2 υ O1 − υ O 2 = V + − I D1 R D1 − V + − I D 2 R D 2 = I D 2 R D 2 − I D1 R D1 (i) υ O1 − υ O 2 = 0 (ii ) υ O1 − υ O 2 = (0.125)(15.5 − 14.5) = 0.125 V R1 =

(

) (

)

(c) I Q = I D1 + I D 2 = (K n1 + K n 2 )(VGS − VTN )

2

0.25 = (0.125 + 0.115)(VGS − VTN ) ⇒ (VGS − VTN ) = 1.04166 I D1 = (0.125)(1.04166) = 0.1302 mA I D 2 = (0.115)(1.04166) = 0.1198 mA (i) υ O1 − υ O 2 = (15)(0.1198 − 0.1302 ) = −0.156 V (ii) υ O1 − υ O 2 = (0.1198)(15.5) − (0.1302 )(14.5) = −0.031 V ______________________________________________________________________________________ 2

2

11.36

(a)

⎛ K Kn i D1 1 = + ⋅υ d 1 − ⎜ n ⎜ 2I Q IQ 2 2I Q ⎝ 0.20 = 0.50 +

⎞ 2 ⎟ ⋅υ d ⎟ ⎠

⎛ 0.20 ⎞ 2 0.20 ⎟⎟ ⋅υ d ⋅υ d 1 − ⎜⎜ 2(0.15) ⎝ 2(0.15) ⎠

− 0.3674 = υ d 1 − (0.6667 )υ d2

[

0.135 = υ d2 1 − (0.6667 )υ d2

]

0.6667υ − υ + 0.135 = 0 ⇒ υ d2 = 0.15 ⇒ υ d = −0.3873 V 4 d

2 d

(b) 0.80 = 0.50 −

⎛ 0.20 ⎞ 2 0.20 ⎟⎟υ d ⋅υ d 1 − ⎜⎜ 2(0.15) ⎝ 2(0.15) ⎠

0.3674 = −υ d 1 − (0.6667 )υ d2 ⇒ υ d = −0.3873 V

(c) υ d = υ d , max =

IQ Kn

=

0.15 = +0.866 V 0.20

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.37 i D1 IQ

Lin

= 0.5 +

g f υd

, Now g f =

IQ

Kn IQ 2

=

(0.2)(0.10) = 0.10 mA/V 2

(0.1) ⋅υ = 0.5 + υ d (0.1) d (0.2) = 1 = 2(0.1)

= 0.5 +

We have

Kn 2I Q

0.5 + υ d − ⎡0.5 + υ d 1 − υ d2 ⎤ ⎥⎦ ⎢⎣ = 0.005 (a) 0.5 + υ d

0.0025 = υ d ⎡0.995 − 1 − υ d2 ⎤ , ⇒ υ d ≅ 0.19 V ⎢⎣ ⎥⎦ 0.5 + υ d − ⎡0.5 + υ d 1 − υ d2 ⎤ ⎢⎣ ⎥⎦ = 0.015 (b) 0.5 + υ d 0.0075 = υ d ⎡0.985 − 1 − υ d2 ⎤ , ⇒ υ d ≅ 0.285 V ⎢⎣ ⎥⎦ ______________________________________________________________________________________

11.38 (b) gm = 2 K p I D = 2

( 0.05)( 0.008696 )

= 0.0417 mA/V Vd = ( 0.0417 )( 0.05 ) = 0.002085 mA 2 ΔvD = ( 0.002085 )( 510 ) = 1.063 vD 2 ↑⇒ vD 2 = 1.063 − 4.565 = −3.502 V ΔI = g m

vD1 = −1.063 − 4.565 = −5.628 V 9 = I S RS + VSG + 1 IS = 2I D 8 = 2 K P RS (VSG + VTP ) + VSG 2

8 = ( 2 )( 0.05 )( 390 )(VSG − 0.8 ) + VSG 2

2 8 = 39 (VSG − 1.6VSG + 0.64 ) + VSG 2 39VSG − 61.4VSG + 16.96 = 0

VSG =

61.4 ± 3769.96 − 4 ( 39 )(16.96 ) 2 ( 39 )

= 1.217 V VS = 2.217 I S = 0.01739 mA I D1 = I D 2 ⇒ 8.696 μ A

vD1 = vD 2 = ( 8.696 )( 0.510 ) − 9 = −4.565 V

(b) g m = 2 K P I DQ = 2

( 0.05)( 0.008696 ) = 0.0417 mA/V

Vd = ( 0.0417 )( 0.05 ) = 0.002085 mA 2 ΔvD = ( 0.002085 )( 510 ) = 1.063 V ΔvD = ΔI D ⋅ RD

ΔI D = g m ⋅

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ v1 ↑, I D1 ↓, vD1 ↓ vD1 = −4.565 − 1.063 = −5.628 V vD 2 = −4.565 + 1.063 = −3.502 V

______________________________________________________________________________________ 11.39 (a) v1 = v2 = 0 I D = K n (VSG + VTP )

2

ID = 6 μA 6 + 0.4 = VSG 30 VSG = 0.847 V VS = +0.847 V vD = I D RD − 3

= ( 6 )( 0.36 ) − 3 = −0.84 V

VSD = VS − vD = 0.847 − ( −0.84 ) vSD = 1.69 V

(b) (i) Ad = g m RD

gm = 2 Kn I D =2

( 30 )( 6 ) = 26.83 μ A/V

Ad = ( 26.83)( 0.36 ) ⇒ Ad = 9.66 Acm = 0

(ii)

g m RD ( 26.83)( 0.36 ) = ⇒ Ad = 4.83 2 2 − ( 26.83)( 0.36 ) − g m RD Acm = = = −0.0448 1 + 2 g m RO 1 + 2 ( 26.83)( 4 ) Ad =

______________________________________________________________________________________ 11.40

(a) g m = 2 K p I DQ = 2 (0.12 )(0.075) = 0.1897 mA/V

(i) υ D1 − υ D 2 = g m R Dυ d = (0.1897 )(30 )(0.1) = 0.5692 V (ii) υ D1 − υ D 2 = g m R Dυ d = (0.1897 )(30 )(0.2) = 1.138 V

(b) (i) Δυ D 2 = g m R D ⋅ (ii) Acm =

υd 2

= (0.1897 )(30 )(0.1) = 0.5692 V

− g m RD − (0.1897 )(30 ) = = −0.003748 1 + 2 g m R o 1 + 2(0.1897 )(4000 )

g m R D (0.1897 )(30 ) = = 2.846 2 2 Δυ D 2 = Ad υ d + Acmυ cm , let υ cm = 1 V, υ d = 0.2 V

Ad =

Then Δυ D 2 = (2.846 )(0.2) − (0.003748)(1) = 0.5655 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.41 For v1 = v2 = 0 0 = VGS + 2 I D RS − 10 10 = VGS + 2 K n RS (VGS − VTN )

2

= VGS + 2 ( 0.15 )( 75 )(VGS − 1)

2

22.5VGS2 − 44VGS + 12.5 = 0 So VGS = 1.61 V and I D = ( 0.15 )(1.61 − 1) ⇒ 55.9 μ A 2

gm = 2 Kn I D = 2

( 0.15 )( 0.0559 )

g m = 0.1831 mA/V

Use Half-circuits – Differential gain ΔR ⎞ ⎛V ⎞⎛ vD1 = − g m ⎜ d ⎟ ⎜ RD + ⎟ 2 ⎠ ⎝ 2 ⎠⎝ ΔR ⎞ ⎛V ⎞⎛ vo 2 = g m ⎜ d ⎟ ⎜ RD − ⎟ 2 ⎠ ⎝ 2 ⎠⎝ vo = vD1 − vD 2 = − g mVd RD v Ad = o = − g m RD Vd

Now – Common-Mode Gain Vi = Vgs + g mVgs ( 2 RS ) = Vcm Vcm Vgs = 1 + g m ( 2 RS )

vD1

vD 2

ΔR ⎞ ⎛ − g m ⎜ RD + D ⎟ Vcm 2 ⎠ ⎝ = 1 + g m ( 2 RS ) ΔR ⎞ ⎛ − gm ⎜ RD − D ⎟ Vcm 2 ⎠ ⎝ = 1 + g m ( 2 RS )

vO = vD1 − vD 2 So vo = Acm =

− g m ( ΔRD )Vcm 1 + g m ( 2 RD )

− g m ( ΔRD ) vo = Vcm 1 + g m ( 2 RS )

Then Ad = − ( 0.1831)( 50 ) = −9.16 Acm =

− ( 0.1831)( 0.5 )

1 + ( 0.1831)( 2 )( 75 )

= −0.003216

C M R R ∫ = 69.1 dB bB

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.42 From 11.41 I D1 = I D 2 = 55.9 μ A g m = 0.183 mA/V

⎛ +V ⎞ ΔvD 2 = + g m 2 ⎜ d ⎟ RD ⎝ 2 ⎠ Vd Vd vO = ΔvD1 − ΔvD 2 = − g m1 RD − g m 2 RD 2 2 −V −V ⎛ Δg Δg ⎞ ⎞ ⎛ vO = d ⋅ RD ( g m 2 + g m1 ) = d ⋅ RD ⎜ g m − m + ⎜ g m − m ⎟ ⎟ 2 2 2 2 ⎠⎠ ⎝ ⎝ Ad = − g m RD = − ( 0.183) ( 50 ) = −9.15 Ad : ΔvD1 = − g m1

Vd ⋅ RD 2

ACM : vO = ΔvD1 − ΔvD 2 Acm = Acm =

Δg ⎞ Δg ⎞ ⎛ ⎛ − ⎜ g m + M ⎟ RD vcm ⎜ g m − M ⎟ RD vCM 2 2 ⎠ ⎝ ⎠ ⎝ = + 1 + g m ( 2 RS ) 1 + g m ( 2 RS )

−Δg m RD vO = vcm 1 + g m ( 2 RS ) − ( 0.00183) ( 50 )

1 + ( 0.183)( 2 ) ( 75 )

Δg m = ( 0.01) ( 0.183) = 0.00183 = −0.003216

C M R R ∫ = 69.1 dB dB

______________________________________________________________________________________ 11.43 2 (a) 5 = I S R S + V SG = 2 K p (V SG + VTP ) R S + V SG

(

)

2 5 = 2(1.2 )(2 ) V SG − 1.2V SG + 0.36 + V SG

4.8V

2 SG

− 4.76V SG − 3.272 = 0 ⇒ V SG = 1.459 V = υ S

I S = 2(1.2)(1.459 − 0.6) = 1.77 mA 2

I D1 = I D 2 = 0.885 mA

υ D1 = υ D 2 = I D R D − 5 = (0.885)(1) − 5 = −4.115 V

(

)

2 − 1.2V SG + 0.36 + V SG + 1 (b) 5 = 2(1.2)(2 ) V SG

4.8V

2 SG

− 4.76V SG − 2.272 = 0 ⇒ V SG = 1.344 V

υ S = V SG + 1 = 2.344 V

I D1 = I D 2 = (1.2)(1.344 − 0.6 ) = 0.664 mA 2

υ D1 = υ D 2 = (0.664 )(1) − 5 = −4.336 V

(c) Using part (a), g m = 2 (1.2 )(0.885) = 2.061 mA/V

υ D2

υd

= −4.115 + (2.061)(1)(0.1) = −3.909 V 2 = −4.115 − (2.061)(1)(0.1) = −4.321 V

υ D1 = −4.115 + g m R D ⋅

(d) Using part (b), g m = 2 (1.2 )(0.664) = 1.785 mA/V

υd

= −4.336 + (1.785)(1)(0.1) = −4.158 V 2 υ D 2 = −4.336 − (1.785)(1)(0.1) = −4.515 V ______________________________________________________________________________________

υ D1 = −4.336 + g m R D ⋅

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.44 (a) gm = 2 Kn I D

( 0.4 )(1)

=2

g m = 1.265 mA/V v 1 Ad = o = = 10 vd 0.1 Ad = g m RD 10 = (1.265 ) RD RD = 7.91 K

(b)

Quiescent v1 = v2 = 0

vD1 = vD 2 = 10 − (1)( 7.91) = 2.09 V ID 1 + VTN = + 0.8 = 2.38 V Kn 0.4

VGS =

VDS ( sat ) = 2.38 − 0.8 = 1.58 So vcm = vD − VDS ( sat ) + VGS = 2.09 − 1.58 + 2.38 vcm = 2.89 V

______________________________________________________________________________________ 11.45 g m RD 2 For vCM = 2.5 V Ad =

I D1 = I D 2 =

IQ 2

= 0.25 mA

Let VD1 = VD 2 = 3 V , then RD = Then 100 =

g m ( 28 )

10 − 3 ⇒ RD = 28 k Ω 0.25

⇒ g m = 7.14 mA / V 2 k′ ⎛ W ⎞ And g m = 2 n ⎜ ⎟ I D 2⎝L⎠

⎛ 0.080 ⎞ ⎛ W ⎞ 7.14 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.25 ) ⇒ ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 1274 (Extremely large transistors to meet the gain requirement.) ⎝ L ⎠1 ⎝ L ⎠ 2 Need ACM = 0.10

From Eq. (11.82(b)) ACM =

g m RD 1 + 2 g m Ro

So 0.10 =

( 7.14 )( 28 ) 1 + 2 ( 7.14 ) Ro

⇒ Ro = 140 k Ω

For the basic 2-transistor current source Ro = ro =

1 1 = = 200 k Ω λ I Q ( 0.01)( 0.5 )

This current source is adequate to meet common-mode gain requirement. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.46 Not in detail, Approximation looks good. a. IS =

−VGS 1 − ( −5 ) RS

and I S = 2 I D = 2 K n (VGS1 − VTN )

5 − VGS 1 2 = 2 ( 0.050 )(VGS 1 − 1) 20 5 − VGS1 = 2 (VGS2 1 − 2VGS 1 + 1) 2VGS2 1 − 3VGS 1 − 3 = 0 VGS 1 =

3±

( 3)

2

+ 4 ( 2 )( 3)

2 ( 2)

⇒ VGS 1 = 2.186 V

5 − 2.186 ⇒ I S = 0.141 mA 20 I I D1 = I D 2 = S ⇒ I D1 = I D 2 = 0.0704 mA 2 v02 = 5 − ( 0.0704 )( 25 ) ⇒ v02 = 3.24 V IS =

b.

g m = 2 K n (VGS − VTN ) = 2 ( 0.05 )( 2.186 − 1) g m = 0.119 mA/V r0 =

1

λ I DQ

=

1

( 0.02 )( 0.0704 )

= 710 kΩ

Vgs1 = v1 − VS , Vgs 2 = v2 − VS v01 v −V + g mVgs1 + 01 S = 0 RD r0 ⎛ 1 V 1⎞ v01 ⎜ + ⎟ + g m ( v1 − VS ) − S = 0 R r r0 ⎝ D 0⎠

(1)

v02 v − VS + g mVgs 2 + 02 =0 RD r0 ⎛ 1 V 1⎞ + ⎟ + g m ( v2 − VS ) − S = 0 v02 ⎜ R r r0 ⎝ D 0⎠

(2)

2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ g mVgs1 +

v01 − VS v02 − VS V + + g mVgs 2 = S r0 r0 RS

g m ( v1 − VS ) +

v01 v02 2VS V + − + g m ( v2 − VS ) = S r0 r0 r0 RS

g m ( v1 + v2 ) +

⎧ v01 v02 2 1 ⎫ + = VS ⎨2 g m + + ⎬ r0 r0 r0 RS ⎭ ⎩

(3)

From (1) ⎛ 1⎞ VS ⎜ g m + ⎟ − g m v1 r0 ⎠ v01 = ⎝ ⎛ 1 1⎞ + ⎟ ⎜ ⎝ RD r0 ⎠

Then ⎛ 1⎞ VS ⎜ g m + ⎟ − g m v1 r0 ⎠ v ⎝ + 02 = VS g m ( v1 + v2 ) + r0 ⎛ 1 1⎞ + ⎟ r0 ⎜ ⎝ RD r0 ⎠

⎧ 2 1 ⎫ ⎨2 g m + + ⎬ r RS ⎭ 0 ⎩

(3)

⎛ 1 ⎛ ⎛ 1 ⎧ 1⎞ 1⎞ 1⎞ 2 1 ⎫ ⎛ 1 1⎞ g m ( v1 + v2 ) r0 ⎜ + ⎟ + VS ⎜ g m + ⎟ − g m v1 + v02 ⎜ + ⎟ = VS ⎨2 g m + + ⎬ ⋅ r0 ⎜ + ⎟ r0 ⎠ r0 RS ⎭ ⎝ RD r0 ⎠ ⎝ RD r0 ⎠ ⎝ ⎝ RD r0 ⎠ ⎩ ⎧⎪⎛ ⎛ 1 ⎛ r ⎞ r0 ⎞ ⎛ 1⎞ 2 1 ⎞⎛ 1 ⎞ ⎪⎫ g m ( v1 + v2 ) ⎜ 1 + 0 ⎟ − g m v1 + v02 ⎜ + ⎟ = VS ⎨⎜ 2 g m + + ⎟ ⎜1 + ⎟ − ⎜ gm + ⎟⎬ R R r r R R r ⎪⎩⎝ ⎪ D ⎠ 0 S ⎠⎝ D ⎠ ⎝ 0 ⎠⎭ ⎝ ⎝ D 0⎠ ⎛ 1 ⎞ ⎧ ⎛ r0 ⎞ r r r0 1 2 1 2 1⎫ g m ⎜ v1 ⋅ + v2 + v2 ⋅ + ⎟ = VS ⎨2 g m + + + 2 gm ⋅ 0 + + 0 − gm − ⎬ ⎟ + v02 ⎜ RD ⎠ r0 RS RD RD RS RD r0 ⎭ ⎝ RD ⎝ RD r0 ⎠ ⎩ ⎫ r0 ⎞ 2 1 1 ⎛ ⎪⎧ (1 + g m r0 )⎪⎬ ⎨2 g m + + ⎜1 + ⎟+ r R R R ⎪⎩ ⎪⎭ (4) 0 S ⎝ D ⎠ D ⎛ 1 ⎛ 1⎞ 1⎞ v02 ⎜ + ⎟ + g m v2 = VS ⎜ g m + ⎟ R r r 0 ⎠ ⎝ Then substituting into (2), ⎝ D 0 ⎠ 710 710 ⎤ 1 ⎤ ⎡1 + v2 + v2 + v02 ⎢ + ( 0.119 ) ⎡⎢v1 ⎥ 25 ⎥⎦ ⎣ 25 ⎣ 25 710 ⎦

⎛ r r g m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 R R ⎝ D D

⎛ 1 ⎞ 1⎞ + ⎟ = VS ⎟ + v02 ⎜ R r ⎠ ⎝ D 0⎠

⎧ ⎫ 1 1 ⎛ 710 ⎞ 2 = VS ⎨0.119 + + ⎜1 + ⎟ + ⎡⎣1 + ( 0.119 )( 710 ) ⎤⎦ ⎬ 710 20 ⎝ 25 ⎠ 25 ⎩ ⎭

Substitute numbers: ( 0.119 ) [ 28.4v1 + 29.4v2 ] + ( 0.0414 ) v02 = VS {0.1204 + 1.470 + 6.8392} = VS ( 8.4296 ) or VS = 0.4010v1 + 0.4150v2 + 0.00491v02 1 ⎞ 1 ⎞ ⎛ 1 ⎛ Then v02 ⎜ + ⎟ + ( 0.119 ) v2 = VS ⎜ 0.119 + ⎟ 710 ⎠ ⎝ 25 710 ⎠ ⎝

(2)

(4)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ v02 ( 0.0414 ) + v2 ( 0.119 ) = ( 0.1204 ) [ 0.401v1 + 0.4150v2 + 0.00491v02 ] v02 ( 0.0408 ) = ( 0.04828 ) v1 − ( 0.0690 ) v2 v02 = (1.183) v1 − (1.691) v2

vd 2 vd v2 = vcm − 2 v ⎞ v ⎞ ⎛ ⎛ So v02 = (1.183) ⎜ vcm + d ⎟ − (1.691) ⎜ vcm − d ⎟ 2⎠ 2⎠ ⎝ ⎝ Or v02 = 1.437vd − 0.508vcm ⇒ Ad = 1.437, Acm = −0.508 Now

v1 = vcm +

⎛ 1.437 ⎞ C M R RdB = 20 log10 ⎜ ⎟ ⇒ C M R RdB = 9.03 dB ⎝ 0.508 ⎠

______________________________________________________________________________________ 11.47

KVL: v1 = Vgs1 − Vgs 2 + v2 So v1 − v2 = Vgs1 − Vgs 2

KCL:

g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2

So Vgs1 =

1 1 ( v1 − v2 ) , Vgs 2 = − ( v1 − v2 ) 2 2

Now

v02 v02 − v01 + = − g mVgs 2 RD RL ⎛ 1 1 ⎞ v01 = v02 ⎜ + ⎟− R R RL L ⎠ ⎝ D v01 v01 − v02 + = − g mVgs1 RD RL ⎛ 1 1 ⎞ v02 = v01 ⎜ + ⎟− R R RL L ⎠ ⎝ D

⎛ R ⎞ v01 = v02 ⎜ 1 + L ⎟ + g m RLVgs 2 R ⎝ D ⎠ From (1):

(1)

(2)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Substitute into (2): ⎛ ⎛ 1 v02 R ⎞⎛ 1 1 ⎞ 1 ⎞ − g mVgs1 = v02 ⎜1 + L ⎟⎜ + + ⎟ + g m RL ⎜ ⎟ Vgs 2 − R R R R R RL ⎝ ⎝ D D ⎠⎝ D L ⎠ L ⎠ ⎛ 1 ⎛ R ⎞⎛ 1 ⎞ R 1 ⎞ − g m ⋅ ( v1 − v2 ) + g m ⎜1 + L ⎟ ⎜ ⎟ ( v1 − v2 ) = v02 ⎜ + L2 + ⎟ ⎝ RD ⎠ ⎝ 2 ⎠ ⎝ RD RD RD ⎠ 1 ⋅ g m RL v02 ⎛ v02 RL ⎞ 1 ⎛ RL ⎞ gm ⎜ = 2 ⎟ ( v1 − v2 ) = ⎜2+ ⎟ ⇒ Ad 2 = 2 ⎝ RD ⎠ RD ⎝ RD ⎠ v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ R ⎝ D ⎠ 1 − ⋅ g m RL v01 = 2 Ad 1 = v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ From symmetry v −v g m RL Av = 02 01 = v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝

Then ______________________________________________________________________________________ 11.48

⎛ −υ ⎞ (a) υ o = − g m (R D R L )⎜⎜ d ⎟⎟ ⎝ 2 ⎠ 1 Ad = g m (R D R L ) 2 (b) υ o = − g m (R D R L )υ gs

υ cm = υ gs + g mυ gs (2 Ro )

υ gs = Acm =

υ cm 1+ 2 g m R o

− g m (R D R L )

1+ 2 g m R o ______________________________________________________________________________________

11.49

(a) Ad =

Δυ o 0.5 = =5 Δυ d 0.1

g m RD , g m = 2 K n I DQ = 2 (0.15)(0.1) = 0.2449 mA/V 2 2(5) RD = = 40.8 k Ω 0.2449 (b) υ CM (max ) = υ O − υ DS (sat ) + υ GS Ad = 5 =

i D = K n (υ GS − VTN )

2

0.1 = 0.15(υ GS − 0.4 ) ⇒ υ GS = 1.216 V 2

υ DS (sat ) = 1.216 − 0.4 = 0.8165 V υ O = 5 − (0.1)(40.8) = 0.92 V υ CM (max ) = 0.92 − 0.8165 + 1.216 = 1.32 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.50

Vd 1 = − g mVgs1 RD = − g m RD (V1 − Vs ) Vd 2 = − g mVgs 2 RD = − g m RD (V2 − Vs ) Now Vo = Vd 2 − Vd 1 = − g m RD (V2 − Vs ) − ( − g m RD (V1 − Vs ) ) Vo = g m RD (V1 − V2 ) Define V1 − V2 ≡ Vd V Then Ad = o = g m RD and Acm = 0 Vd

______________________________________________________________________________________ 11.51 (a) υ CM = υ O − υ DS (sat ) + υ GS i D = K n (υ GS − VTN )

2

⎛ 0.1 ⎞ 2 0.1 = ⎜ ⎟(10 )(υ GS − 0.4 ) ⇒ υ GS = 0.8472 V ⎝ 2 ⎠ υ DS (sat ) = 0.8472 − 0.4 = 0.4472 V

Then υ O = 1.5 + 0.4472 − 0.8472 = 1.1 V υ O = 3 − (0.1)R D = 1.1 ⇒ R D = 19 k Ω Now Ad = Ad =

g m RD ⎛ 0.1 ⎞ , where g m = 2 ⎜ ⎟(10 )(0.1) = 0.4472 mA/V 2 ⎝ 2 ⎠

(0.4472)(19) = 4.248

2 CMRR dB = 50 dB ⇒ CMRR = 316.2

[

]

⎤ 1 1⎡ ⎛ 0.1 ⎞ 1 + 2 K n I Q ⋅ R o ⇒ 316.2 = ⎢1 + 2 ⎜ ⎟(10 )(0.2) ⋅ R o ⎥ ⇒ Ro = 998 k Ω 2 2 ⎢⎣ ⎝ 2 ⎠ ⎥⎦ (b) Use cascode current source similar to Figure 10.18 with υ DS 2 (sat ) = 0.3 V. ______________________________________________________________________________________ CMRR =

11.52 (a) From Problem 11.27, υ d ≅ 12.5 mV. g R Ad = m D , where g m = 2 (1)(0.1) = 0.6325 mA/V 2 ( 0.6325)(20 ) Ad = = 6.325 2 υ O = (6.325)(0.0125) = 0.0791 V So −0.0791 ≤ υ O ≤ 0.0791 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.53 From previous results Ad 1 =

vo 2 − vo1 = g m1 R1 = 2 K n1 I Q1 ⋅ R1 = 20 v1 − v2

and Ad 2 = Set

I Q1 R1 2

vo 3 1 1 2 K n 3 I Q 2 ⋅ R2 = 30 = g m 3 R2 = 2 vo 2 − vo1 2 I Q 2 R2

= 5 V and

2

= 2.5 V

Let I Q1 = I Q 2 = 0.1 mA Then R1 = 100 k Ω, R2 = 50 k Ω 2

⎛ 0.06 ⎞⎛ W ⎞ ⎛ 20 ⎞ ⎛W ⎞ ⎛W ⎞ Then 2 ⎜ ⎟⎜ ⎟ ( 0.1) = ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 6.67 ⎝ 2 ⎠⎝ L ⎠1 ⎝ 100 ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2 ⎛ 2 ( 30 ) ⎞ ⎛ 0.060 ⎞⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ and 2 ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 240 ⎟⎜ ⎟ ( 0.1) = ⎜ 50 ⎝ 2 ⎠⎝ L ⎠3 ⎝ L ⎠3 ⎝ L ⎠ 4 ⎝ ⎠ 2

______________________________________________________________________________________ 11.54 ⎛ v ⎞ iD1 = I DSS ⎜ 1 − GS1 ⎟ VP ⎠ ⎝

a.

⎛ v ⎞ iD 2 = I DSS ⎜1 − GS 2 ⎟ VP ⎠ ⎝

iD1 − iD 2

2

2

⎛ v ⎞ ⎛ v ⎞ = I DSS ⎜1 − GS 1 ⎟ − I DSS ⎜ 1 − GS 2 ⎟ VP ⎠ VP ⎠ ⎝ ⎝ I DSS

=

( vGS 2 − vGS1 )

VP

I DSS

=−

⋅ vd =

VP

I DSS

( −VP )

⋅ vd

iD1 + iD 2 = I Q ⇒ iD 2 = I Q − iD1

(

iD1 − I Q − iD1

)

2

=

I DSS

( −VP )

2

⋅ vd2

iD1 − 2 iD1 ( I Q − iD1 ) + ( I Q − iD1 ) = Then iD1 ( I Q − iD1 ) =

I DSS

( −VP )

2

⋅ vd2

⎤ I 1⎡ ⎢ I Q − DSS 2 ⋅ vd2 ⎥ 2 ⎢⎣ ( −VP ) ⎦⎥

Square both sides 2

⎤ I 1⎡ iD2 1 − iD1 I Q + ⎢ I Q − DSS 2 ⋅ vd2 ⎥ = 0 4 ⎣⎢ ( −VP ) ⎦⎥ ⎤ I ⎛1⎞⎡ I Q ± I − 4 ⎜ ⎟ ⎢ I Q − DSS 2 ⋅ vd2 ⎥ 4 ⎝ ⎠ ⎢⎣ ( −VP ) ⎥⎦

2

2 Q

iD1 =

iD1

2

2 I Q 1 2 ⎡ 2 2 I Q I DSS vd2 ⎛ I DSS vd2 ⎞ ⎤ ⎢ ⎟ ⎥ = ± +⎜ IQ − IQ − 2 ⎜ ( −V )2 ⎟ ⎥ ⎢ 2 2 −VP ) ( P ⎝ ⎠ ⎦ ⎣

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Use + sign iD1

⎞ 1 2 I Q I DSS 2 ⎛ I DSS = + ⋅ vd − ⎜ ⋅ vd2 ⎟ 2 2 ⎜ ( −V ) ⎟ 2 2 ( −VP ) P ⎝ ⎠

2

IQ

2

2

2

2

2

2

2 I DSS ⎛ I DSS 1 IQ vd −⎜ ⎜ IQ 2 ( −VP ) IQ ⎝

⎞ ⎛ vd ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

2 I DSS ⎛ I DSS iD1 1 ⎛ 1 ⎞ = +⎜ −⎜ ⎟ ⋅ vd ⎜ I I Q 2 ⎝ −2VP ⎠ IQ ⎝ Q We had

⎞ ⎛v ⎞ ⎟⎟ ⎜ d ⎟ ⎠ ⎝ VP ⎠

iD1 =

IQ 2

+

Or

iD 2 = I Q − iD1 Then 2 I DSS ⎛ I DSS iD 2 1 ⎛ 1 ⎞ −⎜ = −⎜ ⎟ ⋅ vd ⎜ I IQ I Q 2 ⎝ −2VP ⎠ ⎝ Q

b.

⎞ ⎛ vd ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

If iD1 = I Q , then 1=

1 ⎛ 1 +⎜ 2 ⎝ −2VP

2

2 I DSS ⎛ I DSS −⎜ ⎜ IQ IQ ⎝

VP = vd

2

⎞ 2 I DSS ⎛ I DSS −⎜ ⎟ ⋅ vd ⎜ I IQ ⎠ ⎝ Q ⎞ ⎛ vd ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

⎞ ⎛v ⎞ ⎟⎟ ⎜ d ⎟ ⎠ ⎝ VP ⎠

2

2

Square both sides VP

2

⎛ I DSS ⎜⎜ ⎝ IQ

vd2 =

⎡ 2I ⎛I = v ⎢ DSS − ⎜ DSS ⎜ IQ ⎢ IQ ⎝ ⎣ 2 d

2

2 2 ⎞ ⎛ vd ⎞ ⎤ ⎟⎟ ⎜ ⎟ ⎥ ⎠ ⎝ VP ⎠ ⎥⎦

2

⎞ ⎛ 1 ⎞ 2 2 2 I DSS 2 ⋅ vd + VP ⎟⎟ ⎜ ⎟ ( vd ) − IQ ⎠ ⎝ VP ⎠ ⎛ 2I 2 I DSS ± ⎜ DSS ⎜ IQ IQ ⎝

2

⎞ ⎛I ⎟⎟ − 4 ⎜⎜ DSS ⎠ ⎝ IQ

⎛ 2I 2 ⎜ DSS ⎜ IQ ⎝

2

2

=0 2

⎞ ⎛ 1 ⎞ 2 ⎟⎟ ⎜ ⎟ (VP ) V ⎠ ⎝ P⎠

2

⎞ ⎛ 1 ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

2

2 ⎛ IQ ⎞ vd2 = (VP ) ⎜ ⎟ ⎝ I DSS ⎠ 1/ 2

⎛ IQ ⎞ Or vd = VP ⎜ ⎟ ⎝ I DSS ⎠

c.

For vd small, 2 I DSS 1 IQ + ⋅ ⋅ vd IQ 2 ( −VP )

iD1 ≈

IQ

gf =

diD1 d vd

2

vd → 0

=

2 I DSS 1 IQ ⋅ ⋅ IQ 2 ( −VP )

⎛ 1 ⎞ I Q I DSS Or ⇒ g f ( max ) = ⎜ ⎟ 2 ⎝ −VP ⎠

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.55 a.

I Q = I D1 + I D 2 ⇒ I Q = 1 mA v0 = 7 = 10 − ( 0.5 ) RD ⇒ RD = 6 kΩ

b. ⎛ 1 ⎞ I Q ⋅ I DSS g f ( max ) = ⎜ ⎟ 2 ⎝ −VP ⎠ ⎛1⎞ g f ( max ) = ⎜ ⎟ ⎝ 4⎠

(1)( 2 ) 2

⇒ g f ( max ) = 0.25 mA/V

c. g m RD = g f ( max ) ⋅ RD 2 Ad = ( 0.25 )( 6 ) ⇒ Ad = 1.5

Ad =

______________________________________________________________________________________ 11.56 a. IS =

−VGS − ( −5 ) RS

⎛ V ⎞ = ( 2 ) I DSS ⎜1 − GS ⎟ ⎝ VP ⎠

⎛ V ⎞ 5 − VGS = ( 2 )( 0.8 )( 20 ) ⎜⎜ 1 − GS ⎟⎟ −2 ) ⎠ ( ⎝ 1 ⎛ ⎞ 5 − VGS = ( 2 )16 ⎜1 + VGS + VGS2 ⎟ 4 ⎝ ⎠ 8VGS2 + 33VGS + 27 = 0 VGS =

2

2

−33 ± 1089 − 4 ( 8 )( 27 ) 2 (8)

= −1.125 V IS =

5 − ( −1.125 )

20 = 0.306 mA

I D1 = I D 2 = 0.153 mA vo 2 = 1.17 V

(b) ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.57 Equivalent circuit and analysis is identical to that in problem 11.47. 1 ⋅ g m RL Ad 2 = 2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ 1 − ⋅ g m RL Ad 1 = 2 ⎛ RL ⎞ ⎜2+ ⎟ R D ⎠ ⎝ Av =

v02 − v01 g m RL = vd ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝

______________________________________________________________________________________ 11.58 (a) Ad = g m (ro 2 ro 4 )

I CQ

gm =

VT

=

0.2 = 7.692 mA/V 0.026

ro 2 =

V A2 120 = = 600 k Ω I CQ 0.2

ro 4 =

V A4 80 = = 400 k Ω I CQ 0.2

Ad = (7.692)(600 400) = 1846

(b) R o = ro 2 ro 4 = 600 400 = 240 k Ω

(

(c) Ad = g m ro 2 ro 4 R L

)

Ad = (0.75)(1846) = 1384.5 = (7.692)(240 R L ) ⇒ R L = 720 k Ω

______________________________________________________________________________________ 11.59 (a) ⎛ 2⎞ I Q = 250 μ A I REF = I Q ⎜1 + ⎟ β ⎝ ⎠ 2 ⎞ ⎛ = 250 ⎜1 + ⎟ = 252.8 μ A ⎝ 180 ⎠ R1 =

5 − ( 0.7 ) − ( −5 ) 0.2528

⇒ R1 = 36.8 K

(b) Ad = g m ( ro 2 ro 4 )

Ad = ( 4.808 ) (1200 800 ) Ad = 2308

0.125 = 4.808 mA/V 0.026 150 ro 2 = = 1200 K 0.125 100 ro 4 = = 800 K 0.125 gm =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) Rid = 2rπ =

2 (180 )( 0.026 ) 0.125

⇒ Rid = 74.9 K

Ro = ro 2 ro 4 = 1200 800 = 480 K = Ro

(d)

vcm ( max ) = 5 − 0.7 = 4.3 V vcm ( min ) = 0.7 + 0.7 − 5 = −3.6 V

______________________________________________________________________________________ 11.60 a. ⎛ IQ ⎞ ⎛ 1 ⎞ I 0 = I B3 + I B 4 ≈ 2 ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠⎝ β ⎠ I Q 0.2 I0 = = ⇒ I0 = 2 μ A β 100

b. r02 = r04 = gm =

I CQ

VA 100 = = 1000 kΩ I CQ 0.1

=

VT

0.1 = 3.846 mA/V 0.026

Ad = g m ( r02 r04 ) = ( 3.846 ) (1000 1000 ) ⇒ Ad = 1923

c.

(

Ad = g m r02 r04 RL

)

Ad = ( 3.846 ) (1000 1000 250 ) ⇒ Ad = 641

______________________________________________________________________________________ 11.61 IQ

(a) Ad =

IQ 2V A 2

+

2VT IQ 2V A4

+

1 RL

IQ

1000 =

2(0.026 ) IQ 1 + + 2(90) 2(60) 250 IQ

( )

1000 I Q (0.005556 + 0.008333) + 4 = I Q (19.23) ⇒ I Q = 0.749 mA (b) υ CM (max ) = V + − 2V EB (on ) = 5 − 2(0.6 ) = 3.8 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.62 ⎛υ ⎞ (b) υ o1 = − g m1 (ro1 Ro 3 )⎜⎜ d ⎟⎟ ⎝ 2 ⎠ υ 1 Ad 1 = o1 = − g m1 (ro1 R o3 ) υd 2 I CQ 0.2 = = 7.692 mA/V g m1 = 0.026 VT

ro1 =

V A1 120 = = 600 k Ω I CQ 0.2

R o3 = ro 3 rπ 3

1 , where g m3 = 7.692 mA/V, g m3 rπ 3 =

Then Ro 3 = 400 15.6

(120)(0.026) = 15.6 k Ω , 0.2

ro3 = 400 k Ω

1 = 15.014 0.130 = 0.1289 k Ω 7.692

1 (7.692)(600 0.1289) = −0.4956 2 1 = + g m 2 (ro 2 Ro 4 ) = +0.4956 2 = + g m1 (ro1 Ro 3 ) = 2(0.4956) = 0.9912

Ad 1 = −

(c) Ad 2 (d) Ad 3

______________________________________________________________________________________ 11.63

g m 4V sg 4 = g m3V sg 3 = g m1V gs1

V gs1 = V1 − V s , and V gs 2 = V 2 − V s g m1V gs1 + g m1V gs 2 +

Vo − V s V s = ro 2 Ro

g m1 (V1 + V 2 − 2V s ) +

⎛ 1 Vo 1 ⎞ ⎟ = V s ⎜⎜ + ⎟ ro 2 ⎝ R o ro 2 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ g m1 (V1 + V 2 ) +

⎛ Vo 1 1 ⎞ ⎟ = V s ⎜⎜ 2 g m1 + + ro 2 R o ro 2 ⎟⎠ ⎝

Now g m1 = 2 K n I DQ = 2 (0.5)(0.1) = 0.4472 mA/V ro 2 =

1 = 500 k Ω (0.02)(0.1)

Then (0.4472 )(V1 + V 2 ) +

Vo 1 1 ⎤ ⎡ = V s ⎢2(0.4472 ) + + ⎥ = V s (0.8969) 500 2000 500 ⎣ ⎦

We have (1) V s = (0.4986 )(V1 + V 2 ) + Vo (0.0022299 ) V − V s Vo Also (2) o + + g m1V gs 2 − g m1V gs1 = 0 ro 2 ro 4

⎛ 1 1 ⎞ Vs ⎟− V o ⎜⎜ + ⎟ r + g m1 (V 2 − V s − (V1 − V s )) = 0 r r o4 ⎠ o2 ⎝ o2

⎛ 1 1 Vo ⎜⎜ + ⎝ ro 2 ro 4 We find ro 4 =

⎞ Vs ⎟⎟ − + g m1 (V2 − V1 ) = 0 ⎠ ro 2 1

(0.03)(0.1)

= 333.3 k Ω

1 ⎞ 1 ⎛ 1 [(0.4986)(V1 + V2 ) + Vo (0.0022299)] + (0.4472)(V2 − V1 ) = 0 Then Vo ⎜ + ⎟− ⎝ 500 333.3 ⎠ 500 V o [(0.005) − (0.00000446 )] − (0.0009972 )(V1 + V 2 ) + (0.4472 )(V 2 − V1 ) = 0 (a) Let V1 = V d and V 2 = 0 V o [0.00499554] − V d (0.0009972 ) − V d (0.4472 ) = 0 V Ad = o = 89.72 Vd (b) Let V1 = 0 and V 2 = −V d V o [0.00499554] + V d (0.0009972 ) − V d (0.4472 ) = 0 V Ad = o = 89.32 Vd Vd V and V 2 = − d 2 2 V o [0.00499554] = V d (0.4472 )

(c) Let V1 =

Vo = 89.52 Vd ______________________________________________________________________________________ A=

11.64 a.

From symmetry.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 0.1 +1 0.1

VGS 3 = VGS 4 = VDS 3 = VDS 4 = Or VDS 3 = VDS 4 = 2 V

0.1 +1 = 2 V 0.1 = VSG1 − (VDS 3 − 10 )

VSG1 = VSG 2 = VSD1 = VSD 2

= 2 − ( 2 − 10 )

Or VSD1 = VSD 2 = 10 V

b. r0 n = r0 p =

1

λn I DQ 1

=

1

( 0.01)( 0.1)

⇒ 1 MΩ

1

=

⇒ 0.667 MΩ

( 0.015)( 0.1) g m = 2 K p (VSG + VTP ) = 2 ( 0.1)( 2 − 1) = 0.2 mA / V Ad = g m ( ron rop ) = ( 0.2 ) (1000 667 ) ⇒ Ad λP I DQ

= 80

(c) I D 2 = I D1 = ro 4 = ro 2 =

1

λn I D 4 1

λP I D 2

IQ 2 = =

= 0.1 mA 1

( 0.01)( 0.1) 1

= 1000 k Ω

( 0.015)( 0.1)

= 667 k Ω

Ro = ro 2 ro 4 = 667 1000 = 400 k Ω

______________________________________________________________________________________ 11.65 (a) Ad = g m (ro 2 ro 4 ) ⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞ g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ = 2 ⎜ ⎟(8)(0.06 ) = 0.3098 mA/V 2 L ⎝ 2 ⎠ ⎝ ⎠⎝ ⎠ n 1 ro 2 = = 666.7 k Ω (0.025)(0.06) 1 ro 4 = = 416.7 k Ω (0.04)(0.06)

Ad = (0.3098)(666.7 416.7 ) = 79.44

⎛ k ′p ⎞⎛ W ⎞ 2 (b) i D 3 = ⎜⎜ ⎟⎟⎜ ⎟ (υ SG 3 + VTP ) L 2 ⎝ ⎠⎝ ⎠ p ⎛ 0.04 ⎞ 2 0.06 = ⎜ ⎟(10 )(υ SG 3 − 0.3) ⇒ υ SG 3 = 0.8477 V 2 ⎝ ⎠

υ O = 1.8 − 0.8477 = 0.9523 υ CM (max ) = υ O − υ DS (sat ) + υ GS ⎛ 0.1 ⎞ 2 0.06 = ⎜ ⎟(8)(υ GS − 0.3) ⇒ υ GS = 0.6873 V, υ DS (sat ) = 0.3873 V ⎝ 2 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ υ CM (max ) = 0.9523 − 0.3873 + 0.6873 = 1.25 V ______________________________________________________________________________________

11.66

R04 = r04 ⎡⎣1 + g m 4 ( R rπ 4 ) ⎤⎦ 80 = 800 K 0.1 0.1 gm4 = = 3.846 0.026 (100 )( 0.026 ) rπ 4 = 0.1 = 26 K r04 =

R rπ 4 = 1 26 = 0.963 K Assume β = 100 rπ 3 =

(100 )( 0.026 ) 0.1

= 26 kΩ

0.1 = 3.846 mA/V 0.026 R04 = 800 ⎡⎣1 + ( 3.846 )( 0.963) ⎤⎦ ⇒ 3.763 MΩ

g m3 =

⇒ R0 = 3.763MΩ Then Av = − g m ( r02 R0 ) 120 = 1200 kΩ 0.1 0.1 gm = = 3.846 mA/V 0.026 Av = − ( 3.846 ) ⎣⎡1200 3763⎤⎦ ⇒ Av = −3499 r02 =

b. For R = 0, r04 = Av = − g m ( r02

80 = 800 kΩ 0.1 r04 )

= − ( 3.846 ) ⎡⎣1200 800 ⎤⎦ ⇒ Av = −1846

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

R = ( 3.763 1.2 ) = 0.910 M Ω For part (a), o R = (1.2 0.8 ) = 0.48 M Ω For part (b), o ______________________________________________________________________________________ (c)

11.67 I B5 =

I E5 I +I I +I = B3 B 4 = C 3 C 4 1+ β 1+ β β (1 + β )

Now I C 3 + I C 4 ≈ I Q So I B 5 ≈ I B6 =

IQ

β (1 + β )

I Q1 IE6 = 1 + β β (1 + β )

For balance, we want I B 6 = I B 5 So that I Q1 = I Q

______________________________________________________________________________________

11.68 Resistance looking into drain of M4.

V sg 4 ≅ I X R1 I X + g m 4V sg 4 =

V X − V sg 4 ro 4

⎡ R ⎤ V I X ⎢1 + g m 4 R1 + 1 ⎥ = X ro 4 ⎦ ro 4 ⎣ ⎡ R ⎤ Or R o = ro 4 ⎢1 + g m 4 R1 + 1 ⎥ ro 4 ⎦ ⎣ a.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Ad = g m 2 ( ro 2 Ro )

g m 2 = 2 K n I DQ = 2

( 0.080 )( 0.1)

= 0.179 mA / V 1 1 ro 2 = = = 667 k Ω λn I DQ ( 0.015 )( 0.1) g m 4 = 2 K P I DQ = 2

( 0.080 )( 0.1)

= 0.179 mA / V ro 4 =

1

λ p I DQ

=

1

( 0.02 )( 0.1)

= 500 k Ω

1 ⎤ ⎡ R0 = 500 ⎢1 + ( 0.179 )(1) + = 590.5 kΩ 500 ⎥⎦ ⎣ Ad = ( 0.179 ) ⎡⎣667 590.5⎤⎦ ⇒ Ad = 56.06

b. When R1 = 0, R0 = r04 = 500 kΩ

Ad = ( 0.179 ) ⎡⎣667 500 ⎤⎦ ⇒ Ad = 51.15

(c)

For part (a),

Ro = ro 2 Ro = 667 590.5 ⇒ Ro = 313 k Ω

R =r

r = 667 500 ⇒ R = 286 kΩ

o For part (b), o o 2 o 4 ______________________________________________________________________________________

11.69

Let β = 100, VA = 100 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VA 100 = = 1000 k Ω I CQ 0.1

ro 2 =

Ro 4 = ro 4 [1 + g m RE′ ] where RE′ = rπ RE Now rπ =

(100 )( 0.026 )

= 26 k Ω 0.1 0.1 = 3.846 mA / V gm = 0.026 RE′ = 26 1 = 0.963 k Ω

Then Ro 4 = 1000 ⎣⎡1 + ( 3.846 )( 0.963) ⎦⎤ = 4704 k Ω

Ad = g m ( ro 2 Ro 4 ) = 3.846 (1000 4704 ) ⇒ Ad = 3172

______________________________________________________________________________________ 11.70 (a) For Q2, Q4

(1)

Ix =

Vx − Vπ 4 V + g m 2Vπ 2 + g m 4Vπ 4 + x ro 2 ro 4

g m 2Vπ 2 +

(2) (3)

Vx − Vπ 4 V = π4 ro 2 rπ 4 rπ 2

Vπ 4 = −Vπ 2

From (2)

⎡ 1 ⎤ Vx 1 = Vπ 4 ⎢ + + gm2 ⎥ ro 2 ⎢⎣ rπ 4 rπ 2 ro 2 ⎥⎦

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Now ⎞ ⎛ 120 ⎞ ⎟=⎜ ⎟ ( 0.5 ) = 0.496 mA ⎠ ⎝ 121 ⎠ ⎛ 120 ⎞ ⎛ I Q ⎞ ⎛ 1 ⎞⎛ β ⎞ ⎟ ⇒ I C 2 = 0.0041 mA = ⎜ ⎟⎜ ⎟⎜ ⎟ = ( 0.5 ) ⎜⎜ 2 ⎟ ⎝ 2 ⎠ ⎝ 1 + β ⎠⎝ 1 + β ⎠ ⎝ (121) ⎠

⎛ β IC 4 = ⎜ ⎝ 1+ β IC 2

⎞ ⎛ IQ ⎟⎜ ⎠⎝ 2

So rπ 2 =

(120 )( 0.026 )

= 761 k Ω 0.0041 0.0041 = 0.158 mA/V gm2 = 0.026 100 ⇒ 24.4 M Ω ro 2 = 0.0041 (120 ) ( 0.026 ) = 6.29 k Ω rπ 4 = 0.496 0.496 gm4 = = 19.08 mA / V 0.026 100 ro 4 = = 202 k Ω 0.496 Now ⎡ ⎤ Vx Vx 1 1 = Vπ 4 ⎢ + + 0.158⎥ ⇒ which yields Vπ 4 = ro 2 6.29 761 24400 0.318 ( ) ro 2 ⎣⎢ ⎦⎥

From (1), Ix =

⎛ Vx Vx 1 ⎞ + + Vπ 4 ⎜ g m 4 − g m 2 − ⎟ ro 2 ro 4 ro 2 ⎠ ⎝

⎡ 1 ⎞⎤ ⎛ ⎜ 19.08 − 0.158 − ⎟ Ix ⎢ 1 V 1 24400 ⎠ ⎥ ⎝ ⎥ which yields Ro 2 = x = 135 k Ω =⎢ + + Vx ⎢ 24400 202 Ix ( 0.318 )( 24400 ) ⎥ ⎢ ⎥ ⎣ ⎦ 80 = 160 k Ω Now ro 6 = 0.5 Then Ro = Ro 2 ro 6 = 135 160 ⇒ Ro = 73.2 k Ω

(b) Ad = g mc Ro where g mc =

Δi vd / 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Δi = g m1Vπ 1 + g m 3Vπ 3 and Vπ 1 + Vπ 3 =

vd 2

⎛V ⎞ Also ⎜ π 1 + g m1Vπ 1 ⎟ rπ 3 = Vπ 3 ⎝ rπ 1 ⎠ ⎛1+ β ⎞ So Vπ 1 ⎜ ⎟ rπ 3 = Vπ 3 ⎝ rπ 1 ⎠ ⎛ 121 ⎞ Or Vπ 1 ⎜ ⎟ ( 6.29 ) = Vπ 3 ≅ Vπ 1 ⎝ 761 ⎠ v v Then 2Vπ 1 = d ⇒ Vπ 1 = d 2 4 ⎛v ⎞ ⎛v ⎞ So Δi = ( g m1 + g m 3 )Vπ 1 = ( 0.158 + 19.08 ) ⎜ d ⎟ = 9.62 ⎜ d ⎟ ⎝ 4⎠ ⎝ 2⎠ Δ i = 9.62 ⇒ Ad = ( 9.62 )( 73.2 ) ⇒ Ad = 704 So g mc = vd / 2 Now Rid = 2 Ri where Ri = rπ 1 + (1 + β ) rπ 3

Ri = 761 + (121)( 6.29 ) = 1522 k Ω Then Rid = 3.044 M Ω

______________________________________________________________________________________ 11.71 - Design Problem ______________________________________________________________________________________ 11.72

Input: −8 ≤ V d ≤ 8 mV Output: −0.8 ≤ Vo ≤ 0.8 V V 0.8 = 100 Ad = o = V d 0.008 Ad = g m (ro 2 ro 4 )

Let I Q = 0.5 mA, I DQ = 0.25 mA 1 = 160 k Ω (0.025)(0.25) 1 = = 100 k Ω (0.04)(0.25)

ro 2 = ro 4

100 = g m (160 100 ) ⇒ g m = 1.625 mA/V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ k′ g m = 2 ⎜⎜ n ⎝ 2

⎞⎛ W ⎞ ⎟⎟⎜ ⎟ I DQ ⎠⎝ L ⎠ n

⎛W ⎞ ⎛ 0.1 ⎞⎛ W ⎞ 1.625 = 2 ⎜ ⎟⎜ ⎟ (0.25) ⇒ ⎜ ⎟ = 52.8 2 L ⎝ ⎠⎝ ⎠ n ⎝ L ⎠n ______________________________________________________________________________________

11.73

For current source, I REF = I Q

( )( ) 0.8 = (2 I )(6 ) ⇒ I ≅ 66 μ A A = g (r r ) P = 2I Q V + − V − Q

d

o2

m

ro 2 =

ro 4 =

Q

o4

1 = 1515 k Ω ⎛ 0.066 ⎞ (0.02)⎜ ⎟ ⎝ 2 ⎠ 1 = 1010 k Ω ⎛ 0.066 ⎞ (0.03)⎜ ⎟ ⎝ 2 ⎠

240 = g m (1515 1010 ) ⇒ g m = 0.396 mA/V

⎛ k ′ ⎞⎛ W ⎞ g m = 2 ⎜⎜ n ⎟⎟⎜ ⎟ I DQ ⎝ 2 ⎠⎝ L ⎠ n ⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 0.396 = 2 ⎜ ⎟⎜ ⎟ (0.033) ⇒ ⎜ ⎟ = 23.76 ⎝ 2 ⎠⎝ L ⎠ n ⎝ L ⎠n ______________________________________________________________________________________

11.74

Ad = g m ( ro 2 Ro ) ≈ g m ro 2 1 ro 2 = λn I D = Ad

1

= 666.7 K

( 0.015 )( 0.1) = 400 = g m ( 666.7 )

g m = 0.60 mA/V ⎛ k′ ⎞⎛ W = 2 ⎜ n ⎟⎜ ⎝ 2 ⎠⎝ L

⎞ ⎟ ID ⎠

⎛ 0.08 ⎞⎛ W ⎞ 0.60 = 2 ⎜ ⎟⎜ ⎟ ( 0.1) ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ 0.090 = 0.004 ⎜ ⎟ ⎝L⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 22.5 ⎝ L ⎠1 ⎝ L ⎠ 2

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.75 Ad = g m ( Ro 4 Ro 6 )

where Ro 4 = ro 4 + ro 2 [1 + g m 4 ro 4 ] Ro 6 = ro 6 + ro8 [1 + g m 6 ro 6 ] We have ro 2 = ro 4 = ro 6 = ro8 =

1

( 0.015)( 0.040 ) 1

( 0.02 )( 0.040 )

= 1667 k Ω

= 1250 k Ω

⎛ 0.060 ⎞ gm4 = 2 ⎜ ⎟ (15 )( 0.040 ) = 0.268 mA/V ⎝ 2 ⎠ ⎛ 0.025 ⎞ gm6 = 2 ⎜ ⎟ (10 )( 0.040 ) = 0.141 mA/V ⎝ 2 ⎠ Then Ro 4 = 1667 + 1667 ⎡⎣1 + ( 0.268 )(1667 ) ⎤⎦ ⇒ 748 M Ω

Ro 6 = 1250 + 1250 ⎡⎣1 + ( 0.141)(1250 ) ⎤⎦ ⇒ 222.8 M Ω

(a) Ro = Ro 4 Ro 6 = 748 222.8 ⇒ Ro = 172 M Ω

(b)

Ad = g m 4 ( Ro 4 Ro 6 ) = ( 0.268 )(172000 ) ⇒ Ad = 46096

______________________________________________________________________________________ 11.76 (a) Ad = g m (ro 2 ro 4 ) g m = 2 K n I DQ = 2 (0.2)(0.06 ) = 0.2191 mA/V ro 2 = ro 4 =

1 = 666.7 k Ω (0.025)(0.06)

Ad = (0.2191)(666.7 666.7 ) = 73.0

(b) R o = ro 2 ro 4 = 333.3 k Ω (c) i D 3 = K p (υ SG 3 + VTP )

2

0.06 = 0.2(υ SG 3 − 0.3) ⇒ υ SG 3 = 0.8477 V 2

υ O = V + − υ SG 3 = 2.8 − 0.8477 = 1.9523 V i D1 = K n (υ GS1 − VTN )

2

0.06 = 0.2(υ GS1 − 0.3) ⇒ υ GS1 = 0.8477 V, ⇒ υ DS1 (sat ) = 0.5477 V 3

υ CM (max ) = υ O − υ DS1 (sat ) + υ GS1 = 1.9523 − 0.5477 + 0.8477 = 2.25 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.77 2 (a) i D 3 = K p (υ SG 3 + VTP ) 0.25 = 0.25(υ SG 3 − 0.4) ⇒ υ SG 3 = 1.4 V 2

Then υ GS 1 = 1.4 V, υ DS1 (sat ) = 1.0 V υ CM = υ O − υ DS1 (sat ) + υ GS1

(

)

3 = V + − 1.4 − 1.0 + 1.4 ⇒ V + = 4 V = −V −

(b) Ad = g m (ro 2 ro 4 )

g m = 2 K n I DQ = 2 (0.25)(0.25) = 0.50 mA/V ro 2 = ro 4 =

1

(0.02)(0.25)

= 200 k Ω

Ad = (0.50 )(200 200 ) = 50

______________________________________________________________________________________ 11.78

(a)

+ For vcm = +2V ⇒ V = 2.7 V

If I Q is a 2-transistor current source, V − = vcm − 0.7 − 0.7 V − = −3.4 V ⇒ V + = −V − = 3.4 V

(b) 100 = 1000 K 0.1 60 ro 4 = = 600 K 0.1 0.1 gm = = 3.846 mA/V 0.026 Ad = ( 3.846 ) (1000 600 ) ⇒ Ad = 1442 Ad = g m ( ro 2 ro 4 )

ro 2 =

______________________________________________________________________________________ 11.79 (a) (b)

V + = −V − = 3.4 V 75 = 1250 K 0.06 40 ro 4 = = 666.7 K 0.06 0.06 gm = = 2.308 mA/V 0.026 Ad = ( 2.308 ) (1250 666.7 ) ro 2 =

Ad = 1004

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.80 (a) I E 2 = 0.25 mA, I B 2 = 0.001656 mA, I C 2 = 0.2483 mA I D1 = 0.25 + 0.001656 = 0.25166 mA g m1 = 2 K n I DQ = 2 (0.2 )(0.25166) = 0.4487 mA/V g m2 = rπ 2 = g mC

I C 2 0.2483 = = 9.55 mA/V 0.026 VT

(150)(0.026) = 15.71 k Ω 0.2483

(0.4487 )[1 + (9.55)(15.71)] = 8.42 mA/V = 1 + (0.4487 )(15.71)

(b) I E 2 = 0.45 mA, I B 2 = 0.002980 mA, I C 2 = 0.4470 mA I D1 = 0.05 + 0.00298 = 0.05298 mA g m1 = 2 (0.2 )(0.05298) = 0.2059 mA/V

0.4470 = 17.19 mA/V 0.026 = 8.725 k Ω

g m2 = rπ 2 g mC =

(0.2059)[1 + (17.19)(8.725)] = 11.12 mA/V 1 + (0.2059 )(8.725)

______________________________________________________________________________________ 11.81 r0 ( M 2 ) = r0 ( Q2 ) =

1

λn I DQ

=

1

( 0.01)( 0.2 )

= 500 kΩ

VA 80 = = 400 kΩ I CQ 0.2

g m ( M 2 ) = 2 K n I DQ = 2

( 0.2 )( 0.2 )

= 0.4 mA/V Ad = g m ( M 2 ) ⎡⎣ r0 ( M 2 ) r0 ( Q2 ) ⎤⎦ = 0.4 ⎡⎣500 400 ⎤⎦ ⇒ Ad = 88.9 Acm = 0

C M RR = ∞

dB and If the IQ current source is ideal, ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.82 (a)

0.7 = 0.0875 mA 8 I Q = 0.5 = I R1 + I E 2 = 0.0875 + I E 2 ⇒ I E 2 = 0.4125 mA

(b) I R1 =

I B 2 = 0.002279 mA, I C 2 = 0.41022 mA I D1 = I R1 + I B 2 = 0.0875 + 0.002279 = 0.08978 mA

g m1 = 2 (0.7 )(0.08978) = 0.5014 mA/V g m2 =

0.41022 = 15.78 mA/V 0.026

rπ 2 = 11.41 k Ω

(

)

(c) V o = − g m1V sg + g m 2Vπ R L

(

)

Vπ = g m1V sg (R1 rπ 2 )

[

]

V o = − g m1V sg + g m 2 g m1V sg (R1 rπ 2 ) R L

[

]

− V o = V sg g m1 + g m 2 g m1 (R1 rπ 2 ) R L

Now V sg = Vo − Vi

[

]

[

]

So V o 1 + g m1 (1 + g m 2 (R1 rπ 2 ))R L = Vi g m1 1 + g m 2 (R1 rπ 2 ) R L Aυ = Aυ

[

]

g m1 1 + g m 2 (R1 rπ 2 ) R L Vo = Vi 1 + g m1 1 + g m 2 (R1 rπ 2 ) R L

[

] (0.5014)[1 + (15.78)(8 11.41)]R = 1 + (0.5014 )[1 + (15.78)(8 11.41)]R

=

L

L

37.71R L 1 + 37.71R L

(i) For R L = 10 k Ω , Aυ =

(37.71)(10) = 0.99736 1 + (37.71)(10 )

(ii) For R L = 100 k Ω , Aυ =

(37.71)(100) = 0.99973 1 + (37.71)(100)

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.83 0.7 (a) I R1 = = 0.0875 mA 8 I Q = 0.8 = I R1 + I E 2 = 0.0875 + I E 2 ⇒ I E 2 = 0.7125 mA I B 2 = 0.003936 mA, I C 2 = 0.7086 mA I D1 = I R1 + I B 2 = 0.0875 + 0.003936 = 0.091436 mA g m1 = 2 (0.7 )(0.091436 ) = 0.506 mA/V

0.7086 = 27.25 mA/V 0.026 rπ 2 = 6.605 k Ω (b) I x = g m 2Vπ + g m1V sg g m2 =

(

)

Vπ = g m1V sg (R1 rπ 2 )

Now V sg = V x Ix 1 = = g m1 + g m 2 g m1 (R1 rπ 2 ) = 0.506 + (0.506)(27.25)(8 6.605) = 50.39 mA/V V x Ro

Or R o = 19.8 Ω ______________________________________________________________________________________

11.84 (a)

(1)

g m 2Vπ + g m 2Vπ +

(2) Vπ =

Then From (1)

Vo − ( −Vπ ) ro 2 Vo − ( −Vπ ) ro 2

g m1Vi ⎛ 1 1⎞ ⎜ + ⎟ ⎝ ro1 rπ ⎠

=0 = g m1Vi +

⎛ 1 1⎞ −Vπ −Vπ + or 0 = g m1Vi − Vπ ⎜ + ⎟ ro1 rπ ⎝ ro1 rπ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ Vo 1 ⎞ =0 ⎜ g m 2 + ⎟ Vπ + r ro 2 o2 ⎠ ⎝ ⎛ 1 ⎞ ⎜ gm2 + ⎟ ro 2 ⎠ ⎛ 1 ⎞ Vo = −ro 2 ⎜ g m 2 + ⎟ Vπ = − ro 2 g m1Vi ⎝ r ⎛ 1 1⎞ o2 ⎠ ⎝ ⎜ + ⎟ r r ⎝ o1 π ⎠ ⎛ 1 ⎞ − g m1ro 2 ⎜ g m 2 + ⎟ ro 2 ⎠ V ⎝ Av = o = Vi ⎛ 1 1⎞ ⎜ + ⎟ ⎝ ro1 rπ ⎠ Now

( 0.25 )( 0.025) = 0.158 mA / V

g m1 = 2 K n I Q = 2 gm2 = ro1 =

VT 1

λ IQ

= =

0.025 = 0.9615 mA / V 0.026 1

( 0.02 )( 0.025)

= 2000 k Ω

VA 50 = = 2000 k Ω I Q 0.025

ro 2 = rπ =

IQ

β VT IQ

=

(100 )( 0.026 ) 0.025

= 104 k Ω

Then 1 ⎞ ⎛ − ( 0.158 )( 2000 ) ⎜ 0.9615 + ⎟ 2000 ⎠ ⎝ Av = ⇒ Av = −30039 1 ⎞ ⎛ 1 + ⎜ ⎟ ⎝ 2000 104 ⎠ Vi = 0 ⇒ g m1Vi = 0

To find Ro; set

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I x = g m 2Vπ +

Vx − ( −Vπ )

Vπ = − I x ( ro1 rπ )

ro 2

Then ⎛ V 1 ⎞ I x = ⎜ g m 2 + ⎟ ( − I x ) ( ro1 rπ ) + x ro 2 ⎠ ro 2 ⎝ Combining terms, Ro =

⎡ ⎛ Vx 1 ⎞⎤ = ro 2 ⎢1 + ( ro1 rπ ) ⎜ g m 2 + ⎟ ⎥ Ix ro 2 ⎠ ⎦ ⎝ ⎣

1 ⎞⎤ ⎡ ⎛ = 2000 ⎢1 + ( 2000 104 ) ⎜ 0.9615 + ⎟ ⎥ ⇒ Ro = 192.2 M Ω 2000 ⎝ ⎠⎦ ⎣

(b)

(1)

g m 3Vgs 3 + g m 3Vgs 3 +

(2) (3)

Vo − ( −Vgs 3 ) ro3 Vo − ( −Vgs 3 )

= g m 2Vπ 2 +

−Vgs 3 − ( −Vπ 2 ) ro 2

−Vgs 3 − ( −Vπ 2 ) ( −Vπ 2 ) Vπ 2 + g m 2Vπ 2 + = g m1Vi + rπ 2 ro 2 ro1

Vπ 2 =

From (2), Then (3) or

ro 3

=0

Vgs 3 ⎛ 1 ⎞ ro 2 ⎜ g m 2 + ⎟ r o2 ⎠ ⎝

Vgs 3 ⎛ 1 1 1⎞ Vπ 2 ⎜ + gm2 + + ⎟ = g m1Vi + ro 2 ro1 ⎠ ro 2 ⎝ rπ 2

⎛ 1 ⎞ Vgs 3 or 0 = Vπ 2 ⎜ g m 2 + ⎟ − ro 2 ⎠ ro 2 ⎝

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Vgs 3 ⎡ 1 1 1⎤ + ⎥ = g m1Vi + ⎢ + gm2 + ro 2 ro1 ⎦ ro 2 ⎛ 1 ⎞ r ro 2 ⎜ g m 2 + ⎟ ⎣ π 2 r o2 ⎠ ⎝ Vgs 3 Vgs 3 1 1 ⎤ ⎡ 1 + 0.9615 + + = 0.9615Vi + 1 ⎞ ⎢⎣104 2000 2000 ⎥⎦ 2000 ⎛ 2000 ⎜ 0.9615 + ⎟ 2000 ⎠ ⎝ Vgs 3 = 1.83 × 105 Vi Vgs 3

Then

⎛ −Vo 1 ⎞ 1 ⎞ ⎛ 5 or Vo = −2000 ⎜ 0.158 + ⎜ g m 3 + ⎟ Vgs 3 = ⎟ (1.83 × 10 )Vi 2000 ⎠ ro3 ⎠ ro 3 ⎝ ⎝ From (1), V Av = o = −5.80 × 107 Vi

To find Ro (1) (2) (3)

I x = g m 3Vgs 3 + g m 3Vgs 3 +

Vx − ( −Vgs 3 ) ro 3

Vx − ( −Vgs 3 ) ro3

= g m 2Vπ 2 +

Vπ 2 = − I x ( ro1 rπ 2 ) ⎛ 1 ⎞ V From (1) I x = Vgs 3 ⎜ g m 3 + ⎟ + x r ro 3 o3 ⎠ ⎝ 1 ⎞ Vx ⎛ I x = Vgs 3 ⎜ 0.158 + ⎟+ 2000 ⎠ 2000 ⎝ V Ix − x 2000 So Vgs 3 = 0.1585

−Vgs 3 − ( −Vπ 2 ) ro 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ From (2), ⎡ ⎛ 1 1 ⎤ V 1 ⎞ + ⎥ + x = Vπ 2 ⎜ g m 2 + ⎟ Vgs 3 ⎢ g m 3 + r r r r o3 o2 ⎦ o3 o2 ⎠ ⎣ ⎝ 1 1 ⎤ Vx 1 ⎞ ⎡ ⎛ + + = Vπ 2 ⎜ 0.9615 + Vgs 3 ⎢ 0.158 + ⎟ 2000 2000 ⎥⎦ 2000 2000 ⎠ ⎣ ⎝ V ⎡ I − Vx / 2000 ⎤ Then ⎢ x ( 0.159 ) + x = − I x ( 2000 104 ) ( 0.962 ) ⎥ 2000 ⎣ 0.1585 ⎦ Vx We find Ro = = 6.09 × 1010 Ω Ix

______________________________________________________________________________________ 11.85 Assume emitter of Q1 is capacitively coupled to signal ground. ⎛ 80 ⎞ I CQ = 0.2 ⎜ ⎟ = 0.1975 mA ⎝ 81 ⎠ 0.2 = 0.00247 mA I DQ = 81 ( 80 )( 0.026 ) = 10.5 k Ω rπ = 0.1975 0.1975 = 7.60 mA / V g m ( Q1 ) = 0.026 gm ( M1 ) = 2 K n I D = 2 g m ( M 1 ) = 0.0445 mA / V

( 0.2 )( 0.00247 )

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Vi = Vgs + Vπ and Vπ = g m ( M 1 )Vgs rπ or Vgs = ⎛ 1 Then Vi = Vπ ⎜⎜1 + ⎝ g m ( M 1 ) rπ

Vπ g m ( M 1 ) rπ

⎞ Vi ⎟⎟ or Vπ = ⎛ 1 ⎠ ⎜⎜ 1 + g M ( m 1 ) rπ ⎝ − g m ( Q1 ) RC V Vo = − g m ( Q1 )Vπ RC ⇒ Av = o = Vi ⎛ ⎞ 1 ⎜⎜ 1 + g ( M ) r ⎟⎟ m 1 π ⎠ ⎝ − ( 7.60 )( 20 ) ⇒ Av = −48.4 Then Av = ⎛ ⎞ 1 + 1 ⎜⎜ ( 0.0445 )(10.5 ) ⎟⎟ ⎝ ⎠

⎞ ⎟⎟ ⎠

______________________________________________________________________________________ 11.86 rπ 11 =

(120)(0.026) = 15.6 k Ω

0.2 R E′ = R3 rπ 11 = 0.2 15.6 = 0.1975 k Ω 0.2 = 7.692 mA/V 0.026 V 120 ro11 = A11 = = 600 k Ω I C11 0.2 RC11 = ro11 (1 + g m11 R E′ ) = 600[1 + (7.692 )(0.1975)] = 1512 k Ω g m11 =

RC 7 = ro 7 =

V A7 60 = = 300 k Ω I C 7 0.2

Z = RC 7 RC11 = 300 1512 = 250 k Ω rπ 8 = I C9

(120)(0.026) = 3.12 k Ω

1 1 ⎛ 120 ⎞ = ⎜ ⎟ = 0.008264 mA 120 ⎝ 121 ⎠

rπ 9 =

(120)(0.026) = 377.5 k Ω 0.008264

⎛r +Z ⎞ ⎛ 377.5 + 250 ⎞ rπ 8 + ⎜⎜ π 9 ⎟⎟ 3.12 + ⎜ ⎟ 121 ⎝ 121 ⎠ ⎝ ⎠ =5 Now R o = R 4 = 5 0.06864 121 121

Or R o = 67.7 Ω ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.87 Ri = rπ 1 + (1 + β ) rπ 2 rπ 2 =

(100 )( 0.026 ) 0.5

= 5.2 kΩ

(100 )( 0.026 ) (100 ) ( 0.026 ) = = 520 kΩ 0.5 ( 0.5 /100 ) Ri = 520 + (101)( 5.2 ) ⇒ Ri ≅ 1.05 MΩ (100 )( 0.026 ) rπ 3 + 50 2

rπ 1 =

R0 = 5 R0 = 5

101

, rπ 3 =

= 2.6 kΩ

1

2.6 + 50 = 5 0.521 ⇒ R0 = 0.472 kΩ 101

⎛V ⎞ V0 = − ⎜ π 3 + g m 3Vπ 3 ⎟ ( 5 ) ⎝ rπ 3 ⎠ ⎛ 1+ β ⎞ V0 = −Vπ 3 ⎜ ⎟ ( 5) ⎝ rπ 3 ⎠

(V − V ) Vπ 3 = g m 2Vπ 2 + 0 π 3 rπ 3 50 ⎛ 1 1 ⎞ V g m 2Vπ 2 = Vπ 3 ⎜ + ⎟− 0 r 50 ⎝ π3 ⎠ 50 Vπ 2

Vin = Vπ 1 + Vπ 2 gm2

(2)

⎛V ⎞ = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2 r ⎝ π1 ⎠ ⎛ 1+ β ⎞ = Vπ 1 ⎜ ⎟ rπ 2 ⎝ rπ 1 ⎠

and

(1)

(3) (4)

0.5 = = 19.23 mA/V 0.026

Then ⎛ 101 ⎞ V0 = −Vπ 3 ⎜ ⎟ ( 5 ) ⇒ Vπ 3 = −V0 ( 0.005149 ) ⎝ 2.6 ⎠

(1)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ And 1 ⎞ V ⎛ 1 + ⎟− 0 19.23Vπ 2 = −V0 ( 0.005149 ) ⎜ ⎝ 2.6 50 ⎠ 50 = −V0 ( 0.02208 ) Or Vπ 2 = −V0 ( 0.001148 )

(2)

And

Vπ 1 = Vin − Vπ 2 = Vin + V0 ( 0.001148 )

(4)

So ⎛ 101 ⎞ −V0 ( 0.001148 ) = ⎣⎡Vin + V0 ( 0.001148 ) ⎤⎦ ⎜ ⎟ ( 5.2 ) ⎝ 520 ⎠ (3) −V0 ( 0.001148 ) − V0 ( 0.001159 ) = Vin (1.01) ⇒ Av =

V0 = −438 Vin

______________________________________________________________________________________ 11.88 I2 =

5 = 1 mA 5

1 + 0.8 = 2.21 V 0.5 2.21 − ( −5 ) I1 = = 0.206 mA 35

VGS 2 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V0 = ( g m 2Vgs 2 ) ( R2 r02 )

Vgs 2 = ( g m1Vsg1 ) ( r01 R1 ) − V0 and Vsg1 = −Vin So Vgs 2 = − ( g m1Vin ) ( r01 R1 ) − V0 Then V0 = g m 2 ( R2 r02 ) ⎡⎣ − ( g m1Vin ) ( r01 R1 ) − V0 ⎤⎦ Av =

V0 − g m 2 ( R2 r02 ) g m1 ( r01 R1 ) = Vin 1 + g m 2 ( R2 r02 )

gm2 = 2 Kn2 I D2 = 2

( 0.5 )(1) = 1.414 mA / V

g m1 = 2 K p1 I D1 = 2

( 0.2 )( 0.206 ) = 0.406 mA / V

r01 = r02 =

1

λ1 I D1 1

λ2 I D 2

=

1

( 0.01)( 0.206 )

=

1

( 0.01)(1)

= 485 kΩ

= 100 kΩ

R2 r02 = 5 100 = 4.76 kΩ R1 r01 = 35 485 = 32.6 kΩ Then Av =

− (1.414 )( 4.76 )( 0.406 )( 32.6 ) 1 + (1.414 )( 4.76 )

So ⇒ Av = −11.5

Output Resistance—From the results for a source follower in Chapter 4. R0 =

1 R2 r02 gm2

=

1 5 100 1.414

= 0.707 4.76 So R0 = 0.616 kΩ

______________________________________________________________________________________ 11.89 5−0 = 5kΩ 1 2 2 I 2 = K p (V SG + VTP ) ⇒ 1 = 1(V SG − 0.8) ⇒ V SG = 1.8 V

(a) R 2 =

R1 =

5 − (− 1.8) = 27.2 k Ω 0.25

(b) g m1 = 2 (0.5)(0.25) = 0.7071 mA/V g m 2 = 2 (1)(1) = 2 mA/V ro1 =

1

(0.02)(0.25)

= 200 k Ω , ro 2 =

1 = 50 k Ω (0.02)(1)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c)

V o1 = − g m1Vin (ro1 R1 ) V sg 2 = Vo − V o1

Vo Vo + + g m 2V sg 2 = 0 R 2 ro 2 ⎛ 1 1 ⎞ ⎟ + g m 2 Vo + g m1Vin (ro1 R1 ) = 0 V o ⎜⎜ + ⎟ R r o2 ⎠ ⎝ 2

[

]

⎞ ⎛ 1 1 V o ⎜⎜ + + g m 2 ⎟⎟ = − g m1 g m 2 (ro1 R1 )Vin ⎠ ⎝ R 2 ro 2 − g m1 g m 2 (ro1` R1 ) − (0.7071)(2)(200 27.2 ) V = Aυ = o = = −15.25 Vin ⎛ 1 ⎞ ⎛1 1 ⎞ 1 + 2⎟ ⎜ ⎟ ⎜ + ⎜ R + r + g m2 ⎟ ⎝ 5 50 ⎠ o2 ⎝ 2 ⎠ V V (d) I x = x + x + g m 2V x ro 2 R 2 R o = ro 2 R 2

1 1 = 50 5 = 4.545 0.5 2 g m2

R o = 0.450 k Ω ______________________________________________________________________________________

11.90

5−0 = 20 k Ω 0.25 − 0.7 − (− 5) R E1 = = 17.2 k Ω 0.25 5 − 0.7 = 17.2 k Ω RC = 0.25 0 − (− 5) RE 2 = = 2.5 k Ω 2 υ g (b) Ad 1 = o 2 = m1 (R rπ 3 ) υd 2 (a) R =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 0.25 g m1 = = 9.615 mA/V 0.026 (120)(0.026) = 12.48 k Ω rπ 3 = 0.25 ( 9.615) (20 12.48) = 36.94 Ad 1 = 2 A3 = − g m3 (RC Ri 4 ) Ri 4 = rπ 4 + (1 + β )R E 2

0.25 = 9.615 mA/V 0.026 (120)(0.026) = 1.56 k Ω rπ 4 = 2 ( )( Ri 4 = 1.56 + 121 2.5) = 304 k Ω g m3 =

A3 = −(9.615)(17.2 304 ) = −156.5

A4 =

Now

(1 + β )R E 2 (121)(2.5) = 0.995 = rπ 4 + (1 + β )R E 2 1.56 + (121)(2.5)

Ad =

(c) Acm1 =

υo = Ad 1 ⋅ A3 ⋅ A4 = (36.94 )(− 156.5)(0.995) = −5752 υd

− g m1 (R rπ 3 ) 1+

2(1 + β )R o rπ 1

rπ 1 = Acm1 = Acm

(120)(0.026) = 12.48 k Ω

0.25 − (9.615)(20 12.48)

= −0.01905 2(121)(200 ) 12.48 = Acm1 ⋅ A3 ⋅ A4 = (− 0.01905)(− 156.5)(0.995) = 2.966 1+

⎛ 5752 ⎞ CMRRdB = 20 log 10 ⎜ ⎟ = 65.8 dB ⎝ 2.966 ⎠ ______________________________________________________________________________________ 11.91 a.

b.

RC1 =

10 − v01 10 − 2 = ⇒ RC1 = 80 kΩ I C1 0.1

RC 2 =

10 − v04 10 − 6 = ⇒ RC 2 = 20 kΩ IC 4 0.2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Ad 1 =

v01 − v02 = − g m1 ( RC1 rπ 3 ) vd

0.1 = 3.846 mA/V 0.026 (180 )( 0.026 ) rπ 3 = = 23.4 kΩ 0.2 Ad 1 = − ( 3.846 ) ( 80 23.4 ) ⇒ Ad 1 = −69.6

g m1 =

Ad 2 =

v04 1 = g m 4 RC 2 v01 − v02 2

0.2 = 7.692 mA/V 0.026 1 Ad 2 = ( 7.692 )( 20 ) = 76.9 2 Then Ad = ( 76.9 )( −69.6 ) ⇒ Ad = −5352 gm4 =

______________________________________________________________________________________

11.92 a.

Neglect the effect of r0 in determining the differential-mode gain. v 1 Ad 1 = 02 = g m 2 ( RC Ri 3 ) where Ri 3 = rπ 3 + (1 + β ) RE vd 2 A2 = I1 =

− β RC 2 rπ 3 + (1 + β ) RE

12 − 0.7 − ( −12 ) R1

=

23.3 = 1.94 mA ≈ I C 5 12

1 ⋅ (1.94 ) gm2 = 2 = 37.3 mA/V 0.026 ( 200 )( 0.026 ) rπ 3 = IC 3 1 (1.94 )(8) = 4.24 V 2 4.24 − 0.7 IC 3 = = 1.07 mA 3.3 ( 200 )( 0.026 ) rπ 3 = = 4.86 kΩ 1.07 Ri 3 = 4.86 + ( 201)( 3.3) = 668 kΩ

v02 = 12 −

Ad 1 =

1 ( 37.3) ⎡⎣8 668⎤⎦ = 147.4 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then Ad = Ad 1 ⋅ A2 = (147.4 )( −1.197 ) ⇒ Ad = −176 R0 = r05 = Acm1 =

− g m 2 ( RC Ri 3 ) 1+

rπ 2 =

VA 80 = = 41.2 kΩ I C 5 1.94

2 (1 + β ) R0 rπ 2

( 200 )( 0.026 )

= 5.36 kΩ 1 ⋅ (1.94 ) 2 − ( 37.3) ( 8 668 ) = −0.09539 Acm1 = 2 ( 201)( 41.2 ) 1+ 5.36 A2 = −1.197 Acm = ( −0.09539 )( −1.197 ) ⇒ Acm = 0.114

b. vd = v1 − v2 = 2.015sin ω t − 1.985sin ω t

vd = 0.03sin ω t ( V ) v +v vcm = 1 2 = 2.0sin ω t 2 v03 = Ad vd + Acm vcm

= ( −176 )( 0.03) + ( 0.114 )( 2 )

Or

v03 = −5.052sin ω t

Ideal, Acm = 0 So v03 = Ad vd = ( −176 )( 0.03) v03 = −5.28sin ω t

c.

Rid = 2rπ 2 = 2 ( 5.36 ) ⇒ Rid = 10.72 kΩ 2 Ricm ≅ 2 (1 + β ) R0 (1 + β ) r0 r0 =

VA 80 = = 82.5 kΩ I C 2 1 ⋅ 1.94 ( ) 2

2 Ricm = ⎡⎣ 2 ( 201)( 41.2 ) ⎤⎦ ⎡⎣( 201)( 82.5 ) ⎤⎦ = 16.6 MΩ 16.6 MΩ So ⇒ Ricm = 4.15 MΩ

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.93 a. I1 =

24 − VGS 4 2 = kn (VGS 4 − VTh ) R1

24 − VGS 4 = ( 55 )( 0.2 ) (VGS 4 − 2 )

2

24 − VGS 4 = 11 (VGS2 4 − 4VGS 4 + 4 ) 11VGS2 4 − 43VGS 4 + 20 = 0 VGS 4 =

43 ±

( 43)

2

− 4 (11) ( 20 )

2 (11)

= 3.37 V

24 − 3.37 = 0.375 mA = I Q 55 ⎛ 0.375 ⎞ v02 = 12 − ⎜ ⎟ ( 40 ) = 4.5 V ⎝ 2 ⎠ v02 − VGS 3 2 = I D 3 = kn (VGS 3 − VTh ) R5 I1 =

4.5 − VGS 3 = ( 0.2 )( 6 ) (VGS2 3 − 4VGS 3 + 4 ) 1.2VGS2 3 − 3.8VGS 3 + 0.3 = 0 VGS 3 = I D3 =

3.8 ±

( 3.8 )

2

− 4 (1.2 ) ( 0.3)

2 (1.2 )

4.5 − 3.09 = 0.235 mA 6

= 3.09 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

( 0.2 ) ⎛⎜

0.375 ⎞ ⎟ ⎝ 2 ⎠

gm2 = 2 Kn I D2 = 2

= 0.387 mA/V 1 1 Ad 1 = g m 2 RD = ( 0.387 )( 40 ) ⇒ Ad 1 = 7.74 2 2 − g m 3 RD 2 A2 = 1 + g m 3 R5 g m3 = 2 K n I D3 = 2

( 0.2 )( 0.235 )

= 0.434 mA/V − ( 0.434 ) ( 4 )

A2 =

1 + ( 0.434 )( 6 )

= −0.482

So Ad = Ad 1 ⋅ A2 = ( 7.74 ) ( −0.482 ) ⇒ Ad = −3.73 R0 = r05 =

1 1 = = 133 kΩ λ I Q ( 0.02 )( 0.375 )

− ( 0.387 ) ( 40 ) − g m 2 RD = 1 + 2 g m 2 R0 1 + 2 ( 0.387 ) (133)

Acm1 =

= −0.149 Acm = ( −0.149 )( −0.482 ) ⇒ Acm = 0.0718

b.

vd = v1 − v2 = 0.3sin ω t v1 + v2 = 2sin ω t 2 v03 = Ad vd + Acm vcm = ( −3.73)( 0.3) + ( 0.0718 )( 2 ) ⇒ v03 = −0.975sin ω t ( V )

vcm =

Ideal, Acm = 0

v03 = Ad vd = ( −3.73)( 0.3)

Or

⇒ v03 = −1.12sin ω t ( V )

______________________________________________________________________________________

11.94

(a) Ad =

β RC rπ + R B

Assuming I CQ ≅ I EQ , Ad =

(150)(10) 9.75 + 0.5

rπ =

= 146

(150)(0.026) = 9.75 k Ω 0.4

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) C M ≅ C μ (1 + g m RC ) 0.4 = 15.38 mA/V 0.026 = (0.2)[1 + (15.38)(10 )] = 30.96 pF gm =

CM

(c)

fH =

2π (rπ

1

RB

1

)(Cπ + C ) = 2π (9.75 0.5)×10 (1.2 + 30.96)×10 3

−12

M

f H = 10.4 MHz ______________________________________________________________________________________

11.95

(a)

fZ =

(b) rπ =

R eq

1 1 = ⇒ f Z = 39.8 kHz 6 2πR o C o 2π 10 × 10 0.4 × 10 −12

(

)(

)

(150)(0.026) = 9.75 k Ω 0.4

⎛ R ⎞ R o ⎜⎜1 + B ⎟⎟ (10)⎛⎜1 + 0.5 ⎞⎟ r π ⎝ ⎠ ⎝ 9.75 ⎠ = = 2(1 + β )R o 0.5 2(151)(10,000) R 1+ + 1+ B + 9.75 9.75 rπ rπ

Or R eq = 33.94 Ω fP =

1 1 = 2πReq C o 2π (33.94) 0.4 × 10 −12

(

)

f P = 11.7 GHz ______________________________________________________________________________________

11.96 fT =

a.

From Equation (7.73),

gm

2π ( Cπ + Cμ )

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 = 38.46 mA/V 0.026 38.46 × 10−3 Then 800 × 106 = 2π ( Cπ + Cμ ) gm =

Or Cπ + Cμ = 7.65 × 10 −12 F = 7.65 pF And Cπ = 6.65 pF CM = Cμ (1 + g m RC ) = 1 ⎣⎡1 + ( 38.46 )(10 ) ⎦⎤ = 386 pF 1 fH = 2π ⎡⎣ rπ RB ⎤⎦ ( Cπ + CM ) (120 )( 0.026 ) = 3.12 kΩ rπ = 1 1 fH = 2π ⎡⎣3.12 1⎤⎦ × 103 × ( 6.65 + 386 ) × 10 −12 Or f H = 535 kHz fZ =

b.

From Equation (11.140),

1 1 = 2π R0 C0 2π (10 × 106 )(10−12 )

f Z = 15.9 kHz

Or ______________________________________________________________________________________ 11.97 1 β RC 2 (a) Aυ = rπ + (1 + β )R E rπ =

(120)(0.026) = 12.48 k Ω

0.25 1 (120)(8) 2 = 19.5 Aυ = 12.48 + (121)(0.1) 1 (120)(8) 2 = 11.2 (b) Aυ = 12.48 + (121)(0.25) ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 12 12.1 (a) A f =

A 5 × 10 4 ⇒ β = 9.98 × 10 − 3 ⇒ 100 = 1 + Aβ 1 + 5 × 10 4 β

(

)

A ⇒ A = 2000 1 + A(0.012 ) ______________________________________________________________________________________

(b) 80 =

12.2 (a) A f =

A − 10 5 ⇒ −80 = ⇒ β = −0.01249 1 + Aβ 1 + − 10 5 β

(

)

− 5 × 10 = −66.58 1 + − 5 × 10 4 (− 0.015) ______________________________________________________________________________________ 4

(b) A f =

(

)

A 1 + Aβ

β = 0.15

12.3 Af =

(a)

T = Aβ T =∞

(i) 4 3 (ii) A = 80 dB ⇒ A = 10 ⇒ T = 1.5 × 10 (iii) T = 15 Af =

(i) (ii)

1

β

= 6.667

Af = 6.662 A = 6.25

(iii) f (b) (i) T = ∞ 3 (ii) T = 2.5 × 10 (iii) T = 25 Af =

(i) (ii)

1

β

= 4.00

Af = 3.9984 A = 3.846

(iii) f ______________________________________________________________________________________ 12.4 (a) A 1 ≅ = 125 1 + Aβ β β = 0.0080 Af =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

Af = (125)(0.9975) = 124.6875 124.6875 =

A 1 + (0.008) A

124.6875 [1 + (0.008) A] = A 124.6875 = A [1 − 0.9975] A = 49,875

______________________________________________________________________________________ 12.5 A − 2 × 10 4 ⇒ −80 = ⇒ β = −0.01245 1 + Aβ 1 + − 2 × 10 4 β

(a) A f = (b)

dA f

(

=

Af

)

1 dA ⋅ (1 + Aβ ) A

1

dA dA ⇒ = 2.5% A 1 + − 2 × 10 (− 0.01245) A ______________________________________________________________________________________ 0.01 =

[ (

4

]

)

⋅

12.6 dA f Af

=

1 dA ⋅ 1 + βA A

1 ⋅ (0.10 ) ⇒ 1 + β A = 100 1 + βA 100 − 1 β= ⇒ β = 1.98 × 10 −3 5 × 10 4 5 × 10 4 A Now A f = = ⇒ A f = 500 1 + βA 1 + 1.98 × 10 − 3 5 × 10 4 ______________________________________________________________________________________ 0.001 =

(

)(

)

12.7 ⎛ A1 ⎞⎛ A2 ⎞ ⎟⎟⎜⎜ ⎟⎟ (a) Aυf = ⎜⎜ ⎝ 1 + A1 β 1 ⎠⎝ 1 + A2 β 1 ⎠ ⎛ 200 ⎞⎛ 10 ⎞ ⎟⎟⎜⎜ ⎟⎟ 50 = ⎜⎜ ⎝ 1 + 200 β 1 ⎠⎝ 1 + 10 β 1 ⎠ (1 + 200β 1 )(1 + 10β 1 ) = (200)(10) = 40 50

Then 2000 β 12 + 210 β 1 − 39 = 0 ⇒ β 1 = 0.096685 Aυf =

50 =

A1 A2 1 + A1 A2 β 2

2000 ⇒ β 2 = 0.0195 1 + 2000 β 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) From circuit (a): ⎛ ⎞⎛ ⎞ 200 9 ⎟⎟⎜⎜ ⎟⎟ = 47.33 ⇒ (− 5.43% ) Aυf = ⎜⎜ ⎝ 1 + (200 )(0.096685) ⎠⎝ 1 + (9 )(0.096685) ⎠ For circuit (b): 1800 Aυf = = 49.86 ⇒ (− 0.28% ) 1 + (1800)(0.0195) ______________________________________________________________________________________ 12.8

(a)

VO = (−10)(−15)( −20)Vε = −3000Vε Vε = β VO + VS

So VO = −3000( β VO + VS ) We find Avf =

For

VO −3000 = VS 1 + 3000 β

Avf = −120 =

−3000 ⇒ β = 0.008 1 + 3000 β

(b) Now VO = (−9)(−13.5)(−18)Vε = −2187Vε Then Avf =

−2187 −2187 = = −118.24 1 + 2187 β 1 + 2187(0.008)

% change =

120 − 118.24 × 100 ⇒ 1.47% change 120

______________________________________________________________________________________ 12.9 (a) Aυ f H = Aυ f f C

(5 ×10 )(10) = (25) f ⇒ f = 20 kHz A f (10 )(8) = 40 A = = 4

C

(b)

υ

υf

C

5

H

fC 20 × 10 3 ______________________________________________________________________________________ 12.10 fC =

(5 ×10 )(10) ⇒ f 4

C = 10 kHz 50 10 4 (10 ) ⇒ f C = 2 kHz (b) f C = 50 ______________________________________________________________________________________

(a)

( )

12.11

(a) (i) Aυ = (ii) 75 =

Aυf f C fH

=

(75)(35 ×10 3 ) = 5.25 ×10 5 5

5.25 × 10 ⇒ β = 0.01333 1 + 5.25 × 10 5 β 5

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(

)

(b) A = 5.25 × 10 5 (0.9) = 4.725 × 10 5 Aυf =

4.725 × 10 5 = 74.99 1 + 4.725 × 10 5 (0.01333)

(

(

)

)

Aυ f H 4.725 × 10 5 (5) = ⇒ f C = 31.5 kHz Aυf 74.99

fC =

______________________________________________________________________________________ 12.12

Low freq. 100 =

Af =

A0 1 + A0 β

5000 ⇒ β = 0.0098 1 + (5000) β

Freq. response

Af =

A 1 + Aβ

5000 f ⎞⎛ f ⎞ ⎛ ⎜ 1 + j f ⎟⎜1 + j f ⎟ 1 ⎠⎝ 2 ⎠ = ⎝ (5000)(0.0098) 1+ f ⎞⎛ f ⎞ ⎛ ⎜1 + j f ⎟ ⎜1 + j f ⎟ ⎝ 1 ⎠⎝ 2 ⎠ 5000 f ⎞⎛ f ⎞ ⎛ ⎜1 + j f ⎟⎜1 + j f ⎟ + 49 ⎝ 1 ⎠⎝ 2 ⎠ 5000 = f f ⎛ jf ⎞⎛ jf 1 + j + j + ⎜ ⎟⎜ f1 f 2 ⎝ f1 ⎠⎝ f 2 =

=

⎞ ⎟ + 49 ⎠

5000 f f ⎛ jf ⎞⎛ jf ⎞ 50 + j + j + ⎜ ⎟⎜ ⎟ f1 f 2 ⎝ f1 ⎠⎝ f 2 ⎠

Also Af =

Af 0 f ⎞⎛ f ⎞ ⎛ ⎜ 1 + j f ⎟⎜1 + j f ⎟ A ⎠⎝ B ⎠ ⎝

=

100 f f ⎛ f ⎞⎛ f ⎞ j ⎟ 1+ j +j + j fA f B ⎜⎝ f A ⎟⎜ ⎠⎝ f B ⎠

So 100 100 = 1 ⎛ jf ⎞⎛ jf ⎞ f f ⎛ f ⎞⎛ f ⎞ f f 1+ j j ⎟ 1+ j +j + j +j + ⎜ ⎟⎜ ⎟ fA f B ⎜⎝ f A ⎟⎜ f f f 50 50 50 1 2 ⎠⎝ B ⎠ ⎝ f1 ⎠⎝ f 2 ⎠

Then 1 1 1 1 + = + f A f B 50 f1 50 f 2

and

1 1 = f A f B 50 f1 f 2 f1 = 10

and f 2 = 2000

1 1 1 1 + = + = 0.002 + 0.000010 = 0.002010 f A f B 50(10) 50(2000)

and f 1 1 1 = ⇒ = B f A f B (50)(10)(2000) f A 106

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Then

fB 1 + = 0.002010 106 f B

10−6 f B2 + 1 = 2.01 + 10−3 f B 10−6 f B2 − 2.01× 10−3 f B + 1 = 0

fB =

2.01× 10−3 ± 4.0401× 10−6 − 4(10−6 )(1) 2(10−6 )

fB =

2.01× 10−3 ± 2.0025 × 10−4 2(10−6 ) f B = 1.105 × 103 Hz

+ sign

f A = 9.05 × 10 Hz + sign ______________________________________________________________________________________ 2

12.13 (a)

Fig. P12.7(a) ⎡ ⎤ ⎛ 200 ⎞ ⎢ ⎥ ⎜ f ⎟ ⎢ ⎥ ⎜⎜ 1 + j ⎟⎟ f1 ⎠ 10 ⎤ ⎢ ⎥⎡ ⎝ Af = ⎢ ⎥ ⎢1 + (10)(0.1126) ⎥ ⎛ 200 ⎞ ⎣ ⎦ ⎢1 + ⎜ ⎟ (0.1126) ⎥ ⎢ ⎜1+ j f ⎟ ⎥ ⎢⎣ ⎜⎝ ⎥⎦ f1 ⎟⎠ 200 ⎡ ⎤ =⎢ ⎥ (4.704) f ⎛ ⎞ ⎢ ⎜ 1 + j ⎟ + 22.52 ⎥ f1 ⎠ ⎣⎢ ⎝ ⎦⎥ =

=

940.73 f 23.52 + j f1

=

40 jf 1+ (23.52) f1

940.73 1 ⋅ f 23.52 1 + j (23.52) f1 f −3dB = (23.52)(100) ⇒ 2.352 kHz

Fig P12.7(b) (200)(10) f 1+ j f1 2000 Af = = f (0.0245)(200)(10) 1+ 1 + j + 49 f f 1 1+ j f1 =

2000 ⋅ 50

1

f −3dB = (50)(100) ⇒ 5 KHz f (50) f1 Overall feedback ⇒ wider bandwidth. 1+ j

(b) ______________________________________________________________________________________ 12.14

v0 = A1 A2 vi + A1vn v0 = (100)vi + (1)vn = (100)(10) + (1)(1) ⇒

S0 1000 = = 1000 N0 1

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.15 (a)

(b)

Circuit (b) – less distortion ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.16 Aυ f =

Aυ 1 + β υ Aυ

=

Rif = Ri (1 + β υ Aυ

5 × 10 3 ⇒ Aυf = 121.95 1 + (0.0080 ) 5 × 10 3

(

) ) = (10)[1 + (0.0080)(5 ×10 )] ⇒ R 3

if

= 410 k Ω

1× 10 ⇒ R of = 24.4 Ω 1 + β υ Aυ 1 + (0.0080 ) 5 × 10 3 ______________________________________________________________________________________ R of =

12.17

Ro

3

=

(

)

V fb = β υ Vo = (0.0096)(2.5) ⇒ V fb = 24 mV V∈ = Vi − V fb = 25 − 24 = 1 mV V o = Aυ V∈ ⇒ Aυ =

Vo 2.5 = = 2.5 × 10 3 V/V V∈ 0.001

Vo 2.5 = = 100 V/V Vi 0.025 ______________________________________________________________________________________ Aυf =

12.18

⎛ R ⎞ R Avf ≈ ⎜ 1 + 2 ⎟ = 20 ⇒ 2 = 19 R1 ⎠ R1 ⎝ vd = iS Ri

vS − vd (vs − vd ) − v0 + R1 R2

(1)

v0 − A0 L vd v0 − (vs − vd ) + =0 R0 R2

(2)

is =

⎛ 1 ( v − vd ) 1 ⎞ A v v0 ⎜ + ⎟ = 0 L d + S R R R R2 2 ⎠ 0 ⎝ 0 A0 L vd (vS − vd ) + R0 R2 v0 = ⎛ 1 1 ⎞ ⎜ + ⎟ R R 2 ⎠ ⎝ 0

From (1):

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 ⎡ A0 L vd (vS − vd ) ⎤ ⋅⎢ + R R R2 ⎥⎦ v −v v −v iS = S d + S d − 2 ⎣ 0 R1 R2 ⎛ 1 1 ⎞ ⎜ + ⎟ R R 2 ⎠ ⎝ 0 A0 L 1 1 ⎞ ⎛ ⎛ − ⎜ 1 ⎜ 1 R0 R2 R2 ⎟ 1 1 ⎜ ⎜ ⎟ + − − vd + + iS = vS R ⎜ R1 R2 ⎜ R1 R2 1 + R2 ⎟ 1+ 2 ⎜ ⎟ ⎜ R R0 0 ⎠ ⎝ ⎝ vd = iS Ri ⎧ ⎡⎛ 1 1 ⎞⎛ R ⎞ A 1 ⎤⎫ ⎪ Ri ⎢⎜ + ⎟ ⎜1 + 2 ⎟ + 0 L − ⎥ ⎪ R R R R R ⎪ 2 ⎠⎝ 0 ⎠ 0 2 ⎦⎪ ⎝ 1 iS ⎨1 + ⎣ ⎬ = vS R 2 ⎪ ⎪ 1+ R0 ⎪ ⎪ ⎩ ⎭

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

⎡⎛ 1 1 ⎞ ⎛ R2 ⎞ 1 ⎢ ⎜ + ⎟ ⎜1 + ⎟ − ⎢ ⎝ R1 R2 ⎠ ⎝ R0 ⎠ R2 ⎢ R 1+ 2 ⎢ R0 ⎣⎢

⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥

⎡1 R 1 ⎡ 1 R2 1 1 A0 L ⎤ ⎪⎫ 1⎤ ⎪⎧ R iS ⎨1 + 2 + Ri ⎢ + 2 ⋅ + + ⎥ ⎬ = vS ⎢ + ⋅ + ⎥ ⎣ R1 R1 R0 R0 R0 ⎦ ⎭⎪ ⎣ R1 R1 R0 R0 ⎦ ⎩⎪ R0

Let

⎡R ⎛ R ⎞ ⎤ ⎪⎫ ⎪⎧ iS ⎨ R0 + R2 + Ri ⎢ 0 + ⎜ 1 + 2 ⎟ + A0 L ⎥ ⎬ = vS R R 1 ⎠ ⎣ 1 ⎝ ⎦ ⎭⎪ ⎩⎪ R2 = 190 kΩ , R1 = 10 kΩ

⎡ R0 ⎛ R2 ⎞ ⎤ ⎢ + ⎜ 1 + ⎟ ⎥ (1) R1 ⎠ ⎦ ⎣ R1 ⎝

⎧ ⎡ 0.1 ⎤⎫ ⎡ 0.1 ⎤ iS ⎨0.1 + 190 + 100 ⋅ ⎢ + 20 + 105 ⎥ ⎬ = vS ⎢ + 20 ⎥ ⎣ 10 ⎦⎭ ⎣ 10 ⎦ ⎩ 7 iS (1.000219 × 10 ) = vS (20.01) Rif =

vS ≅ 5 × 105 kΩ ⇒ Rif ≅ 500 MΩ iS

Output Resistance

IX =

VX − A0 L vd VX + R0 R2 + R1 || Ri

vd =

− R1 || Ri ⋅ VX R1 || Ri + R2

A0 L ⋅ R1 || Ri IX 1 1 1 = = + + VX R0 f R0 R0 ( R1 || Ri + R2 ) R2 + R1 || Ri R1 || Ri = 10 ||100 = 9.09 1 1 105 ⎛ 9.09 ⎞ 1 = + ⋅⎜ ⎟+ R0 f 0.1 0.1 ⎝ 9.09 + 190 ⎠ 190 + 9.09 = 10 + 4.566 × 104 + 0.00502 R0 f = 2.19 × 10−5 kΩ ⇒ R0 f = 0.0219 Ω

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.19 a.

vS − vd v0 − (vS − vd ) = R1 R2

and

vd =

v0 A

⎛ 1 vS vS v 1 ⎞ + = 0 + vd ⎜ + ⎟ R1 R2 R2 R R ⎝ 1 2 ⎠ v0 v0 ⎛ 1 1 ⎞ = + ⎜ + ⎟ R2 A ⎝ R1 R2 ⎠ ⎛ 1 1 ⎞ v ⎡ 1 ⎛ R ⎞⎤ vS ⎜ + ⎟ = 0 ⎢1 + ⎜ 1 + 2 ⎟ ⎥ R R R2 ⎣ A ⎝ R1 ⎠ ⎦ 2 ⎠ ⎝ 1 ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ v0 = ⎝ vS 1 ⎛ R2 ⎞ 1 + ⎜1 + ⎟ A⎝ R1 ⎠

which can be written as Avf =

v0 = vS

β=

b.

1 R 1+ 2 R1

20 =

c.

A ⎡ ⎛ R2 ⎞ ⎤ 1 + ⎢ A / ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ ⎝

105 1 + (105 ) β

105 −1 β = 20 5 ⇒ β = 0.04999 10 So R2 1 R 1 = −1 = − 1 ⇒ 2 = 19.004 R1 β 0.04999 R1

Then A → 9 × 104 d. Af =

ΔAf Af

9 × 104 = 19.99956 1 + (9 × 10 4 )(0.04999)

=

ΔAf −4.444 × 10−4 = −2.222 × 10−3 % ⇒ = −0.005% 20 Af

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.20

I ∈ = I i − I fb = 20 − 19 = 1 μ A I fb = β i I o ⇒ I o =

19 ⇒ I o = 2 mA 0.0095

I o 2 × 10 −3 = = 2 × 10 3 A/A I ∈ 1× 10 − 6 I 2 = 100 A/A Aif = o = I i 0.020

Ai =

Rif =

Ri

1 + β i Ai

=

500 = 25 Ω 1 + (0.0095) 2 × 10 3

(

[

)

(

)](

)

R of = (1 + β i Ai )Ro = 1 + (0.0095) 2 × 10 3 20 × 10 3 ⇒ R of = 400 k Ω

______________________________________________________________________________________ 12.21

I fb = I i − I ∈ = 25 − 0.8 = 24.2 μ A

I o = Ai f I i = (125)(25) ⇒ I o = 3.125 mA

βi = Aif =

I fb

=

Io

24.2 = 0.007744 A/A 3125

Ai

1 + β i Ai

Ai ⇒ Ai = 3906 A/A 1 + (0.007744 )Ai ______________________________________________________________________________________

125 =

12.22 a.

Assume that V1 is at virtual ground. V0 = − I fb RF

Now I fb = I 0 +

I fb RF V0 = I0 − R3 R3

I fb = I S − I ε

and I 0 = Ai I ε =

so

I0 Ai

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I fb = I S −

I0 Ai

From above ⎛ R ⎞ I fb ⎜ 1 + F ⎟ = I 0 R3 ⎠ ⎝ ⎛ I 0 ⎞⎛ RF ⎞ ⎜ I S − ⎟⎜ 1 + ⎟ = I0 Ai ⎠⎝ R3 ⎠ ⎝ ⎛ R I S ⎜1 + F R3 ⎝

⎡ ⎞ 1 ⎛ RF ⎞ ⎤ ⎟ = I 0 ⎢1 + ⎜ 1 + ⎟⎥ R3 ⎠ ⎦ ⎠ ⎣ Ai ⎝

or I Aif = 0 IS

βi =

b. 25 =

c.

⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ ⎝ = ⎡ 1 ⎛ RF ⎞ ⎤ ⎢1 + ⎜1 + ⎟⎥ A R3 ⎠ ⎦ i ⎝ ⎣ Ai = = Aif Ai 1+ ⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ ⎝

1 ⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ ⎝ 105 1 + (105 ) β i

105 −1 β i = 25 5 ⇒ β i = 0.03999 10 so RF R 1 1 = −1 = − 1 ⇒ F = 24.0 R3 β i R3 0.03999

so

Ai = 105 − (0.15)(105 ) = 8.5 × 104

d. so

Aif =

ΔAif A

8.5 × 104 = 24.9989 1 + (8.5 × 104 )(0.03999)

=−

1.10 × 10−3 = −4.41× 10−5 ⇒ −4.41× 10−3 % 25

so if ______________________________________________________________________________________ 12.23

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I S = I ε + I fb , V1 = I ε Ri I fb = I 0 +

V0 R3

and

I 0 = Ai I ε ⇒ I ε =

V0 = V1 − I fb RF

I0 Ai

Now I fb = Ai I ε +

1 (V1 − I fb RF ) R3

⎡ R ⎤ V I fb ⎢1 + F ⎥ = Ai I ε + 1 R3 ⎣ R3 ⎦ I fb = I S − I e ⎡ R ⎤ V ( I S − I e ) ⎢1 + F ⎥ = Ai I e + 1 R3 ⎣ R3 ⎦ ⎡⎛ R ⎞ ⎤ V ⎡ R ⎤ I S ⎢1 + F ⎥ = I e ⎢⎜1 + F ⎟ + Ai ⎥ + 1 R R 3 ⎦ 3 ⎠ ⎣ ⎣⎝ ⎦ R3 V Iε = 1 Ri

⎡ R I S ⎢1 + F ⎣ R3

The

⎧⎪ 1 ⎡⎛ RF ⎤ ⎥ = V1 ⎨ ⋅ ⎢⎜ 1 + R3 ⎦ ⎪⎩ Ri ⎣⎝

⎤ 1 ⎫⎪ ⎞ ⎟ + Ai ⎥ + ⎬ ⎠ ⎦ R3 ⎪⎭

⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ V ⎝ Rif = 1 = I S ⎧⎪ 1 ⎡⎛ RF ⎞ ⎤ 1 ⎫⎪ ⎨ ⋅ ⎢⎜ 1 + ⎟ + Ai ⎥ + ⎬ R3 ⎠ ⎪⎩ Ri ⎣⎝ ⎦ R3 ⎪⎭ 1/ R3

term in the denominator will be negligible. Then

25 ⎧1 ⎡ 5 ⎫ ⎨ ⎣(25) + 10 ⎤⎦ ⎬ ⎩2 ⎭ Rif ≅ 5 × 10−4 kΩ ⇒ Rif = 0.5 Ω Rif =

Output Resistance (Let Z L = 0)

IX =

VX VX + Ai I ε + R3 RF + Ri

Iε =

VX RF + Ri

so A + 1 RF IX 1 1 = = + i , = 24 VX R0 f R3 RF + Ri R3

Let RF = 240 kΩ, R3 = 10 k Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 1 105 + 1 = + Rof 10 240 + 2 R0 f ≈

RF + Ri 240 + 2 = 5 ⇒ R0 f ≈ 2.42 × 10−3 kΩ or R0 f ≈ 2.42 Ω 10 + 1 Ai + 1

so ______________________________________________________________________________________ 12.24

V∈ = Vi − V fb = 0.2 − 0.195 ⇒ V∈ = 5 μ V

βz =

V fb Io

=

0.195 × 10 −3 = 0.039 V/A 5 × 10 −3

I o 5 × 10 −3 = = 1000 A/V V∈ 5 × 10 −6 Ag 1000 = = = 25 A/V 1 + Ag β z 1 + (1000)(0.039)

Ag = Agf

( ) ( ) = R (1 + β A ) = (10 × 10 )[1 + (0.039)(1000 )] ⇒ R

Rif = Ri 1 + β z Ag = 20 × 10 3 [1 + (0.039 )(1000 )] ⇒ Rif = 800 k Ω R of

3

o

z

g

of

= 400 k Ω

______________________________________________________________________________________ 12.25 Agf =

Ag

1 + β z Ag

=

(

2000 = 40 A/V 1 + (2000)(0.0245)

)

I o = Agf Vi = (40 ) 150 × 10 −6 ⇒ I o = 6 mA

(

)

V fb = β z I o = (0.0245) 6 × 10 −3 ⇒ V fb = 147 μ V V∈ = Vi − V fb = 150 − 147 ⇒ V∈ = 3 μ V

______________________________________________________________________________________ 12.26

IE =

Also Then

V − Vε (1 + hFE ) ⋅ I0 = S hFE RE

I 0 = hFE ( AgVε )

Vε =

so

I0 hFE Ag

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V I0 1 + hFE ⋅ I0 = S − hFE RE hFE Ag RE ⎡1 + hFE ⎤ VS 1 + ⎢ ⎥ I0 = hFE Ag RE ⎥⎦ RE ⎢⎣ hFE ⎡ Ag (1 + hFE ) RE + 1 ⎤ VS ⎢ ⎥ I0 = hFE Ag RE RE ⎣⎢ ⎦⎥ ⎤ hFE Ag RE hFE Ag I0 I0 1 ⎡ = ⋅⎢ ≈ ⎥⇒ VS RE ⎣⎢1 + Ag (1 + hFE ) RE ⎦⎥ VS 1 + (hFE Ag ) RE

β z = RE

b.

10 =

c.

d.

5 × 105 1 + (5 × 105 ) β z

5 × 105 −1 β z = 10 5 ⇒ β z = RE = 0.099998 kΩ 5 × 10 Ag → 5.5 × 105

If

then

5.5 × 105 = 10.0000182 Agf = 1 + (5.5 × 105 )(0.099998) ΔAgf

Agf

=

1.82 × 10 −5 ⇒ 1.82 × 10−4 % 10

______________________________________________________________________________________ 12.27

I E = (1 + hFE ) AgVε ,

Now

(1 + hFE ) Ag I S Ri =

IE =

Vε − IS RE

and Vε = I S Ri , Vε = VS − Vε = VS − I S Ri

1 ⋅ (VS − I S Ri ) − I S RE

⎡ ⎤ V Ri + 1⎥ I S = S ⎢(1 + hFE ) Ag Ri + R RE E ⎣ ⎦ Rif =

⎡ ⎤ VS R = RE ⎢(1 + hFE ) Ag Ri + i + 1⎥ IS RE ⎦ ⎣

We have

(1 + hFE ) Ag

≈ hFE Ag = 5 × 105 mS

RE ≈ 0.1 kΩ 20 ⎤ ⎡ Rif = (0.1) ⎢(5 × 105 )(20) + + 1⎥ 0.1 ⎣ ⎦ so

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or

Rif = 106 kΩ

Vπ = AgVε rπ I X = g mVπ +

VX − (−Vε ) R0

(1)

Vε = −( I X + AgVε )( RE || Ri )

(2)

V ⎡1 + Ag ( Rε || Ri ) ⎦⎤ = − I X ( RE || Ri )

or ε ⎣ Now:

I X = g m Ag rπ Vε +

VX Vε + R0 R0

(1)

⎛ 1 ⎞ ⎡ − I X ( RE || Ri ) ⎤ VX I X = ⎜ g m Ag rπ + ⎟ ⎢ ⎥+ R ⎢1 + Ag ( RE || Ri ) ⎦⎥ R0 0 ⎠⎣ ⎝ V R0 f = X IX

⎧⎪ ⎛ 1 ⎞ ⎡ ( RE || Ri ) ⎤ ⎫⎪ = R0 ⎨1 + ⎜ g m Ag rπ + ⎟ ⎢ ⎥⎬ R0 ⎠ ⎢⎣1 + Ag ( RE || Ri ) ⎥⎦ ⎭⎪ ⎩⎪ ⎝ g m rπ Ag = hFE Ag = 5 × 105 mS

A = 5× 103 mS Let hFE = 100 so g

RE || Ri = 0.1|| 20 ≈ 0.1 kΩ

Then ⎤ ⎪⎫ 1 ⎞⎡ 0.1 ⎪⎧ ⎛ R0 f = 50 ⎨1 + ⎜ 5 × 105 + ⎟ ⎢ ⎥⎬ 3 50 + × 1 (5 10 )(0.1) ⎠⎣ ⎦ ⎭⎪ ⎩⎪ ⎝ R0 f = 5.04 MΩ

or ______________________________________________________________________________________ 12.28 Azf =

Az

1 + β g Az

0.20 × 10 6 = Rif =

Rof =

(

Az

)

1 + 4.25 × 10 − 6 Az

⇒ Az = 1.333 V/ μ A

Ri 500 = ⇒ Rif = 75 Ω 1 + β g Az 1 + 4.25 × 10 − 6 1.333 × 10 6

(

Ro

1 + β g Az

)(

)

= 75 Ω

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.29 I ε = I i − I fb = 40 − 38 = 2 μA Vo 8 V = =4 Iε 2 μA

Az =

βg = Azf =

I fb Vo

=

38 μA = 4.75 8 V

Vo 8 V = = 0.2 I i 40 μA

______________________________________________________________________________________ 12.30 a.

Assuming V1 is at virtual ground (−V0 ) = − I fb RF

and

(−V0 ) = − Az I ε ⇒ I ε =

I fb = I S − I ε ⎛V V0 = ( I S − I ε ) RF = I S RF − ⎜ 0 ⎝ Az So ⎡ R ⎤ V0 ⎢1 + F ⎥ = I S RF ⎣ Az ⎦ Azf =

V0 RF AR = = z F I S ⎡ RF ⎤ Az + RF ⎢1 + A ⎥ z ⎦ ⎣

Azf =

Az Az = ⎛ 1 ⎞ 1 + Az β g 1 + Az ⎜ ⎟ ⎝ RF ⎠

so

or b.

βg =

⎞ ⎟ RF ⎠

1 RF

5 × 104 =

c.

5 × 106 1 + (5 × 106 ) β g

5 × 106 −1 4 β g = 5 × 10 6 ⇒ β g = 1.98 × 10−5 5 × 10 1 RF = ⇒ RF = 50.5 kΩ

βg

V0 Az

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ d.

Az = (0.9)(5 × 106 ) = 4.5 × 106 Azf = ΔAzf

Azf

4.5 × 106 = 4.994 × 10 4 1 + (4.5 × 106 )(1.98 × 10 −5 ) =−

55.4939 = −1.11× 10−3 ⇒ −0.111% 5 × 104

______________________________________________________________________________________ 12.31

V1 = Iε Ri , − V0 = − Az I ε ⇒ V0 = Ax Iε I fb = I S − I ε −V0 = V1 − I fb RF

and

− Az I ε = V1 − ( I S − Iε ) RF ⎛V ⎞ ⎛V ⎞ − Az ⎜ 1 ⎟ = V1 − I S RF + ⎜ 1 ⎟ RF R ⎝ i⎠ ⎝ Ri ⎠ ⎡ A R ⎤ I S RF = V1 ⎢1 + z + F ⎥ ⎣ Ri Ri ⎦ Rif =

V1 RF = I S ⎡ Az RF ⎤ + ⎢1 + ⎥ ⎣ Ri Ri ⎦

We have Rif =

50.5 × 103

⎡ 5 × 106 50.5 × 103 ⎤ ⎢1 + 10 × 103 + 10 × 103 ⎥ ⎣ ⎦ 50.5 × 103 = ⇒ Rif = 99.79 Ω [1 + 500 + 5.05]

______________________________________________________________________________________ 12.32 (a)

Low input R ⇒ Shunt input Low output R ⇒ Shunt output Or a Shunt-Shunt circuit (b) High input R ⇒ Series input High output R ⇒ Series output Or a series-Series circuit (c) Shunt-Series circuit (d) Series-Shunt circuit ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.33 Ri (max) = Ri (1 + T ) = 10(1 + 104 ) ⇒ Ri (max) ≅ 105 k Ω

(a)

Ri 10 = ≅ 10−3 k Ω 1 + T 1 + 104 Ri (min) = 1Ω Ri (min) =

Or (b)

Ro (max) = Ro (1 + T ) = 1(1 + 104 ) ⇒ Ro (max) ≅ 104 k Ω Ro 1 = ≅ 10−4 k Ω 1 + T 1 + 104 Ro (min) = 0.1Ω Ro (min) =

Or ______________________________________________________________________________________ 12.34 Ag =

io vi

Overall Transconductance Amplifier, Series output = current signal and Shunt input = current signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions are shown.

______________________________________________________________________________________ 12.35

I CQ = 0.2 mA, rπ =

(140)(0.026) = 18.2 k Ω , 0.2

(a)

(1)

⎛ 1 + h FE Vπ V −VA + g m Vπ = o = Vπ ⎜⎜ rπ R2 ⎝ rπ

(2)

Vπ A V − Vo = υ ∈ rπ Ro + rπ

(3)

V A − V o V A V A − Vi + + =0 R2 R1 Ri

(4) V∈ = Vi − V A

⎞ ⎟ ⎟ ⎠

gm =

0.2 = 7.692 mA/V 0.026

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎞ ⎡ ⎛ Aυ V∈ − V o ⎟ ⎢rπ ⎜ ⎟ ⎜ R +r ⎠ ⎣⎢ ⎝ o π

From (1),

V o − V A ⎛ 1 + h FE = ⎜⎜ R2 ⎝ rπ

Using (4),

⎛ A V − Vo V o − Vi + V∈ = (1 + h FE )⎜⎜ υ ∈ R2 ⎝ R o + rπ

⎛ A V − Vo ⎞⎤ ⎟⎥ = (1 + h FE )⎜ υ ∈ ⎜ R +r ⎟ ⎝ o π ⎠⎦⎥ ⎞ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ ⎠

(Eq. 5)

⎛ 1 V 1 1 ⎞ V From (3), V A ⎜⎜ + + ⎟⎟ = o + i R R R R Ri 1 i ⎠ 2 ⎝ 2

⎛ 1 V 1 1 ⎞ V + + ⎟⎟ = o + i R R R R Ri 1 i ⎠ 2 ⎝ 2 V V (Vi − V∈ )⎛⎜ 1 + 1 + 1 ⎞⎟ = o + i 10 1 30 10 30 ⎝ ⎠

(Vi − V∈ )⎜⎜

(Vi − V∈ )(1.1333) = Vo (0.10) + Vi (0.0333) We find, V∈ = Vi (0.9706 ) − V o (0.08824 ) From Eq. (5) above, Vo − Vi (1 + h FE )Vo (1 + h FE ) V + = ⋅ Aυ V∈ − ∈ R2 Ro + rπ R o + rπ R2 V o − Vi V 141 141 + ⋅Vo = 10 5 V∈ − ∈ 10 0.5 + 18.2 0.5 + 18.2 10

( )

7.640V o − Vi (0.1) = 7.540 × 10 5 V∈ = 7.540 × 10 5 [Vi (0.9706) − Vo (0.08824 )]

(

6.653705 × 10 4 V o = Vi 7.318235 × 10 5

)

V Then o = 11.0 Vi ⎛ A V − Vo ⎞ ⎟ (b) From (2), Vπ = rπ ⎜⎜ υ ∈ ⎟ ⎝ R o + rπ ⎠ ⎛ 1 + h FE From (1), V o = V A + R 2Vπ ⎜⎜ ⎝ rπ

⎞ ⎟ ⎟ ⎠

⎛ A V − Vo V o = (Vi − V∈ ) + R 2 (1 + h FE )⎜⎜ υ ∈ ⎝ R o + rπ

⎞ ⎟ ⎟ ⎠

⎤ ⎡ R (1 + h FE ) ⎤ ⎡ R 2 (1 + h FE )Aυ V o ⎢1 + 2 − 1⎥ (Eq. 6) ⎥ = Vi + V∈ ⎢ R o + rπ ⎦ ⎦ ⎣ ⎣ R o + rπ From (3), ⎛ 1 V 1 1 ⎞ V V A ⎜⎜ + + ⎟⎟ = o + i ⎝ R 2 R1 Ri ⎠ R 2 Ri ⎛ 1 V 1 1 ⎞ V + + ⎟⎟ = o + i R R R R Ri 1 i ⎠ 2 ⎝ 2 Then, using Eq. (6),

(Vi − V∈ )⎜⎜

⎧ ⎡ R 2 (1 + h FE )Aυ ⎤⎫ − 1⎥ ⎪ ⎪Vi + V∈ ⎢ ⎛ 1 ⎛ 1 1 ⎞ 1 1 ⎞ 1 ⎪ ⎣ R o + rπ ⎦⎪ ⎟⎟ − V∈ ⎜ Vi ⎜⎜ + + ⎟⎟ = + ⎨ ⎬ ⎜ ( 1 ) R h + FE ⎝ R 2 R1 ⎠ ⎝ R 2 R1 Ri ⎠ R 2 ⎪ ⎪ 1+ 2 ⎪ ⎪ R o + rπ ⎩ ⎭

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Vi (1.10 ) − V∈ (1.1333) =

(

(

Vi (1.098674 ) = V∈ 1.1333 + 1.0 × 10 4

So

)

6 1 ⎧⎪Vi + V∈ 7.540106 × 10 ⎪⎫ ⎨ ⎬ 10 ⎪⎩ 75.401 ⎪⎭

)

Vi = 9.103 × 10 3 V∈

⎛V ⎞ Then Rif = Ri ⎜⎜ i ⎟⎟ = 30 × 10 3 9.103 × 10 3 ⇒ Rif = 273 M Ω ⎝ V∈ ⎠

(

(c) I x + g mVπ +

)(

)

Vπ V x − V A = rπ R2

⎛ A V −Vx ⎞ ⎟ From (2), Vπ = rπ ⎜⎜ υ ∈ ⎟ ⎝ R o + rπ ⎠ V −V (1 + h FE ) ( Aυ V∈ − V x ) = x A Then, I x + R o + rπ R2

Now, V A = −V∈ and V∈ = −V x (0.08824) So, I x +

V V (0.08824) 141 10 5 (− 0.08824V x ) − V x = x − x 0.5 + 18.2 10 10

[

(

]

)

I x = V x 0.091176 + 6.654 × 10 4 V R of = x = 15 μ Ω Ix ______________________________________________________________________________________

12.36 a.

Neglecting base currents I C 2 = 0.5 mA, VC 2 = 12 − (0.5)(22.6) = 0.7 V I C1 = 0.5 mA ⇒ v0 = 0

Then I C 3 = 2 mA b.

rπ 1 = rπ 2 =

hFE ⋅ VT (100)(0.026) = = 5.2 kΩ 0.5 I C1

0.5 = 19.23 mA / V 0.026 (100)(0.026) rπ 3 = = 1.3 kΩ 2 2 g m3 = = 76.92 mA / V 0.026 g m1 = g m 2 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Vπ 1 V + g m1Vπ 1 + g m1Vπ 2 + π 2 = 0 rπ 1 rπ 1 ⎛ 1 ⎞ (Vπ 1 + Vπ 2 ) ⎜ + g m1 ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 r ⎝ π1 ⎠ Vπ 1 ( RS + rπ 1 ) − Vπ 2 + Vb 2 Vi = rπ 1

(1)

⎛ R ⎞ Vi = Vπ 1 ⎜1 + S ⎟ − Vπ 2 + Vb 2 ⎝ rπ 1 ⎠ or Vπ 2 = −Vπ 1

But so

⎛ R ⎞ Vi = Vπ 1 ⎜ 2 + S ⎟ + Vb 2 rπ 1 ⎠ ⎝

(2)

V02 V −V + g m1Vπ 2 + 02 0 = 0 RC rπ 3

(3)

Vπ 3 V V −V + g m 3Vπ 3 = 0 + 0 b 2 rπ 3 RL R2 Vπ 3 = V02 − V0

so ⎛ 1 + hFE ⎞ ⎛ 1 1 ⎞ V (V02 − V0 ) ⎜ + ⎟ − b2 ⎟ = V0 ⎜ r R R R2 2 ⎠ ⎝ L ⎝ π3 ⎠

(4)

Vb 2 − V0 Vb 2 Vπ 2 + + =0 R2 R1 rπ 1

(5) Substitute numbers into (2), (3), (4) and (5): 1 ⎞ ⎛ Vi = −Vπ 2 ⎜ 2 + ⎟ + Vb 2 5.2 ⎠ ⎝ Vi = −Vπ 2 (2.192) + Vb 2

(2)

1 ⎞ ⎛ 1 ⎛ 1 ⎞ + V02 ⎜ ⎟ + (19.23)Vπ 2 − V0 ⎜ ⎟ = 0 ⎝ 22.6 1.3 ⎠ ⎝ 1.3 ⎠ V02 (0.8135) + (19.23)Vπ 2 − (0.7692)V0 = 0 ⎛ 101 ⎞ ⎛ 101 1 1 ⎞ ⎛ 1 ⎞ V02 ⎜ + + ⎟ − Vb 2 ⎜ ⎟ ⎟ = V0 ⎜ ⎝ 1.3 ⎠ ⎝ 1.3 4 50 ⎠ ⎝ 50 ⎠ V02 (77.69) = V0 (77.96) − Vb 2 (0.02)

(4)

⎛ 1 1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ Vb 2 ⎜ + ⎟ − V0 ⎜ ⎟ + Vπ 2 ⎜ ⎟=0 ⎝ 50 10 ⎠ ⎝ 50 ⎠ ⎝ 5.2 ⎠ Vb 2 (0.120) − V0 (0.020) + Vπ 2 (0.1923) = 0

(5)

(3)

From (2): Vb 2 = Vi + Vπ 2 (2.192). Substitute in (4) and (5) to obtain: V02 (77.69) = V0 (77.96) − [Vi + Vπ 2 (2.192)](0.02) (4′) [Vi + Vπ 2 (2.192)](0.120) − V0 (0.020) + Vπ 2 (0.1923) = 0 (5′) So we now have the following three equations: V02 (0.8135) + (19.23)Vπ 2 − (0.7692)V0 = 0

(3)

V02 (77.69)

= V0 (77.96) − Vi (0.02) − Vπ 2 (0.04384) (4′) (0.120)Vi + Vπ 2 (0.4553) − V0 (0.020) = 0 (5′) From (3): V02 = V0 (0.9455) − Vπ 2 (23.64). Substitute for V02 in (4′) to obtain:

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (77.69)[Vo (0.9455) − Vπ 2 (23.64)] = V0 (77.96) − Vi (0.02) − Vπ 2 (0.04384)

or 0 = V0 (4.504) − Vi (0.02) + Vπ 2 (1836.5) (5′) Vπ 2 :

Next, solve

for

(0.120)Vi + Vπ 2 (0.4553) − V0 (0.020) = 0 Vπ 2 = V0 (0.04393) − Vi (0.2636)

Finally, 0 = V0 (4.504) − Vi (0.02) + (1836.5)[V0 (0.04393) − Vi (0.2636)] 0 = V0 (85.18) − Vi (484.12)

So Avf =

V0 484.12 = ⇒ Avf = 5.68 Vi 85.18

______________________________________________________________________________________ 12.37

RTH = R1 || R2 = 400 || 75 = 63.2 kΩ

a.

⎛ R2 ⎞ ⎛ 75 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10) = 1.579 V ⎝ 75 + 400 ⎠ ⎝ R1 + R2 ⎠ 1.579 − 0.7 = 0.007106 mA I BQ1 = 63.2 + (121)(0.5) I CQ1 = 0.853 mA VC1 = 10 − (0.853)(8.8) = 2.49 V 2.49 − 0.7 = 0.497 mA 3.6 = 10 − (0.497)(13) = 3.54 V

IC 2 ≈ VC 2

IC 3 ≈

3.54 − 0.7 = 2.03 mA 1.4

Then (120)(0.026) = 3.66 kΩ 0.853 0.853 g m1 = = 32.81 mA / V 0.026 (120)(0.026) rπ 2 = = 6.28 kΩ 0.497 0.497 gm2 = = 19.12 mA / V 0.026 (120)(0.026) rπ 3 = = 1.54 kΩ 2.03 2.03 g m3 = = 78.08 mA / V 0.026 rπ 1 =

b.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Vi = Vπ 1 + Vε 1 ⇒ Vε 1 = Vi − Vπ 1

(1)

Vπ 1 V V −V + g m1Vπ 1 = ε 1 + ε 1 0 rπ 1 RE1 RF

(2)

Vπ 2 = −( g m1Vπ 1 )( RC1 || rπ 2 )

(3)

V +V V g m 2Vπ 2 + π 3 0 + π 3 = 0 RC 2 rπ 3

(4)

Vπ 3 V V −V + g m 3Vπ 3 = 0 + 0 ε 1 rπ 3 RE 3 RF

(5)

Substitute numbers in (2), (3), (4) and (5):

or

1⎞ V ⎛ 1 ⎞ ⎛ 1 Vπ 1 ⎜ + 32.81⎟ = (Vi − Vπ 1 ) ⎜ + ⎟− 0 ⎝ 3.66 ⎠ ⎝ 0.5 10 ⎠ 10 Vπ 1 (35.18) = Vi (2.10) − V0 (0.10)

(2)

Vπ 2 = −(32.81)Vπ 1 (88 || 6.28) V or π 2 = −Vπ 1 (120.2) (3) Vπ 3 V0 Vπ 3 (19.12)Vπ 2 + + + =0 13 13 1.54

or

Vπ 2 (19.12) + Vπ 3 (0.7263) + V0 (0.07692) = 0

(4)

1 ⎞ V −V ⎛ 1 ⎞ ⎛ 1 Vπ 3 ⎜ + 78.08 ⎟ = V0 ⎜ + ⎟ − i π1 10 ⎝ 1.54 ⎠ ⎝ 1.4 10 ⎠

or

Vπ 3 (78.73) = V0 (0.8143) − Vi (0.10) + Vπ 1 (0.10)

Now substituting

Vπ 2 = −Vπ 1 (120.2)

(5)

in (4):

(19.12)[−Vπ 1 (120.2)] + Vπ 3 (0.7263) + V0 (0.07692) = 0

or −Vπ 1 (2298.2) + Vπ 3 (0.7263) + V0 (0.07692) = 0

Then Vπ 3 = Vπ 1 (3164.3) − V0 (0.1059) Substituting Vπ 3 = Vπ 1 (3164.3) − V0 (0.1059) in (5): (78.73)[Vπ 1 (3164.3) − V0 (0.1059)] = V0 (0.8143) − Vi (0.10) + Vπ 1 (0.10) 5 or Vπ 1 (2.49 × 10 ) − V0 (9.152) = −Vi (0.10) Then

Vπ 1 = V0 (3.674 × 10−5 ) − Vi (4.014 × 10−7 ) −5 Now substituting Vπ 1 = V0 (3.674 × 10 )

−Vi (4.014 × 10−7 )

in (2):

(35.18)[V0 (3.674) × 10−5 ) − Vi (4.014 × 10−7 )]

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ = Vi (2.10) − V0 (0.10)

or V0 (0.1013) = Vi (2.10) V0 = 20.7 Vi

So

Rif =

c.

I RB1 =

I b1 =

Vi Ii

and I i = I RB1 + I b1

Vi RB1

Vπ 1 rπ 1

Now Vπ 1 = (20.7Vi )(3.674 × 10−5 ) − Vi (4.014 × 10−7 ) Vπ 1 = Vi (7.60 × 10 −4 )

Then

Vi Vi V (7.60 × 10−4 ) + i 63.2 3.66 1 = 0.01582 + 2.077 × 10 −4 Rif = 62.4 kΩ Rif =

or d.

To determine

R0 f :

Equation (1) is modified to Vπ 1 + Ve1 = 0 (Vi = 0) Equation (5) is modified to: Vπ 3 (78.73) + I X = V0 (0.8143) + Vπ 1 (0.10)

(5)

Now

Vπ 1 (35.18) = −V0 (0.10)

Vπ 2 = −Vπ 1 (120.2)

(2)

(3)

Vπ 2 (19.12) + Vπ 3 (0.7263) + V0 (0.07692) = 0

(4)

Now Vπ 1 = −V0 (0.002843)

so Vπ 2 = −(−V0 )(0.002843)(120.2) Vπ 2 = V0 (0.3417)

Then V0 (0.3417)(19.12) + Vπ 3 + (0.7263) + V0 (0.07692) = 0 Vπ 3 = −V0 (9.101)

or So then

(4)

−V0 (9.101)(78.73) + I X

= V0 (0.8143) + (0.10)(−V0 )(0.002843)

or

I X = V0 (717.3)

(5)

or R0 f =

V0 = 0.00139 kΩ ⇒ R0 f = 1.39 Ω IX

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 12.38 (a)

(1)

Vi − VA V −V V + g m1Vπ 1 = A + A O rπ 1 RE RF

(2)

VB V + g m1Vπ 1 + B = 0 RC1 rπ 2

(3)

VC V − Vo + g m 2Vπ 2 + C =0 RC 2 rπ 3

(4)

g m 3Vπ 3 +

VC − VO VO − VA = rπ 3 RF

Vπ 1 = Vi − VA Vπ 2 = VB Vπ 3 = VC − VO

(1)

⎛ 1 ⎞ V V −V + g m1 ⎟ = A + A O RF ⎝ rπ 1 ⎠ RE

(Vi − VA ) ⎜

(2)

⎛ 1 1 ⎞ VB ⎜ + ⎟ + g m1 (Vi − VA ) = 0 ⎝ RC1 rπ 2 ⎠

(3)

⎛ 1 VO 1 ⎞ + =0 VC ⎜ ⎟ + g m 2VB − R r rπ 3 2 3 π C ⎝ ⎠ ⎛

(4)

(VC − VO ) ⎜ g m3 + ⎝

1 ⎞ VO − VA ⎟= rπ 3 ⎠ RF

(100)(0.026) 14.3 rπ 1 = = 0.182 K g m1 = = 550 mA/V 14.3 0.026 (100)(0.026) 4.62 rπ 2 = = 0.563 K g m 2 = = 178 mA/V 4.62 0.026 (100)(0.026) 4.47 rπ 3 = = 0.582 K g m 3 = = 172 mA/V 4.47 0.026 V −V 1 ⎞ V + 550 ⎟ = A + A O (Vi − VA ) ⎛⎜ 1.2 ⎝ 0.182 ⎠ 0.05 (1)

(2)

1 ⎞ ⎛ 1 VB ⎜ + ⎟ + (550)(Vi − VA ) = 0 ⎝ 0.3 0.563 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VO 1 ⎞ ⎛ 1 VC ⎜ + =0 ⎟ + 178 VB − 0.582 ⎝ 0.65 0.582 ⎠

(3)

(VC − VO ) ⎛⎜172 + ⎝

(4)

1 ⎞ VO − VA ⎟= 0.582 ⎠ 1.2

(Vi − VA )(555.5) = VA (20) + (VA − VO )(0.8333)

(1)

VB (5.109) + 550(Vi − VA ) = 0

(2)

VC (3.257) + 178VB − VO (1.718) = 0

(3) (4)

(VC − VO )(173.7) = (VO − VA )(0.8333)

(1)

Vi (555.5) + VO (0.8333) = VA (576.3) VB (5.109) + 550Vi − VA (550) = 0

(2)

VC (3.257) + 178VB − VO (1.718) = 0

(3)

VC (173.7) + VA (0.8333) = VO (174.5)

(4)

VB = VA (107.7) − Vi (107.7)

From (2)

VC = VO (1.0046) − VA (0.004797)

From (4) Substitute into (3)

(3.257) [VO (1.0046) − VA (0.004797) ]

+(178)[VA (107.7) − Vi (107.7)] − Vo (1.718) = 0 VO (3.272) − VA (0.01562) + VA (19170.6) − Vi (19170.6) − VO (1.718) = 0 VA (19170.6) = Vi (19170.6) − VO (1.554) VA = Vi (1.00) − VO (0.00008106)

Substitute into (1)

Vi (555.5) + Vo (0.8333) = (576.3) [Vi (1.00) − Vo (0.00008106)] = Vi (576.3) − Vo (0.0467)

Vo (0.880) = Vi (20.8) Vo = Avf = 23.6 Vi

Ideal Avf =

RF + RE 1.2 + 0.05 = = 25.0 0.05 RE

Rif =

(b) We have

Vi Ii

and

Ii =

Vπ 1 Vi − VA = rπ 1 rπ 1

VA = Vi (1.00) − Vo (0.00008106) = Vi (1.00) − (23.6)Vi (0.00008106) VA = Vi (0.99809)

Then

Ii =

Vi (1 − 0.99809) = Vi (0.01051) 0.182

Vi ⇒ Rif = 95.1 K Vi (0.01051) Rof , Vi = 0

Rif =

To find

set

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I X + g m 3Vπ 3 +

Vπ 3 Vx − VA = rπ 3 RF

Vπ 3 = VC − VX I X + (VC − VX )( g m 3 +

V − VA 1 )= X rπ 3 RF

For Vi = 0, we have VC = VX (1.0046) − VA (0.004797) VA (576.3) = VX (0.8333) VA = VX (0.001446) VC = VX (1.0046) − VX (0.001446)(0.004797) VC = VX (1.0046)

1 ⎞ VX (1 − 0.004797) ⎛ I X + VX (1.0046 − 1.0) ⎜172 + ⎟= 0.582 1.2 ⎝ ⎠ I X + VX (0.7991) = VX (0.8293) I X = VX (0.03024) Rof =

VX = 33.1 K IX

______________________________________________________________________________________

12.39

(

)

g m υ gs1 + υ gs 2 = 0 ⇒ υ gs 2 = −υ gs1 ⎛ R2 ⎞ ⎟⎟ ⋅ V o Vi = υ gs1 − υ gs 2 + ⎜⎜ ⎝ R1 + R 2 ⎠ 1 Vi = −2υ gs 2 + Vo 2 Vo V + g mυ gs 2 + o = 0 Also R1 + R 2 RD

υ gs 2 = −

Vo gm

⎛ 1 1 ⎜⎜ + ⎝ R D R1 + R 2

⎞ ⎟⎟ ⎠

⎛ 1 1 ⎜⎜ + ⎝ R D R1 + R 2 We have R1 + R 2 >> R D

Then Vi =

2Vo gm

⎞ 1 ⎟⎟ + V o ⎠ 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ 2 1⎞ Vi ≅ Vo ⎜⎜ + ⎟⎟ ⎝ g m RD 2 ⎠ V g m RD 1 = Aυf = o = Vi ⎛ 2 1 ⎞ ⎛2 + 1 g R ⎞ ⎟ ⎜ ⎜ m D ⎟ ⎜ g R + 2⎟ ⎝ 2 ⎠ ⎠ ⎝ m D

Now g m = 2 K n I DQ = 2 (0.5)(0.5) = 1.0 mA/V Aυf =

(1)(7 )

= 1.273 1 ⎡ ⎤ ( )( ) 2 + 1 7 ⎢ ⎥ 2 ⎣ ⎦ ______________________________________________________________________________________

12.40 (a) Neglect base currents VGG − VGS = I D1 + I C 2 RL I D1 = I C2 =

5 − VD 5 − VD = 5 R D1

5 − (V D + 0.7 ) 4.3 − V D = 1.6 RE 2

V D = 5 − 5 I D1 ⇒ I C 2 =

Then

4.3 − (5 − 5 I D1 ) = −0.4375 + 3.125I D1 1.6

2.5 − VGS = I D1 + (− 0.4375 + 3.125 I D1 ) 1.2

3.025 = VGS + 4.95 I D1 = VGS + 4.95(K n )(VGS − 0.5)

2

We find 7.425VGS2 − 6.425VGS − 1.16875 = 0 ⇒ VGS = 1.0197 V Then I D1 = (1.5)(1.0197 − 0.5) = 0.405 mA I C 2 = −0.4375 + 3.125(0.405) = 0.828 mA 2

(b)

g m1 = 2 K n I D1 = 2 (1.5)(0.405) = 1.559 mA/V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I 0.828 = 31.85 mA/V g m2 = C 2 = 0.026 VT rπ 2 =

β VT

=

I C2

(120)(0.026) = 3.768 k Ω 0.828

(1) V o = ( g m1V gs + g m 2Vπ ) R L (2) Vi = V gs + Vo ⇒ V gs = Vi − Vo (3)

V VA + g m1V gs = π R D1 rπ

(4)

V A + Vπ Vπ + + g m 2 Vπ = 0 RE 2 rπ

So

−VA 1 ⎞ ⎛ 1 = Vπ ⎜ + + 31.85 ⎟ = Vπ (32.74 ) 1.6 ⎠ ⎝ 1.6 3.768 V A = −52.385Vπ

−52.385Vπ V + 1.559V gs = π 5 3.768 or 1.559V gs = Vπ (0.2654 + 10.477 )

(3)

so Vπ = 0.1451V gs = 0.1451(Vi − V o )

(1) V o = [1.559(Vi − V o ) + 31.85(0.1451)(Vi − V o )](1.2) V Then Aυ = o = 0.8812 Vi (c)Set Vi = 0 I x + g m 2Vπ + g m1V gs =

Vx RL

V gs = −V x

Vπ = 0.1451V gs = −0.1451V x ⎡ 1 ⎤ I x = Vx ⎢ + (31.85)(0.1451) + (1.559 )⎥ = V x (7.014 ) ⎣1.2 ⎦ V R o = x = 143 Ω Ix ______________________________________________________________________________________

12.41 g m RS 1 + g m RS

(a) (i) Aυ f =

g m = 2 K n I DQ = 2 (1.5)(1.2) = 2.683 mA/V Aυ f =

(2.683)(1.5) = 0.801 1 + (2.683)(1.5)

(ii) R o f =

1 1 RS = 1.5 = 0.3727 1.5 gm 2.683

Ro f = 299 Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) (i) g m = 2 (2.25)(1.2 ) = 3.286 mA/V Aυ f = ΔAυ f Aυ f

(3.286)(1.5) = 0.8313 mA/V 1 + (3.286)(1.5)

× 100% = +3.78%

1 1.5 = 0.3043 1.5 3.286 = 253 Ω

(ii) Ro f = Ro f ΔR o f Ro f

× 100% = −15.4%

______________________________________________________________________________________ 12.42 dc analysis: RTH 1 = 150 || 47 = 35.8 kΩ , ⎛ 47 ⎞ VTH 1 = ⎜ ⎟ (25) = 5.96 V ⎝ 47 + 150 ⎠ RTH 2 = 33 || 47 = 19.4 kΩ , ⎛ 33 ⎞ VTH 2 = ⎜ ⎟ (25) = 10.3 V ⎝ 33 + 47 ⎠ 5.96 − 0.7 I B1 = = 0.0187 mA 35.8 + (51)(4.8) I C1 = (50)(0.0187) = 0.935 mA

10.3 − 0.7 = 0.03705 mA 19.4 + (51)(4.7) = (50)(0.03705) = 1.85 mA

IB2 = IC 2

(50)(0.026) = 1.39 kΩ; 0.935 (50)(0.026) rπ 2 = = 0.703 kΩ 1.85 0.935 g m1 = = 35.96 mA / V 0.026 1.85 gm2 = = 71.15 mA / V 0.026 rπ 1 =

VS = Vπ 1 + Ve Vπ 1 V V − V0 + g m1Vπ 1 = e + e rπ 1 R1 RF

(1) (2)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ g m1Vπ 1 +

Vπ 2 Vπ 2 Vπ 2 + + =0 RC1 RB 2 rπ 2

(3)

V V −V g m 2Vπ 2 + 0 + 0 e = 0 RC 2 RF

(4) Substitute numerical values in (2), (3) and (4): Ve = VS − Vπ 1

(1)

Vπ 1 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + (35.96)Vπ 1 = (VS − Vπ 1 ) ⎜ + ⎟ − V0 ⎜ ⎟ 1.39 ⎝ 0.1 4.7 ⎠ ⎝ 4.7 ⎠

or

Vπ 1 (46.89) = VS (10.213) − V0 (0.2128)

(2)

1 1 ⎞ ⎛1 (35.96)Vπ 1 + Vπ 2 ⎜ + + ⎟=0 ⎝ 10 19.4 0.703 ⎠

or

(35.96)Vπ 1 + Vπ 2 (1.574) = 0

(3)

1 ⎞ ⎛ 1 ⎛ 1 ⎞ (71.15)Vπ 2 + V0 ⎜ + ⎟ − (VS − Vπ 1 ) ⎜ ⎟=0 4.7 4.7 ⎝ ⎠ ⎝ 4.7 ⎠

or

(71.15)Vπ 2 + V0 (0.4255) − VS (0.2128) + Vπ 1 (0.2128) = 0 Vπ 2 = −Vπ 1 (22.85)

(4)

From (3): Then substitute in (4):

−(71.15)Vπ 1 (22.85) + V0 (0.4255) − VS (0.2128) + Vπ 1 (0.2128) = 0

or −Vπ 1 (1625.6) + V0 (0.4255) − VS (0.2128) = 0 Vπ 1 = VS (0.2178) − V0 (0.004538)

From (2): Then

−(1625.6)[VS (0.2178) − V0 (0.004538)] + V0 (0.4255) − VS (0.2128) = 0 −VS (354.3) + V0 (7.802) = 0

or Finally ⇒

V0 = 45.4 VS

______________________________________________________________________________________ 12.43

For example, use a 2-stage amplifier. Each stage is shown in Fig. 12.29. ______________________________________________________________________________________ 12.44

⎛ k ′ ⎞⎛ W ⎞ 2 I DQ1 = ⎜⎜ n ⎟⎟⎜ ⎟(VGS − VTN ) 2 L ⎝ ⎠ ⎝ ⎠ ⎛W ⎞ Let all ⎜ ⎟ = 20 ⎝L⎠ ⎛ 0.1 ⎞ 2 0.5 = ⎜ ⎟(20)(VGS − 1.5) ⇒ VGS1, 2 = 2.207 V ⎝ 2 ⎠

⎛ 0.1 ⎞ 2 I DQ 3 = 2 = ⎜ ⎟(20)(VGS 3 − 1.5) ⇒ VGS 3 = 2.914 V 2 ⎝ ⎠ Want VG 3 = 2.914 V, then

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12 − 2.914 RD = = 18.2 k Ω 0.5

g m1υ gs1 + g m1υ gs 2 = 0 ⇒ υ gs 2 = −υ gs1

Vi = υ gs1 − υ gs 2 + V A = −2υ gs 2 + V A ⎛ R1 ⎞ ⎟⎟ ⋅ V o V A = ⎜⎜ ⎝ R1 + R 2 ⎠ Vo Vo + = g m3υ gs 3 R L R1 + R 2

υ gs 3 = −(g m1υ gs 2 R D ) − Vo ⎛ 1 1 V o ⎜⎜ + R R + R2 1 ⎝ L

υ gs 2 =

⎞ ⎟⎟ = g m3 − g m1υ gs 2 R D − V o ⎠

(

⎛ R 1 (V A − Vi ) = 1 ⎜⎜ 1 2 2 ⎝ R1 + R 2

⎛ 1 1 V o ⎜⎜ + ⎝ R L R1 + R 2

)

⎞ 1 ⎟⎟ ⋅ V o − ⋅ Vi 2 ⎠

⎡ 1 ⎛ R1 ⎞ ⎟⎟ = − g m1 g m3 R D ⎢ ⎜⎜ ⎢⎣ 2 ⎝ R1 + R 2 ⎠

⎤ ⎞ 1 ⎟⎟ ⋅ Vo − ⋅ Vi ⎥ − g m3Vo 2 ⎥⎦ ⎠

⎡ 1 ⎤ 1 ⎛ R1 ⎞ 1 1 ⎟⎟ + g m3 ⎥ = g m1 g m3 R DVi + + g m1 g m3 R D ⎜⎜ Vo ⎢ ⎢⎣ R L R1 + R 2 2 ⎥⎦ 2 ⎝ R1 + R 2 ⎠ 1 g m1 g m 3 R D Vo 2 Aυ f = = Vi ⎡ 1 ⎤ ⎛ R1 ⎞ 1 1 ⎟⎟ + g m3 ⎥ + + g m1 g m3 R D ⎜⎜ ⎢ ⎢⎣ R L R1 + R 2 2 ⎥⎦ ⎝ R1 + R 2 ⎠

⎛ 0.1 ⎞ g m1 = 2 ⎜ ⎟(20)(0.5) = 1.414 mA/V ⎝ 2 ⎠ ⎛ 0.1 ⎞ g m3 = 2 ⎜ ⎟(20 )(2 ) = 2.828 mA/V ⎝ 2 ⎠ 1 (1.414)(2.828)(18.2) 2 Then 8 = ⎡1 ⎛ 15 1 1 + (1.414 )(2.828)(18.2 )⎜⎜ ⎢ + ⎢⎣10 R1 + R 2 2 ⎝ R1 + R 2

⎤ ⎞ ⎟⎟ + 2.828⎥ ⎠ ⎦⎥

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 36.389 8= ⎡ 546.8 ⎤ ⎥ ⎢2.928 + R1 + R 2 ⎦ ⎣ ⇒ R1 + R 2 = 337 k Ω ⇒ R 2 = 322 k Ω ______________________________________________________________________________________

12.45 Io R R 5 = = 83.33 = 1 + 1 ⇒ 1 = 82.33 I s 0.06 R2 R2

For example, let R 2 = 3 k Ω , R 2 = 247 k Ω ______________________________________________________________________________________ 12.46 (a) (1) V A = (I D1 + I D 2 )R D 2 (2) I D1 R D1 = V SG 2 (V − VGS1 ) − V A = I D1 (3) G RF

Now VG − VGS1 − (I D1 + I D 2 )R D 2 = I D1 R F And VGS 2 =

I D2 − VTP = I D1 R D1 Kp

I D 2 (0.3162 ) + 1 = I D1 (0.525) I D 2 = I D1 (1.660 ) − 3.162 I D 2 = [I D1 (1.660 ) − 3.162]

2

Then VG − VGS 1 − [I D1 (1.660 ) − 3.162] R D 2 = I D1 (R D 2 + R F ) 2

I D1 = K n (VGS1 − VTN )

2

7.6 − VGS1 − [I D1 (1.660 ) − 3.162] (0.25) = I D1 (0.75) 2

I D1 = 10(VGS1 − 1) By trial and error, I D1 ≅ 3.98 mA Then V SG 2 = I D1 R D1 = (3.98)(0.525) = 2.0895 V 2

I D1 = (10)(2.0895 − 1) = 11.87 mA 2

(b)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I o = g m 2V sg 2 V sg 2 = g m1V gs1 R D1

(

)

V A V A − − V gs1 + RD2 RF V A + V gs1 =0 I i + g m1V gs1 + RF

Io =

V gs1 ⎤ ⎡ or V A = − R F ⎢ I i + g m1V gs1 + ⎥ RF ⎦ ⎣ ⎛ 1 1 ⎞ V gs1 ⎟⎟ + + I o = V A ⎜⎜ ⎝ RD2 RF ⎠ RF V gs1 ⎞⎛ 1 ⎛ 1 ⎞ V gs1 ⎟⎜ ⎟+ + I o = − R F ⎜⎜ I i + g m1V gs1 + ⎜ ⎟ R F ⎠⎝ R D 2 R F ⎟⎠ R F ⎝ V sg 2 I 1 Now V gs1 = = o ⋅ g m1 R D1 g m 2 g m1 R D1 ⎡⎛ ⎛ 1 1 ⎞ 1 ⎞⎛ 1 1 ⎞ 1 ⎤ ⎟⎟ − R F V gs1 ⎢⎜⎜ g m1 + ⎟⎟⎜⎜ ⎟− + + I o = − R F I i ⎜⎜ ⎥ R F ⎠⎝ R D 2 R F ⎟⎠ R F2 ⎥⎦ ⎝ RD2 RF ⎠ ⎣⎢⎝ ⎡ RF Then I o ⎢1 + ⎣⎢ g m1 g m 2 R D1

⎛ g m1 g m1 ⎛ R 1 ⎞⎤ ⎜⎜ ⎟⎟⎥ = − I i ⎜⎜1 + F + + ⎝ R D 2 R F R F R D 2 ⎠⎦⎥ ⎝ RD2 I − g m 2 R D1 So Ai = o = g R R 1 Ii 1+ + m 2 D1 D 2 g m1 (R F + R D 2 ) R F + R D 2

⎞ ⎟⎟ ⎠

(c) g m1 = 2 K n I D1 = 2 (10 )(3.98) = 12.62 mA/V g m 2 = 2 K p I D 2 = 2 (10 )(11.87 ) = 21.79 mA/V − (21.79 )(0.525) (21.79)(0.525)(0.25) 1 1+ + (12.62)(0.75) 0.75 Ai = −2.33 ______________________________________________________________________________________ Ai =

12.47 (a) I Q = I D1 + I D 2 V SG 2 = I D1 R D

I D 2 = K p (V SG 2 + VTP ) = K p (I D1 R D + VTP ) 2

[

]

I Q = I D1 + K p I D2 1 R D2 − 2 I D1 R D + 1

We find 2.756 I

2 D1

− 9.5 I D1 − 6 = 0

⇒ I D1 = 3.99 mA and I D 2 = 12.01 mA

2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

I i + g m1V gs1 + I o = 0

V sg 2 = g m1V gs1 R D ⇒ V gs1 =

V sg 2 g m1 R D

I o = g m 2V sg 2

Then V gs1 =

Io g m1 g m 2 R D

⎛ ⎞ Io ⎟ + Io = 0 I i + g m1 ⎜⎜ ⎟ ⎝ g m1 g m 2 R D ⎠ − g m2 R D I −1 = So that Ai = o = Ii ⎛ 1 ⎞ 1 + g m2 R D ⎜1 + ⎟ ⎜ g R ⎟ m2 D ⎠ ⎝

We find g m 2 = 2 K p I D 2 = 2 (10 )(12.01) = 21.92 mA/V − (21.92)(0.525) = −0.920 1 + (21.92 )(0.525) ______________________________________________________________________________________ Ai =

12.48 (a) Neglect base currents I Q = I C1 + I C 2 V EB 2 = I C1 RC ⎛I R I Q = I C1 + I S 2 exp⎜⎜ C1 C ⎝ VT

⎞ ⎟⎟ ⎠

⎡ I (200 ) ⎤ 16 × 10 −3 = I C1 + 10 −15 exp ⎢ C1 ⎥ ⎣ 0.026 ⎦ By trial and error, I C1 = 3.92 mA and I C 2 = 12.08 mA

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

I o = g m 2 Vπ 2

Vπ 2 = g m1Vπ 1 (rπ 2 RC )

Then I o = g m1 g m 2Vπ 1 (rπ 2 RC ) I i + I o + g m1Vπ 1 +

Vπ 1 =0 rπ 1

⎛ 1 + β1 ⎞ ⎟=0 I i + I o + Vπ 1 ⎜⎜ ⎟ ⎝ rπ 1 ⎠ Io Now Vπ 1 = g m1 g m 2 (rπ 2 RC ) ⎡ ⎛ 1 + β1 1 ⎜ Then I i + I o ⎢1 + ⎜ ⎢⎣ g m1 g m 2 (rπ 2 RC ) ⎝ rπ 1

⎞⎤ ⎟⎥ = 0 ⎟ ⎠⎥⎦

⎛ β ⎞ − ⎜⎜ 1 ⎟⎟(g m 2 )(rπ 2 RC ) I ⎝ 1 + β1 ⎠ So Ai = o = Ii ⎛ β ⎞ 1 + ⎜⎜ 1 ⎟⎟(g m 2 )(rπ 2 RC ) ⎝ 1 + β1 ⎠ I 12.08 = 464.6 mA/V (c) g m 2 = C 2 = 0.026 VT rπ 2 =

β 2 VT I C2

=

(180)(0.026) = 0.3874 k Ω 12.08

rπ 2 RC = 387.4 200 = 131.9 Ω ⎛ 180 ⎞ −⎜ ⎟(464.6 )(0.1319 ) ⎝ 181 ⎠ Ai = = −0.984 ⎛ 180 ⎞ ( )( ) 1+ ⎜ 464 . 6 0 . 1319 ⎟ ⎝ 181 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.49 ⎛ h ⎞ ⎛ 100 ⎞ I C1 = ⎜ FE ⎟ I E1 = ⎜ ⎟ (0.2) = 0.198 mA 1 + h ⎝ 101 ⎠ FE ⎠ ⎝ VC1 = 10 − (0.198)(40) = 2.08 V

2.08 − 0.7 = 1.38 mA 1 ⎛ 100 ⎞ =⎜ ⎟ (1.38) = 1.37 mA ⎝ 101 ⎠

IE2 =

(a)

IC 2

For Q1 : (100)(0.026) = 13.1 k Ω 0.198 0.198 = = 7.62 mA / V 0.026

rπ 1 = g m1

For Q2 : (100)(0.026) = 1.90 k Ω 1.37 1.37 = = 52.7 mA / V 0.026

rπ 2 = gm2

(b)

IS =

Vπ 1 Vπ 1 Vπ 1 − Ve + + RS rπ 1 RF

V + Ve Vπ 2 + =0 g m1Vπ 1 + π 2 RC1 rπ 2

(1) (2)

Vπ 2 V V −V + g m 2Vπ 2 = e + e π 1 rπ 2 RE RF

(3) Substitute numerical values in (1), (2), and (3): 1 1⎞ ⎛1 ⎛ 1⎞ I S = Vπ 1 ⎜ + + ⎟ − Ve ⎜ ⎟ ⎝ 10 13.1 10 ⎠ ⎝ 10 ⎠ I S = Vπ 1 (0.2763) − Ve (0.10)

(1)

1 ⎞ ⎛ 1 ⎛ 1 ⎞ (7.62)Vπ 1 + Vπ 2 ⎜ + ⎟ + Ve ⎜ ⎟ = 0 40 1.90 ⎝ ⎠ ⎝ 40 ⎠ (7.62)Vπ 1 + Vπ 2 (0.5513) + Ve (0.025) = 0

(2)

⎛ 1 ⎞ ⎛1 1 ⎞ ⎛ 1⎞ Vπ 2 ⎜ + 52.7 ⎟ = Ve ⎜ + ⎟ − Vπ 1 ⎜ ⎟ ⎝ 1.90 ⎠ ⎝ 1 10 ⎠ ⎝ 10 ⎠ Vπ 2 (53.23) = Ve (1.10) − Vπ 1 (0.10)

From (3), Ve = Vπ 2 (48.39) + Vπ 1 (0.0909)

Substituting into (1),

(3)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I S = Vπ 1 (0.2763) − (0.10) [Vπ 2 (48.39) + Vπ 1 (0.0909) ]

or I S = Vπ 1 (0.2672) − Vπ 2 (4.839)

(1′)

and substituting into (2),

(7.62)Vπ 1 + Vπ 2 (0.5513) + ( 0.025 ) ⎣⎡Vπ 2 ( 48.39 ) + Vπ 1 ( 0.0909 ) ⎦⎤ = 0

or

(7.622)Vπ 1 + Vπ 2 (1.761) = 0 ⇒ Vπ 1 = −Vπ 2 (0.2310) (2′)

Then substituting (2′) into (1′), we obtain I S = (0.2672)(−Vπ 2 )(0.2310) − Vπ 2 (4.839)

or I S = −Vπ 2 (4.901)

Now ⎛ RC 2 ⎞ I O = − g m 2Vπ 2 ⎜ ⎟ ⎝ RC 2 + RL ⎠ ⎛ 2 ⎞ = −(52.7) ⎜ ⎟ Vπ 2 = −(42.16)Vπ 2 ⎝ 2 + 0.5 ⎠

Then ⎛ −IS ⎞ I O = −(42.16) ⎜ ⎟ ⎝ 4.901 ⎠

or Aif =

IO = 8.60 Is

Ri =

(c) We had

Vπ 1 IS

and

Ri = RS || Rif

Vπ 1 = −Vπ 2 (0.2310)

and I S = −Vπ 2 (4.901)

so ⎛ −Vπ 1 ⎞ IS = − ⎜ ⎟ (4.901) = Vπ 1 (21.22) ⎝ 0.2310 ⎠

Then Ri =

Vπ 1 1 = = 0.04713 21.22 IS

Finally 0.04713 =

10 Rif 10 + Rif

⇒ Rif = 47.4 Ω

______________________________________________________________________________________ 12.50 (a) Ii =

(1)

Vπ 1 V −V + π 1 e2 RF RS RB1 rπ 1

g m1Vπ 1 +

(2) (3)

VC1 V + π2 = 0 RC1 RB 2 rπ 2

Vπ 2 V V −V + g m 2Vπ 2 = e 2 + e 2 π 1 rπ 2 RE 2 RF

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(4) Now

⎛ RC 2 ⎞ I o = −( g m 2Vπ 2 ) ⎜ ⎟ ⎝ RC 2 + RL ⎠ Ii =

(1)′

Vπ 1 V − e2 RS RB1 rπ 1 RF RF

⎛ ⎞ RF ⎟V − I R Ve 2 = ⎜ ⎜ RS RB1 rπ 1 RF ⎟ π 1 i F ⎝ ⎠ So

Now, from (2) g m1Vπ 1 +

(2)′ Also (3)′ And

Vπ 2 + Ve 2 Vπ 2 + =0 Rc1 RB 2 rπ 2

⎛ Vπ 2 Ve 2 1 ⎞ + =0 ⎜ g m1 + ⎟ Vπ 1 + r R R R r π2 ⎠ C1 RB 2 ⎝ C1 B2 π 2

⎛ ⎛ 1 Vπ 1 1 ⎞ 1 ⎞ = Ve 2 ⎜ + ⎜ gm2 + ⎟ Vπ 2 + ⎟ r R R R F F ⎠ π2 ⎠ ⎝ E2 ⎝ ⎛ I ⎞⎛ R + RL ⎞ Vπ 2 = − ⎜ O ⎟⎜ C 2 ⎟ ⎝ g m 2 ⎠⎝ RC 2 ⎠

(4)′ Substitute (1)′ into (2)′ and (3)′

(2)″

⎛ Vπ 2 1 ⎞ 1 + ⎜ g m1 + ⎟ Vπ 1 + rπ 2 ⎠ RC1 RB 2 rπ 2 RC1 RB 2 ⎝

⎡ ⎤ RF ⎢ − Vπ 1 − I i RF ⎥ = 0 ⎢ RS RB1 rπ 1 RF ⎥ ⎣ ⎦

⎤ ⎛ Vπ 1 ⎛ 1 RF 1 ⎞ 1 ⎞⎡ =⎜ + − Vπ 1 − I i RF ⎥ ⎜ gm2 + ⎟ Vπ 2 + ⎟⎢ rπ 2 ⎠ RF ⎝ RE 2 RF ⎠ ⎢ RS RB1 rπ 1 RF ⎥ ⎝ ⎣ ⎦ (3)″ V Solve for π 1 from (2)″ and substitute into (3)″. Also use Equation (4)′.

(b)

RB1 = R1 R2 = 20 80 = 16 K ⎛ 20 ⎞ VTH 1 = ⎜ ⎟ (10) = 2 V ⎝ 100 ⎠ 2 − 0.7 I BQ1 = = 0.0111 mA 16 + (101)(1) I CQ1 = 1.11 mA RTH 2 = 15 85 = 12.75 K ⎛ 15 ⎞ VTH 2 = ⎜ ⎟ (10) = 1.5 V ⎝ 100 ⎠ 1.5 − 0.7 I BQ 2 = = 0.01265 mA 12.75 + (101)(0.5) I CQ 2 = 1.265 mA 1.11 = 42.69 mA/V 0.026 1.265 = = 48.65 mA/V 0.026

g m1 = gm2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (100)(0.026) = 2.34 K 1.11 (100)(0.026) = 2.06 K rπ 2 = 1.265 RC1 RB 2 = 2 12.75 = 1.729 K rπ 1 =

Now

RS RB1 rπ 1 RF ≅ RB1 rπ 1 RF = 16 2.34 10 = 1.695 K RC1 RB 2 rπ 2 = 1.729 2.06 = 0.940 K

Now

(2)″

Vπ 2 1 ⎞ 1 ⎡ 10 ⎛ ⎤ + ⎜ 42.69 + ⎟ Vπ 1 + ⎢1.695 ⋅ Vπ 1 − I i (10) ⎥ 2.06 0.940 1.729 ⎝ ⎠ ⎣ ⎦ 46.587Vπ 1 + 1.064Vπ 2 − 5.784 I i = 0

(3)″

Vπ 1 ⎛ 1 1 ⎞ ⎡ 10 ⎛ 101 ⎞ ⎤ =⎜ + ⎟ ⋅ Vπ 1 − I i (10) ⎥ ⎜ ⎟ Vπ 2 + 10 ⎝ 0.5 10 ⎠ ⎣⎢1.695 ⎝ 2.06 ⎠ ⎦ 49.03Vπ 2 = 12.29Vπ 1 − 21I i

From (2)″ Vπ 1 = (0.1242) I i − (0.02284)Vπ 2 Then (3)″

49.03Vπ 2 = 12.29 [ (0.1242) I i − (0.02284)Vπ 2 ] − 21I i

49.31Vπ 2 = −19.47 I i

From (4)′ Then

⎛ I ⎞⎛ 4 + 4 ⎞ Vπ 2 = − ⎜ o ⎟ ⎜ ⎟ = −(0.0411) I o ⎝ 48.65 ⎠ ⎝ 4 ⎠

(49.31) [ −(0.0411) I o ] = −19.47 I i Io = Ai = 9.61 Ii

______________________________________________________________________________________ 12.51

a.

RTH = 13.5 || 38.3 = 9.98 kΩ ⎛ 13.5 ⎞ VTH = ⎜ ⎟ (10) = 2.606 V ⎝ 13.5 + 38.3 ⎠ (120)(2.606 − 0.7) I C1 = = 1.75 mA 9.98 + (121)(1) VC1 = 10 − (1.75 )( 3) = 4.75 V 4.75 − 0.7 = 0.50 mA 8.1 (120)(0.026) rπ 1 = = 1.78 kΩ 1.75 1.75 g m1 = = 67.31 mA / V 0.026

IC 2 ≈

(120)(0.026) = 6.24 kΩ 0.50 0.50 gm2 = = 19.23 mA / V 0.026

rπ 2 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b.

and

VS − Vπ 1 Vπ 1 V −V = + π 1 e2 RS RB || rπ 1 RF

(1)

V + Ve 2 Vπ 2 g m1Vπ 1 + π 2 + =0 RC1 rπ 2

(2)

Vπ 2 V V −V + g m 2Vπ 2 = ε 2 + ε 2 π 1 rπ 2 RE 2 RF

(3)

V0 = −( g m 2Vπ 2 ) RC 2

(4) Substitute numerical values in (1), (2), and (3) VS Vπ 1 ⎡ 1 1 ⎤ Ve 2 = Vπ 1 ⎢ + + ⎥− 0.6 ⎣ 0.6 9.98 || 1.78 1.2 ⎦ 1.2 VS (1.67) = Vπ 1 (4.011) − Ve 2 (0.8333)

(1)

1 ⎞ Ve 2 ⎛1 (67.31)Vπ 1 + Vπ 2 ⎜ + =0 ⎟+ ⎝ 3 6.24 ⎠ 3

or

Vπ 1 (67.31) + Vπ 2 (0.4936) + Ve 2 (0.3333) = 0

(2)

V V ⎛ 1 ⎞ V Vπ 1 ⎜ + 19.23 ⎟ = e 2 + e 2 − π 2 ⎝ 6.24 ⎠ 8.1 1.2 1.2

or

Vπ 2 (19.39) = Ve 2 (0.9568) − Vπ 1 (0.8333)

(3)

From (1) Ve 2 = Vπ 1 (4.813) − VS (2.00)

Then Vπ 1 (67.31) + Vπ 2 (0.4936) + (0.3333)[Vπ 1 (4.813) − VS (2.00)] = 0

or and or

Vπ 1 (68.91) + Vπ 2 (0.4936) − VS (0.6666) = 0

(2′)

Vπ 2 (19.39) = (0.9568) [Vπ 1 (4.813) − VS (2.00) ] − Vπ 1 (0.8333) Vπ 2 (19.39) = Vπ 1 (3.772) − VS (1.914)

(3′)

We find Vπ 1 = VS (0.009673) − Vπ 2 (0.007163)

Then Vπ 2 (19.39) = (3.772)[VS (0.009673) − Vπ 2 (0.007163)] − VS (1.914)

Vπ 2 (19.42) = VS (−1.878) or Vπ 2 = −VS (0.09670)

so that

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V0 = −(19.23)(4)(−VS )(0.09670)

Then V0 = 7.44 VS

______________________________________________________________________________________ 12.52

Using the circuit from Problem 12.51, we have

Rif =

Vπ 1 IS

.

V − Vπ 1 IS = S RS

Where From Problem 12.51

Vπ 1 = VS (0.009673) − Vπ 2 (0.007163) = VS (0.009673) − (0.007163)(−VS )(0.09670) = VS (0.01037)

So Rif =

VS (0.01037) ⋅ (0.6) = 0.00629 kΩ VS − VS (0.01037)

or Rif = 6.29 Ω

______________________________________________________________________________________ 12.53 RTH = 1.4 || 17.9 = 1.298 kΩ ⎛ 1.4 ⎞ VTH = ⎜ ⎟ (10) = 0.7254 V ⎝ 1.4 + 17.9 ⎠ 0.7254 − 0.7 I B1 = = 0.0196 mA 1.298 I C1 = (50)(0.0196) = 0.98 mA

Neglecting dc base currents, VB 2 = 10 − (0.98)(7) = 3.14 V 3.14 − 0.7 = 3.25 mA 0.25 + 0.5 ⎛ 50 ⎞ I C 2 = ⎜ ⎟ (3.25) = 3.19 mA ⎝ 51 ⎠ (50)(0.026) rπ 1 = = 1.33 kΩ 0.98 0.98 g m1 = = 37.7 mA / V 0.026 (50)(0.026) rπ 2 = = 0.408 kΩ 3.19 3.19 gm2 = = 123 mA / V 0.026 IE2 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

IS =

Vπ 1 V −V + π1 1 R1 || R2 || rπ 1 RF

V V + Ve 2 g m1Vπ 1 + π 2 + π 2 =0 rπ 2 RC1

(1) (2)

Vπ 2 V −V + g m 2Vπ 2 = e 2 1 rπ 2 RE1

(3)

Ve 2 − Vπ 1 V −V V = 1 + 1 π1 RE1 RE 2 RF

(4) Enter numerical values in (1), (2), (3) and (4): IS =

or

Vπ 1 V −V + π1 1 17.9 ||1.4 || 1.33 5

I S = Vπ 1 (1.722) − V1 (0.20)

(1)

V V + Ve 2 (37.7)Vπ 1 + π 2 + π 2 =0 0.408 7

or

or

Vπ 1 (37.7) + Vπ 2 (2.594) + Ve 2 (0.1429) = 0 (2) Vπ 2 V −V + (123)Vπ 2 = e 2 1 0.408 0.25 Vπ 2 (125.5) = Ve 2 (4) − V1 (4)

(3)

Ve 2 − V1 V −V V = 1 + 1 π1 0.25 0.50 5

or

Ve 2 (4) = V1 (6.20) − Vπ 1 (0.20)

(4)

From (4): Ve 2 = V1 (1.55) − Vπ 1 (0.05)

Then substituting in (3): Vπ 2 (125.5) = (4)[V1 (1.55) − Vπ 1 (0.05)] − V1 (4)

or

Vπ 2 (125.5) = V1 (2.20) − Vπ 1 (0.20)

(3′)

and substituting in (2):

Vπ 1 ( 37.7 ) + Vπ 2 ( 2.594 ) + ( 0.1429 ) ⎡⎣V1 (1.55 ) − Vπ 1 ( 0.05 ) ⎤⎦ = 0

or Vπ 1 (37.69) + Vπ 2 (2.594) + V1 (0.2215) = 0

Now V1 = −Vπ 1 (170.16) − Vπ 2 (11.71)

Then substituting in (1): I S = Vπ 1 (1.722) − (0.20)[−Vπ 1 (170.16) − Vπ 2 (11.71)]

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or I S = Vπ 1 (35.75) + Vπ 2 (2.342)

and substituting in (3′): Vπ 2 (125.5) = (2.20)[−Vπ 1 (170.16) − Vπ 2 (11.71)] − Vπ 1 (0.20) Vπ 2 (151.3) = −Vπ 1 (374.55)

or Then

Vπ 1 = −Vπ 2 (0.4040) so that

I S = (35.75)[−Vπ 2 (0.4040)] + Vπ 2 (2.342) I S = −Vπ 2 (12.10)

⎛ RC 2 ⎞ I 0 = −( g m 2Vπ 2 ) ⎜ ⎟ ⎝ RC 2 + RL ⎠ ⎛ 2.2 ⎞ = −(123) ⎜ ⎟ Vπ 2 = −(64.43)Vπ 2 ⎝ 2.2 + 2 ⎠ Vπ 2 = −(0.01552) I 0

or Then

I0 I 1 = ⇒ 0 = 5.33 I S (0.01552)(12.10) IS

______________________________________________________________________________________ 12.54 For example, use the circuit shown in Figure P12.49 ______________________________________________________________________________________ 12.55

(a)

Vi − V L V L − V O = R1 RF ⎛ 1 VO 1 ⎞ Vi ⎟⎟ − = V L ⎜⎜ + RF ⎝ R1 R F ⎠ R1

⎛ R So (1) VO = V L ⎜⎜1 + F R1 ⎝ VO − V L V L V L = + R3 R L R2

⎞ ⎛ RF ⎟⎟ − ⎜⎜ ⎠ ⎝ R1

⎞ ⎟⎟Vi ⎠

⎛ R R ⎞ So (2) VO = V L ⎜⎜1 + 3 + 3 ⎟⎟ ⎝ R L R2 ⎠ Then, from (1) = (2) ⎛ ⎛ R ⎞ ⎛R ⎞ R R ⎞ V L ⎜⎜1 + F ⎟⎟ − ⎜⎜ F ⎟⎟Vi = V L ⎜⎜1 + 3 + 3 ⎟⎟ R1 ⎠ ⎝ R1 ⎠ ⎝ R L R2 ⎠ ⎝ ⎛ R R R ⎞ ⎛R V L ⎜⎜1 + F − 1 − 3 − 3 ⎟⎟ = ⎜⎜ F R R R L 1 2 ⎠ ⎝ R1 ⎝ Now V L = I O R L ⎛R R R ⎞ ⎛R I O R L ⎜⎜ F − 3 − 3 ⎟⎟ = ⎜⎜ F R R R 2 ⎠ L ⎝ R1 ⎝ 1

⎞ ⎟⎟Vi ⎠

⎞ ⎟⎟Vi ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ RF R1

IO = Vi ⎛ R L R F R R ⎜⎜ − R3 − L 3 R R2 1 ⎝ R R (b) Set F = 3 R1 R2 Ag f =

⎞ ⎟⎟ ⎠

RF

Then Ag f = Or Ag f = (c) For

R1 − R F = R1 R3 − R3

I O −1 = Vi R2

IO −1 = −0.5 mA/V = ⇒ R2 = 2 k Ω Vi R2

R RF = 3 , set R3 = 2 k Ω and R1 = R F = 10 k Ω R1 R2 ______________________________________________________________________________________

For

12.56 (a) V D1 = 3 − (1)(1.6) = 1.40 V, V SG 3 = (1)(1.6) = 1.6 V

I DQ 3 = K p (V SG 3 + VTP ) = (10 )(1.6 − 0.5) = 12.1 mA 2

2

VG = (12.1)(0.248) − 3 ≅ 0 (b) g m1V gs1 + g m1V gs 2 = 0 ⇒ V gs 2 = −V gs1 V sg 3 = g m1V gs1 R D I o = g m3V sg 3 = g m1 g m3 R DV gs1

Vi = V gs1 − V gs 2 + I o R L = 2V gs1 + I o R L Vi − I o R L 2 ⎛ V − I o RL ⎞ ⎟⎟ I o = g m1 g m3 R D ⎜⎜ i 2 ⎝ ⎠ ⎛ g g R R ⎞ g g R I o ⎜⎜1 + m1 m3 D L ⎟⎟ = m1 m3 D ⋅ Vi 2 2 ⎝ ⎠ V gs1 =

Ag f =

Io g m1 g m3 R D = Vi 2 + g m1 g m 3 R D R L

(c) g m1 = 2 K n I DQ1 = 2 (2 )(1) = 2.828 mA/V g m3 = 2 K p I DQ 3 = 2 (10 )(12.1) = 22 mA/V

(2.828)(22)(1.6) = 3.73 mA/V 2 + (2.828)(22)(1.6)(0.248) ______________________________________________________________________________________ Ag f =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.57

rπ 1 = 6.24 kΩ, rπ 2 = 3.12 kΩ, rπ 3 = 1.56 kΩ g m1 = 19.23mA / V , g m 2 = 38.46 mA / V, g m 3 = 76.92 mA / V

VS = Vπ 1 + Ve1

(1)

Vπ 1 V V −V + g m1Vπ 1 = e1 + e1 e3 rπ 1 RE1 RF

(2)

Vπ 2 = − g m1Vπ 1 ( RC1 || rπ 2 )

(3)

V +V V g m 2Vπ 2 + π 3 e 3 + π 3 = 0 RC 2 rπ 3

(4)

Vπ 3 V V −V + g m 3Vπ 3 = e3 + e3 e1 rπ 3 RE 2 RF

(5) Enter numerical values in (2)-(5): Vπ 1 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + (19.23)Vπ 1 = Ve1 ⎜ + ⎟ − Ve3 ⎜ ⎟ 6.24 ⎝ 0.1 0.8 ⎠ ⎝ 0.8 ⎠

or

Vπ 1 (19.39) = Ve1 (11.25) − Ve 3 (1.25) Vπ 2 = −(19.23)Vπ 1 (5 || 3.12) = −(36.94)Vπ 1

(2) (3)

1 ⎞ ⎛1 ⎛1⎞ (38.46)Vπ 2 + Vπ 3 ⎜ + ⎟ + Ve 3 ⎜ ⎟ = 0 2 1.56 ⎝ ⎠ ⎝ 2⎠

or

Vπ 2 (38.46) + Vπ 3 (1.141) + Ve3 (0.5) = 0

(4)

1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + 76.92 ⎟ = Ve 3 ⎜ + Vπ 3 ⎜ ⎟ − Ve3 ⎜ ⎟ ⎝ 1.56 ⎠ ⎝ 0.1 0.8 ⎠ ⎝ 0.8 ⎠

or

Vπ 3 (77.56) = Ve3 (11.25) − Ve1 (1.25)

From (1) Then or

Vπ 1 = VS − Ve1

(VS − Ve1 )(19.39) = Ve1 (11.25) − Ve3 (1.25) VS (19.39) = Ve1 (30.64) − Ve3 (1.25) Vπ 2 = −VS (36.94) + Ve1 (36.94)

(5)

(2′)

(3′)

(38.46)[−VS (36.94) + Ve1 (36.94)] + Vπ 3 (1.141) + Ve3 (0.5) = 0

From (5): Then

Ve 3 = Vπ 3 (6.894) + Ve1 (0.1111)

VS (19.39) = Ve1 (30.64) − (1.25)[Vπ 3 (6.894) + Ve1 (0.1111)]

or

VS (19.39) = Ve1 (30.50) − Vπ 3 (8.6175)

(2″)

(4′)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ and −VS (1420.7) + Ve1 (1420.7) + Vπ 3 (1.141) + (0.5)[Vπ 3 (6.894) + Ve1 (0.1111)] = 0

or

−VS (1420.7) + Ve1 (1420.76) + Vπ 3 (4.588) = 0

(4″)

From (2″): Ve1 = VS (0.6357) + Vπ 3 (0.2825)

Then substituting in (4″): −VS (1420.7) + (1420.76)[VS (0.6357) + Vπ 3 (0.2825)] + Vπ 3 (4.588) = 0 −VS (517.5) + Vπ 3 (405.95) = 0

Now

I 0 = g m 3Vπ 3 = 76.92Vπ 3

or Vπ 3 = I 0 (0.0130)

Then −VS (517.5) + I 0 (0.0130)(405.95) = 0 or I0 = 98.06 mA / V VS

______________________________________________________________________________________ 12.58

Computer Analysis ______________________________________________________________________________________ 12.59 (100)(0.026) = 5.2 kΩ 0.5 0.5 g m1 = g m 2 = = 19.23 mA / V 0.026 (100)(0.026) rπ 3 = = 1.3 kΩ 2 2 g m3 = = 76.92 mA / V 0.026 rπ 1 = rπ 2 =

Vπ 1 V + g m1Vπ 1 + g m 2Vπ 2 + π 2 = 0 rπ 1 rπ 2

Since

rπ 1 = rπ 2

and

g m1 = g m 2 ,

(1)

then Vπ 1 = −Vπ 2

VS = Vπ 1 − Vπ 2 + Ve3 = −2Vπ 2 + Ve3 V V +V g m 2Vπ 2 + π 3 + π 3 e 3 = 0 rπ 3 RC 2

(3)

Vπ 3 V V + g m 3Vπ 3 = e3 + π 2 rπ 3 RF rπ 2

(4)

(2)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ RC 3 ⎞ I0 = − ⎜ ⎟ ( g m 3Vπ 3 ) ⎝ RC 3 + RL ⎠ Ve 3 = VS + 2Vπ 2

From (2): (19.23)

Vπ 2 +

(5)

Vπ 3 Vπ 3 1 (VS + 2Vπ 2 ) = 0 + + 1.3 18.6 18.6

or (19.23)Vπ 2 + (0.8230)Vπ 3 + (0.05376)VS = 0

(3′)

V ⎛ 1 ⎞ ⎛1⎞ + 76.92 ⎟ = ⎜ ⎟ (VS + 2Vπ 2 ) + π 2 Vπ 3 ⎜ 5.2 ⎝ 1.3 ⎠ ⎝ 10 ⎠

or

(77.69)Vπ 3 = (0.3923)Vπ 2 + (0.1)VS

(4′)

⎛ 2 ⎞ I0 = − ⎜ ⎟ (76.92)Vπ 3 = −(51.28)Vπ 3 ⎝ 2 +1⎠

(5′)

From (3′): Vπ 2 = −(0.04255)Vπ 3 − (0.002780)VS

Then (77.69)Vπ 3 = (0.3923)[ −(0.04255)Vπ 3 − (0.002780)VS ] + (0.1)VS (77.71)Vπ 3 = (0.0989)VS

or Vπ 3 = (0.001273)VS

so that I 0 = −(51.28)(0.001273)VS

or I0 = −(0.0653) mA/V VS

______________________________________________________________________________________ 12.60 Io 1 = = 3 mA/V ⇒ R E = 333.3 Ω Vi R E h FE Ag I (80)(1000) = o = = = 2.9998875 mA/V Vi 1 + (h FE )Ag R E 1 + (80)(1000 )(0.33333)

(a) Ag f = (b) Ag f

I o = Ag f (5) = 14.9994377

14.9994377 − 15.0 × 100% = −0.00375% 15.0 ______________________________________________________________________________________

Error =

12.61 (a) 5 = (1 + h FE )I BQ R E + V EB (on ) + I BQ R B 5 − 0.7 = 0.030605 mA 100 + (81)(0.5) = 2.448 mA, and I EQ = 2.479 mA

I BQ = I CQ

V ECQ = 5 − (2.479)(0.5) − (2.448)(1) = 1.31 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

V − Vπ Vo Vo + + h FE I b + o =0 RF RC ro ⎛ 1 V 1 1 ⎞ ⎟ + h FE I b − π = 0 + + V o ⎜⎜ ⎟ RF ⎝ RC ro R F ⎠ Vπ = I b rπ V V V − Vo Is = π + π + π R B rπ RF ⎛ r r ⎞ V I s = I b ⎜⎜1 + π + π ⎟⎟ − o ⎝ RB RF ⎠ RF V Is + o RF Ib = ⎛ r r ⎞ ⎜⎜1 + π + π ⎟⎟ ⎝ RB RF ⎠ ⎛ 1 1 1 ⎞ ⎛ ⎟ + ⎜⎜ h FE + + V o ⎜⎜ ⎟ R r R o F ⎠ ⎝ ⎝ C

V ⎡ ⎤ Is + o ⎥ ⎢ r ⎞ RF ⎥ − π ⎟⎟ ⎢ =0 rπ rπ ⎥ R F ⎠⎢ + ⎢1 + ⎥ ⎣ RB RF ⎦

⎛ 1 ⎛ r ⎞ V r ⎞ r r ⎞ ⎛ 1 1 ⎞⎛ ⎟⎜⎜1 + π + π ⎟⎟ + ⎜⎜ h FE − π ⎟⎟ ⋅ I s + ⎜⎜ h FE − π ⎟ ⋅ o = 0 + + V o ⎜⎜ ⎟ R F ⎟⎠ R F RF ⎠ ⎝ ⎝ RC ro R F ⎠⎝ R B R F ⎠ ⎝ V 100 = 40.85 k Ω Now ro = A = I CQ 2.448

rπ =

h FE VT (80)(0.026) = = 0.8497 k Ω 2.448 I CQ

1 1 ⎞⎛ 0.8497 0.8497 ⎞ ⎛ 0.8497 ⎞ 0.8497 ⎞ V o ⎛ ⎛1 =0 Then V o ⎜ + + + ⎟⎜1 + ⎟⋅ ⎟ ⋅ I s + ⎜ 80 − ⎟ + ⎜ 80 − 100 10 ⎠ ⎝ 10 ⎠ 10 ⎠ 10 ⎝ ⎝ 1 40.85 10 ⎠⎝ V o (9.22114 ) + I s (79.915) = 0 V Az f = o = −8.666 V/mA Is

(c) Ri f

Ri f

V ⎛ 1 ⎜ 1+ o ⋅ I s RF V I r ⎜ = π = b π = rπ ⎜ rπ r Is Is ⎜⎜ 1 + + π RB RF ⎝ = 0.1037 k Ω

(d) I x =

⎞ 8.666 ⎛ ⎞ ⎟ 1− ⎜ ⎟ ⎟ 10 ⎜ ⎟ ( ) = 0 . 8497 ⎟ 0.8497 0.8497 ⎟ ⎜ + ⎟⎟ ⎜1+ ⎟ 100 10 ⎠ ⎝ ⎠

Vx Vx Vx + + h FE I b + RC ro R F + R B rπ

⎛ R B rπ ⎞ Vπ ⎟ ⋅V x , and Vπ = ⎜ ⎜ ⎟ rπ ⎝ R B rπ + R F ⎠ ⎞ I 1 1 1 ⎛ h ⎞⎛ R B rπ 1 ⎟+ Then x = = + + ⎜⎜ FE ⎟⎟⎜ ⎜ ⎟ V x Ro f RC ro ⎝ rπ ⎠⎝ R B rπ + R F ⎠ R B rπ + R F Ib =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ R B rπ = 100 0.8497 = 0.8425 k Ω 1 1 1 1 ⎛ 80 ⎞⎛ 0.8425 ⎞ = + +⎜ ⎟+ ⎟⎜ Ro f 1 40.85 ⎝ 0.8497 ⎠⎝ 0.8425 + 10 ⎠ 0.8425 + 10

Then Ro f = 0.1186 k Ω ______________________________________________________________________________________ 12.62 2 (a) (i) V + = I DQ R D + VGS = K n R D (VGS − VTN ) + VGS

(

)

2 3 = (0.5)(5) VGS − 0.8VGS + 0.16 + VGS

We find 2.5V

2 GS

− VGS − 2.6 = 0 ⇒ VGS = 1.239 V

I DQ = (0.5)(1.239 − 0.4) = 0.3522 mA 2

(ii) g m = 2 K n I DQ = 2 (0.5)(0.3522 ) = 0.8392 mA/V Vo − V gs Vo + g mV gs + =0 RF RD V gs − Vo (2) I s = ⇒ V gs = I s R F + V o RF

(b) (1)

⎛ 1 1 ⎞ ⎛ 1 ⎞ ⎟⎟ + ⎜⎜ g m − ⎟( I s R F + V o ) = 0 + Then V o ⎜⎜ R F ⎟⎠ ⎝ RD RF ⎠ ⎝ ⎛ 1 ⎞ + g m ⎟⎟ + I s (g m R F − 1) = 0 V o ⎜⎜ ⎝ RD ⎠ V o 1 − g m R F 1 − (0.8392)(25) Az f = = = 1 1 Is + gm + 0.8392 RD 5 Az f = −19.23 V/mA

(c) I x =

Vx + g mV gs and V gs = V x RD

Ix 1 1 1 = = + g m = + 0.8392 V x Ro f RD 5

Ro f = 0.962 k Ω ______________________________________________________________________________________ 12.63 1 − RF 1 − g m RF gm = (a) Az f = 1 1 + gm +1 g m RD RD As g m → ∞ , Az f = − R F

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 − g m RF 1 + gm RD 1 − g m (25) − 23.75 = 0.2 + g m

(b) − 0.95 R F =

So that g m = 4.6 mA/V ______________________________________________________________________________________ 12.64 dc analysis RTH = 24 || 150 = 20.7 kΩ

so

⎛ 24 ⎞ VTH = ⎜ ⎟ (12) = 1.655 V ⎝ 24 + 150 ⎠ 1.655 − 0.7 I BQ = = 0.00556 mA 20.7 + (151)(1) I CQ = 0.834 mA (150)(0.026) = 4.68 kΩ 0.834 0.834 gm = = 32.08 mA / V 0.026

rπ =

VS − Vπ Vπ V − V0 = + π RS RB || rπ RF

(1)

V V −V g mVπ + 0 + 0 π = 0 RC RF

(2)

From (1): VS = Vπ 5

⎡1 1 1 ⎤ V0 + ⎢ + ⎥− 5 20.7 || 4.68 R RF F ⎦ ⎣

or ⎛ 1 ⎞ V0 VS (0.20) = Vπ ⎜ 0.4620 + ⎟− R RF F ⎠ ⎝

From (2): ⎛ ⎛1 1 ⎞ 1 ⎞ ⎜ 32.08 − ⎟ Vπ + V0 ⎜ + ⎟=0 RF ⎠ ⎝ ⎝ 6 RF ⎠

so ⎛ 1 ⎞ −V0 ⎜ 0.1667 + ⎟ RF ⎠ ⎝ Vπ = ⎛ 1 ⎞ ⎜ 32.08 − ⎟ RF ⎠ ⎝

(2)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then ⎛ 1 VS (0.20) = ⎜ 0.4620 + RF ⎝

⎡ ⎛ 1 ⎢ −V0 ⎜ 0.1667 + R ⎞⎢ F ⎝ ⎟⎢ ⎛ 1 ⎞ ⎠ ⎢ ⎜ 32.08 − ⎟ RF ⎠ ⎢⎣ ⎝

⎞⎤ ⎟⎥ ⎠ ⎥ − V0 ⎥ RF ⎥ ⎥⎦

Neglect the RF in the denominator term. Now V0 V = −5 ⇒ VS = − 0 = −V0 (0.20) VS 5

⎡ −V ( 0.1667 RF + 1) ⎤ −V0 (0.20)(0.20) RF = (0.4620 RF + 1) ⎢ 0 ⎥ − V0 32.08 RF ⎣ ⎦ −1.283RF2 = −(0.4620 RF + 1)(0.1667 RF + 1) − 32.08 RF 1.206 RF2 − 32.71RF − 1 = 0 RF =

32.71 ± (32.71) 2 + 4(1.206)(1) 2(1.206)

so that RF = 27.2 kΩ

______________________________________________________________________________________ 12.65 dc analysis RTH = 4 ||15 = 3.16 kΩ = RB ⎛ 4 ⎞ VTH = ⎜ ⎟12 = 2.526 V ⎝ 4 + 15 ⎠ 2.526 − 0.7 = 0.00251 I BQ = 3.16 + (181)(4) I CQ = 0.452 mA (180)(0.026) = 10.4 kΩ 0.452 0.452 gm = = 17.4 mA/V 0.026

rπ =

Vi − Vπ 1 Vπ 1 V −V = + π1 0 RS RB || rπ RF

(1)

Vπ 2 g mVπ 1 + =0 RC || RB || rπ

(2)

Vπ 3 g mVπ 2 + =0 RC || RB || rπ

(3)

V V V −V g mVπ 3 + 0 + 0 + 0 π 1 = 0 RC RL RF

(4)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Now RC || RB || rπ = 8 || 3.16 ||10.4 = 1.86 kΩ RB || rπ = 3.16 ||10.4 = 2.42 kΩ

Now substituting in (2): (17.4)Vπ 1 +

Vπ 2 =0 1.86 or Vπ 2 = −(32.36)Vπ 1

and substituting in (3): (17.4)Vπ 2 +

Vπ 3 =0 1.86

(17.4)[−(32.36)Vπ 1 ] +

Vπ 3 =0 1.86

or Vπ 3 = (1047.3)Vπ 1 Substitute numerical values in (1): ⎛1 Vi 1 1 ⎞ V0 = Vπ 1 ⎜ + + ⎟− 10 10 2.42 R RF F ⎠ ⎝

or ⎛ 1 ⎞ V0 Vi (0.10) = Vπ 1 ⎜ 0.513 + ⎟− RF ⎠ RF ⎝

Substitute numerical values in (4): ⎛ 1 1 1 ⎞ Vπ 1 (17.4)(1047.3)Vπ 1 + V0 ⎜ + + =0 ⎟− ⎝ 8 4 RF ⎠ RF ⎛ 1 Vπ 1 ⎜1.822 × 104 − R F ⎝

⎞ ⎛ 1 ⎞ ⎟ + V0 ⎜ 0.375 + ⎟=0 R F ⎠ ⎠ ⎝

⎛ 1 ⎞ −V0 ⎜ 0.375 + ⎟ RF ⎠ ⎝ Vπ 1 = 1 1.822 × 104 − RF

so that ⎛ 1 Vi (0.10) = ⎜ 0.513 + R F ⎝

We have

V0 = −80 Vi

or

⎡ ⎛ 1 ⎞⎤ ⎢ −V0 ⎜ 0.375 + ⎟⎥ R ⎞⎢ V F ⎠⎥ ⎝ − 0 ⎟⎢ ⎥ 1 R F ⎠ 1.822 × 104 − ⎢ ⎥ RF ⎥ ⎢⎣ ⎦

Vi = −

V0 80

⎡ ⎛ 1 ⎞ ⎢ − ⎜ 0.375 + ⎟ RF ⎠ (0.10) ⎛ 1 ⎞⎢ ⎝ − = ⎜ 0.513 + ⎟ 1 80 RF ⎠ ⎢ 4 ⎝ ⎢1.822 × 10 − R F ⎣⎢ 1/ RF

Neglect that

⎤ ⎥ ⎥− 1 ⎥ RF ⎥ ⎦⎥

term in the denominator.

−(0.00125RF ) = −

(0.513RF + 1)(0.375 RF + 1) −1 1.822 × 104 RF

22.775 RF2 = (0.513RF + 1)(0.375 RF + 1) + 1.822 × 104 RF

We find 22.58 RF2 − 1.822 × 104 RF − 1 = 0

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ RF =

1.822 × 104 ± (1.822 × 10 4 ) 2 + 4(22.58)(1) 2(22.58)

or RF = 0.807 MΩ

______________________________________________________________________________________ 12.66

VS − (−Vd ) −Vd − V1 = RS RF

a. or

⎛ 1 1 ⎞ VS V1 + + =0 Vd ⎜ ⎟+ R R RS RF F ⎠ ⎝ S

(1)

V0 − V1 V1 V1 − ( −Vd ) = + R1 R2 RF

or ⎛ 1 V0 1 1 ⎞ Vd = V1 ⎜ + + ⎟+ R1 ⎝ R1 R2 RF ⎠ RF

(2)

V Vd = 0 A0 L

and V0 = A0 LVd or Substitute numerical values in (1) and (2): V0 ⎛ 1 1 ⎞ VS V1 ⋅⎜ + ⎟ + + = 0 104 ⎝ 5 10 ⎠ 5 10

or V0 (0.3 × 10−4 ) + VS (0.20) + V1 (0.10) = 0 V0 ⎛ 1 1 1⎞ V = V1 ⎜ + + ⎟ + 04 50 ⎝ 50 10 10 ⎠ 10

(1)

⎛ 1⎞ ⋅⎜ ⎟ ⎝ 10 ⎠

or V0 (0.02 − 10−5 ) = V1 (0.22)

(2)

⎛ 0.02 − 10−5 ⎞ V1 = V0 ⎜ ⎟ ⎝ 0.22 ⎠ Then

and ⎡ ⎛ 0.02 − 10−5 ⎞ ⎤ V0 (0.3 × 10−4 ) + VS (0.20) + (0.10) ⎢V0 ⎜ ⎟⎥ = 0 ⎣ ⎝ 0.22 ⎠ ⎦ V0 ⎡⎣0.3 × 10 −4 − 0.4545 × 10−5 + 0.00909 ⎤⎦ + VS (0.20) = 0

Then

V0 V −0.20 = ⇒ 0 = −21.94 VS 9.115 × 10 −3 VS

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ −Vd −V ⋅ R = d S VS − (−Vd ) VS + Vd RS

Rif =

b. Now Then or

Vd =

V0 21.94VS =− A0 L 104

Rif =

(21.94 × 10−4 )(5) 1 − 21.94 × 10−4

Rif = 1.099 × 10−2 kΩ ⇒ Rif = 10.99 Ω

Because of the A0LVd source,

c.

R0 f = 0

______________________________________________________________________________________ 12.67 For example, use the circuit shown in Figure 12.41 ______________________________________________________________________________________ 12.68 Break the loop

It = Iε Ai I t +

Now

V0 V0 + =0 R1 RF + RS Ri

⎛ RS Ir = ⎜ ⎝ RS + Ri

⎞ V0 ⎟⋅ ⎠ RF + RS Ri

⎛ R + Ri ⎞ V0 = I r ⎜ S ⎟ ⋅ ( RF + RS Ri ) ⎝ RS ⎠ or

Then ⎛ 1 ⎞ ⎡ ⎛ RS + Ri ⎞ ⎤ 1 Ai I t + ⎜ + × I ⎟ ( RF + RS Ri ) ⎥ = 0 ⎜ R R + R R ⎟⎟ ⎢ r ⎜ R F S i ⎠ ⎣ S ⎝ ⎠ ⎦ ⎝ 1 Ai I T = − r ⇒T = It ⎡1 ⎤ ⎛ RS + Ri ⎞ 1 ⎢ + ⎥⎜ ⎟ ( RF + RS Ri ) ⎣ R1 RF + RS Ri ⎦ ⎝ RS ⎠

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.69

Vπ 1 V V −V + g m1Vπ 1 = ε 1 + ε 1 0 rπ 1 RE1 RF g m1Vπ 1 +

Vπ 2 = Vt

Vr RC1 rπ 2

(1)

= 0 ⇒ Vr = −( g m1Vπ 1 )( RC1 rπ 2 )

(2)

so that

V +V V g m 2Vt + π 3 0 + π 3 = 0 RC 2 rπ 3

(3)

Vπ 3 V V −V + g m 3Vπ 3 = 0 + 0 ε 1 rπ 3 RE 3 RF

(4)

From (4):

But

⎛ 1 ⎛ 1 ⎞ V 1 ⎞ V0 ⎜ + + g m3 ⎟ + ε 1 ⎟ = Vπ 3 ⎜ R R r F ⎠ ⎝ E3 ⎝ π3 ⎠ RF Vε 1 = −Vπ 1

⎛ 1 ⎞ V Vπ 3 ⎜ + g m3 ⎟ − π 1 rπ 3 ⎝ ⎠ RF V0 = ⎛ 1 1 ⎞ + ⎜ ⎟ ⎝ RE 3 RF ⎠

so Then

⎡⎛ 1 ⎞ ⎛ 1 1 ⎞⎤ + Vπ 1 ⎢⎜ + g m1 ⎟ − ⎜ ⎟⎥ = ⎠ ⎝ RE1 RF ⎠ ⎦ ⎣⎝ rπ 1

⎛ 1 ⎞ V −Vπ 3 ⎜ + g m3 ⎟ + π 1 ⎝ rπ 3 ⎠ RF ⎛ 1 1 ⎞ RF ⋅ ⎜ + ⎟ R R F ⎠ ⎝ E3

(1′)

and ⎛ 1 1 ⎞ + g m 2Vt + Vπ 3 ⎜ ⎟+ ⎝ RC 2 rπ 3 ⎠

⎛ 1 ⎞ V Vπ 3 ⎜ + g m3 ⎟ − π 1 r ⎝ π3 ⎠ RF = 0 ⎛ 1 1 ⎞ + RC 2 ⋅ ⎜ ⎟ R R F ⎠ ⎝ E3

(3′) From (3′), solve for Vπ 3 and substitute into (1′). Then from (1′), solve for Vπ 1 and substitute into (2). T =−

Vr . Vt

Then ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.70

Vr Vr Vr − Ve + + =0 RS rπ 1 RF

(1)

V + Ve Vπ 2 g m1Vt + π 2 + =0 RC1 rπ 2

(2)

Vπ 2 V V − Vr + g m 2Vπ 2 = e + e rπ 2 RE RF

(3) Assume rπ 1 = 15.8 k Ω , g m1 = 7.62 mA/V, rπ 2 = 2.28 k Ω , g m 2 = 52.7 mA/V 1 1⎞ V ⎛1 Vr ⎜ + + ⎟− e = 0 ⎝ 10 15.8 10 ⎠ 10

or

Vr (0.2633) = Ve (0.10)

(1)

1 ⎞ Ve ⎛ 1 (7.62)Vt + Vπ 2 ⎜ + =0 ⎟+ ⎝ 40 2.28 ⎠ 40

or

Vt (7.62) + Vπ 2 (0.4636) + Ve (0.025) = 0

(2)

⎛ 1 ⎞ ⎛1 1 ⎞ V + 52.7 ⎟ = Ve ⎜ + ⎟ − r Vπ 2 ⎜ 2.28 ⎝ ⎠ ⎝ 1 10 ⎠ 10

or Vπ 2 (53.14) = Ve (1.10) − Vr (0.10)

Then

Vπ 2 = Ve (0.0207) − Vr (0.001882)

(3)

Substituting in (2): Vt (7.62) + (0.4636)[Ve (0.0207) − Vr (0.001882)] + Ve (0.025) = 0

or Vt (7.62) + Ve (0.03460) − Vr (0.0008725) = 0 From (1) Ve = Vr (2.633)

Then Vt (7.62) + Vr (2.633)(0.03460) − Vr (0.0008725) = 0 Vt (7.62) + Vr (0.09023) = 0

Vr = −84.45 Vt

or Now

T =−

Vr ⇒ T = 84.45 Vt

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.71 RTH = R1 R 2 = 24 150 = 20.69 k Ω ⎛ R2 ⎞ ⎛ 24 ⎞ ⎟⎟ ⋅ VCC = ⎜ VTH = ⎜⎜ ⎟(12 ) = 1.655 V + R R ⎝ 24 + 150 ⎠ 2 ⎠ ⎝ 1 VTH − V BE (on ) 1.655 − 0.7 = 0.0133 mA I BQ = = RTH + (1 + h FE )R E 20.69 + (51)(1) I CQ = 0.666 mA

0.666 = 25.62 mA/V 0.026 (50)(0.026) = 1.951 k Ω rπ = 0.666 100 ro = = 150 k Ω 0.666 From Problem 12.64, let R F = 27.2 k Ω gm =

We see that Vπ = Vt . Let R B = R1 R 2 R S (1) g mVt +

Vo Vo + =0 ro RC R F + rπ R B

⎛ R B rπ ⎞ ⎟ ⋅V (2) V r = ⎜ ⎜R r +R ⎟ o π B F ⎝ ⎠

Now R B = 24 150 5 = 20.69 5 = 4.027 k Ω ; rπ R B = 1.951 4.027 = 1.314 k Ω ro RC = 150 6 = 5.769 k Ω

1 ⎤ ⎡ 1 (1) 25.62Vt + Vo ⎢ + ⎥ = 0 ⇒ Vo = −122.93Vt ⎣ 5.769 27.2 + 1.314 ⎦ ⎞ ⎛ 1.314 (2) V r = ⎜ ⎟ ⋅ V o = (0.04608)(− 122.93)Vt ⎝ 1.314 + 27.2 ⎠ V T = − r = 5.66 Vt ______________________________________________________________________________________

12.72

⎛ f ⎞ ⎛ f ⎞ (a) φ = − tan −1 ⎜ 3 ⎟ − 2 tan −1 ⎜ ⎟ 4 ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ At f 180 = 5.1× 10 4 Hz, φ ≅ −180°

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

β (5 × 10 4 )

(b) T = 0.25 =

⎞ ⎟ ⎟ ⎠

⎛ 5.1× 10 4 1 + ⎜⎜ 3 ⎝ 10

0.25 =

β (5 × 10 4 )

2

⎡ ⎛ 5.1× 10 4 ⎢1 + ⎜ ⎢ ⎜⎝ 5 × 10 4 ⎣

⎞ ⎟ ⎟ ⎠

2

⎤ ⎥ ⎥ ⎦

(51.0098)(2.0404)

β = 5.2 × 10 −4 ______________________________________________________________________________________ 12.73 ⎛ f ⎞ ⎛ f ⎞ (a) φ = −2 tan −1 ⎜ ⎟ ⎟ − tan −1 ⎜ 3 5 ⎝ 5 × 10 ⎠ ⎝ 5 × 10 ⎠

At f 180 = 7.1× 10 4 Hz, φ ≅ −180°

β (10 4 )

(b) T = 1 =

⎡ ⎛ 7.1× 10 4 ⎞ 2 ⎤ ⎛ 7.1× 10 4 ⎢1 + ⎜ ⎟ ⎥ 1+ ⎜ 3 ⎟ ⎜ 5 × 10 5 ⎜ ⎢ ⎝ 5 × 10 ⎠ ⎥ ⎝ ⎦ ⎣ 4 β 10 1= ⇒ β = 0.0205 (202.64)(1.01)

( )

(c) Aυ f (0 ) =

⎞ ⎟ ⎟ ⎠

2

10 4 = 48.54 1 + (0.0205) 10 4

( )

(d) Smaller ______________________________________________________________________________________ 12.74 ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ − tan −1 ⎜ − tan −1 ⎜ 5 ⎟ 4 ⎟ 4 ⎟ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ f = 8.1× 104 Hz, φ = −180.28°

φ = − tan −1 ⎜

At

Determine

T( f )

T = β (10 ) ×

at this frequency. 1

3

⎛ 8.1× 10 ⎞ 1+ ⎜ ⎟ 4 ⎝ 10 ⎠ 4

=

a. b.

2

1

×

⎛ 8.1× 10 ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠ 4

2

×

1 ⎛ 8.1× 10 4 ⎞ 1+ ⎜ ⎟ 5 ⎝ 10 ⎠

2

β (103 ) (8.161)(1.904)(1.287) β = 0.005

For

T ( f ) = 0.250 < 1 ⇒ Stable β = 0.05

For

T ( f ) = 2.50 > 1 ⇒ Unstable

______________________________________________________________________________________ 12.75 (b) Phase margin = 80° ⇒ φ = −100° ⎛ f ⎞ ⎛ f ⎞ − tan −1 ⎜ 3 ⎟ 4 ⎟ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ 3 f = 1.16 × 10 Hz

φ = −100 = −2 tan −1 ⎜

By trial and error, Then

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ β (5 × 103 ) T =1=

2

⎛ 3 2 ⎞ 3 2 ⎜ 1 + ⎛⎜ 1.16 × 10 ⎞⎟ ⎟ ⋅ 1 + ⎛⎜ 1.16 × 10 ⎞⎟ 3 4 ⎜ ⎝ 10 ⎠ ⎟⎠ ⎝ 5 × 10 ⎠ ⎝ β (5 × 103 ) = ⇒ β = 4.7 × 10−4 (2.35)(1.00)

______________________________________________________________________________________

12.76

For β = 0.005,

c.

T ( f ) = 1(0 dB)

4 at f ≈ 2.10 ×10 Hz

Then 4 4 ⎛ 2.10 × 104 ⎞ −1 ⎛ 2.10 × 10 ⎞ −1 ⎛ 2.10 × 10 ⎞ − − tan tan ⎟ ⎜ ⎟ ⎜ ⎟ 3 4 5 ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠

φ = − tan −1 ⎜

= −87.27 − 64.54 − 11.86

or

φ = −163.7 System is stable. Phase margin = 16.3° For β = 0.05, T ( f ) = 1 (0 dB)

4 at f ≈ 6.44 × 10 Hz

Then 4 4 ⎛ 6.44 × 104 ⎞ −1 ⎛ 6.44 × 10 ⎞ −1 ⎛ 6.44 × 10 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 3 4 5 ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠

φ = − tan −1 ⎜

= −89.11 − 81.17 − 32.78

or

φ = −203.1° ⇒ System is unstable.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 12.77 ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ (a) φ = − tan −1 ⎜ 3 ⎟ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 7 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠

At f 180 = 10 6 Hz, φ ≅ −180°

⎛f (b) φ = −135° = − tan⎜⎜ 1353 ⎝ 10

⎛ f ⎞ ⎟⎟ − tan −1 ⎜⎜ 1355 ⎝ 10 ⎠

f 135 ≅ 10 5 Hz

T ( f 135 ) = 1 =

1=

⎛ f ⎞ ⎟⎟ − tan −1 ⎜⎜ 1357 ⎝ 10 ⎠

⎞ ⎟⎟ ⎠

β (5 × 10 4 ) ⎛ 10 5 1 + ⎜⎜ 3 ⎝ 10

β (5 × 10 4 )

(100)(1.414)(1)

⎞ ⎟ ⎟ ⎠

2

⎛ 10 5 1 + ⎜⎜ 5 ⎝ 10

⎞ ⎟ ⎟ ⎠

2

⎛ 10 5 1 + ⎜⎜ 7 ⎝ 10

⎞ ⎟ ⎟ ⎠

2

⇒ β = 2.83 × 10 −3

5 × 10 4 = 351 1 + 2.83 × 10 − 3 5 × 10 4 ______________________________________________________________________________________

(c) Aυ f (0 ) =

(

)(

)

12.78

(a) 100 =

10 5 ⇒ β = 9.99 ± 10 −3 1 + 10 5 β

( )

T = 1 = β Av =

1=

(9.99 × 10−3 )(105 ) ⎛ f ⎞ 1+ ⎜ 3 ⎟ ⎝ 10 ⎠ 999 2

2

⎛ f ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠

⎛ f ⎞ ⎛ f ⎞ 1+ ⎜ 3 ⎟ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ f = 3.08 × 105 Hz

2

2

Phase f f − tan −1 5 103 10 3.08 × 105 3.08 × 105 = − tan −1 − tan −1 3 10 105 = −89.81 − 72.01

φ = − tan −1

φ = −161.8 Stable (b) Phase Margin = 180 − 161.8 = 18.2° ______________________________________________________________________________________

12.79

⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ (a) φ = − tan −1 ⎜ 4 ⎟ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 6 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ At f 180 = 3.33 × 10 5 Hz, φ ≅ −180°

(0.019)(10 3 ) 2 2 2 1 + (33.3) 1 + (3.33) 1 + (0.333) (0.019)(10 3 ) T ( f 180 ) = = 0.156 (33.315)(3.477 )(1.054)

(b) (i) T ( f 180 ) =

(0.019)(10 3 )

(ii) T = 1 = ⎛ f ⎞ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠

2

⎛ f ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠

2

⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠

5 ⎛ 1.2 × 10 5 ⎞ −1 ⎛ 1.2 × 10 ⎟ ⎜ − tan 4 ⎟ ⎜ 10 5 ⎝ 10 ⎠ ⎝ = −85.236° − 50.194° − 6.843° φ = −142.3°

φ = − tan −1 ⎜⎜

2

⇒ f = 1.2 × 10 5 Hz

⎞ ⎛ 1.2 × 10 5 ⎟ − tan −1 ⎜ ⎟ ⎜ 10 6 ⎠ ⎝

⎞ ⎟ ⎟ ⎠

10 3 = 50 1 + (0.019) 10 3 ______________________________________________________________________________________ (c) A f (0 ) =

( )

12.80 f180 f − 2 tan −1 1805 5 × 103 10 = 1.05 × 105 Hz

φ = −180 = − tan −1

(a)

f180

(0.0045)(2 × 103 )

T =

⎛ 1.05 × 105 ⎞ 1+ ⎜ 3 ⎟ ⎝ 5 × 10 ⎠ =

2

⎡ ⎛ 1.05 × 105 ⎞ 2 ⎤ ⎢1 + ⎜ ⎟ ⎥ 5 ⎢⎣ ⎝ 10 ⎠ ⎥⎦

9 (21.02)(2.1025)

T = f 0 = 0.204

180 (b) System is stable

T =1=

9 2 2 ⎛ f ⎞ ⎡ ⎛ f ⎞ ⎤ 1+ ⎜ 1 + ⎢ ⎥ ⎟ ⎜ ⎟ 3 5 ⎝ 5 × 10 ⎠ ⎣⎢ ⎝ 10 ⎠ ⎦⎥

f = 3.88 × 104 Hz 4 3.88 × 104 −1 3.88 × 10 − 2 tan 5 × 103 105 = −82.66 − 42.41 φ = −125.1°

φ = − tan −1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (0.15)(2 × 103 )

T =

⎛ 1.05 × 105 ⎞ 1+ ⎜ 3 ⎟ ⎝ 5 × 10 ⎠ =

2

⎡ ⎛ 1.05 × 105 ⎞ 2 ⎤ ⎢1 + ⎜ ⎟ ⎥ 5 ⎢⎣ ⎝ 10 ⎠ ⎥⎦

300 (21.02)(2.1025)

T = 6.79

(c) System is unstable 300

T = 1=

⎛ f ⎞ 1+ ⎜ 3 ⎟ ⎝ 5 × 10 ⎠

2

⎡ ⎛ f ⎞2 ⎤ ⎢1 + ⎜ 5 ⎟ ⎥ ⎢⎣ ⎝ 10 ⎠ ⎥⎦

f = 2.33 × 105 Hz

2.33 × 105 2.33 × 105 − 2 tan −1 3 5 × 10 105 = −88.77 − 133.54 φ = −222.3°

φ = − tan −1

______________________________________________________________________________________ 12.81 Phase Margin = 45° ⇒ φ = −135° φ = −135°

At

⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ = − tan −1 ⎜ 3 ⎟ − tan −1 ⎜ 4 ⎟ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 6 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ f = 104 Hz, φ = −135.6°

T =1 1

= β (103 ) ×

⎛ 104 ⎞ 1+ ⎜ 3 ⎟ ⎝ 10 ⎠ ×

1 ⎛ 104 ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠

1=

or

2

2

1

×

⎛ 104 ⎞ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠

2

×

1

×

⎛ 10 4 ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠

2

β (103 ) (10.05)(1.414)(1.005)(1.00)

β = 0.01428

______________________________________________________________________________________ 12.82 ⎛ f 120 ⎝ f PD

φ = −120° = − tan −1 ⎜⎜ f 120 ≅ 2.31× 10 5 Hz

⎞ ⎛ f ⎟⎟ − tan −1 ⎜⎜ 120 5 ⎝ 4 × 10 ⎠

⎞ ⎛ f ⎟⎟ − tan −1 ⎜⎜ 120 6 ⎠ ⎝ 4 × 10

⎞ ⎛ f ⎟⎟ − tan −1 ⎜⎜ 120 7 ⎠ ⎝ 4 × 10

⎞ ⎟⎟ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4000 T ( f 120 ) = 1 = 2 2 2 2 ⎛ 2.31× 10 5 ⎞ ⎛ 2.31× 10 5 ⎞ ⎛ 2.31× 10 5 ⎞ ⎛ 2.31× 10 5 ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ 1+ ⎜ 1 + ⎜⎜ ⎜ 4 × 10 5 ⎟ 1 + ⎜ 4 × 10 6 ⎟ 1 + ⎜ 4 × 10 7 ⎟ ⎟ f PD ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 4000 1= 2 ⎛ 2.31× 10 5 ⎞ ⎟ (1.1548)(1.00167 )(1.0) 1 + ⎜⎜ ⎟ f PD ⎝ ⎠ 2.31× 10 5 4000 = (1.1548)(1.00167 )(1.09) f PD f PD = 66.8 Hz ______________________________________________________________________________________

12.83

(a)

⎛ f f 180 ⇒ −180° = −2 tan −1 ⎜⎜ 1804 ⎝ 10

⎞ ⎛ f ⎟⎟ − tan −1 ⎜⎜ 1806 ⎠ ⎝ 10

⎞ ⎟⎟ ⎠

f 180 = 1.42 × 10 5 Hz T ( f 180 ) =

10 3

[1 + (14.2) ] 1 + (0.142) 2

2

=

10 3 = 4.89 (202.64)(1.01)

T ( f 180 ) > 1 ⇒ Unstable

⎛ f (b) φ = −135° = − tan −1 ⎜⎜ 135 ⎝ f PD

⎞ ⎛ f ⎞ ⎛ f ⎞ ⎟⎟ − 2 tan −1 ⎜⎜ 1354 ⎟⎟ − tan −1 ⎜⎜ 1356 ⎟⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎠

f 135 ≅ 0.414 × 10 4 Hz T ( f 135 ) = 1 =

10 3 ⎛ 0.414 × 10 4 1 + ⎜⎜ f PD ⎝

⎞ ⎟ ⎟ ⎠

2

⎡ ⎛ 0.414 × 10 4 ⎢1 + ⎜ 10 4 ⎢ ⎜⎝ ⎣

⎞ ⎟ ⎟ ⎠

2

⎤ ⎛ 0.414 × 10 4 ⎥ 1+ ⎜ ⎜ 10 6 ⎥ ⎝ ⎦

⎞ ⎟ ⎟ ⎠

2

0.414 × 10 4 10 3 = f PD (1.171)(1) f PD = 4.85 Hz ______________________________________________________________________________________

12.84

(a)

⎛ f180 ⎞ ⎛ f ⎞ ⎛ f ⎞ − tan −1 ⎜ 180 4 ⎟ − tan −1 ⎜ 1805 ⎟ 4 ⎟ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠

φ = −180 = − tan −1 ⎜ f180 ≅ 8.06 × 104 Hz

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 500

T =

2

⎛ 8.06 × 10 ⎞ ⎛ 8.06 × 104 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ 4 4 ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ 500 = (8.122)(1.897)(1.284) 4

(b)

2

⎛ 8.06 × 104 ⎞ 1+ ⎜ ⎟ 5 ⎝ 10 ⎠

2

T = 25.3 T=

500 ⎛ f ⎞⎛ f ⎞⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟ ⎜1 + j 4 ⎟ ⎜1 + j ⎟⎜1 + j 5 ⎟ f PD ⎠ ⎝ 10 ⎠ ⎝ 5 × 104 ⎠⎝ 10 ⎠ ⎝ = 60° ⇒ φ = −120°

(c) Phase Margin

−120 = − tan −1 tan −1

Assume

f f PD

f f f f − tan −1 4 − tan −1 − tan −1 5 f PD 10 5 × 104 10 ≅ 90°

3 Then f ≅ 4.2 × 10 Hz

500

T =1=

1=

2

⎛ 4.2 × 10 ⎞ ⎛ 4.2 × 10 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ 4 ⎝ 10 ⎠ ⎝ f PD ⎠ 500 3

3

2

⎛ 4.2 × 103 ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠

2

⎛ 4.2 × 103 ⎞ 1+ ⎜ ⎟ 5 ⎝ 10 ⎠

2

2

⎛ 4.2 × 103 ⎞ 1+ ⎜ ⎟ (1.085)(1.004)(1.0) ⎝ f PD ⎠ 4.2 × 103 500 ≅ (1.0846)(1.0035)(1.0) f PD f PD = 9.14 Hz

______________________________________________________________________________________ 12.85 50 =

(a) T=

104 ⇒ β = 0.0199 1 + (104 ) β

(0.0199)(104 ) ⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟ ⎜1 + j 5 ⎟ f 10 ⎝ ⎠ PD ⎠ ⎝

Phase margin = 45° ⇒ φ = −135° −135 = − tan −1

f = 105 Hz

f f − tan −1 5 f PD 10

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (0.0199)(104 )

T =1=

⎛ 105 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠

2

⎛ 105 ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠

2

105 (0.0199)(104 ) = 1.414 f PD f PD = 711 Hz 20 =

(b) T=

104 ⇒ β = 0.0499 1 + (104 ) β

(0.0499)(104 ) f ⎞⎛ f ⎞ ⎛ ⎜1 + j ⎟⎜1 + j 5 ⎟ 711 ⎠⎝ 10 ⎠ ⎝ (0.0499)(104 )

T =1=

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 711 ⎠

2

⎛ f ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠

2

f = 1.76 × 105 Hz 5 ⎛ 1.76 × 105 ⎞ −1 ⎛ 1.76 × 10 ⎞ ⎟ − tan ⎜ ⎟ 5 ⎝ 711 ⎠ ⎝ 10 ⎠

φ = − tan −1 ⎜

= −89.77 − 60.40

φ = −150.2

Phase Margin = 180 − 150.2 = 29.8° ______________________________________________________________________________________

12.86 AO = 100 dB ⇒ AO = 105

(a)

20 =

105 ⇒ β = 0.04999 1 + (105 ) β

(0.04999)(105 ) ⎛ f ⎞⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟ ⎜1 + j 6 ⎟⎜ 1 + j 7 ⎟ f 10 10 ⎝ ⎠⎝ ⎠ ⎝ PD ⎠ Margin = 45° ⇒ φ = −135°

T=

Phase

−135 = − tan −1

f f f − tan −1 6 − tan −1 7 f PD 10 10

f ≈ 106 Hz

(0.04999)(105 )

T =1=

⎛ 106 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠ 1=

2

⎛ 106 ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠

(0.04999)(105 ) 2

⎛ 106 ⎞ 1+ ⎜ ⎟ (1.414)(1.005) ⎝ f PD ⎠ 106 (0.04999)(105 ) = f PD (1.414)(1.005) f PD = 2.84 Hz

2

⎛ 106 ⎞ 1+ ⎜ 7 ⎟ ⎝ 10 ⎠

2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5=

(b)

105 ⇒ β = 0.19999 1 + (105 ) β

(0.19999)(105 )

T =1=

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 284 ⎠

2

⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠

2

⎛ f ⎞ 1+ ⎜ 7 ⎟ ⎝ 10 ⎠

2

f = 2.25 × 106 Hz 6 6 ⎛ 2.25 × 106 ⎞ −1 ⎛ 2.25 × 10 ⎞ −1 ⎛ 2.25 × 10 ⎞ − tan ⎟ − tan ⎜ ⎟ ⎜ ⎟ 6 7 ⎝ 284 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠

φ = − tan −1 ⎜

= −89.99 − 66.04 − 12.68 φ = −168.7 Phase Margin = 180 − 168.7 = 11.3°

______________________________________________________________________________________ 12.87 T (0) = 100 dB ⇒ T (0) = 105

a.

T( f ) =

105 f ⎞⎛ f ⎞⎛ f ⎛ ⎞ 1+ j ⎜1 + j ⎟⎜1 + j ⎟ 6 ⎟⎜ 10 ⎠⎝ 5 × 10 ⎠⎝ 10 × 106 ⎠ ⎝

T =1=

1

= 105 ×

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠

2

1

×

⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 5 × 10 ⎠

2

1

×

f ⎛ ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 × 10 ⎠

By trial and error f = 0.976 MHz

⎛ 0.976 × 106 ⎞ −1 ⎛ 0.976 ⎞ −1 ⎛ 0.976 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 10 ⎝ 5 ⎠ ⎝ 10 ⎠ ⎝ ⎠

φ = − tan −1 ⎜

= −90° − 11.05° − 5.574° = −106.6° = 180° − 106.6° = 73.4°

Phase Margin b. or

f P′1 ∝

1 CF

10 75 = f P′1 20

so

f P′1 = 2.67 Hz

Now T =1=

1

= 105 ×

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 2.67 ⎠

2

×

1 ⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 5 × 10 ⎠

×

2

1 f ⎛ ⎞ 1+ ⎜ 6 ⎟ 10 × 10 ⎝ ⎠

By trial and error f ≈ 2.66 × 105 Hz

then

2

2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ 2.66 × 105 ⎞ −1 ⎛ 0.266 ⎞ −1 ⎛ 0.266 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ ⎝ 5 ⎠ ⎝ 10 ⎠ ⎝ 2.67 ⎠

φ = − tan −1 ⎜

= −90° − 3.045° − 1.524° = −94.57° = 180° − 94.57° = 85.4°

Phase Margin ______________________________________________________________________________________ 12.88

(a)

fP =

2π (R

1 o1

Ri 2

)C

= i

1 2π (2 0.75)× 10 6 × 1.2 × 10 −12

(

)

f P = 243 kHz

(b)

f PD = 6 =

1 1 = 2π (Ro1 Ri 2 )(C i + C M ) 2π (2 0.75)× 10 6 × (C i + C M

)

C i + C M = 4.863 × 10 −8 = 1.2 × 10 −12 + C M

C M = 0.0486 μ F

(c) C M = C F (1 + A )

0.0486 × 10 −6 = C F (1001)

C F = 48.6 pF ______________________________________________________________________________________

12.89 ⎛ f 120 ⎝ f PD

φ = −120° = − tan −1 ⎜⎜

⎞ ⎛ f ⎟⎟ − tan −1 ⎜⎜ 120 6 ⎝ 4 × 10 ⎠

⎞ ⎛ f ⎟⎟ − tan −1 ⎜⎜ 120 7 ⎠ ⎝ 4 × 10

⎞ ⎟⎟ ⎠

f 120 ≅ 2.05 × 10 6 Hz T ( f 120 ) = 1 =

4000 ⎛ 2.05 × 10 6 1 + ⎜⎜ f PD ⎝

⎞ ⎟ ⎟ ⎠

2

⎛ 2.05 × 10 6 1 + ⎜⎜ 6 ⎝ 4 × 10

⎞ ⎟ ⎟ ⎠

2

⎛ 2.05 × 10 6 1 + ⎜⎜ 7 ⎝ 4 × 10

⎞ ⎟ ⎟ ⎠

2

2.05 × 10 6 4000 = f PD (1.1237 )(1.0013) f PD = 577 Hz ______________________________________________________________________________________

12.90

(a) 40 =

5 × 10 5 ⇒ β = 0.024998 1 + β 5 × 10 5

(

)

⎛ f (b) φ = −120° = − tan −1 ⎜⎜ 120 ⎝ f PD

⎞ ⎛ f ⎟⎟ − tan −1 ⎜⎜ 120 5 ⎝ 5 × 10 ⎠

⎛ f ⎞ ⎟⎟ − tan −1 ⎜⎜ 1207 ⎠ ⎝ 10

⎞ ⎟⎟ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ f 120 ≅ 2.71× 10 5 Hz T ( f 120 ) = 1 =

(0.024998)(5 ×10 5 )

⎛ 2.71× 10 5 1 + ⎜⎜ f PD ⎝

(

⎞ ⎟ ⎟ ⎠

2.71× 10 5 (0.024998) 5 × 10 5 = f PD (1.1374)(1)

2

)

⎛ 2.71× 10 5 1 + ⎜⎜ 5 ⎝ 5 × 10

⎞ ⎟ ⎟ ⎠

2

⎛ 2.71× 10 5 1 + ⎜⎜ 7 ⎝ 10

⎞ ⎟ ⎟ ⎠

2

f PD = 24.66 Hz ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 13 13.1

0 − (− 3) = 15 k Ω 0 .2 ⎛ k ′p ⎞⎛ W ⎞ ⎟⎜ ⎟ (V SG 3 + VTP )2 I D 3 = ⎜⎜ ⎟⎝ L ⎠ 2 3 ⎝ ⎠ ⎛ 0.04 ⎞ 2 0. 2 = ⎜ ⎟(40 )(V SG 3 − 0.4 ) ⇒ V SG 3 = 0.9 V ⎝ 2 ⎠

(a) R D 2 =

0 .9 = 9kΩ 0. 1 υ g (b) (i) Ad = o1 = m1 ⋅ R D1 υd 2 R D1 =

⎛ 0. 1 ⎞ g m1 = 2 ⎜ ⎟(20 )(0.1) = 0.6325 mA/V ⎝ 2 ⎠ ⎛ 0.6325 ⎞ Ad = ⎜ ⎟(9 ) = 2.846 ⎝ 2 ⎠

(ii) A2 =

υo = − g m3 R D 2 υ o1

⎛ 0.04 ⎞ g m3 = 2 ⎜ ⎟(40 )(0.2 ) = 0.8 mA/V ⎝ 2 ⎠ A2 = −(0.8)(15) = −12

(c) A = Ad ⋅ A2 = (2.846)(− 12 ) = −34.15 ______________________________________________________________________________________ 13.2 3−0 = 7.5 k Ω 0.4 = 0.7 + (0.4 )(0.5) − 3 = −2.1 V

(a) RC 2 = V B3

− 2.1 − (− 3) = 3. 6 k Ω 0.25 υ g (b) (i) Ad = o1 = m1 (RC1 Ri 3 ) υd 2 R C1 =

(180)(0.026 ) = 11.7 k Ω 0.25 = 9.615 mA/V, rπ 3 = 0.026 0.4 Ri 3 = rπ 3 + (1 + β n )R E = 11.7 + (181)(0.5) = 102.2 k Ω

g m1 =

⎛ 9.615 ⎞ Ad = ⎜ ⎟(3.6 102.2 ) = 16.72 ⎝ 2 ⎠ υ − β n RC 2 − (180)(7.5) (ii) A2 = o = = = −13.21 υ o1 rπ 3 + (1 + β n )R E 11.7 + (181)(0.5)

(c) A = Ad ⋅ A2 = (16.72 )(− 13.21) = −220.9 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 13.3 Computer Analysis ______________________________________________________________________________________ 13.4 Computer Analysis ______________________________________________________________________________________ 13.5

(

Ad = g m1 ro 2 ro 4 Ri 6

(a)

)

g m1 =

I C1 20 = ⇒ 0.769 mA / V VT 0.026

ro 2 =

VA 2 80 = = 4 MΩ I C 2 20

ro 4 =

VA 4 80 = = 4 MΩ I C 2 20

Ri 6 = rπ 6 + (1 + β n ) ⎡⎣ R1 rπ 7 ⎤⎦ (120)(0.026) rπ 7 = = 15.6 k Ω 0.2 V (on) 0.6 I C 6 ≅ BE = = 0.030 mA R1 20 rπ 6 =

(120)(0.026) = 104 k Ω 0.030

Then Ri 6 = 104 + (121) ⎡⎣ 20 15.6 ⎤⎦ ⇒ 1.16 M Ω

Then

(

)

Ad = 769 4 4 1.16 ⇒ Ad = 565

Now ⎛ R1 ⎞ Vo = − I c 7 ro 7 = −( β n I b 7 )ro 7 = − β n ro 7 ⎜ ⎟ Ic6 ⎝ R1 + rπ 7 ⎠ ⎛ R1 ⎞ Vo1 = − β n (1 + β n )ro 7 ⎜ ⎟ I b 6 and I b 6 = Ri 6 ⎝ R1 + rπ 7 ⎠

Then Av 2 =

Vo − β n (1 + β n ) ro 7 = Vo1 Ri 6

ro 7 =

VA 80 = = 400 k Ω I C 7 0.2

⎛ R1 ⎞ ⎜ ⎟ ⎝ R1 + rπ 7 ⎠

So −(120)(121)(400) ⎛ 20 ⎞ ⎜ ⎟ ⇒ Av 2 = −2813 1160 ⎝ 20 + 15.6 ⎠ Overall gain = Ad ⋅ Av 2 = (565)(−2813) ⇒ A = −1.59 × 106 Av 2 =

(b) Rid = 2rπ 1 and Rid = 208 k Ω

rπ 1 =

(80)(0.026) = 104 k Ω 0.020

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ f PD =

(c)

1 and CM = (10)(1 + 2813) = 28,140 pF 2π Req CM

Req = ro 2 ro 4 Ri 6 = 4 4 1.16 = 0.734 M Ω f PD =

1 = 7.71 Hz 2π (0.734 × 10 )(28,140 × 10−12 ) 6

Gain-Bandwidth Product = (7.71)(1.59 × 106 ) ⇒ 12.3 MHz

______________________________________________________________________________________ 13.6 Q3 a. acts as the protection device. b. Same as part (a). ______________________________________________________________________________________

13.7

If we assume VBE (on) = 0.7 V, then Vin = 0.7 + 0.7 + 50 + 5

So breakdown voltage ≈ 56.4 V. ______________________________________________________________________________________ 13.8 ⎛ 0.5 × 10 −3 ⎞ ⎟ = 0.7184 V (a) V EB12 = V BE11 = (0.026) ln⎜⎜ −16 ⎟ ⎝ 5 × 10 ⎠ 15 − 0.7184 − 0.7184 − (− 15) R5 = = 57.1 k Ω 0.5 (0.026 ) ln⎛ 0.5 ⎞ = 2.438 k Ω R4 = ⎜ ⎟ (0.03) ⎝ 0.03 ⎠

V BE10 = V BE11 − I C10 R 4 = 0.7184 − (0.03)(2.438) = 0.6453 V

(b) I REF =

15 − 0.6 − 0.6 − (− 15) = 0.5044 mA 57.1

⎛ 0.5044 ⎞ ⎟ ⇒ I C10 ≅ 30.1 μ A I C10 (2.438) = (0.026) ln⎜⎜ ⎟ ⎝ I C10 ⎠ ⎛ 0.5044 − 0.5 ⎞ (c) I REF : ⎜ ⎟ × 100% = 0.88% 0.5 ⎝ ⎠ ⎛ 30.1 − 30 ⎞ I C10 : ⎜ ⎟ × 100% = 0.33% ⎝ 30 ⎠ ______________________________________________________________________________________

13.9

5 − 0.7184 − 0.7184 − (− 5) = 17.13 k Ω 0. 5 (0.026 ) ln⎛ 0.5 ⎞ = 2.438 k Ω R4 = ⎜ ⎟ (0.03) ⎝ 0.03 ⎠

(a) R5 =

V EB12 = V BE11 = 0.7184 V V BE10 = 0.6453 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5 − 0.6 − 0.6 − (− 5) (b) I REF = = 0.5137 mA 17.13 ⎛ 0.5137 ⎞ ⎟ ⇒ I C10 ≅ 30.22 μ A I C10 (2.438) = (0.026 ) ln⎜⎜ ⎟ ⎝ I C10 ⎠ ⎛ 0.5137 − 0.5 ⎞ (c) I REF : ⎜ ⎟ × 100% = 2.74% 0. 5 ⎝ ⎠ ⎛ 30.22 − 30 ⎞ I C10 : ⎜ ⎟ × 100% = 0.733% 30 ⎝ ⎠ ______________________________________________________________________________________

13.10

3 − V EB 2 − V BE1 − (− 3) 80 ⎛ I ⎞ V EB 2 = V BE1 = (0.026 ) ln⎜⎜ REF−15 ⎟⎟ ⎝ 3 × 10 ⎠ By trial and error, I REF ≅ 59.6 μ A

(a) I REF =

⎛ 0.0596 ⎞ ⎟ ⇒ I 3 ≅ 11.9 μ A I 3 (3.5) = (0.026) ln⎜⎜ ⎟ ⎝ I3 ⎠ I 3 × 10 −15 I 4 = S 4 ⋅ I REF = ⋅ (59.6 ) = 35.76 μ A I S2 5 × 10 −15 I5 =

I S5 ⋅ I REF I S2

( ( (1×10 = (5 ×10

−15 −15

) ) ) ⋅ (59.6) = 11.92 μ A )

(b) I REF = 59.6 μ A, I 3 = 11.9 μ A I4 =

( ( (2 ×10 = (5 ×10

) ) ) ⋅ (59.6) = 23.84 μ A )

I S4 8 × 10 −15 ⋅ I REF = ⋅ (59.6 ) = 95.36 μ A I S2 5 × 10 15

−15 I S5 ⋅ I REF −15 I S2 ______________________________________________________________________________________

I5 =

13.11 5 − 0.6 − 0.6 − (−5) ⇒ I REF = 0.22 mA 40 ⎛I ⎞ I C10 R4 = VT ln ⎜ REF ⎟ ⎝ I C10 ⎠

I REF =

⎛ 0.22 ⎞ I C10 (5) = (0.026) ln ⎜ ⎟ ⎝ I C10 ⎠

By trial and error; I C10 ≅ 14.2 μ A I C10 ⇒ I C 6 = 7.10μ A 2 = 0.75I REF ⇒ I C17 = 0.165 mA

IC 6 ≅ I C17

I C13 A = 0.25I REF ⇒ I C13 A = 0.055 mA

_____________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 13.12 I C8 = I C9

⎛ 2 ⎞⎟ = I C 9 ⎜1 + ⎜ βp ⎟ βp ⎝ ⎠ ⎛ 1+ β n ⎞ ⎟ = I ⎜⎜ ⎟ ⎝ βn ⎠

2I = I C8 + I E1 = I E 2

2I C 9

⎞⎛ 1 ⎞ ⎟ ⎟⎜ ⎟⎜ 1 + β ⎟ p ⎠ ⎠⎝ ⎛ 1 + β n ⎞⎛ 1 ⎞ ⎟ ⎟⎜ + 2 I ⎜⎜ ⎟⎜ ⎟ ⎝ β n ⎠⎝ 1 + β p ⎠ ⎛ 2 ⎞⎟⎛ 1 + β n ⎞⎛⎜ 1 ⎜ ⎟ + I C 9 ⎜1 + ⎜ β p ⎟⎜ β n ⎟⎜ 1 + β p ⎠⎝ ⎝ ⎠⎝

⎛ 1+ β n I B 3 = I B 4 = I ⎜⎜ ⎝ βn

I C10 = I C 9 I C10 = I C 9

⎞ ⎟ ⎟ ⎠

⎡ ⎛ 2 ⎞⎛ 91 ⎞⎛ 1 ⎞⎤ 50 = I C 9 ⎢1 + ⎜1 + ⎟⎜ ⎟⎜ ⎟⎥ = I C 9 (1.0259 ) ⇒ I C 9 = 48.738 μ A 40 ⎠⎝ 90 ⎠⎝ 41 ⎠⎦ ⎣ ⎝

⎛ ⎞ ⎜1 + 2 ⎟ = 48.738 ⎛⎜1 + 2 ⎞⎟ = 25.587 μ A ⎜ βp ⎟ 2 ⎝ 40 ⎠ ⎝ ⎠ ⎛ β p ⎞⎛ 1 + β n ⎞ 40 91 ⎟⎜ ⎟ = (25.587 )⎛⎜ ⎞⎟⎛⎜ ⎞⎟ = 25.240 μ A = I C2 ⎜ ⎜ 1 + β p ⎟⎜ β n ⎟ ⎝ 41 ⎠⎝ 90 ⎠ ⎠ ⎝ ⎠⎝ I 48.738 = C9 = = 1.218 μ A 40 βp

I C2 = I = I C4 I B9

I C4

I C9 2

25.240

= = 0.631 μ A βp 40 ______________________________________________________________________________________ I B4 =

13.13 VB 5 − V − = VBE (on) + I C 5 (1) = 0.6 + (0.0095)(1) = 0.6095 0.6095 IC 7 = ⇒ I C 7 = 12.2 μ A 50 I C 8 = I C 9 = 19 μ A

I REF = 0.72 mA I E13 = I REF = 0.72 mA I C14 = 138 μ A

Power = (V + − V − ) [ I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 ] = 30[0.0122 + 0.019 + 0.019 + 0.72 + 0.72 + 0.138] ⇒ Power = 48.8 mW Current supplied by V + and V − = I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 = 1.63 mA

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 13.14 (a)

vcm (min) = −15 + 0.6 + 0.6 + 0.6 + 0.6 = −12.6 V

vcm (max) = +15 − .6 = 14.4 V So − 12.6 ≤ vcm ≤ 14.4 V

(b)

vcm (min) = −5 + 4(0.6) = −2.6 V vcm (max) = 5 − 0.6 = 4.4 V So − 2.6 ≤ vcm ≤ 4.4 V

______________________________________________________________________________________ 13.15

I C13 A = (0.25)I REF = (0.25)(0.5) = 0.125 mA 0. 6 = 0.012 mA 50 ≅ I C13 A − I R10 = 0.125 − 0.012 = 0.113 mA

I R10 ≅

I C19

I B19 =

I C19

βn

=

0.113 ⇒ I B19 = 0.565 μ A 200

I C18 = I R10 + I B19 = 12 + 0.565 = 12.565 μ A ⎛ 12.565 × 10 −6 V BE18 = (0.026 ) ln⎜⎜ 10 −14 ⎝

⎞ ⎟ = 0.54474 V ⎟ ⎠

⎛ 0.113 × 10 −3 ⎞ ⎟ = 0.60185 V V BE19 = (0.026) ln⎜⎜ −14 ⎟ ⎝ 10 ⎠ V BB V BE18 + V Be19 = = 0.57329 V 2 2 ⎛ 0.57329 ⎞ I C14 = 3 × 10 −14 exp⎜ ⎟ ⇒ I C14 = 113 μ A ⎝ 0.026 ⎠ ______________________________________________________________________________________

(

)

13.16 ⎛I (a) V BB = 2VT ln⎜⎜ Bias ⎝ I SD I CN = I CP = I SQ

⎞ ⎛ 80 × 10 −6 ⎞ ⎟ = 2(0.026) ln⎜ ⎟ ⎟ ⎜ 5 × 10 −15 ⎟ = 1.22178 V ⎠ ⎝ ⎠ ⎡1.22178 2 ⎤ exp ⎢ ⎥ ⇒ I CN = I CP = 128 μ A ⎣ 0.026 ⎦

(b) For υ I = 3 V, υ O ≅ 3 V, i L ≅

3 = 0.3 mA 10

First approximation: 0.3 I BN ≅ ⇒ I BN = 2.5 μ A 120 I D = 80 − 2.5 = 77.5 μ A ⎛ 77.5 × 10 −6 V BB = 2(0.026 ) ln⎜⎜ −15 ⎝ 5 × 10

⎞ ⎟ = 1.22013 V ⎟ ⎠

⎛ 0.3 × 10 −3 ⎞ ⎟ = 0.63304 V V BEN = (0.026) ln⎜⎜ −15 ⎟ ⎝ 8 × 10 ⎠ V EBP = V BB − V BEN = 0.58709 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(

)

⎛ 0.58709 ⎞ I CP = 8 × 10 −15 exp⎜ ⎟ ⇒ I CP = 51.24 μ A ⎝ 0.026 ⎠ Second approximation: ⎛ 120 ⎞ I CN = ⎜ ⎟(0.3 + 0.05124 ) = 0.34834 mA, I BN = 2.903 μ A ⎝ 121 ⎠

I D = 80 − 2.903 = 77.1 μ A ⎛ 77.1× 10 −6 V BB = 2(0.026) ln⎜⎜ −15 ⎝ 5 × 10

⎞ ⎟ = 1.219864 V ⎟ ⎠

⎛ 0.34834 × 10 −3 ⎞ ⎟ = 0.636922 V V BEN = (0.026) ln⎜⎜ −15 ⎟ ⎝ 8 × 10 ⎠ V EBP = 1.219864 − 0.636922 = 0.582942 V

(

)

⎛ 0.582942 ⎞ I CP = 8 × 10 −15 exp⎜ ⎟ ⇒ I CP = 43.7 μ A ⎝ 0.026 ⎠ ____________________________________________________________________________________

13.17 R1 + R 2 = V BE = VT

V BB

1.160

=

(0.1)I Bias

= 96.67 k Ω

(0.1)(0.12) ⎛ I CQ ⎞ ⎛ (0.9)(120 × 10 −6 ) ⎞ ⎟ = (0.026) ln⎜ ⎟ = 0.6187 V ln⎜

⎜ ⎟ ⎜ I ⎟ 5 × 10 −15 ⎝ ⎠ ⎝ S ⎠ ⎛ R2 ⎞ ⎛ R ⎞ ⎟⎟(V BB ) ⇒ 0.6187 = ⎜⎜ 2 ⎟⎟(1.16 ) V BE = ⎜⎜ ⎝ 96.67 ⎠ ⎝ R1 + R 2 ⎠ So R 2 = 51.56 k Ω , R1 = 45.11 k Ω ____________________________________________________________________________________

13.18 A = −g

(r

r

d m o4 o6 (a) From example 13.4

gm =

Ri 2

)

9.5 = 365 μ A / V , ro 4 = 5.26 M Ω 0.026

Now ro 6 = ro 4 = 5.26 M Ω Assuming R8 = 0, we find

Ri 2 = rπ 16 + (1 + β n ) RE′ = 329 + (201) ( 50 9.63 ) ⇒ 1.95 M Ω

Then

(

)

Ad = −(365) 5.26 5.26 1.95 ⇒ Ad = −409

(b)

From Equation (13.20), − β n (1 + β n ) R9 ( Ract 2 Ri 3 R017 )

Av 2 =

For

{

}

Ri 2 R9 + ⎡⎣ rπ 17 + (1 + β n ) Rg ⎤⎦

Rg = 0, Ri 2 = 1.95 M Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Using the results of Example 13.5 Av 2 =

(

−200(201)(50) 92.6 4050 92.6 (1950){50 + 9.63}

)⇒A

v2

= −792

______________________________________________________________________________________ 13.19

Let I C10 = 40 μ A, then I C1 = I C 2 = 20 μ A. Use the procedure in Example 13.4: Ri 2 = 4.07 MΩ (200)(0.026) = 260 kΩ 0.020 0.020 gm6 = = 0.769 mA/V 0.026 50 r06 = ⇒ 2.5 MΩ 0.02 rπ 6 =

Then Ract1 = 2.5[1 + (0.769)(1 260)] = 4.42 MΩ r06 =

50 ⇒ 2.5 MΩ 0.02

Then ⎛ I CQ ⎞ Ad = − ⎜ ⎟ (r04 Ract1 Ri 2 ) ⎝ VT ⎠ ⎛ 20 ⎞ = −⎜ ⎟ (2.5 4.42 4.07) ⎝ 0.026 ⎠

So Ad = −882

______________________________________________________________________________________ 13.20 From Problem 13.11

I1 = I 2 = 7.10 μ A, I C17 = 0.165 mA, I C13 A = 0.055 mA I C16 ≈ I B17 +

I E17 R8 + VBE17 R9

=

0.165 (0.165)(0.1) + 0.6 + 200 50

= 0.000825 + 0.01233

I C16 = 0.0132 mA (200)(0.026) = 31.5 K 0.165 RE1 = R9 [ rπ 17 + (1 + β ) R8 ] = 50 [31.5 + (201)(0.1)]

rπ 17 =

rπ 16

= 50 51.6 = 25.4 K (200)(0.026) = = 394 K 0.0132

Then Ri 2 = rπ 16 + (1 + β ) RE1 = 394 + (201)(25.4) ⇒ 5.50 MΩ

Now

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (200)(0.026) = 732 K 0.0071 0.0071 gm6 = = 0.273 mA/V 0.026 50 ro 6 = ⇒ 7.04 MΩ 0.0071 Ract1 = ro 6 [1 + g m 6 ( R rπ 6 )] = 7.04[1 + (0.273)(1 732)] = 8.96 MΩ rπ 6 =

ro 4 =

50 ⇒ 7.04 MΩ 0.0071

Then Ad = − g m1 (ro 4 Ract1 Ri 2 ) ⎛ 7.10 ⎞ = −⎜ ⎟ (7.04 8.96 5.5) ⎝ 0.026 ⎠ Ad = −627

Now

Ract 2 =

50 ⇒ 303 K 0.165

Ro17 =

50 = 303 K 0.165

From Eq. (13.20), assuming Ri 3 → ∞ Av 2 ≅ − =

β (1 + β ) R9 ( Ract 2 R017 ) Ri 2 { R9 + [rπ 17 + (1 + β ) R8 ]}

−(200)(201)(50)(303 303)

(5500)[50 + 31.5 + (201)(0.1)] Av 2 = −545

=

−3.045 × 108 5.588 × 105

Overall gain Av = (−627)(−545) = 341, 715 ______________________________________________________________________________________ 13.21 Using results from Problem 13.20 ⎛ 100 ⎞ Ri 2 = 5.50 MΩ, Ract1 ⎜ ⎟ [1 + (0.273)(1 732)] ⇒ 17.93 MΩ ⎝ 0.0071 ⎠ 100 ro 4 = ⇒ 14.08 MΩ 0.0071 ⎛ 7.10 ⎞ Ad = − ⎜ ⎟ (14.08 17.93 5.50) ⎝ 0.026 ⎠ Ad = −885

Now 100 100 = 606 K Ro17 = = 606 K 0.165 0.165 −(200)(201)(50)(606 606) −6.09 × 108 Av 2 = = (5500)[50 + 31.5 + (201)(0.1)] 5.588 × 105 Av 2 = −1090

Ract 2 =

Overall gain Av = ( −885)(−1090) = 964, 650

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 13.22 (a) I C18 + I C19 = I Bias (0.25)I C19 + I C19 = 0.18 mA, ⇒ I C19 = 0.144 mA I C18 = 0.036 mA ⎛ 0.144 × 10 −3 V BE19 = (0.026) ln⎜⎜ −14 ⎝ 10 0.60815 R10 = = 16.9 k Ω 0.036 (b) V BE19 = 0.60815 V

⎞ ⎟ = 0.60815 V ⎟ ⎠

⎛ 0.036 × 10 −3 ⎞ ⎟ = 0.5721 V V BE18 = (0.026) ln⎜⎜ −14 ⎟ ⎝ 10 ⎠ ______________________________________________________________________________________

13.23

(b) I C19 = 0.168 mA, g m19 = rπ 19 =

(200 )(0.026 ) = 30.95 k Ω , 0.168

I C18 = 12.84 μ A, g m18 =

rπ 18 =

0.168 = 6.462 mA/V, 0.026 50 = 298 k Ω 0.168

0.01284 = 0.4938 mA/V 0.026

(200 )(0.026 ) = 405 k Ω , 0.01284

ro19 =

ro18 =

50 = 3894 k Ω 0.01284

V x = Vπ 18 + Vπ 19

(1) I x = (2)

Vπ 18 V V + g m18Vπ 18 + π 18 + g m19Vπ 19 + x rπ 18 ro18 ro19

Vπ 18 V V V + g m18Vπ 18 + π 18 = π 19 + π 19 rπ 18 ro18 R10 rπ 19

1 ⎞ 1 ⎞ ⎛ 1 ⎛ 1 Then Vπ 18 ⎜ + 0.4938 + ⎟ = Vπ 19 ⎜ + ⎟ 405 3894 50 30 .95 ⎠ ⎝ ⎠ ⎝ Vπ 18 (0.49653) = Vπ 19 (0.05231) ⇒ Vπ 18 = Vπ 19 (0.10535) V Then (1) I x = Vπ 18 (0.49653) + (6.462 )Vπ 19 + x 298 I x = Vπ 19 (0.10535)(0.49653) + (6.462)Vπ 19 + (0.003356)V x I x = Vπ 19 (6.5143) + V x (0.003356) Now V x = Vπ 18 + Vπ 19 = Vπ 19 (0.10535) + Vπ 19 = (1.10535)Vπ 19 Or Vπ 19 = (0.90469 )V x Then I x = (0.90469 )V x (6.5143) + V x (0.003356 ) I 1 So x = = 5.8968 ⇒ R eq = 170 Ω V x R eq

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 13.24 Now Re14 =

rπ 14 + R01 1+ βP

and R0 = R6 + Re14

Assume series resistance of Q18 and Q19 is small. Then R01 = r013 A Re 22 Re 22 =

rπ 22 + R017 r013 B

where R

1+ βP

= r [1 + g

( R r )]

m17 8 π 17 and 017 017 Using results from Example 13.6,

rπ 17 = 9.63 kΩ rπ 22 = 7.22 kΩ g m17 = 20.8 mA/V r017 = 92.6 kΩ

Then R017 = 92.6[1 + (20.8)(0.1 9.63)] = 283 kΩ r013 B =

50 = 92.6 kΩ 0.54

Then 7.22 + 283 92.6

= 1.51 kΩ 51 R01 = r013 A Re 22 = 278 1.51 = 1.50 kΩ (50)(0.026) rπ 14 = = 0.65 kΩ 2 Re 22 =

Then Re14 =

0.65 + 1.50 = 0.0422 kΩ 51

or Re14 = 42.2 Ω

Then R0 = 42.2 + 27 ⇒ R0 = 69.2 Ω

______________________________________________________________________________________ 13.25 ⎡ ⎛ r Rid = 2 ⎢ rπ 1 + (1 + β n ) ⎜ π 3 ⎝ 1+ βP ⎣ β n = 200, β P = 10

(a) I C1 = 9.5 μ A (200)(0.026) = 547 K 0.0095 (10)(0.026) rπ 3 = = 27.4 K 0.0095 rπ 1 =

Then (201)(27.4) ⎤ ⎡ Rid = 2 ⎢547 + ⎥ 11 ⎣ ⎦ Rid ⇒ 2.095 MΩ

⎞⎤ ⎟⎥ ⎠⎦

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) I C1 = 7.10 μ A (200)(0.026) = 732 K 0.0071 (10)(0.026) rπ 3 = = 36.6 K 0.0071 (201)(36.6) ⎤ ⎡ Rid = 2 ⎢ 732 + ⎥ 11 ⎣ ⎦ Rid ⇒ 2.80 MΩ rπ 1 =

______________________________________________________________________________________ 13.26 We can write A0 ⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟ ⎜1 + j ⎟ f f 1 ⎠ PD ⎠ ⎝ ⎝ 181, 260 = f ⎞⎛ f ⎞ ⎛ ⎜1 + j ⎟ ⎜1 + j ⎟ f1 ⎠ 10.7 ⎠ ⎝ ⎝

A( f ) =

Phase: ⎛ f ⎞ −1 ⎛ f ⎞ ⎟ − tan ⎜ ⎟ ⎝ 10.7 ⎠ ⎝ f1 ⎠ φ = −110° = 70 ° , For a Phase margin

φ = − tan −1 ⎜

So ⎛ f ⎞ −1 ⎛ f ⎞ −110° = − tan −1 ⎜ ⎟ − tan ⎜ ⎟ ⎝ 10.7 ⎠ ⎝ f1 ⎠

Assuming f >> 10.7 Hz, we have ⎛ f ⎞ f tan −1 ⎜ ⎟ = 20° ⇒ = 0.364 f1 ⎝ f1 ⎠ A( f ) = 1,

so

At this frequency, 1=

=

181, 260 2

⎛ f ⎞ 2 1+ ⎜ ⎟ ⋅ 1 + (0.364) ⎝ 10.7 ⎠ 170,327 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 10.7 ⎠

2

f = 170,327 ⇒ f = 1.82 MHz 10.7 or

Then, second pole at f1 =

f ⇒ f1 = 5 MHz 0.364

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 13.27 (a) 10 × 10 6 = f PD 10 6 ⇒ f PD = 10 Hz

( )

(b) f PD =

1 2πR eq C M

⇒ CM =

1 = 1.326 × 10 −8 F 2π 1.2 × 10 6 (10 )

(

)

−8

1.326 ×10 ⇒ C F = 13.25 pF 1001 ______________________________________________________________________________________ C M = C F (1 + A ) ⇒ C F =

13.28

⎛ f ⎝ f PD

φ = −110 = − tan −1 ⎜⎜

⎛ f ⎞ ⎛ ⎞ ⎞ ⎟ = −90 − 2 tan −1 ⎜ f ⎟ ⎟⎟ − 2 tan −1 ⎜ ⎜f ⎟ ⎜f ⎟ ⎠ ⎝ 2,3 ⎠ ⎝ 2, 3 ⎠

⎛ f ⎞ ⎟ = 0.1763 So ⎜ ⎜f ⎟ ⎝ 2,3 ⎠ 200,000

2

≅

200,000

⇒ f = 1.9397 MHz 2 f ⎛ f ⎞ ⎡ ⎛ f ⎞ ⎤ ⎛⎜ ⎞⎟ 1 + (0.1763)2 ⎟ ⎥ ⎟⎟ ⎢1 + ⎜ 1 + ⎜⎜ ⎝ 10 ⎠ ⎜ ⎟ ⎝ f PD ⎠ ⎢⎣ ⎝ f 2,3 ⎠ ⎥⎦ 1.9397 × 10 6 ⇒ f 2,3 = 11.0 MHz Then f 2,3 = 0.1763 ______________________________________________________________________________________ A =1=

[

]

13.29 ⎛ k ′p ⎞⎛ W ⎞ 2 (a) I D 3 = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 3 + VTP ) L 2 ⎝ ⎠⎝ ⎠ 3 ⎛ 40 ⎞ 2 150 = ⎜ ⎟(50 )(V SG 3 − 0.4 ) ⇒ V SG 3 = 0.7873 V ⎝ 2 ⎠ 0.7873 R D1 = = 7.87 k Ω 0.1 ⎛ k ′ ⎞⎛ W ⎞ 2 I D 4 = ⎜⎜ n ⎟⎟⎜ ⎟ (VGS 4 − VTN ) ⎝ 2 ⎠⎝ L ⎠ 4 ⎛ 100 ⎞ 2 200 = ⎜ ⎟(40 )(VGS 4 − 0.4 ) ⇒ VGS 4 = 0.7162 V 2 ⎝ ⎠ 0.7162 − (− 3) RD2 = = 24.8 k Ω 0.15 0 − (− 3) RS = = 15 k Ω 0.15 ⎛g ⎞ ⎛ 0. 1 ⎞ (b) (i) Ad 1 = ⎜⎜ m1 ⎟⎟ ⋅ R D1 , g m1 = 2 ⎜ ⎟(20 )(0.1) = 0.6325 mA/V 2 ⎝ 2 ⎠ ⎝ ⎠ ⎛ 0.6325 ⎞ Ad 1 = ⎜ ⎟(7.87 ) = 2.49 ⎝ 2 ⎠ ⎛ 0.04 ⎞ (ii) A2 = − g m3 R D 2 , g m 3 = 2 ⎜ ⎟(50 )(0.15) = 0.7746 mA/V ⎝ 2 ⎠ A2 = −(0.7746 )(24.8) = −19.21

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (iii) A3 =

g m4 RS ⎛ 0. 1 ⎞ , g m4 = 2 ⎜ ⎟(40 )(0.2 ) = 1.265 mA/V 1+ g m 4 R S ⎝ 2 ⎠

(1.265)(15) = 0.950 1 + (1.265)(15) A = Ad 1 ⋅ A2 ⋅ A3 = (2.49 )(− 19.21)(0.95) = −45.4 A3 =

(c) ______________________________________________________________________________________ 13.30 ⎛ 100 ⎞ 2 (a) I D 3 = 100 = ⎜ ⎟(25)(VGS 3 − 0.4 ) ⇒ VGS 3 = 0.6828 V ⎝ 2 ⎠ 0.6828 R D1 = = 13.66 k Ω 0.05 ⎛ 40 ⎞ 2 I D 4 = 200 = ⎜ ⎟(100 )(V SG 4 − 0.4 ) ⇒ V SG 4 = 0.7162 V ⎝ 2 ⎠ 3 − (− 0.7162 ) = 37.16 k Ω 0. 1 3−0 RS = = 15 k Ω 0.2 R D3 =

⎛g ⎞ ⎛ 0.04 ⎞ (b) (i) Ad = ⎜⎜ m1 ⎟⎟ ⋅ R D1 , g m1 = 2 ⎜ ⎟(80 )(0.05) = 0.5657 mA/V ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 0.5657 ⎞ Ad = ⎜ ⎟(13.66 ) = 3.864 ⎝ 2 ⎠ ⎛ 0. 1 ⎞ (ii) A2 = − g m3 R D 3 , g m 3 = 2 ⎜ ⎟(25)(0.1) = 0.7071 mA/V ⎝ 2 ⎠ A2 = −(0.7071)(37.16 ) = −26.28

(iii) A3 =

g m4 RS ⎛ 0.04 ⎞ , g m4 = 2 ⎜ ⎟(100 )(0.2 ) = 1.265 mA/V 1+ g m 4 R S ⎝ 2 ⎠

(1.265)(15) = 0.950 1 + (1.265)(15) A = Ad ⋅ A2 ⋅ A3 = (3.864)(− 26.28)(0.95) = −96.5 A3 =

(c) ______________________________________________________________________________________ 13.31

a.

Original g m1 and g m 2 ⎛W ⎞⎛ μ C ⎞ K p1 = K p 2 = ⎜ ⎟ ⎜ P ox ⎟ = (12.5)(10) ⎝ L ⎠⎝ 2 ⎠ = 125 μ A / V 2

So ⎛ IQ ⎞ g m1 = g m 2 = 2 K p1 ⎜ ⎟ ⎝ 2⎠

= 2 (0.125)(10) = 0.09975 mA/V

⎛W ⎞ ⎜ ⎟ If ⎝ L ⎠ is increased to 50, then

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ K p1 = K p 2 = (50)(10) = 500 μ A / V 2

So g m1 = g m 2 = 2 (0.5)(0.0199) = 0.1995 mA/V

b.

Gain of first stage Ad = g m1 (r02 r04 ) = (0.1995)(5025 5025)

or Ad = 501

Voltage gain of second stage remains the same, or Av 2 = 251 Then Av = Ad ⋅ Av 2 = (501)(251)

or Ad = 125, 751

______________________________________________________________________________________ 13.32

υ CM (max) = V + − υ SD 6 (min ) − υ SG1 υ SD 6 (min ) = υ SD (sat ) + 0.2 = 0.9022 − 0.5 + 0.2 = 0.6022 V i D1 = K p1 (υ SG1 + VTP )

2

20.2 = 250(υ SG1 − 0.5) ⇒ υ SG1 = 0.7843 V υ CM (max ) = 5 − 0.6022 − 0.7843 = 3.61 V 2

υ CM (min ) = V − + υ GS 3 + υ SD1 (min ) − υ SG1

⎛ 0. 1 ⎞ 2 2 i D 3 = K n (υ GS 3 − VTN ) , K n = ⎜ ⎟(6.25) = 0.3125 mA/V ⎝ 2 ⎠

20.2 = 312.5(υ GS 3 − 0.5) ⇒ υ GS 3 = 0.7542 V υ SD1 (min ) = 0.7843 − 0.5 + 0.2 = 0.4843 V υ CM (min ) = −5 + 0.7542 + 0.4843 − 0.7843 = −4.55 V So −4.55 ≤ υ CM ≤ 3.61 V ______________________________________________________________________________________ 2

13.33 ⎛ k ′p ⎞⎛ W ⎞ ⎛ 0.04 ⎞ 2 (a) K p 5 = ⎜⎜ ⎟⎟⎜ ⎟ = ⎜ ⎟(50 ) = 1.0 mA/V 2 L 2 ⎠ ⎝ ⎠⎝ ⎠ 5 ⎝ + V − V SG 5 − V − 2 K p 5 (V SG 5 + VTP ) = R set

(1)(50)(V SG2 5 − 1.4VSG5 + 0.49) = 10 − VSG5

2 50V SG 5 − 69V SG 5 + 14.5 = 0 ⇒ V SG 5 = 1.121 V

I set = I Q =

10 − 1.121 = 0.1776 mA = I D 7 50

⎛ 0.04 ⎞ ⎛ 0.1776 ⎞ (b) Ad = g m1 (ro 2 ro 4 ) , g m1 = 2 ⎜ ⎟(50 )⎜ ⎟ = 0.5960 mA/V 2 ⎝ ⎠ ⎝ 2 ⎠

ro 2 =

1 1 = 281.5 k Ω , ro 4 = = 563.1 k Ω ⎛ 0.1776 ⎞ ⎛ 0.1776 ⎞ (0.04)⎜ (0.02)⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Ad = (0.5960)(281.5 563.1) = 111.9 ⎛ 0. 1 ⎞ A2 = − g m 7 (ro 7 ro8 ) , g m 7 = 2 ⎜ ⎟(30 )(0.1776 ) = 1.032 mA/V ⎝ 2 ⎠

ro 7 =

1

= 281.5 k Ω , ro8 =

(0.02)(0.1776) A2 = −(1.032)(281.5 140.8) = −96.86 A = Ad ⋅ A2 = (111.9 )(− 96.86 ) = −10,839

1

(0.04)(0.1776)

= 140.8 k Ω

______________________________________________________________________________________ 13.34 ⎛ 0.04 ⎞ Ad = g m1 (ro 2 ro 4 ) , g m1 = 2 ⎜ ⎟(50 )(0.1) = 0.6325 mA/V ⎝ 2 ⎠

ro 2 =

1 1 = 400 k Ω , ro 4 = = 666.7 k Ω (0.025)(0.1) (0.015)(0.1)

Ad = (0.6325)(400 666.7 ) = 158.1

⎛ 0. 1 ⎞ A2 = − g m 7 (ro 7 ro8 ) , g m 7 = 2 ⎜ ⎟(30 )(0.2 ) = 1.095 mA/V ⎝ 2 ⎠

ro 7 =

1 1 = 333.3 k Ω , ro8 = = 200 k Ω (0.015)(0.2) (0.025)(0.2)

A2 = −(1.095)(333.3 200) = −136.9

A = Ad ⋅ A2 = (158.1)(− 136.9) = −21,644 ______________________________________________________________________________________

13.35 f PD =

1 2π Req Ci

R =r

r

where eq 04 02 and We can find that

Ci = C1 (1 + Av 2 )

Av 2 = 251 and r04 = r02 = 5.025 MΩ

Now Req = 5.025 5.025 = 2.51 MΩ

and Ci = 12(1 + 251) = 3024 pF

So f PD =

1 2π (2.51× 106 )(3024 × 10−12 )

or f PD = 21.0 Hz

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 13.36

From Problem 13.33, A2 = 96.86 , ro 2 = 281.5 k Ω , ro 4 = 563.1 k Ω f PD =

1 2π (ro 2 ro 4 )C M

Then C M =

1 = 1.06 × 10 − 7 F 3 2π (8)(281.5 563.1)× 10

C M = C1 (1 + A2 ) ⇒ 1.06 × 10 −7 = C1 (97.86 )

Or C1 = 1.08 ×10 −9 F ______________________________________________________________________________________ 13.37 R0 = r07 r08

We can find that r07 = r08 = 2.52 MΩ

Then R0 = 2.52 2.52

or R0 = 1.26 MΩ

______________________________________________________________________________________ 13.38 6 =2V 3 ⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 2 I D = 0.5 = ⎜ = 3.91 ⎟⎜ ⎟(2 − 0.4 ) ⇒ ⎜ ⎟ 2 L ⎝ ⎠⎝ ⎠ ⎝ L ⎠ 2−5 g m1 (ro1 ro 2 ) 1 = 80 k Ω , ro1 ro 2 = 40 k Ω (b) Aυ = , ro1 = ro 2 = (0.025)(0.5) 1 + g m1 (ro1 ro 2 )

(a) VGS 2 =

0.98 =

g m1 (40 ) ⇒ g m1 = 1.225 mA/V 1 + g m1 (40 )

⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ g m1 = 1.225 = 2 ⎜ ⎟⎜ ⎟ (0.5) ⇒ ⎜ ⎟ = 15 2 L ⎝ ⎠⎝ ⎠ 1 ⎝ L ⎠1

(c) R o =

1 1 ro1 ro 2 = 40 = 0.8163 40 g m1 1.225

R o = 800 Ω ______________________________________________________________________________________

13.39

(a)

2 ⎛ 80 ⎞ I Q 2 = ⎜ ⎟ (20) [1.1737 − 0.7 ] 2 ⎝ ⎠ I Q 2 = 180 μ A

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(b)

2 ⎛ 80 ⎞ I D 6 = ⎜ ⎟ (25) (VGS 6 − 0.7 ) = 25 ⇒ VGS 6 = 0.8581 V ⎝ 2⎠ 2 ⎛ 40 ⎞ I D 7 = ⎜ ⎟ (50) (VSG 7 − 0.7 ) = 25 ⇒ VSG 7 = 0.8581 V ⎝ 2 ⎠

Set VSG 8 P = VGS 8 N = 0.8581 V

⎛ 40 ⎞⎛ W ⎞ ⎛W ⎞ 180 = ⎜ ⎟⎜ ⎟ (0.8581 − 0.7)2 ⇒ ⎜ ⎟ = 360 L 2 ⎝ ⎠⎝ ⎠8 P ⎝ L ⎠8 P ⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ 180 = ⎜ ⎟⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 180 ⎝ 2 ⎠⎝ L ⎠8 N ⎝ L ⎠8 N

______________________________________________________________________________________ 13.40 ⎛ 0.1 ⎞⎛ W ⎞ ⎛ 0. 1 ⎞ 2 2 I REF = I Q1= 0.150 = ⎜ ⎟⎜ ⎟ (VGS 11 − 0.5) = ⎜ ⎟(20 )(VGS 11 − 0.5) ⇒ VGS 11 = 0.8873 V 2 L 2 ⎝ ⎠⎝ ⎠ 11 ⎝ ⎠ V GS , REF = 5 − 0.8873 = 4.1127 V

For three NMOS transistors in series: 4.1127 = 1.3709 V VGS = 3 ⎛ 0.1 ⎞⎛ W ⎞ ⎛W ⎞ 2 I REF = 0.15 = ⎜ ⎟⎜ ⎟(1.3709 − 0.5) ⇒ ⎜ ⎟ = 3.96 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ ______________________________________________________________________________________ 13.41 (a) 2 ⎛ 80 ⎞ I Q 2 = 250 μ A = ⎜ ⎟ (5) (VGS 8 − 0.7 ) 2 ⎝ ⎠ ⇒ VGS 8 = 1.818 V

⇒ VGS 6 = VSG 7 =

1.818 = 0.909 V 2

⎛ 80 ⎞ I D 6 = I D 7 = ⎜ ⎟ (25)(0.909 − 0.7) 2 = 43.7 μ A ⎝ 2⎠

(b) ⎛ 80 ⎞ ⎛ 250 ⎞ g m1 = 2 ⎜ ⎟ (15) ⎜ ⎟ ⇒ 0.5477 mA/V ⎝ 2⎠ ⎝ 2 ⎠ 1 = 800 K ro 2 = ( 0.01)( 0.125) r04 =

1

= 533.3K

( 0.015)( 0.125) Ad 1 = g m1 ( ro 2 ro 4 ) = ( 0.5477 ) ( 800 Ad 1 = 175

Second stage:

533.3)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ A2 = − g m 5 ( ro 5 ro9 ) ⎛ 40 ⎞ g m 5 = 2 ⎜ ⎟ (80)(250) ⇒ 1.265 mA/V ⎝ 2 ⎠ 1 r05 = = 266.7 K (0.015)(0.25) 1 r09 = = 400 K (0.01)(0.25) A2 = −(1.265)(266.7 400) A2 = −202

Assume the gain of the output stage ≈ 1, then Av = Ad 1 ⋅ A2 = (175)(−202) Av = −35,350

______________________________________________________________________________________ 13.42 Ad = g m1 ( Ro 6 Ro8 )

(a)

g m1 = 2 K n I DQ = 2 (0.5)(0.025) ⇒ 224 μ A / V g m1 = g m8 g m 6 = 2 (0.5)(0.025) ⇒ 224 μ A / V 1 1 ro1 = ro 6 = ro8 = ro10 = = = 2.67 M Ω λ I DQ (0.015)(25) ro 4 =

1

λ I D4

=

1 ⇒ 1.33 M Ω 0.015 ( )( 50 )

Now Ro 8 = g m8 (ro8 ro10 ) = (224)(2.67)(2.67) = 1597 M Ω Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) = (224)(2.67)(2.67 1.33) ⇒ Ro 6 = 531 M Ω

Then Ad = (224)(531 1597) ⇒ Ad = 89, 264

(b)

Ro = Ro 6 Ro8 = 531 1597 ⇒ Ro = 398 M Ω f PD =

(c)

1 1 = ⇒ f PD = 80 Hz 2π Ro CL 2π ( 398 × 106 )( 5 × 10−12 )

GBW = (89, 264)(80) ⇒ GBW = 7.14 MHz

______________________________________________________________________________________ 13.43 (a)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ro1 = ro8 = ro10 = ro 6 = ro 4 =

1

λn I D 1

λn I D 4

=

1

λp ID

=

1 = 2 MΩ (0.02)(25)

1 = 2.67 M Ω (0.015)(25)

=

1 = 1.33 M Ω (0.015)(50)

⎛ 35 ⎞⎛ W ⎞ ⎛W ⎞ g m1 = 2 ⎜ ⎟⎜ ⎟ (25) = 41.8 ⎜ ⎟ = g m8 ⎝ 2 ⎠⎝ L ⎠1 ⎝ L ⎠1 ⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ g m 6 = 2 ⎜ ⎟⎜ ⎟ (25) = 63.2 ⎜ ⎟ ⎝ 2 ⎠⎝ L ⎠6 ⎝ L ⎠6 Ro = Ro 6 Ro8 = [ g m 6 ( ro 6 )(ro 4 ro1 )] [ g m8 (ro8 ro10 )] ⎛W ⎞ ⎛W ⎞ X1 = ⎜ ⎟ X6 = ⎜ ⎟ ⎝ L ⎠1 and ⎝ L ⎠6 Define

Then

Ro = ⎣⎡63.2 X 6 ( 2.67 ) (1.33 2 ) ⎦⎤ ⎡⎣ 41.8 X 1 ( 2 )( 2 ) ⎤⎦ 22,539 X 1 X 6 = 134.8 X 6 167.2 X 1 = 134.8 X 6 + 167.2 X 1

Ad = g m1 Ro

⎛ 22,539 X 1 X 6 ⎞ = (41.8 X 1 ) ⎜ ⎟ 134.8 167.2 X + X 6 1 ⎠ ⎝ = 10, 000

⎛W ⎞ X6 = ⎜ ⎟ = ⎝ L ⎠6 Now

1 ⎛W ⎞ ⎜ ⎟ = 0.674 X 1 2.2 ⎝ L ⎠1

We then find ⎛W ⎞ ⎛W ⎞ X 12 = ⎜ ⎟ = 4.06 = ⎜ ⎟ L ⎝ ⎠1 ⎝ L ⎠p

and ⎛W ⎞ ⎜ ⎟ = 1.85 ⎝ L ⎠n

______________________________________________________________________________________ 13.44 + − Let V = 5V , V = −5V

P = IT (10) = 3 ⇒ IT = 0.3 mA ⇒ I REF = 0.1 mA = 100 μ A 1 ro1 = ro8 = ro10 = = 1 MΩ (0.02)(50) 1 ro 6 = = 1.33 MΩ (0.015)(50) 1 ro 4 = = 0.667 M Ω (0.015)(100) ⎛ 35 ⎞⎛ W ⎞ g m1 = 2 ⎜ ⎟⎜ ⎟ (50) = 59.2 X 1 = g m8 ⎝ 2 ⎠⎝ L ⎠1

⎛W ⎞ X1 = ⎜ ⎟ ⎝ L ⎠1 where

Assume all width-to-length ratios are the same.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ 80 ⎞⎛ W ⎞ g m 6 = 2 ⎜ ⎟⎜ ⎟ (50) = 89.4 X 1 ⎝ 2 ⎠⎝ L ⎠

Now

Ro = Ro 6 Ro8 = ⎡⎣ g m 6 ( ro 6 ) ( ro 4 ro1 ) ⎤⎦ ⎡⎣ g m8 ( ro8 ro10 ) ⎤⎦ = ⎡⎣89.4 X 1 (1.33) ( 0.667 1) ⎤⎦ ⎡⎣59.2 X 1 (1)(1) ⎤⎦ ( 47.6 X 1 )( 59.2 X 1 ) = [ 47.6 X 1 ] [59.2 X 1 ] = 47.6 X 1 + 59.2 X 1

So Ro = 26.4 X 1 Now

Ad = g m1 Ro = ( 59.2 X 1 )( 26.4 X 1 ) = 25, 000 X 12 =

W = 16 L for all transistors

So that ______________________________________________________________________________________

13.45 (a) Ad = Bg m1 ro 6 ro8

(

)

⎛ 0.1 ⎞ g m1 = 2 ⎜ ⎟(20 )(0.06 ) = 0.4899 mA/V ⎝ 2 ⎠

1 = 138.9 k Ω (0.04)(0.06)(3) 1 ro8 = = 277.8 k Ω (0.02)(0.06)(3)

ro 6 =

Ad = (3)(0.4899)(138.9 277.8) = 136.1 (b) Ro = ro 6 ro8 = 138.9 277.8 = 92.6 k Ω (c)

f PD =

1 1 = ⇒ f PD = 343.7 kHz 2πR o C 2π 92.6 × 10 3 5 × 10 −12

(

(

)

)(

)

GBW = (136.1) 343.7 × 10 ⇒ GBW = 46.8 MHz ______________________________________________________________________________________ 3

13.46 1 = 0.5 M Ω (0.02)(2.5)(40) 1 ro8 = = 0.667 M Ω (0.015)(2.5)(40) Ad = Bg m1 ( ro 6 ro8 )

ro 6 =

400 = (2.5) g m1 ( 0.5 0.667 ) ⇒ g m1 = 560 μ A / V ⎛ 80 ⎞ ⎛ W g m1 = 560 = 2 ⎜ ⎟ ⎜ ⎝ 2 ⎠⎝ L

⎞ ⎛W ⎟ (40) ⇒ ⎜ ⎠ ⎝L

⎞ ⎟ = 49 ⎠

(a) Assume all (W/L) ratios are the same except for

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛W ⎞ ⎛W ⎞ M 5 and M 6 . ⎜ ⎟ = ⎜ ⎟ = 122.5 ⎝ L ⎠5 ⎝ L ⎠ 6

(b)

Assume the bias voltages are V + = 5V , V − = −5V .

⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 49 Assume ⎝ L ⎠ A ⎝ L ⎠ B ⎛ 80 ⎞ I Q = ⎜ ⎟ (49)(VGSA − 0.5) 2 = 80 ⇒ VGSA = 0.702 V ⎝ 2⎠

Then ⎛ 80 ⎞ ⎛ W ⎞ I REF = 80 = ⎜ ⎟ ⎜ ⎟ (VGSC − 0.5) 2 ⎝ 2 ⎠ ⎝ L ⎠C

For four transistors 10 − 0.702 = 2.325 V 4 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 80 = ⎜ ⎟ ⎜ ⎟ (2.325 − 0.5) 2 ⇒ ⎜ ⎟ = 0.60 ⎝ 2 ⎠ ⎝ L ⎠C ⎝ L ⎠C

VGSC =

f 3− dB =

(c)

1 2π Ro C

Ro = 0.5 0.667 = 0.286 M Ω

1 = 185 kHz 2π (286 × 103 )(3 × 10−12 ) GBW = (400)(185 × 103 ) ⇒ 74 MHz f 3− dB =

______________________________________________________________________________________ 13.47 (a)

From previous results, we can write Ro10 = g m10 ( ro10 ro 6 ) Ro12 = g m12 ( ro12 ro8 ) Ad = Bg m1 ( Ro10 Ro12 )

Now 1

λP B ( I Q / 2 )

=

1 = 0.5 M Ω (0.02)(2.5)(40)

1

=

1 = 0.667 M Ω (0.015)(2.5)(40)

ro10 = ro 6 = ro12 = ro8 =

λn B ( I Q / 2 )

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Assume all transistors have the same width-to-length ratios except for M 5 and M 6 . ⎛W ⎜ Let ⎝ L

⎞ 2 ⎟= X ⎠

Then ⎛ k p′ ⎞ ⎛ W ⎞ ⎛ 35 ⎞ g m10 = 2 ⎜ ⎟ ⎜ ⎟ ( I DQ10 ) = 2 ⎜ ⎟ X 2 (2.5)(40) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠10 = 83.67 X ⎛ k′ ⎞⎛ W ⎞ ⎛ 80 ⎞ g m12 = 2 ⎜ n ⎟ ⎜ ⎟ ( I DQ12 ) = 2 ⎜ ⎟ X 2 (2.5)(40) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠12 = 126.5 X ⎛ 80 ⎞ g m1 = 2 ⎜ ⎟ X 2 (40) = 80 X ⎝ 2⎠

Then Ro10 = (83.67 X )(0.5)(0.5) = 20.9 X M Ω Ro12 = (126.5 X )(0.667)(0.667) = 56.3 X M Ω

We want 20, 000 = (2.5)(80 X )[20.9 X 56.3 X ] ⎡ (20.9 X )(56.3 X ) ⎤ 2 = 200 X ⎢ ⎥ = 3048 X ⎣ 20.9 X + 56.3 X ⎦

Then ⎛W ⎞ X 2 = 6.56 = ⎜ ⎟ ⎝ L⎠

Then ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = (2.5)(6.56) = 16.4 ⎝ L ⎠ 6 ⎝ L ⎠5

(b)

+ − Assume bias voltages are V = 5V , V = −5V

⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 6.56 L Assume ⎝ ⎠ A ⎝ L ⎠ B ⎛ 80 ⎞ I Q = 80 = ⎜ ⎟ (6.56)(VGSA − 0.5) 2 ⇒ VGSA = 1.052 V ⎝ 2⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Need 5 transistors in series VGSC =

10 − 1.052 = 1.79 V 5

Then ⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ I REF = 80 = ⎜ ⎟⎜ ⎟ (1.79 − 0.5) 2 ⇒ ⎜ ⎟ = 1.20 2 L ⎝ ⎠⎝ ⎠C ⎝ L ⎠C f 3− dB =

(c) Now

1 2π Ro C

where

Ro = Ro10 Ro12

Ro10 = 20.9 6.56 = 53.5 M Ω Ro12 = 56.3 6.56 = 144 M Ω

Then Ro = 53.5 144 = 39 M Ω 1 f 3− dB = = 1.36 kHz 2π (39 ×106 )(3 × 10−12 ) GBW = (20, 000)(1.36 x103 ) ⇒ GBW = 27.2 MHz

______________________________________________________________________________________

13.48 (a) ΔVO1 = 0.7 + (0.3)(0.4) = 0.82 V 0.82 = 5.47 k Ω 0.15 3−0 = 10 k Ω RC = 0.3

RD =

(b) Ad =

g m1 (R D RiC ) , g m1 = 2 ⎛⎜ 0.04 ⎞⎟(50)(0.15) = 0.7746 mA/V 2 ⎝ 2 ⎠

RiC = rπ + (1 + β )R E , rπ =

(120 )(0.026) = 10.4 k Ω

0. 3 RiC = 10.4 + (121)(0.4 ) = 58.8 k Ω

⎛ 0.7746 ⎞ Ad = ⎜ ⎟(5.47 58.8) = 1.938 ⎝ 2 ⎠ − β RC − (120)(10) (c) A2 = = = −20.41 rπ + (1 + β )R E 10.4 + (121)(0.4)

(d) Aυ = Ad ⋅ A2 = (1.938)(− 20.41) = −39.6 ______________________________________________________________________________________ 13.49

(a) R S =

0 − (− 3) = 10 k Ω 0. 3

I D1 = K n (VGS1 − VTN )

2

0.3 = 3(VGS1 − 0.4) ⇒ VGS1 = 0.7162 V 2

RC 2 =

ΔVO1

0.7162 − (− 3) = 12.39 k Ω 0. 3 = (0.3)(0.5) + 0.7 = 0.85 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 0.85 R C1 = = 3.4 k Ω 0.25 (120 )(0.026 ) = 12.48 k Ω 0.25 (b) g m1 = = 9.615 mA/V, rπ 1 = 0.026 0.25 ( 0.3 120 )(0.026 ) g m3 = = 11.54 mA/V, rπ 3 = = 10.4 k Ω 0.026 0. 3 g Ad = m1 (RC1 Ri 3 ) 2 Ri 3 = rπ 3 + (1 + β )R E = 10.4 + (121)(0.5) = 70.9 k Ω 9.615 (3.4 70.9) = 15.60 2 − βRC 2 − (120 )(12.39 ) = = −20.97 (c) A2 = rπ 3 + (1 + β )R E 10.4 + (121)(0.5) Ad =

(d) A3 =

g mD R S , g mD = 2 (3)(0.3) = 1.897 mA/V 1 + g mD R S

(1.897)(10) = 0.950 1 + (1.897)(10) = Ad ⋅ A2 ⋅ A3 = (15.6 )(− 20.97 )(0.95) = −310.8

A3 =

(e) Aυ ______________________________________________________________________________________ 13.50 (a) For PMOS: ⎛ 0.04 ⎞ g m1 = g m 2 = 2 ⎜ ⎟(40 )(0.125) = 0.6325 mA/V ⎝ 2 ⎠

roP =

1 1 = = 228.6 k Ω λ I D1 (0.035)(0.125)

For BJT: ro 2 =

VA 150 = = 1200 k Ω I C 0.125

(b) Ad = g m 2 (roP ro 2 ) = (0.6325)(228.6 1200) = 121.5 ______________________________________________________________________________________ 13.51 (a) For NMOS: ⎛ 0. 1 ⎞ g mN = 2 ⎜ ⎟(40 )(0.125) = 1.0 mA/V ⎝ 2 ⎠

roN =

1 = 400 k Ω (0.02)(0.125)

For BJT: 100 = 800 k Ω 0.125 (b) Ad = g mN roN ro 2 = (1.0) 400 800 = 266.7 ro 2 =

(

)

(

)

______________________________________________________________________________________ 13.52

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I REF = 200 μ A K n = K p = 0.5 mA / V 2 λn = λ p = 0.015 V −1 (a) where

Ad = g m1 ( Ro 6 Ro8 )

Ro8 = g m8 (ro8 ro10 )

Ro 6 = g m 6 (ro 6 ) ( ro 4 ro1 )

Now g m8 = 2 K P I D 8 = 2 (0.5)(0.1) = 0.447 mA/V 1 1 ro8 = = = 667 k Ω λP I D 8 (0.015)(0.1) ro10 =

1 = 667 k Ω λP I D 8

gm6 =

IC 6 0.1 = = 3.846 mA/V VT 0.026

ro 6 = ro 4 = ro1 =

VA 80 = = 800 k Ω I C 6 0.1 1

λn I D 4

=

1 = 333 k Ω (0.015)(0.2)

1 1 = = 667 k Ω λ p I D1 (0.015)(0.1)

g m1 = 2 K P I D1 = 2 (0.5)(0.1) = 0.447 mA/V

So Ro8 = (0.447)(667)(667) ⇒ 198.9 M Ω Ro 6 = (3.846)(800)(333 667) ⇒ 683.4 M Ω

Then Ad = 447(198.9 683.4) ⇒ Ad = 68,865

______________________________________________________________________________________ 13.53 + − Assume biased at V = 10V , V = −10V .

P = 3I REF (20) = 10 ⇒ I REF = 167 μ A Ad = g m1 ( Ro 6 Ro8 ) = 25, 000 kn′ = 80 μ A / V 2 , k ′p = 35 μ A / V 2

λn = 0.015V −1 , λ p = 0.02 V −1 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = 2.2 ⎜ ⎟ L ⎝ L ⎠n Assume ⎝ ⎠ p Ro8 = g m8 ( ro8 ro10 )

Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) ro8 = ro10 =

1

λP I D 8

=

1 = 0.60 M Ω (0.02)(83.3)

1 = 0.60 M Ω λP I D 8

⎛ k ′p ⎞ ⎛ W ⎞ ⎛ 35 ⎞ g m8 = 2 ⎜ ⎟ ⎜ ⎟ I D 8 = 2 ⎜ ⎟ (2.2) X 2 (83.3) L 2 ⎝ ⎠ ⎝ 2⎠ 8 ⎝ ⎠ = 113.3 X

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛W ⎞ X2 =⎜ ⎟ ⎝ L ⎠n where V 80 = 0.960 M Ω ro 6 = A = I C 6 83.3 ro 4 = ro1 = gm6 =

1

λn I D 4

=

1 = 0.40 M Ω (0.015)(167)

1 1 = = 0.60 M Ω λ p I D1 (0.02)(83.3) IC 6 83.3 = = 3204 μ A / V 0.026 VT

⎛ k ′p ⎞ ⎛ W ⎞ ⎛ 35 ⎞ g m1 = 2 ⎜ ⎟ ⎜ ⎟ I D1 = 2 ⎜ ⎟ (2.2) X 2 (83.3) 2 L ⎝ ⎠ ⎝ 2⎠ 1 ⎝ ⎠ = 113.3 X

Now Ro 6 = (3204)(0.960) ⎡⎣ 0.40 0.60 ⎤⎦ = 738 M Ω Ro8 = (113.3 X )(0.60)(0.60) = 40.8 X M Ω

Then Ad = 25, 000 = (113.3 X ) ⎡⎣738 40.8 X ⎤⎦ ⎡ 30,110 X ⎤ = (113.3 X ) ⎢ ⎥ ⎣ 738 + 40.8 X ⎦

which yields X = 2.48 or ⎛W ⎞ X 2 = 6.16 = ⎜ ⎟ ⎝ L ⎠n

and ⎛W ⎞ ⎜ ⎟ + (2.2)(6.16) = 12.3 ⎝ L ⎠P

______________________________________________________________________________________ 13.54

For vcm (max), assume VCB (Q5 ) = 0. Then VS = 15 − 0.6 − 0.6 = 13.8 V I D 9 = I D10 =

0.236 = 0.118 mA 2

Using parameters given in Example 13.12 VSG =

I D9 0.118 − VTP = + 1.4 = 2.17 V 0.20 KP

Then

vcm (max) = 13.8 − 2.17 ⇒ vcm (max) = 11.6 V

For vcm (min) , assume VSD ( M 9 ) = VSD ( sat ) = VSG + VTP = 2.17 − 1.4 = 0.77 V

Now VD10 = I D10 (0.5) + 0.6 + I D10 (0.5) − 15 = 0.118 + 0.6 − 15 ⇒ VD10 = −14.28 V

Then

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ vcm (min) = −14.28 + VSD (sat) − VSG = −14.28 + 0.77 − 2.17 = −15.68 V

Then, common-mode voltage range −15.68 ≤ vcm ≤ 11.6

Or, assuming the input is limited to ±15 V, then −15 ≤ vcm ≤ 11.6 V

______________________________________________________________________________________ 13.55

I1 = I 2 K p (V SG + VTP ) = 2

I 2 = 0.15 =

V SG − V BE 7 R1

V SG − 0.6 ⇒ V SG = 1.8 V 8

I 1 = 0.15 = K p (1.8 − 1) ⇒ K p = 0.234 mA/V 2 2

______________________________________________________________________________________

13.56

(a) K p (V SG + VTP ) =

V SG − V BE 7 R1

0.15(V SG − 1.2 ) =

V SG − 0.6 8

2

2

2 − 3.88V SG + 2.328 = 0 , ⇒ V SG = 2.437 V We find 1.2V SG

I1 = I 2 =

2.437 − 0.6 = 0.2296 mA 8

(b) VC 7 = V + − 0.6 − 0.6

VC 6 = V − + 2.437 Set VC 6 = VC 7 Then V S − 1.2 = −V S + 2.437 ⇒ V S = 1.82 V ______________________________________________________________________________________ 13.57

I C 5 = I C 4 = 300 μ A

We have Ri 2 = rπ 13 =

β nVT I C13

=

(200)(0.026) = 17.3 kΩ 0.3

Ad = 2 K n I Q 5 ⋅ ( Ri 2 ) = 2(0.6)(0.3) ⋅ (17.3)

or Ad = 10.38

Now

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ I C13 0.3 = = 11.5 mA/V 0.026 VT

g m13 = r013 =

VA 50 = = 167 kΩ I C13 0.3

Then | Av 2 | = g m13 ⋅ r013 = (11.5)(167)

or Av 2 = 1917

Overall gain: Av = (10.38)(1917) = 19,895

_____________________________________________________________________________ 13.58

Assuming the resistances looking into Q4 and into the output stage are very large, we have β R013 | Av 2 | = rπ 13 + (1 + β ) RE13 where

R013 = r013 ⎡⎣1 + g m13 ( RE13 rπ 13 ) ⎤⎦

I C13 = 300 μ A, r013 =

50 = 167 kΩ 0.3

0.3 = 11.5 mA / V 0.026 (200)(0.026) = = 17.3 kΩ 0.3

g m13 = rπ 13

So

R013 = (167) ⎡⎣1 + (11.5) (1 17.3) ⎤⎦ ⇒ 1.98 MΩ

Then | Av 2 | =

(200)(1980) = 1814 17.3 + (201)(1)

Now

Ci = C1 (1 + Av 2 ) = 12 [1 + 1814 ] ⇒ Ci = 21, 780 pF

f PD =

1 2π Req Ci

Req = Ri 2 r012 r010

Neglecting R3 , r010 =

1

λ I D10

=

1 = 333 kΩ (0.02)(0.15)

Neglecting R5 , 50 = 333 kΩ 0.15 Ri 2 = rπ 13 + (1 + β ) RE13

r012 =

= 17.3 + (201)(1) = 218 kΩ

Then f PD =

1 2π ⎡⎣ 218 333 333⎤⎦ ×103 × ( 21, 780 ) × 10−12

or f PD = 77.4 Hz

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Unity-Gain Bandwidth Gain of first stage: Ad = 2 K n I Qs ⋅ ( R12 ro12 ro10 ) = 2(0.6)(0.3) ⋅ (218 333 333) = (0.6)(218 333 333) Ad = 56.6 or Overall gain: Av = (56.6)(1814) = 102, 672 Then unity-gain bandwidth = (77.4)(102, 672) ⇒ 7.95 MHz

______________________________________________________________________________________ 13.59

Since VGS = 0 in J 6 , I REF = I DSS ⇒ I DSS = 0.8 mA

______________________________________________________________________________________

13.60

Ri 2 = rπ 5 + (1 + β ) [ rπ 6 + (1 + β ) RE ]

a.

(100)(0.026) = 13 kΩ 0.2 I 200 μ A ≅ C6 = = 2 μA β 100

rπ 6 = IC 5

So rπ 5 =

(100)(0.026) = 1300 kΩ 0.002

Then

Ri 2 = 1300 + (101) [13 + (101)(0.3)]

or Ri 2 = 5.67 MΩ Av = g m 2 ( r02 r04 Ri 2 )

b.

gm2 =

2 ⋅ I D ⋅ I DSS VP

=

2 ⋅ (0.1)(0.2) 3

= 0.0943 mA / V r02 = r04 =

Then

1

λ ID

=

1 = 500 kΩ (0.02)(0.1)

VA 5.0 = = 500 kΩ I C 4 0.1

Av = (0.0943)[500 || 500 || 5670]

or

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Av = 22.6

______________________________________________________________________________________ 13.61

a. Set

Need

VSD (QE ) ≥ VSD ( sat ) = VP

VP = 3 V I REF 2 =

so that

and

VZK = 3 V

For minimum bias ±3 V

VZK − VD1 R3

R3 =

3 − 0.6 ⇒ R3 = 24 kΩ 0.1

Set bias in QE = I REF 2 + I Z 2 = 0.1 + 0.1 = 0.2 mA Therefore, I DSS = 0.2 mA

b.

Neglecting base currents I 01 = I REF 1 = 0.5 mA =

12 − 0.6 R4

so that R4 = 22.8 kΩ

______________________________________________________________________________________

13.62 a.

We have gm2 =

2 ⋅ I D ⋅ I DSS | VP |

=

2 ⋅ (0.5)(1) 4

= 0.354 mA/V

1 1 r02 = = = 100 kΩ λ I D (0.02)(0.5) r04 =

VA 100 = = 200 kΩ I D 0.5

0.5 = 19.23 mA/V 0.026 (200)(0.026) rπ 4 = = 10.4 kΩ 0.5

gm4 =

So

R04 = r04 ⎡⎣1 + g m 4 ( rπ 4 R2 ) ⎤⎦

= 200 ⎡⎣1 + (19.23) (10.4 0.5 ) ⎤⎦ = 2035 kΩ

Ad = g m 2 ( r02 R04 RL ) RL → ∞

For

Ad = 0.354 (100 || 2035 ) = 33.7

With these parameter values, gain can never reach 500. b. Similarly for this part, gain can never reach 700. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 14 14.1 Ad =

vo = −80 vi

vo (max) = 4.5 ⇒ vi (max) = 56.25 mV

vi (max) rms =

56.25

= 39.77 mV

2 So ______________________________________________________________________________________

14.2

(a) 4.5 = 0.028125 mA 160 4.5 iL = = 4.5 mA 1 Output Circuit = 4.528 mA i2 =

vi = −

vo −4.5 = ⇒ vi = −0.05625 V 80 A

(b) io ≈ 15 mA =

vo 4.5 = RL RL

⇒ RL (min) = 300 Ω

______________________________________________________________________________________ 14.3 (1)

vo = 2 V

(2)

v2 = 12.5 mV

(3)

AOL = 2 × 10 4

(4)

v1 = 8 μ V

(5) AOL = 1000 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 14.4 R 120 = −21.42857 (a) ACL (∞ ) = − 2 = − R1 5.6 −21.42857 = −21.42376 22.42857 1+ 10 5 − 21.42376 − (− 21.42857 ) × 100% = −0.0224% − 21.42857 R 120 (b) ACL (∞ ) = − 2 = − = −14.634146 R1 8.2 ACL =

−14.634146 = −14.63186 15.634146 1+ 10 5 − 14.63186 − (− 14.634146 ) × 100% = −0.0156% − 14.634146 ______________________________________________________________________________________ ACL =

14.5 47 7.91176 6.8 = = 7.90863 (a) (i) ACL = 7.91176 47 ⎞ ⎛ 1 + ⎜1 + ⎟ 2 × 10 4 ⎝ 6.8 ⎠ 1+ 2 × 10 4 7.90863 − 7.91176 (ii) × 100% = −0.03956% 7.91176 7.91176 (b) (i) ACL = = 7.84966 7.91176 1+ 10 3 7.84966 − 7.91176 (ii) × 100% = −0.785% 7.91176 ______________________________________________________________________________________ 1+

14.6 − R2 R R1 R1 (a) − 15.0 = ⇒ 2 = 15.12091 = R1 ⎛1 + R 2 ⎞ 1.0005 + 5 × 10 − 4 ⎛ R 2 ⎞ ⎜ R ⎟ ⎜ R1 ⎟⎠ 1⎠ ⎝ ⎝ 1+ 2 × 10 3 −15.12091 (b) ACL = = −15.1160 16.12091 1+ 5 × 10 4 ______________________________________________________________________________________ − R2

14.7 90 ⇒ AOL = 8.9991 × 10 5 90 1+ AOL ______________________________________________________________________________________

(1 − 0.0001)(90) =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 14.8 ACL = (1 − 0.0002 )(1) =

1

⇒ AOL = 4999 1 1+ AOL ______________________________________________________________________________________

14.9 A =+

(a)

R 2 210(1 ± 0.001) = R1 21(1 ± 0.001)

210.21 = 10.02 20.979 209.79 A min = = 9.98 21.021 So 9.98 ≤ A ≤ 10.02 A max =

10.02 = 10.009 11.02 1+ 10 4 9.98 A min = = 9.969 10.98 1+ 10 4 So 9.969 ≤ A ≤ 10.009 A max =

(b)

______________________________________________________________________________________ 14.10

vI − v1 v1 − v0 v1 = + and v0 = − A0 L v1 R1 R2 Ri so that v1 = −

v0 A0 L

⎛ 1 vI v0 1 1⎞ + = v1 ⎜ + + ⎟ R1 R2 ⎝ R1 R2 Ri ⎠

So ⎡1 vI 1 ⎛ 1 1 1 ⎞⎤ = −v0 ⎢ + + ⎟⎥ ⎜ + R1 ⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦

Then v0 −(1/ R1 ) = = ACL vI ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ + ⎟⎥ ⎢ + ⎜ + ⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦ From Equation (14.20) for RL = ∞ and R0 = 0

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 1 1 (1 + A0 L ) = + ⋅ 1 Rif Ri R2

a.

For Ri = 1 kΩ −(1/ 20) ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎢100 + 103 ⎜ 20 + 100 + 1 ⎟ ⎥ ⎝ ⎠⎦ ⎣ −0.05 = [0.01 + 1.06 × 10−3 ]

ACL =

or ⇒ ACL = −4.52 1 1 1 + 103 = + ⇒ Rif = 90.8 Ω Rif 1 100

b.

For Ri = 10 kΩ −(1/ 20) ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎢100 + 103 ⎜ 20 + 100 + 10 ⎟ ⎥ ⎝ ⎠⎦ ⎣ −0.05 = [0.01 + 1.6 × 10−4 ]

ACL =

or ⇒ ACL = −4.92 1 1 1 + 103 = + ⇒ Rif = 98.9 Ω Rif 10 100

c.

For Ri = 100 kΩ ACL =

=

−(1/ 20) ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎢100 + 103 ⎜ 20 + 100 + 100 ⎟ ⎥ ⎝ ⎠⎦ ⎣ −0.05 [0.01 + 7 × 10−5 ]

or ⇒ ACL = −4.965 1 1 1 + 103 = + ⇒ Rif = 99.8 Ω Rif 100 100

______________________________________________________________________________________ 14.11

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ vo ⎝ ACL = = vi ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ A R1 ⎠ ⎦ OL ⎝ ⎣

For the ideal: ⎛ R2 ⎞ 0.10 = 50 ⎜1 + ⎟ = R1 ⎠ 0.002 ⎝ vo (actual ) = (0.10)(1 − 0.001) = 0.0999

So 0.0999 50 = = 49.95 0.002 1 + 1 (50) AOL

which yields AOL = 1000

______________________________________________________________________________________ 14.12 From Equation (14.18) Avf 1

⎛A 1 ⎞ − ⎜ OL − ⎟ R R v 2 ⎠ ⎝ o = o1 = v1 ⎛ 1 1 1 ⎞ + + ⎟ ⎜ R R R o 2 ⎠ ⎝ L

Or ⎛ 5 × 103 1 ⎞ −⎜ − ⎟ 3 1 100 ⎠ ⋅ v = −(4.99999 × 10 ) ⋅ v vo1 = ⎝ 1 1 1.11 ⎛1 1 1 ⎞ ⎜ + + ⎟ ⎝ 10 1 100 ⎠ vo1 = −4.504495 × 103 ⋅ v1

Now i1 vi − v1 = ≡K v1 R1v1

Then vi − v1 = KR1v1

which yields v1 =

vi KR1 + 1

Now, from Equation (14.20) 1 ⎤ ⎡ 1 + 5 × 103 + ⎥ 1 1 ⎢ 10 K= + ⎢ ⎥ 10 100 ⎢ 1 + 1 + 1 ⎥ 10 100 ⎦⎥ ⎣⎢ ⎡ 5.0011× 103 ⎤ = (0.1) + (0.01) ⎢ ⎥ = 45.15495 1.11 ⎣ ⎦

Then v1 =

We find

vi vi = ( 45.15495 )(10 ) + 1 452.5495

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ vi ⎡ ⎤ vo1 = −4.504495 × 103 ⎢ ⎥ ⎣ 452.5495 ⎦

Or Avf 1 =

vo1 = −9.9536 vi

For the second stage, RL = ∞ ⎛ 5 × 103 1 ⎞ −⎜ − ⎟ 1 100 ⎠ ⋅ v ′ = −4.950485 × 103 ⋅ v ′ vo 2 = ⎝ 1 1 ⎛1 1 ⎞ ⎜ + ⎟ ⎝ 1 100 ⎠ ⎡ ⎤ 1 1 ⎢1 + 5 × 103 ⎥ K≡ + = 49.61485 ⎢ 1 ⎥⎥ 10 100 ⎢ 1+ ⎢⎣ 100 ⎥⎦ vo1 vo1 vo1 = = v1′ = KR1 + 1 (49.61485)(10) + 1 497.1485

Then vo 2 −4.950485 × 103 = = −9.95776 497.1485 vo1

So Avf =

vo 2 = (−9.9536)(−9.95776) ⇒ Avf = 99.12 vi

______________________________________________________________________________________ 14.13 a.

v1 − vI v v −v + 1 + 1 0 =0 R3 + Ri R1 R2

⎡ 1 vI 1 1⎤ v v1 ⎢ + + ⎥= 0 + R R R R R R + i 1 2 ⎦ 2 3 + Ri ⎣ 3 v0 v0 − A0 L vd v0 − v1 =0 + + RL R0 R2

(1)

(2)

or ⎡ 1 A v 1 1⎤ v v0 ⎢ + + ⎥ = 1 + 0L d R0 ⎣ RL R0 R2 ⎦ R2 ⎛ v −v ⎞ vd = ⎜ I 1 ⎟ ⋅ Ri ⎝ R3 + Ri ⎠

So substituting numbers:

(3)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ vI 1⎤ v 1 ⎡ 1 v1 ⎢ + + ⎥= 0 + ⎣10 + 20 10 40 ⎦ 40 10 + 20

(1)

or v1[0.15833] = v0 [0.025] + vI [0.03333] 1 ⎤ v (10 4 )vd ⎡1 1 v0 ⎢ + + ⎥= 1 + 0.5 ⎣1 0.5 40 ⎦ 40

or

So or

(2)

v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) vd ⎛ v −v ⎞ vd = ⎜ I 1 ⎟ ⋅ 20 = 0.6667 ( vI − v1 ) ⎝ 10 + 20 ⎠

(3)

v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) ( 0.6667 )( vI − v1 )

(2)

v0 [3.025] = 1.333 ×104 vI − 1.333 ×104 v1

From (1):

v1 = v0 ( 0.1579 ) + vI ( 0.2105)

Then

v0 [3.025] = 1.333 × 104 vI − 1.333 × 104 ⎡⎣v0 ( 0.1579 ) + vI ( 0.2105 ) ⎤⎦ v0 ⎡⎣ 2.1078 ×103 ⎤⎦ = vI ⎡⎣1.0524 × 104 ⎤⎦

or ACL =

To find

v0 = 4.993 vI

Rif :

Use Equation (14.27)

⎛ 0.5 0.5 ⎞ iI ⎜1 + + ⎟ 1 40 ⎠ ⎝

3 ⎧⎛ 1 1 ⎞⎛ 0.5 0.5 ⎞ 0.5 ⎫ (10 ) vd = v1 ⎨⎜ + ⎟⎜1 + + − ⎬ ⎟− 1 40 ⎠ (40) 2 ⎭ 40 ⎩⎝ 10 40 ⎠⎝ iI (1.5125) = v1{(0.125)(1.5125) − 0.0003125} − 25vd

or iI (1.5125) = vI {0.18875} − 25vd

Now

vd = iI Ri = iI (20) and v1 = vI − iI (20)

So iI (1.5125) = [vI − iI (20)] ⋅ [0.18875] − 25iI (20) iI [505.3] = vI (0.18875)

or vI = 2677 kΩ iI

Now

Rif = 10 + 2677 ⇒ Rif = 2.687 MΩ

To determine

R0 f :

Using Equation (14.36)

⎡ 1 1 ⎢⎢ A0 L = ⋅ R2 R0′ f R0 ⎢ ⎢1 + R R i 1 ⎣

⎤ ⎡ ⎤ ⎥ ⎢ 103 ⎥ 1 ⎥= ⎥ ⋅⎢ ⎥ 0.5 ⎢1 + 40 ⎥ ⎥ ⎢ 10 20 ⎥ ⎣ ⎦ ⎦

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or

R0′ f = 3.5 Ω

Then

R0 f = 1 kΩ 3.5 Ω

⇒ R0 f = 3.49 Ω

b.

Using Equation (14.16) dACL dA ⎛ 5 ⎞ = (−10) ⎜ 3 ⎟ ⇒ CL = −(0.05)% ACL ACL ⎝ 10 ⎠

______________________________________________________________________________________ 14.14 (a)

(b)

(i)

υ I −υO Ri

υI

+

Ri

=

υ O − AOL (υ I − υ O ) Ro

⎛ 1 AOLυ I A ⎞ 1 = υ O ⎜⎜ + + OL ⎟⎟ Ro Ro ⎠ ⎝ Ri R o

⎛ 1 5 × 10 3 + 1 ⎝ 10

υ I ⎜⎜

⎞ ⎛ 1 1 5 × 10 3 ⎟ = υO ⎜ + + ⎟ ⎜ 10 1 1 ⎠ ⎝

υ I (5.0001×10 3 ) = υ O (5.0011×10 3 ) υO = 0.9998 υI (ii) I x =

⎞ ⎟ ⎟ ⎠

V x − AOL (− V x ) V x + Ro Ri

Ix 1 + AOL 1 1 1 + 5 × 10 3 1 = = + = + V x Rof Ro Ri 1 10 R of ≅ 0.2 Ω

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 14.15

and

vI 1 − v1 vI 2 − v1 v1 − v0 + = 20 10 40 vI 1 vI 2 v0 1 1⎤ ⎡1 + + = v1 ⎢ + + ⎥ 20 10 40 ⎣ 20 10 40 ⎦ v v1 = − 0 A0L v0 = − A0L v1

so that

Then ⎧1 1 ⎛ 7 ⎞⎫ vI 1 (0.05) + vI 2 (0.10) = −v0 ⎨ + ⋅ ⎟⎬ 3 ⎜ ⎩ 40 2 × 10 ⎝ 40 ⎠ ⎭ = −v0 [2.50875 × 10−2 ] ⇒ v0 = −1.993vI 1 − 3.986vI 2 Δv0 2 − 1.993 Δv = ⇒ 0 = 0.35% v0 v0 2

______________________________________________________________________________________ 14.16

⎛ 40 ⎞ ⎛ 4⎞ vB = ⎜ ⎟ v2 = ⎜ ⎟ v2 = 0.8v2 ⎝ 40 + 10 ⎠ ⎝5⎠

(1)

v1 − v A v A − v0 = 10 40 v1 v0 1 ⎞ ⎛1 + = vA ⎜ + ⎟ 10 40 10 40 ⎝ ⎠ v1 (0.1) + v0 (0.025) = v A (0.125)

(2)

v0 = A0 L vd = A0 L (vB − v A )

or

(3)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ v0 = A0 L [0.8v2 − v A ] v0 − 0.8v2 = −v A A0 L ⇒ v A = 0.8v2 −

v0 A0 L

Then ⎡ v ⎤ v1 (0.1) + v0 (0.025) = (0.125) ⎢ 0.8v2 − 0 ⎥ A 0L ⎦ ⎣ 0.125 ⎤ ⎡ v1 (0.1) − v2 (0.1) = −v0 ⎢0.025 + 103 ⎥⎦ ⎣ = −v0 [2.5125 × 10−2 ] ⇒ Ad = ⇒

v0 = 3.9801 v2 − v1

ΔAd 0.0199 = ⇒ 0.4975% Ad 4

______________________________________________________________________________________

14.17 a.

Considering the second op-amp and Equation (14.20), we have ⎡ ⎤ 1 1 1 ⎢1 + 100 ⎥ 101 = + ⋅⎢ ⎥ = 0.10 + (0.1)(11) Rif 2 10 0.1 ⎢ 1 + 1 ⎥ ⎢⎣ 0.1 ⎥⎦ Rif 2 = 0.0109 kΩ

So The effective load on the first op-amp is then RL1 = 0.1 + Rif 2 = 0.1109 kΩ

Again using Equation (14.20), we have 1 1 + 100 + 1 1 1 110.017 0.1109 = + ⋅ = 0.10 + 11.017 Rif 10 1 1 + 1 + 1 0.1109 1

so that Rif = 99.1 Ω R

:

b. To determine 0 f For the first op-amp, we can write, using Equation (14.36) ⎡ 1 1 ⎢ A0 L = ⋅⎢ R2 R0 f 1 R0 ⎢ 1+ ⎢⎣ R1 || Ri R

⎤ ⎡ ⎤ ⎥ 1 ⎢ 100 ⎥ ⎥ = ⋅⎢ ⎥ ⎥ 1 ⎢1 + 40 ⎥ ⎥⎦ ⎢⎣ 1||10 ⎥⎦ = 0.021 kΩ

which yields 0 f 1 For the second op-amp, then

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎡ ⎢ A0 L 1 1 ⎢ = ⋅ R2 R0 f R0 ⎢ ⎢1 + ( R + R ) || R 1 0 f1 i ⎣ ⎡ ⎤ ⎥ 1 ⎢ 100 ⎥ = ⋅⎢ 0.10 1 ⎢1 + ⎥ ⎢⎣ (0.121) ||10 ⎥⎦

or c.

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

R0 f = 18.4 Ω

To find the gain, consider the second op-amp.

v01 − ( −vd 2 ) vd 2 −vd 2 − v02 + = 0.1 Ri 0.1

(1)

v01 v02 1 1 ⎞ ⎛ 1 + vd 2 ⎜ + + ⎟=− 0.1 0.1 10 0.1 0.1 ⎝ ⎠

or v01 (10) + vd 2 (20.1) = −v02 (10)

v02 − A0 L vd 2 v02 − ( −vd 2 ) + =0 R0 0.1 v02 ⎛ 100 1 ⎞ v02 =0 − vd 2 ⎜ − ⎟+ 1 0.1 ⎠ 0.1 ⎝ 1 v02 (11) − vd 2 (90) = 0

or vd 2 = v02 (0.1222)

Then Equation (1) becomes v01 (10) + v02 (0.1222)(20.1) = −v02 (10)

or v01 = −v02 (1.246)

Now consider the first op-amp.

(2)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ vI − ( −vd 1 ) vd 1 −vd 1 − v01 + = 1 Ri 1

(1)

⎛1 1 1⎞ vI (1) + vd 1 ⎜ + + ⎟ = −v01 (1) ⎝ 1 10 1 ⎠

or vI (1) + vd 1 (2.1) = −v01 (1) v01 v − A0 L vd 1 v01 − ( −vd 1 ) + 01 + =0 0.1109 R0 1

(2)

1 1⎞ ⎛ 1 ⎛ 100 1 ⎞ v01 ⎜ + + ⎟ − vd 1 ⎜ − ⎟=0 ⎝ 0.1109 1 1 ⎠ ⎝ 1 1⎠ v01 (11.017) − vd 1 (99) = 0

or vd 1 = v01 (0.1113)

Then Equation (1) becomes vI (1) + v01 (0.1113)(2.1) = −v01 or vI = −v01 (1.234) We had v01 = −v02 (1.246)

So vI = v02 (1.246)(1.234) or

v02 = 0.650 vI v02 =1 vI

d. Ideal So ratio of actual to ideal = 0.650. ______________________________________________________________________________________

14.18

(a)

6 For the op-amp. A0 L ⋅ f 3dB = 10

f 3dB =

106 = 50 Hz 2 × 104

For the closed-loop amplifier. f 3dB =

(b)

106 = 40 kHz 25

Open-loop amplifier. A=

2 × 104 2 × 104 ⇒| A | = 2 f ⎛ f ⎞ 1+ j 1+ ⎜ f 3dB ⎟ ⎝ f 3dB ⎠

f = 0.25 f 3dB ⇒ A = f = 5 f 3− dB ⇒ A =

2 × 104 1 + (0.25) 2

2 × 104 1 + (5)

2

= 1.94 × 104

= 3.92 × 103

Closed-loop amplifier f = 0.25 f 3dB ⇒ A = f = 5 f 3− dB ⇒ A =

25 1 + (0.25) 2 25

1 + (5) 2

= 24.25

= 4.90

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 14.19 Ao = 100 dB, ⇒ Ao = 10 5 A = 38 dB, A = 79.43 10 5 Then 79.43 = 2 ⎛ 10 4 ⎞ ⎟ 1 + ⎜⎜ ⎟ ⎝ f PD ⎠ 10 4 10 5 ≅ ⇒ f PD = 7.94 Hz f PD 79.43

( )

GBW = 10 5 (7.94) = 7.94 ×10 5 Hz ______________________________________________________________________________________ 14.20 ⎛ R ⎞ ⎛ 150 ⎞ (a) ACLO = ⎜⎜1 + 2 ⎟⎟ = ⎜1 + ⎟ = 11 R1 ⎠ ⎝ 15 ⎠ ⎝

f T = 1.2 × 10 6 = (11) f 3− dB ⇒ f 3− dB = 109 kHz

⎡ (150 )(1 ± 0.05) ⎤ (b) ACLO = ⎢1 + (15)(1 ± 0.05) ⎥⎦ ⎣ 157.5 ACLO (max ) = 1 + = 12.05 14.25 142.5 ACLO (min ) = 1 + = 10.05 15.75 Then 10.05 ≤ ACLO ≤ 12.05

f T = 1.2 ×10 6 = (12.05) f 3− dB ⇒ f 3− dB = 99.6 kHz

f T = 1.2 × 10 6 = (10.05) f 3− dB ⇒ f 3− dB = 119.4 kHz Then 99.6 ≤ f 3− dB ≤ 119.4 kHz ______________________________________________________________________________________ 14.21 The open loop gain can be written as A0

A0 L ( f ) =

⎛ f ⎞⎛ f ⎞ ⎜1 + j ⋅ ⎟ ⎜1 + j ⋅ ⎟ × 5 106 ⎠ f ⎝ ⎝ PD ⎠ 5 where A0 = 2 × 10 .

The closed-loop response is ACL =

A0 L 1 + β A0 L

At low frequency, 100 =

2 × 105 1 + β (2 × 105 )

So that β = 9.995 × 10 . Assuming the second pole is the same for both the open-loop and closed-loop, then −3

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ f ⎞ f ⎞ −1 ⎛ ⎟ − tan ⎜ 6 ⎟ ⎝ 5 × 10 ⎠ ⎝ f PD ⎠ For a phase margin of 80°, φ = −100°.

φ = − tan −1 ⎜

So ⎛ f ⎞ −100 = −90 − tan −1 ⎜ 6 ⎟ ⎝ 5 × 10 ⎠

or f = 8.816 × 105 Hz

Then A0 L = 1 2 × 105

=

⎛ 8.816 × 105 ⎞ 1+ ⎜ ⎟ f PD ⎝ ⎠

2

⎛ 8.816 × 105 ⎞ 1+ ⎜ ⎟ 6 ⎝ 5 × 10 ⎠

2

or 8.816 × 105 ≅ 1.9696 × 105 f PD

or f PD = 4.48 Hz

______________________________________________________________________________________ 14.22 (a) nd

1st stage (10) f 3− dB = 1 MHz ⇒ f 3− dB = 100 kHz

2 stage (50) f 3− dB = 1 MHz ⇒ f 3− dB = 20 kHz

Bandwidth of overall system ≅ 20 kHz

(b)

If each stage has the same gain, so K 2 = 500 ⇒ K = 22.36

Then bandwidth of each stage (22.36) f 3− dB = 1 MHz ⇒ f 3− dB = 44.7 kHz

______________________________________________________________________________________ 14.23

(a) ACLO =

−

R2

R1 − 10.0 = = −9.9978 ⎛1 + R 2 ⎞ 1 + 11 ⎜ ⎟ R1 ⎠ ⎝ 5 × 10 4 1+ AO

f T = 1.5 ×10 6 = (9.9978) f 3− dB ⇒ f 3− dB = 150.033 kHz

(b) ACLO = (− 9.9978) = −999.34 3

At f 3− dB ; ⇒ ACL =

999.34 2

= 706.64

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 999.34 Then 706.64 = 3 2 ⎤ ⎡ f ⎞ ⎛ −dB 3 ⎢ 1+ ⎜ ⎟ ⎥ ⎜ 3 ⎟ ⎢ ⎝ 150.033 × 10 ⎠ ⎥ ⎣ ⎦ 3

2 2 ⎡ ⎛ f 3− dB ⎞ ⎤ ⎛ 999.34 ⎞ ⎢1 + ⎜⎜ ⎟ ⎥ = ⇒ f 3− dB = 76.49 kHz ⎜ ⎟ 3 ⎟ ⎝ 706.64 ⎠ ⎢⎣ ⎝ 150.033 × 10 ⎠ ⎥⎦ ______________________________________________________________________________________

14.24 (5 × 10 4 ) f PD = 106 ⇒ f PD = 20 Hz (25) f 3− dB 106 ⇒ f 3− dB = 40 kHz Av =

Avo 1+ j

f

⇒ Av =

f 3− dB

25 f ⎛ ⎞ 1+ ⎜ 3 ⎟ ⎝ 40 × 10 ⎠

2

At f = 0.5 f 3− dB = 20 kHz Av =

25 1 + (0.5) 2

= 22.36

At f = 2 f 3− dB = 80 kHz Av =

25 1 + (2)2

= 11.18

______________________________________________________________________________________ 14.25 (20 × 103 ) ⋅ Avf

MAX

= 106 ⇒ Avf

MAX

= 50

______________________________________________________________________________________

14.26

(a)

f max =

(b) f max =

SR 5 × 10 6 = ⇒ f max = 159 kHz 2πV PO 2π (5)

5 × 10 6 ⇒ f max = 530.5 kHz 2π (1.5)

5 × 10 6 ⇒ f max = 1.99 MHz 2π (0.4 ) ______________________________________________________________________________________

(c)

14.27 a.

f max =

Using Equation (14.55), VP 0 =

8 × 106 2π (250 × 103 )

or VP 0 = 5.09 V

b.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

T=

1 1 = = 4 × 10−6 s f 250 × 103

Period One-fourth period = 1 μ s Slope =

VP 0 = SR = 8 V/μ s 1μ s ⇒ VP 0 = 8 V

______________________________________________________________________________________ 14.28 f max =

SR 2πV PO

(

)

SR = 2π (10) 12 ×10 3 = 7.54 ×10 5 V/s Or SR = 0.754 V/ μ s ______________________________________________________________________________________ 14.29

(a)

f max = 20 × 10 3 =

0.63 × 10 6 ⇒ V PO = 5.0 V 2πV PO

3 × 10 6 = 23.87 V 2π 20 × 10 3 ______________________________________________________________________________________

(b) V PO =

(

)

14.30 For input (a), maximum output is 5 V. S R = 1 V/μs

so

For input (b), maximum output is 2 V.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

For input (c), maximum output is 0.5 V so the output is

______________________________________________________________________________________ 14.31

For input (a),

Then

v02

max

max v01 = 3 V.

= 3(3) = 9 V

For input (b),

max v01 = 1.5 V.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Then

v02

max

= 3 (1.5) = 4.5V

______________________________________________________________________________________ 14.32 ⎛V ⎞ ⎛V ⎞ I1 = I S1 exp ⎜ BE1 ⎟ , I 2 = I S 2 exp ⎜ BE 2 ⎟ V ⎝ T ⎠ ⎝ VT ⎠ Want I1 = I 2 , so I1 =1= I2

=

⎛V ⎞ 5 × 10−14 (1 + x) exp ⎜ BE1 ⎟ ⎝ VT ⎠ ⎛ V ⎞ 5 × 10−14 (1 − x) exp ⎜ BE 2 ⎟ ⎝ VT ⎠

⎛ V −V ⎞ (1 + x) exp ⎜ BE1 BE 2 ⎟ VT (1 − x) ⎝ ⎠

Or ⎛ V − VBE1 ⎞ ⎛ VOS ⎞ 1+ x = exp ⎜ BE 2 ⎟ ⎟ = exp ⎜ 1− x V T ⎝ ⎠ ⎝ VT ⎠ ⎛ 0.0025 ⎞ = exp ⎜ ⎟ = 1.10 ⎝ 0.026 ⎠

Now 1 + x = (1 − x)(1.10) ⇒ x = 0.0476 ⇒ 4.76%

______________________________________________________________________________________ 14.33 (a) Balanced circuit, I S 4 = 5 ×10 −15 A (b) From Eq. (14.62), υ CE1 = 5 V, υ CE 2 = 5.6 − 1.2 = 4.4 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5 120 = I S 3 0.6 I S 4 1+ 80

1+

4 .4 ⎞ ⎛ ⎟ ⎜1 + ⎝ 120 ⎠ ⋅ ⎛ 1 .2 ⎞ ⎟ ⎜1 + ⎝ 80 ⎠

1.041667 I S 3 (1.036667 ) = ⋅ (1.015) 1.0075 I S4 I S3 = 1.0123 ⇒ I S 4 = 4.939 × 10 −15 A I S4

(c) υ CE1 = 5 V, υ CE 2 = 5.6 − 2.5 = 3.1 V

5 120 = I S 3 0.6 I S 4 1+ 80

1+

3 .1 ⎞ ⎛ ⎜1 + ⎟ 120 ⎠ ⎝ ⋅ ⎛ 2.5 ⎞ ⎜1 + ⎟ 80 ⎠ ⎝

1.041667 I S 3 (1.025833) = ⋅ 1.0075 I S 4 (1.03125) I S3 = 1.03937 ⇒ I S 4 = 4.811× 10 −15 A I S4 ______________________________________________________________________________________

14.34 K n = 150 μ A/V 2

ΔK n = 150(1 + x ) − 150(1 − x ) = 300 x μ A/V 2

VOS =

1 2

I Q ⎛ ΔK n ⎜ 2 K n ⎜⎝ K n

⎞ ⎟ ⎟ ⎠

1 200 ⎛ 300 x ⎞ ⎜ ⎟ = 0.8165x ⇒ x = 0.01837 2 2(150) ⎝ 150 ⎠ ______________________________________________________________________________________ 15 ×10 −3 =

14.35 (a) υ O = −30(10 ± 2)× 10 −3 = (− 300 ± 60)×10 −3 V So −0.360 ≤ υ O ≤ −0.240 V

(b) υ O = −30(100 ± 2)×10 −3 = −3 ± 0.06 V So −3.06 ≤ υ O ≤ −2.94 V ______________________________________________________________________________________

14.36

υ O = −30(25 sin ω t ± 2) mV υ O = −0.75 sin ω t ± 0.06 V So (− 0.75 sin ω t − 0.06 ) ≤ υ O ≤ (− 0.75 sin ω t + 0.06 ) V ______________________________________________________________________________________ 14.37

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

I=

0.5 × 10 −3 = 5 × 10−8 A 104

Also I =C

i dVo I 1 ⇒ Vo = ∫ Idt = ⋅ t dt C0 C

Then 5=

5 × 10−8 t ⇒ t = 1 03 s 10 × 10−6

______________________________________________________________________________________ 14.38 ⎛ 100 ⎞ (a) υ O1 = ⎜1 + ⎟(± 3) mV, −33 ≤ υ O1 ≤ 33 mV 10 ⎠ ⎝ ⎛ 50 ⎞ ⎟(± 33 ± 3) mV, −180 ≤ υ O 2 ≤ 180 mV ⎝ 10 ⎠

υ O 2 = −⎜

(b) υ O1 = (11)(10 ± 3) mV, ⇒ 77 ≤ υ O1 ≤ 143 mV υ O 2 = −5(143 + 3) = −730 mV υ O 2 = −5(77 − 3) = −370 mV So −0.73 ≤ υ O 2 ≤ −0.37 V (c) υ O1 = (11)(100 ± 3) mV 1.067 ≤ υ O1 ≤ 1.133 V υ O 2 = −5(1.133 + 0.003) = −5.68 V υ O 2 = −5(1.067 − 0.003) = −5.32 V So −5.68 ≤ υ O 2 ≤ −5.32 V ______________________________________________________________________________________

14.39 v0 due to vI 1 ⎞ ⎛ v0 = (0.5) ⎜ 1 + ⎟ = 0.9545 V ⎝ 1.1 ⎠ + Wiper arm at V = 10 V, (using superposition)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ R1 || R5 ⎞ ⎛ 0.0909 ⎞ v1 = ⎜ ⎟ (10) = ⎜ ⎟ (10) + R || R R ⎝ 0.0909 + 10 ⎠ 4 ⎠ ⎝ 1 5 = 0.090 ⎛1⎞ v01 = − ⎜ ⎟ (0.090) = −0.090 ⎝1⎠ Then Wiper arm in center, v1 = 0 and v02 = 0

− Wiper arm at V = −10 V, v1 = −0.090 So

v03 = 0.090

Finally, total output v0 : (from superposition) + Wiper arm at V ,

v0 = 0.8645 V

Wiper arm in center, v0 = 0.9545 V − Wiper arm at V ,

v0 = 1.0445 V

______________________________________________________________________________________ 14.40

a. or

R1′ = R2′ = 0.5 || 25 = 0.490 kΩ

R1′ = R2′ = 490 Ω

b.

From Equation (14.75), ⎛ 125 × 10−6 ⎞ + (0.125) R1′ (0.026) ln ⎜ −14 ⎟ ⎝ 2 × 10 ⎠ ⎛ 125 × 10−6 ⎞ = (0.026) ln ⎜ + (0.125) R2′ −14 ⎟ ⎝ 2.2 × 10 ⎠ 0.586452 + (0.125) R1′ = 0.583974 + (0.125) R2′ 0.002478 = (0.125)( R2′ − R1′)

′

′

So R2 − R1 = 0.0198 kΩ ⇒ 19.8 Ω Then R2 (1 − x) Rx R × Rx − 1 = 0.0198 R2 + (1 − x) Rx R1 + xRx (0.5)(1 − x)(50) (0.5)(50) x − = 0.0198 (0.5) + (1 − x)(50) (0.5) + x(50) 25(1 − x) 25 x − = 0.0198 50.5 − 50 x 0.5 + 50 x (0.5 + 50 x)(25 − 25 x) − (25 x)(50.5 − 50 x) = 0.0198 (50.5 − 50 x)(0.5 + 50 x)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 25 {0.5 − 0.5 x + 50 x − 50 x 2 − 50.5 x + 50 x 2 } = 0.0198{25.25 + 2525 x − 25 x − 2500 x 2 } 25 {0.5 − x} = 0.0198 {25.25 + 2500 x − 2500 x 2 } 0.5 − x = 0.019998 + 1.98 x − 1.98 x 2 1.98 x 2 − 2.98 x + 0.48 = 0 x=

2.98 ± (2.98) 2 − 4(1.98)(0.48) 2(1.98)

So x = 0.183

and 1 − x = 0.817

______________________________________________________________________________________ 14.41

R1′ = R1 || 15 = 0.5 || 15 = 0.4839 kΩ R2′ = R2 || 35 = 0.5 || 35 = 0.4930 kΩ

From Equation (14.75), ⎛i ⎞ ⎛i ⎞ (0.026) ln ⎜ C1 ⎟ + iC1 R1′ = (0.026) ln ⎜ C 2 ⎟ + iC 2 R2′ I ⎝ S3 ⎠ ⎝ IS 4 ⎠ ⎛i ⎞ (0.026) ln ⎜ C1 ⎟ = iC 2 R2′ − iC1 R1′ ⎝ iC 2 ⎠ ⎛i ⎞ ⎡ i R′ ⎤ (0.026) ln ⎜ C1 ⎟ = iC 2 R2′ ⎢1 − C1 ⋅ 1 ⎥ ′ i i R 2 ⎦ C2 ⎝ C2 ⎠ ⎣ ⎡ ⎛i ⎞ ⎛ i ⎞⎤ (0.026) ln ⎜ C1 ⎟ = iC 2 (0.4930) ⎢1 − (0.9815) ⎜ C1 ⎟ ⎥ i ⎝ C2 ⎠ ⎝ iC 2 ⎠ ⎦ ⎣

By trial and error: iC1 = 252 μ A i = 248 μ A and C 2 or iC1 = 1.0155 iC 2

______________________________________________________________________________________ 14.42 (a) υ O ( 1 μA ) = 10 −6 200 × 10 3 = 0.2 V Insert resistor R3

(

υO (b) υ O

υO

( 2 μA )

)(

)

(

)

⎛ 200 ⎞ = −0.2 = − 2 × 10 − 6 R3 ⎜1 + ⎟ ⇒ R 3 = 9.09 k Ω 20 ⎠ ⎝

)( ) ⎛ 200 ⎞ ⎟⇒ R ) = −0.16 = − (0.5 × 10 )R ⎜1 + 20

( 0 . 8 μA )

(

= 0.8 × 10 −6 200 × 10 3 = 0.16 V

3 = 29.09 k Ω ⎠ ⎝ ______________________________________________________________________________________

( 0.5 μA

−6

3

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 14.43 (a) υ O = − I B R 2 = − 2 ×10 −6 150 × 10 3 = −0.3 V

(

)(

)

150 (0.02 ) − 0.3 = −0.5 V 15 150 (c) υ O = − (− 0.02) − 0.3 = −0.1 V 15 150 (d) υ O = − (0.1) − 0.3 = −1.3 V 15 ______________________________________________________________________________________

(b) υ O = −

14.44 (a) υ O = 0.6 ×10 −6 250 ×10 3 = 0.15 V (b) υ O = (41)(0.008) + 0.15 = 0.478 V (c) υ O = (41)(− 0.0035) + 0.15 = 0.0065 V (d) υ O = (41)(0.005 sin ω t ) + 0.15 = 0.205 sin ω t + 0.15 (V) ______________________________________________________________________________________

(

)(

)

14.45 a.

v = − (10−6 )(104 ) For I B 2 = 1 μ A, then 0

or b.

v0 = −0.010 V

If a 10 kΩ resistor is included in the feedback loop

Now v0 = − I B 2 (10) + I B1 (10) = 0

Circuit is compensated if I B1 = I B 2 . ______________________________________________________________________________________

14.46 From Equation (14.83), we have

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ v0 = R2 I 0S where R2 = 40 kΩ and I 0 S = 3 μ A.

Then

v0 = ( 40 × 103 )( 3 × 10−6 )

or v0 = 0.12 V

______________________________________________________________________________________ 14.47 a.

Assume all bias currents are in the same direction and into each op-amp.

v01 = I B1 (100 kΩ ) = (10−6 )(105 ) ⇒ v01 = 0.1 V

Then

v02 = v01 ( −5 ) + I B1 ( 50 kΩ )

= ( 0.1)( −5 ) + (10−6 )( 5 × 10 4 ) = −0.5 + 0.05

or v02 = −0.45 V

b.

Connect R3 = 10 ||100 = 9.09 kΩ resistor to noninverting terminal of first op-amp, and R3 = 10 || 50 = 8.33 kΩ resistor to noninverting terminal of second op-amp.

______________________________________________________________________________________ 14.48 a.

For a constant current through a capacitor.

1 t I dt C ∫0 0.1× 10−6 v0 = ⋅ t ⇒ v0 = (0.1)t 10 −6 or v0 = 1 V b. At t = 10 s, v0 =

c.

Then 100 × 10−12 ⋅ t ⇒ v0 = (10 −4 )t 10−6 v0 = 1 mV t = 10 s, v0 =

At ______________________________________________________________________________________ 14.49 (a) υ O1 = 3 ×10 −6 50 × 10 3 = 0.15 V υ O 2 = 0.15 V

(

)(

)

(

)(

)

20 (0.15) + 3 ×10 − 6 20 ×10 3 = −0.09 V 20 (b) R A = 10 50 = 8.33 k Ω

υ O3 = −

R B = 20 20 = 10 k Ω

(

)(

)

(c) υ O1 = ± 50 × 10 3 0.3 × 10 −6 = ±0.015 V υ O 2 = ±0.015 V

υ O 3 = ±(20 ×10 3 )(0.3 ×10 −6 ) ± 0.015 = ±0.021 V

______________________________________________________________________________________ 14.50

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ a. Using Equation (14.79), Circuit (a), ⎛ 50 ⎞ v0 = ( 0.8 × 10 −6 )( 50 × 103 ) − ( 0.8 × 10−6 )( 25 × 103 ) ⎜ 1 + ⎟ ⎝ 50 ⎠

or v0 = 0

Circuit (b), ⎛ 50 ⎞ v0 = ( 0.8 × 10−6 )( 50 × 103 ) − ( 0.8 × 10 −6 )(103 ) ⎜1 + ⎟ ⎝ 50 ⎠ −2 = 4 × 10 − 1.6

or v0 = −1.56 V

b. Assume I B1 = 0.7 μ A and I B 2 = 0.9 μ A, then using Equation (14.79): Circuit (a), ⎛ 50 ⎞ v0 = ( 0.7 × 10 −6 )( 50 × 103 ) − ( 0.9 × 10−6 )( 25 × 103 ) ⎜1 + ⎟ ⎝ 50 ⎠ = 0.035 − 0.045

v = −0.010 V

0 or Circuit (b),

⎛ 50 ⎞ v0 = ( 0.7 × 10 −6 )( 50 × 103 ) − ( 0.9 × 10 −6 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ = 0.035 − 1.8

v = −1.765 V

0 or ______________________________________________________________________________________

14.51 ⎛ 100 ⎞ (a) For V OS : υ O = ⎜1 + ⎟(± 3) = ±33 mV 10 ⎠ ⎝

( )( ) (max) = (0.37 ×10 )(100 ×10 ) = 0.037 V

For I B : υ O (max ) = 0.43 ×10 −6 100 ×10 3 = 0.043 V

υO

−6

So 4 ≤ υ O ≤ 76 mV (b) For V OS : υ O = ±33 mV

(

)(

3

)

For I OS : υ O = ± 0.06 × 10 −6 100 ×10 3 = ±0.006 V So −39 ≤ υ O ≤ 39 mV ⎛ 100 ⎞ (c) υ O = ⎜1 + ⎟(0.2 ) ± 0.039 10 ⎠ ⎝ So 2.161 ≤ υ O ≤ 2.239 V ______________________________________________________________________________________

14.52

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ Ri ⎞ ⎜ ⎟ (15) = 0.010 V ⎝ Ri + R2 ⎠

a.

15 = 0.0006667 15 + R2 15(1 − 0.0006667) = 0.0006667 R2

Then R2 = 22.48 MΩ R = R || R = 15 ||10 ⇒ R = 6 kΩ

1 i F 1 b. ______________________________________________________________________________________

14.53 a.

Assume the offset voltage polarities are such as to produce the worst case values, but the bias currents are in the same direction. Use superposition: Offset voltages ⎛ 100 ⎞ | v01 | = ⎜ 1 + ⎟ (10) = 110 mV =| v01 | 10 ⎠ ⎝ ⎛ 50 ⎞ | v02 | = (5)(110) + ⎜1 + ⎟ (10) ⎝ 10 ⎠ ⇒ | v02 | = 610 mV

Bias Currents: v01 = I B (100 kΩ) = (2 × 10 −6 )(100 × 103 ) = 0.2 V

Then v02 = (−5)(0.2) + (2 × 10 −6 )(50 × 103 ) = −0.9 V v01 v02

Worst case:

v01 = 0.31 V

b.

is positive and

v02 = −1.51 V

is negative, then

and Compensation network:

If we want ⎛ RB ⎞ + + ⎜ ⎟ V = 20 mV and V = 10 V R R + C ⎠ ⎝ B ⎛ 8.33 ⎞ ⎜ ⎟ (10) = 0.020 ⎝ 8.33 + RC ⎠

R ≅ 4.15 MΩ

C or ______________________________________________________________________________________

14.54

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) Offset voltage: ⎛ 50 ⎞ υ O1 = ⎜1 + ⎟(± 2 ) = ±12 mV ⎝ 10 ⎠ υ O 2 = ±12 ± 2 = ±14 mV ⎛ 20 ⎞ ⎟(± 12 ) + (2 )(± 2 ) = ±16 mV ⎝ 20 ⎠ Bias current: υ O1 = 0.21×10 −6 50 ×10 3 = 0.0105 V

υ O3 = ⎜ −

or υ O1

( )( ) = (0.19 ×10 )(50 ×10 ) = 0.0095 V −6

3

υ O 2 = υ O1

υ O 3 = (− 1)(υ O1 ) + (0.21×10 −6 )(20 ×10 3 ) = −υ O1 + 0.0042

(

)(

)

or υ O3 = −υ O1 + 0.19 × 10 −6 20 × 10 3 = −υ O1 + 0.0038 By superposition −2.5 ≤ υ O1 ≤ 22.5 mV −4.5 ≤ υ O 2 ≤ 24.5 mV −22.3 ≤ υ O 3 ≤ 10.7 mV (b) Bias currents: υ O1 = ± I OS 50 ×10 3 = ± 0.02 ×10 −6 50 ×10 3 ⇒ υ O1 = ±1 mV

υ O 3 = ± I OS

( ) ( )( ) (20 ×10 ) = ±(0.02 ×10 )(20 ×10 ) ⇒ υ 3

−6

3

υ O 3 = ±υ O1 ± 2(2 ) ± 0.4

O3

= ±0.4 mV

By superposition: −13 ≤ υ O1 ≤ 13 mV −15 ≤ υ O 2 ≤ 15 mV −17.4 ≤ υ O 3 ≤ 17.4 mV ______________________________________________________________________________________ 14.55 For circuit (a), effect of bias current: v0 = (50 × 103 )(100 × 10 −9 ) ⇒ 5 mV

Effect of offset voltage ⎛ 50 ⎞ v0 = (2) ⎜ 1 + ⎟ = 4 mV ⎝ 50 ⎠ v0 = 9 mV

So net output voltage is For circuit (b), effect of bias current:

Let I B 2 = 550 nA, I B1 = 450 nA, then from Equation (14.79), ⎛ 50 ⎞ v0 = (450 × 10−9 )(50 × 103 ) − (550 × 10−9 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ −2 = 2.25 × 10 − 1.1

or v0 = −1.0775 V

If the offset voltage is negative, then v0 = ( −2)(2) = −4 mV

So the net output voltage is v0 = −1.0815 V

_____________________________________________________________________________________ 14.56

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ At T = 25°C, V0 S = 2 mV so the output voltage for each circuit is

a.

v0 = 4 mV

For T = 50°C, the offset voltage for is

b.

V0 S = 2 mV + (0.0067)(25) = 2.1675 mV

so the output voltage for each circuit is v0 = 4.335 mV

______________________________________________________________________________________ 14.57

At T = 25°C,

a.

V0 S = 1 mV, then

⎛ 50 ⎞ v01 = (1) ⎜ 1 + ⎟ ⇒ v01 = 6 mV ⎝ 10 ⎠

and ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜1 + ⎟ + (1) ⎜ 1 + ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = 6(4) + (1)(4) ⇒ v02 = 28 mV

b.

At T = 50°C, V0 S = 1 + (0.0033)(25) = 1.0825 mV, then v01 = (1.0825)(6) ⇒ v01 = 6.495 mV

and v02 = (6.495)(4) + (1.0825)(4)

or v02 = 30.31 mV

______________________________________________________________________________________ 14.58

a.

25°C; I B = 500 nA, I 0 S = 200 nA 50°C, I B = 500 nA + (8 nA / °C)(25°C) = 700 nA I 0 S = 200 nA + (2 nA / °C)(25°C) = 250 nA Circuit (a): For I B , bias current cancellation, v0 = 0

Circuit (b): For I B , Equation (14.79), ⎛ 50 ⎞ v0 = (500 × 10 −9 )(50 × 103 ) − (500 × 10−9 )(106 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.025 − 1.00 ⇒ v0 = −0.975 V

b. Due to offset bias currents. Circuit (a): v0 = (200 × 10−9 )(50 × 103 ) ⇒ v0 = 0.010 V

Circuit (b): Let I B 2 = 600 nA I B1 = 400 nA

Then ⎛ 50 ⎞ v0 = (400 × 10−9 )(50 × 103 ) − (600 × 10−9 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ = 0.020 − 1.20 ⇒ v0 = −1.18 V

c.

Circuit (a): Due to I B , v = 0

Circuit (b): Due to I B ,

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ 50 ⎞ v0 = (700 × 10−9 )(50 × 103 ) − (700 × 10 −9 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ = 0.035 − 1.40 ⇒ v0 = −1.365 V

Circuit (a): Due to

I0S ,

v0 = (250 × 10−9 )(50 × 103 ) ⇒ v0 = 0.0125 V

Circuit (b): Due to I 0 S , Let I B 2 = 825 nA I B1 = 575 nA

Then ⎛ 50 ⎞ v0 = (575 × 10−9 )(50 × 103 ) − (825 × 10−9 )(106 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.02875 − 1.65 ⇒ v0 = −1.62 V

______________________________________________________________________________________ 14.59

a.

25°C; I B = 2 μ A, I 0 S = 0.2 μ A 50°C, I B = 2 μ A + (0.020 μ A / °C)(25°C ) = 2.5 μ A I 0 S = 0.2 μ A + (0.005 μ A / °C)(25°C) = 0.325 μ A Due to I B : (Assume bias currents into op-amp).

v01 = I B (50 kΩ) = (2 × 10 −6 )(50 × 103 ) ⇒ v01 = 0.10 V ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜ 1 + ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = (0.1)(4) + (2 × 10−6 )(60 × 103 ) − (2 × 10 −6 )(60 × 103 )4

or b.

v02 = 0.12 V

Due to I 0 S :

1st op-amp. Let I B1 = 2.1 μ A 2nd op-amp. Let I B1 = 2.1 μ A I B 2 = 1.9 μ A

v01 = I B1 (50 kΩ) = (2.1× 10 −6 )(50 × 103 ) ⇒ v01 = 0.105 V ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜1 + ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = (0.105)(4) + (2.1× 10−6 )(60 × 103 ) − (1.9 × 10−6 )(50 × 103 )(4)

or v02 = 0.166 V

c.

Due to I B : v01 = (2.5 × 10−6 )(50 × 103 ) ⇒ v01 = 0.125 V

or

⎛ 60 ⎞ ⎛ 60 ⎞ v01 = v02 ⎜1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜ 1 + ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = (0.125)(4) + (2.5 × 10−6 )(60 × 103 ) − (2.5 × 106 )(50 × 103 (4) v02 = 0.15 V

Due to I 0 S :

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Let I B1 = 2.625 μ A I B 2 = 2.3375 μ A

v01 = I B1 (50 kΩ) = (2.6625 × 10−6 )(50 × 103 ) ⇒ v01 = 1.133 V ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜1 + ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = (0.133)(4) + (2.6625 × 10−6 )(60 × 103 ) − (2.3375 × 10−6 )(50 × 103 )(4)

or v02 = 0.224 V

______________________________________________________________________________________ 14.60 50 = 5.0 10 For common-mode, υ I 1 = υ I 2 From Chapter 9, ⎛ R2 ⎞ ⎜⎜1 + ⎟ R1 ⎟⎠ R 2 ⎝ Acm = − ⎛ R3 ⎞ R1 ⎜⎜1 + ⎟⎟ ⎝ R4 ⎠ If R 2 = 50(1.015 ) = 50.75 , R1 = 10(1 − 0.015 ) = 9.85

(a) Ad =

R3 = 10(1 − 0.015) = 9.85 , R 4 = 50(1.015 ) = 50.75

50.75 9.85 − 50.75 = 6.15228 − 5.15228 = 5.046 × 10 − 6 Then Acm = 9.85 9.85 1.19409 1+ 50.75 If R3 = 10(1.015) = 10.15 , R 4 = 50(1 − 0.015) = 49.25 1+

50.75 9.85 − 50.75 = 6.15228 − 5.15228 = −0.051268 Acm = Then 10.15 9.85 1.20609 1+ 49.25 If R 2 = 49.25 , R1 = 10.15 1+

49.25 10 .15 − 49.25 = 5.85222 − 4.85222 = +0.04877 Acm = Then 9.85 10.15 1.19409 1+ 50.75 5 ⎛ ⎞ Now CMRR dB (min ) = 20 log 10 ⎜ ⎟ = 39.8 dB ⎝ 0.051268 ⎠ (b) R 2 = 50(1.03) = 51.5 , R1 = 10(0.97 ) = 9.70 R 4 = 50(0.97 ) = 48.5 , R3 = 10(1.03) = 10.3 1+

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 51.5 1+ 9.70 − 51.5 = 6.30928 − 5.30928 = −0.10519 Acm = 10.3 9.70 1.21237 1+ 48.5 5 ⎛ ⎞ CMRR dB = 20 log 10 ⎜ ⎟ = 33.5 dB 0 . 10519 ⎝ ⎠ ______________________________________________________________________________________ 14.61 (a) CMRR dB = 50 dB ⇒ CMRR = 316.2 5 = 0.01581 316.2 From Problem 14.60, 50(1 + x ) 1+ 10(1 − x ) 50(1 + x ) − Acm = −0.01581 = 10(1 + x ) 10(1 − x ) 1+ 50(1 − x ) Acm =

1 ≅ 1+ x 1− x 1 + 5(1 + x )(1 + x ) − 5(1 + x )(1 + x ) Then − 0.01581 = 1 + 0.2(1 + x )(1 + x )

x is small, so that

Neglect x 2 ,

1 + 5(1 + 2 x ) − 5(1 + 2 x ) 1 + 0.2(1 + 2 x ) (− 0.01581)[1 + 0.2(1 + 2 x )] = 1 + 5(1 + 2 x ) − 5(1 + 2 x )[1 + 0.2(1 + 2 x )] − 0.01581 ≅

We find (1 + 2 x ) − 0.003162(1 + 2 x ) − 1.01581 = 0 ⇒ (1 + 2 x ) = 1.009456 Then x = 0.004728 ⇒ x = 0.4728% (b) CMRR dB = 75 dB, ⇒ CMRR = 5623.4 2

Acm =

5 = 0.00088914 5623.4

Then (− 0.00088914)[1 + 0.2(1 + 2 x )] = 1 − (1 + 2 x )

2

(1 + 2 x )2 − 0.000177828(1 + 2 x ) − 1.00088914 = 0 ⇒ (1 + 2 x ) = 1.0005334 ⇒ x = 0.0267% ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 15 15.1 (a) Noninverting amplifier R R 8 = 1 + 2 ⇒ 2 = 7 ⇒ R 2 = 210 k Ω , R1 = 30 k Ω R1 R1 At noninverting terminal 1 1 = = 5.305 × 10 − 6 RC = 2π f 2π 30 × 10 3 Let input C = 0.001 μ F, then R = 5.305 k Ω (b) Set R1 = 15 k Ω , R 2 = 300 k Ω for inverting amplifier

(

)

1 1 = = 7.958 × 10 − 6 2π f 2π 20 × 10 3 Put C in series with R1

R1C =

(

)

7.958 × 10 −6 ⇒ C = 530.5 pF 15 × 10 3 ______________________________________________________________________________________ C=

15.2 (a) T = (b) T = (c) T =

1 1 + (1.5)

= 0.4061 ⇒ −7.83 dB 4

1 1 + (1.5)

= 0.2841 ⇒ −10.93 dB 6

1 1 + (1.5)

= 0.1938 ⇒ −14.25 dB

8

1

(d) T =

= 0.1306 ⇒ −17.68 dB 10 1 + (1.5) ______________________________________________________________________________________

15.3 T = −6 dB, ⇒ T = 0.50 1

0.50 =

=

⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ 0.9 ⎠

or (1.111)

2N

1 1 + (1.111)

2N

2

2N

⎛ 1 ⎞ =⎜ ⎟ −1 = 3 ⎝ 0.5 ⎠

For N=6, (1.111) = 3.54 , ⇒ 6-pole filter ______________________________________________________________________________________ (2 )(6 )

15.4 (a) From Figure 15.8(a) 1 1 RC = = = 6.366 × 10 − 6 2π f 3− dB 2π 25 × 10 3

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Let C = 0.001 μ F, then R = 6.366 k Ω And R3 = (0.707 )R = 4.50 k Ω R 4 = (1.414 )R = 9.0 k Ω

(b) (i) T =

1 ⎛ 25 ⎞ 1+ ⎜ ⎟ ⎝ 22 ⎠

(ii) T =

1 ⎛ 25 ⎞ 1+ ⎜ ⎟ ⎝ 25 ⎠

(iii) T =

= 0.6123 ⇒ −4.26 dB 4

= 0.707 ⇒ −3 dB 4

1

= 0.7819 ⇒ −2.14 dB 4 ⎛ 25 ⎞ 1+ ⎜ ⎟ ⎝ 28 ⎠ ______________________________________________________________________________________

15.5 1 1 = = 7.958 × 10 − 6 2π f 3− dB 2π 20 × 10 3 Let R = 20 k Ω , then C = 397.9 pF And C1 = (3.546 )C = 1411 pF

(a) RC =

(

)

C 2 = (1.392 )C = 553.9 pF

C 3 = (0.2024 )C = 80.5 pF

(b) (i) T =

1 ⎛ 10 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠

(ii) T =

1 ⎛ 15 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠

(iii) T =

= 0.7071 ⇒ −3.0 dB 6

1 ⎛ 25 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠

(v) T =

= 0.9214 ⇒ −0.711 dB 6

1 ⎛ 20 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠

(iv) T =

= 0.9923 ⇒ −0.0673 dB 6

1

= 0.4557 ⇒ −6.83 dB 6

= 0.2841 ⇒ −10.9 dB 6 ⎛ 30 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.6 From Equation (15.7). T (s) =

Y1Y2 Y1Y2 + Y4 (Y1 + Y2 + Y3 ) Y1 = Y2 = sC ,

For a high-pass filter, let Y3 =

1 , R3

Y4 =

and

1 R4

Then s 2C 2 1 ⎛ 1 ⎞ s 2 C 2 + ⎜ sC + sC + ⎟ R4 ⎝ R3 ⎠ 1 = 1 ⎛ 1 ⎞ 1+ ⎜2+ ⎟ sR4 C ⎝ sR3C ⎠

T (s) =

Define τ 3 = R3 C and τ 4 = R 4 C T (s ) =

Set s = jω

1 1 ⎛ 1 ⎞ ⎜2 + ⎟ 1+ ⎜ sτ 4 ⎝ sτ 3 ⎟⎠

T ( jω ) =

1 ⎛ 1 1 ⎜2 + 1+ jωτ 4 ⎜⎝ jωτ 3 1 = j ⎛ j ⎞ ⎟ ⎜2 − 1− ⎜ ωτ 4 ⎝ ωτ 3 ⎟⎠ 1 = ⎛ 1 ⎞⎟ 2 j ⎜1 − ⎜ ω 2τ τ ⎟ − ωτ 4 3 4 ⎠ ⎝

⎧⎛ 1 ⎪ T ( jω ) = ⎨⎜⎜1 − 2 ⎪⎩⎝ ω τ 3τ 4

⎞ ⎟ ⎟ ⎠

2 ⎫ ⎞ ⎟ + 4 ⎪⎬ ⎟ ω 2τ 42 ⎪ ⎠ ⎭

−1 / 2

For a maximally flat filter, we want dT =0 dω ω →∞ Taking the derivative, we find d T ( jω ) dω

or

1 1 ⎧⎪⎛ = − ⎨⎜⎜1 − 2 2 ⎪⎝ ω τ 3τ 4 ⎩

2 ⎫ ⎞ ⎟ + 4 ⎪⎬ ⎟ ω 2τ 42 ⎪ ⎠ ⎭

−3 / 2

⎡ ⎛ 1 × ⎢2⎜⎜1 − 2 ⎢⎣ ⎝ ω τ 3τ 4

⎞⎛ 2 ⎟⎜ ⎟⎜ ω 3τ τ 3 4 ⎠⎝

⎞ 4(− 2) ⎤ ⎟+ ⎟ ω 3τ 2 ⎥ 4 ⎥ ⎠ ⎦

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ d T ( jω ) ω →∞

dω

=0

⎡⎛ 4 = ⎢⎜⎜ 3 ⎣⎢⎝ ω τ 3τ 4 =

⎞⎛ 1 ⎟⎜1 − ⎟⎜ ω 2τ τ 3 4 ⎠⎝

4 ⎡ 1 ⎛⎜ 1 1− ⎢ ω 3 ⎣⎢τ 3τ 4 ⎜⎝ ω 2τ 3τ 4

⎤ ⎞ ⎟− 8 ⎥ ⎟ ω 3τ 2 ⎥ 4 ⎦ ⎠

⎞ 2⎤ ⎟− ⎥ ⎟ τ2 ⎠ 4 ⎦⎥

Then ⎡ 1 ⎛ 1 ⎞⎟ 2 ⎤ ⎜1 − =0 ⎢ 2 ⎜ ⎟− 2 ⎥ ⎢⎣τ 3τ 4 ⎝ ω τ 3τ 4 ⎠ τ 4 ⎥⎦ ω →∞ 1 2 = ⇒ 2τ 3 = τ 4 So that

τ3

τ4

Then the transfer function can be written as: 2 ⎧⎡ ⎤ 1 4 ⎪ T ( jω ) = ⎨⎢1 − 2 + 2 2 ⎥ ω 2 τ ω 4τ 32 ⎥ 3 ⎦ ⎪⎩⎣⎢

( )

(

(

−1 / 2

( )

⎧⎪ 1 1 = ⎨1 − 2 2 + 2 ⎪⎩ ω τ 3 4 ω τ 32 ⎧⎪ 1 = ⎨1 + 2 2 ⎪⎩ 4 ω τ 3

⎫ ⎪ ⎬ ⎪⎭

⎫⎪ 2 ⎬ ⎪⎭

⎫⎪ + 2 2⎬ ω τ 3 ⎪⎭

−1 / 2

1

)

2

−1 / 2

)

3 − dB frequency

1

2ω 2τ 32 = 1 or ω =

2 (τ 3 )

=

1 2 (R 3 C )

Define ω=

1 RC

So that R3 =

R 2

We had 2τ 3 = τ 4 or 2( R3C ) = R4 C ⇒ R4 = 2 R3 R = 2⋅R So that 4 ______________________________________________________________________________________

15.7 −14 dB ⇒ T = 0.1995

0.1995 =

1

⇒ (1.2 )

2

2N

⎛ 1 ⎞ =⎜ ⎟ − 1 = 24.1 ⎝ 0.1995 ⎠

1 + (1.2) th N=9, 9 order filter ______________________________________________________________________________________ 2N

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.8 −12 dB ⇒ T = 0.2512 ⇒ (1.333)

1

0.2512 =

2N

2

2N

⎛ 1 ⎞ =⎜ ⎟ − 1 = 14.85 ⎝ 0.2512 ⎠

⎛4⎞ 1+ ⎜ ⎟ ⎝3⎠ th N=5, 5 order filter ______________________________________________________________________________________

15.9 1

T =

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠

2N

T = 0.9 At f = 12 kHz, 1 1 0.9 = = 2N 2N ⎛ f ⎞ ⎛ 12 ⎞ 1+ ⎜ 1 + ⎟ ⎜ ⎟ ⎝ f 3dB ⎠ ⎝ f 3dB ⎠

⎛ 12 ⎞ ⎜ ⎟ ⎝ f3dB ⎠ Also

2N

=

1 − 1 = 0.2346 (0.9) 2 1

0.01 =

⎛ 14 ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠ ⎛ 14 ⎞ ⎜ ⎟ ⎝ f3dB ⎠

2N

=

⎛ 14 ⎞ ⎜ ⎟ ⎝ f3dB ⎠

2N

⎛ 12 ⎞ ⎜ ⎟ ⎝ f3dB ⎠

2N

2N

1 − 1 = 9999 (0.01) 2

⎛ 14 ⎞ =⎜ ⎟ ⎝ 12 ⎠

2N

=

9999 = 4.262 × 104 0.2346

(1.16667) 2 N = 4.262 × 104 N = 35 Then 1

0.9 =

⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠ ⎛ 12 ⎞ ⎜ ⎟ ⎝ f3dB ⎠

2N

2N

= 0.2346

⎛ ⎞ ⎜ ⎟ ⎛ 12 ⎞ ⎝ 2N ⎠ = (0.2346)0.014286 ⎜ ⎟ = (0.2346) f ⎝ 3dB ⎠ = 0.9795 1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ So f 3dB = 12.25 kHz

______________________________________________________________________________________ 15.10

(a) T =

1

1

=

⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ 0.8 ⎠

1 + (1.25)

2N

1

For N=3, T =

1 + (1.25)

= 0.4557 ⇒ −6.83 dB 6

1

(b) For N=5, T =

2N

= 0.3114 ⇒ −10.1 dB

1 + (1.25)

10

1

(c) For N=7, T =

= 0.2053 ⇒ −13.8 dB 14 1 + (1.25) ______________________________________________________________________________________ 15.11

(a) T =

1 1 + (1.4)

2N

For N=3, T = (b) For N=5, T =

1 1 + (1.4 )

= 0.3424 ⇒ −9.31 dB 6

1 1 + (1.4 )

= 0.1828 ⇒ −14.8 dB

10

(c) For N=7, T =

1

= 0.0944 ⇒ −20.5 dB 14 1 + (1.4) ______________________________________________________________________________________

15.12 Consider

For low-frequency:

vo R2 + R3 = vi R1 + R2 + R3

For high-frequency:

vo R2 = vi R1 + R2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ So we need ⎛ R2 ⎞ R2 + R3 = 25 ⎜ ⎟ R1 + R2 + R3 ⎝ R1 + R2 ⎠

R = 1.5 k Ω ⇒ R1 = 48.5 k Ω Let R1 + R2 = 50 k Ω and 2 Then 1.5 + R3 ⎛ 1.5 ⎞ = 25 ⎜ ⎟ ⇒ R3 = 144 k Ω 50 + R3 ⎝ 50 ⎠ Connect the output of this circuit to a non-inverting op-amp circuit.

At low-frequency: vo1 =

R2 + R3 1.5 + 144 ⋅ vi = ⋅ vi = 0.75vi R1 + R2 + R3 48.5 + 1.5 + 144 vo = 25.

Need to have

⎛ R ⎞ ⎛ R ⎞ R vo = 25 = ⎜1 + 5 ⎟ ⋅ vo1 = ⎜ 1 + 5 ⎟ (0.75)vi ⇒ 5 = 32.3 R R R ⎝ ⎝ 4 ⎠ 4 ⎠ 4 To check at high-frequency. R2 1.5 vo1 = vi = vi = 0.03vi R1 + R2 1.5 + 48.5 vo = (1 + 32.3)vo1 = (33.3)(0.03)vi = (1.0)vi which meets the design specification Consider the frequency response. vo1 = vi

R2 + R3

1 sC

R1 + R2 + R3

Now R3

R3 1 = sC 1 + sR3C

1 sC

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then, we find vo1 R3 + R2 (1 + sR3C ) = vi R3 + ( R1 + R2 )(1 + sR3C ) which can be rearranged as ( R2 + R3 ) (1 + s( R2 || R3 )C ) vo1 = vi ( R1 + R2 + R3 ) 1 + s ( R3 || ( R1 + R2 ) ) C

(

)

So fL ≅

1 1 1 = = 2π ( R2 || R3 ) C 2π (1.5 ||144 ) × 103 C ( 9.33 × 103 ) C

fH ≅

1 1 = π 2 144 || 50 ) × 103 C 2π ( R3 || ( R1 + R2 ) ) C (

=

1

( 2.33 ×10 ) C 5

Set 25 kHz =

⎤ fL + fH 1 ⎡ 1 1 ⎥ = ⎢ + 2 2 ⎢ ( 9.33 × 103 ) C ( 2.33 × 105 ) C ⎥ ⎣ ⎦

Which yields C = 2.23 nF ______________________________________________________________________________________ 15.13

Av =

1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠

− 100 dB ⇒ 10−5

2N

So 10−5 =

1 ⎛ 770 ⎞ 1+ ⎜ ⎟ ⎝ 12 ⎠

2N

or 2

⎛ 1 ⎞ 1 + (64.2) 2 N = ⎜ −5 ⎟ = 1010 ⎝ 10 ⎠

or (64.2) 2 N ≅ 1010

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Now

N 1 2 3

Left Side 4.112 × 103 1.7 × 107 7 × 1010

So, we need a 3rd order filter. ______________________________________________________________________________________ 15.14

−3 Low-pass: −50 dB ⇒ 3.16 × 10 Then

3.16 × 10−3 =

1

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ fL ⎠ f L = 3.37 Hz

4

=

1 ⎛ 60 ⎞ 1+ ⎜ ⎟ ⎝ fL ⎠

4

We find High Pass:

3.16 × 10−3 =

We find

1

⎛ f ⎞ 1+ ⎜ H ⎟ ⎝ f ⎠ f H = 1067 Hz

4

=

1 ⎛f ⎞ 1+ ⎜ H ⎟ ⎝ 60 ⎠

4

Bandwidth: BW = f H − f L = 1067 − 3.37 ⇒ BW ≅ 1064 Hz ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.15 a. v vI = − 02 − R4 R3

v0 ⎛ 1 ⎞ R1 ⎜ ⎟ ⎝ sC ⎠

v0 v = − 01 R2 ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ v01 v = − 02 ⇒ v01 = −v02 R5 R5 Then v0 v = + 02 ⎛ 1 ⎞ R2 ⎛ 1 ⎞ v02 = v0 ⎜ ⎟ ⎜ ⎟ ⎝ sR2 C ⎠ ⎝ sC ⎠ or

(1)

(2)

(3)

(2)

And v ⎛ 1 ⎞ v0 vI = − 0 ⋅⎜ ⎟− R4 R3 ⎝ sR2 C ⎠ ⎛ 1 ⎞ R1 ⎜ ⎟ ⎝ sC ⎠ ⎡ ⎤ ⎢ ⎥ 1 1 ⎥ = −v0 ⎢ + ⎢ R3 ( sR2 C ) R1 ⋅ (1/ sC ) ⎥ ⎢⎣ R1 + (1/ sC ) ⎥⎦ ⎡ 1 + sR1C ⎤ 1 = −v0 ⎢ + ⎥ R1 ⎦ ⎣ R3 ( sR2 C ) ⎡ R + (1 + sR1C )( sR2 R3C ) ⎤ = −v0 ⎢ 1 ⎥ ( sC ) R1 R2 R3 ⎣ ⎦

Then v0 1 =− vI R4

⎡ ⎤ ( sC )( R1 R2 R3 ) ⎢ 2 2 ⎥ R + sR R C + s R R R C 2 3 1 2 3 ⎣ 1 ⎦

or 1 − v0 R4 Av ( s ) = = 1 1 vI + sC + R1 sCR2 R3 − Av ( jω ) =

b.

1 R4

1 1 + jω C + R1 jω CR2 R3

(1)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or − Av ( jω ) =

1 R4

⎡ ⎤ 1 1 + j ⎢ω C − ω CR2 R3 ⎥⎦ R1 ⎣ R 1 =− 1⋅ R4 ⎧⎪ ⎡ R1 ⎤ ⎫⎪ ⎨1 + j ⎢ω R1C − ⎬ ω CR2 R3 ⎥⎦ ⎭⎪ ⎣ ⎩⎪ R 1 Av ( jω ) = 1 ⋅ 2 −1/ 2 R4 ⎧ R1 ⎤ ⎫⎪ ⎪ ⎡ ⎨1 + ⎢ω R1C − ⎬ ω CR2 R3 ⎥⎦ ⎭⎪ ⎩⎪ ⎣ Av

⎡ R1 ⎤ ⎢ω R1C − ⎥=0 ω CR 2 R3 ⎦ when ⎣

max

Then Av

max

R1 85 = ⇒ Av 3 R4

=

max

= 28.3

Now ⎡

1

⎣

ω 2 C 2 R2 R3 ⎦

ω R1C ⎢1 −

⎤ ⎥=0

ω=

or

1

C R2 R3

Then f =

1 2π C R2 R3

=

1 2π (0.1× 10−6 ) (300) 2

So f = 5.305 kHz

To find the two 3 − dB frequencies, ⎡ R1 ⎤ ⎢ω R1C − ⎥ = ±1 ω CR2 R3 ⎦ ⎣

ω 2 R1 R2 R3 C 2 − R1 = ±ω R2 R3 C ω 2 (85 × 103 )(300) 2 (0.1× 10−6 ) 2 − 85 × 103 = ±ω (300) 2 (0.1× 10−6 ) ω 2 (7.65× 10 −5 ) − 85 ×103 = ±ω (9 × 10−3 ) ω 2 (7.65× 10−5 ) ± ω (9 × 10−3 ) − 85 × 103 = 0 ω=

(9 × 10−3 ) 2 + 4(7.65 × 10−5 )(85 × 10−3 ) ± (9 × 10−3 ) ± 2(7.65 × 10 −5 ) 2(7.65 × 10−5 )

We find f = 5.315 kHz and f = 5.296 kHz ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.16 a. vI − v A v = A R ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ vI − vB vB − v0 = R R

(1)

(2)

and vA = vB So

vI ⎛1 ⎞ ⎛ 1 + sRC ⎞ = v A ⎜ + sC ⎟ = v A ⎜ ⎟ R ⎝R ⎠ ⎝ R ⎠

(1)

or vA =

vI 1 + sRC

Then 2vI (2) 1 + sRC ⎡ 2 ⎤ ⎡1 − sRC ⎤ v0 = vI ⎢ − 1⎥ = vI ⎢ ⎥ ⎣1 + sRC ⎦ ⎣1 + sRC ⎦ Now v0 1 − jω RC = A( jω ) = 1 + jω RC vI vI + v0 = 2vB = 2v A =

A=

1 + ω 2 R2C 2 1 + ω 2 R2C 2

⇒ A =1

Phase: φ = −2 tan −1 (ω RC ) b.

RC = (104 )(15.9 × 10−9 ) = 1.59 × 10 −4

f 0 102 5 × 103 1/ 2π RC = 103 Hz 5 × 103 104

φ

0

−11.4 −53.1 −90° −157 −169

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.17 a. Vi Vi − V0 + =0 R1 R2 || (1/ sC ) Vi Vi − V0 + =0 R1 ⎡ R2 ⎤ ⎢1 + sR C ⎥ ⎣ ⎦ 2 R2 1 ⋅ (Vi ) + Vi = V0 R1 1 + sR2 C

V0 R2 + R1 (1 + sR2 C ) ( R2 + R1 ) [1 + s ( R1 || R2 )C ] = = Vi R1 (1 + sR2 C ) R1 (1 + sR2 C ) ⇒

V0 ⎛ R2 ⎞ ⎡1 + s ( R1 || R2 )C ⎤ = ⎜1 + ⎟ ⎢ ⎥ Vi ⎝ R1 ⎠ ⎣ (1 + sR2 C ) ⎦

⇒ f 3dB1 =

1 2π R2 C

⇒ f 3dB2 =

1 2π ( R1 || R2 )C

b. Vi V − V0 + i =0 R1 || (1/ sC ) R2 Vi

+

Vi V0 = R2 R2

⎛ R1 ⎞ ⎜ ⎟ ⎝ 1 + sR1C ⎠ ⎡R ⎤ Vi ⎢ 2 ⋅ (1 + sR1C ) + 1⎥ = V0 ⎣ R1 ⎦ Vi ⋅ [ R2 + R1 + sR1 R2 C ] = V0 R1 V0 R2 + R1 V ⎛ R ⎞ 1 = ⋅ [1 + s ( R1 || R2 )C ] ⇒ 0 = ⎜1 + 2 ⎟ [1 + s ( R1 || R2 )C ] ⇒ f 3dB = Vi R1 Vi ⎝ R1 ⎠ 2π ( R1 || R2 )C ______________________________________________________________________________________ 15.18 a. −V0 Vi = R1 + (1/ sC1 ) R2 || (1/ sC2 ) ⎛ sC1 ⎞ ⎛ 1 + sR2 C2 ⎞ Vi ⎜ ⎟ = −V0 ⎜ ⎟ ⎝ 1 + sR1C1 ⎠ ⎝ sC2 ⎠ V0 − sR2 C1 − sR2 C1 = = Vi (1 + sR1C1 )(1 + sR2 C2 ) 1 + sR1C1 + sR2 C2 + s 2 R1 R2 C1C2 ⎡ ⎤ ⎢ ⎥ V0 R sC1 ⎥ = − 2 ×⎢ ⎥ Vi R1 ⎢ 1 ⎛ R2 C2 ⎞ 2 ⎢ + sC1 ⎜ 1 + ⋅ ⎟ + s R2 C1C2 ⎥ R1 C1 ⎠ ⎢⎣ R1 ⎥⎦ ⎝

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or ⎡ ⎢ V0 R2 ⎢ 1 =− ⋅ T (s) = Vi R1 ⎢ 1 ⎛ R2 C2 + ⎜1 + ⋅ ⎢ R1 C1 ⎣⎢ sR1C1 ⎝

⎤ ⎥ ⎥ ⎥ ⎞ ⎟ + sR2 C2 ⎥ ⎠ ⎦⎥

b. R2 1 × 2 2 1/ 2 R1 ⎧ 1 ⎞ ⎪⎫ ⎪⎛ R2 C2 ⎞ ⎛ ⎨⎜ 1 + . ⎟ + ⎜ ω R2 C2 − ⎟ ⎬ R1 C1 ⎠ ⎝ ω R1C1 ⎠ ⎪ ⎪⎩⎝ ⎭ ⎛ 1 ⎞ ⎜ ω R2 C2 − ⎟ = 0, ω R1C1 ⎠ ⎝

T ( jω ) = −

when

we want

R 1 T ( jω ) = 50 = 2 ⋅ R1 ⎛ R2 C2 ⎞ ⎜1 + ⋅ ⎟ R1 C1 ⎠ ⎝ At the 3 − dB frequencies, we want

For

⎛ ⎛ R2 C2 ⎞ 1 ⎞ ⎜ ω R2 C2 − ⎟ = ± ⎜1 + ⋅ ⎟ ω R1C1 ⎠ R1 C1 ⎠ ⎝ ⎝ f = 5 kHz, f = 200 Hz, +

use

sign and for

ω1 = 2π (200) = 1257 ω 2 = 2π (5 × 103 ) = 3.142 × 104 Define τ 2 = R 2 C 2 and τ 1 = R1C1 Then 50 =

R2 ⋅ R1

1

(1)

τ 1+ 2 τ1 ⎞ ⎛ τ ⎟⎟ = +⎜⎜1 + 2 ⎠ ⎝ τ1

⎞ ⎟⎟ ⎠

(2)

⎛ ⎛ τ 1 ⎞ ⎜⎜ ω1τ 2 − ⎟⎟ = −⎜⎜1 + 2 ω1τ 1 ⎠ ⎝ ⎝ τ1 From (2) ω 22τ 1τ 2 − 1 τ 1 + τ 2 =

⎞ ⎟⎟ ⎠

(3)

⎛ 1 ⎜⎜ ω 2τ 2 − ω 2τ 1 ⎝

ω 2τ 1

τ1

or

ω 2τ 1τ 2 −

1

ω2

= τ 1 +τ 2

τ 1 (ω 2τ 2 − 1) =

1

ω2

+τ 2

So 1

+τ 2 ω2 τ1 = ω 2τ 2 − 1

Substituting into (3), we find

use − sign.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

ω1τ 2 −

1 ⎛ 1 ⎞ ⎜⎜ + τ 2 ⎟⎟ ω 2 ⎠ ω1 ⋅ ⎝ (ω 2τ 2 − 1)

⎡ ⎤ ⎢ τ (ω τ − 1) ⎥ ⎥ = − ⎢1 + 2 2 2 1 ⎢ ⎥ +τ 2 ⎥ ⎢ ω2 ⎣ ⎦

⎡⎛ ⎤ ⎞ ⎤ 1 (ω 2τ 2 − 1) = − ⎢⎜⎜ 1 + τ 2 ⎟⎟ + τ 2 (ω 2τ 2 − 1)⎥ +τ 2 ⎥ − ⎥⎦ ⎠ ⎣ω2 ⎦ ω1 ⎣⎢⎝ ω 2 ω1 ω 1 1 ⋅τ 2 + ω1τ 22 − 2 ⋅τ 2 + =− − τ 2 − ω 2τ 22 + τ 2 ⎡ 1

ω1τ 2 ⎢

ω2

ω1

⎛ ω1

(ω1 + ω 2 )τ 22 + ⎜⎜

⎝ ω2

(3.2677 ×10 )τ 4

τ2 =

24.96 ±

2 2

−

ω1

ω2 ω1

ω2

⎛ 1 ⎞ 1 ⎞ ⎟⎟ = 0 ⎟⎟τ 2 + ⎜⎜ + ω ω 2 ⎠ ⎝ 1 ⎠

− 24.96τ 2 + 8.273 × 10 −4 = 0

(24.96)2 − 4(3.2677 × 10 4 )(8.273 ×10 − 4 )

(

)

2 3.2677 × 10 4 Since ω 2 is large, τ 2 should be small so use minus sign:

τ 2 = 3.47 × 10 −5 s Then

τ1 =

3.18 × 10 −5 + 3.47 × 10 −5 ⇒ τ 1 = 7.32 × 10 − 4 s 9.09 × 10 − 2

Now 50 =

R2 ⋅ R1

1 3.47 × 10−5 1+ 7.32 × 10−4

Then R2 = 52.37 R1

or

R2 = 524 kΩ

Also

τ 1 = R1C1 so that C1 = 0.0732 μ F τ 2 = R 2 C 2 so that C 2 = 66.3 pF

______________________________________________________________________________________ 15.19

Two noninverting amplifiers, 2

⎛ R2 ⎞ ⎜⎜1 + ⎟ = 30 dB ⇒= 31.62 R1 ⎟⎠ ⎝ R which gives 2 = 4.62 , then R 2 = 250 k Ω , R1 = 54.1 k Ω R1 For high-pass filter: 1 RC = = 7.958 × 10 − 6 2π 20 × 10 3 Set R = 250 k Ω , then C = 31.8 pF For low-pass filter:

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 = 1.061× 10 −3 2π (150 ) Set R = 250 k Ω , then C = 0.00424 μ F ______________________________________________________________________________________

RC =

15.20 1 ⇒ R eq = 40 M Ω 50 × 10 0.5 × 10 −12

(a) R eq =

(

(b) R eq =

1 ⇒ R eq = 10 M Ω 50 × 10 2 × 10 −12

3

(

3

)(

)

)(

)

1 ⇒ R eq = 2 M Ω (c) R eq = 3 50 × 10 10 × 10 −12 ______________________________________________________________________________________

(

15.21 a.

)(

)

From Equation (15.28), Q=

and Now

V1 − V2 ⋅ TC Req

f C = 100 kHz

Req =

so that

TC =

1 ⇒ 10 μ s 100 × 103

1 1 = ⇒ 1 MΩ f C C (100 × 103 )(10 × 10−12 )

So Q=

(2 − 1)(10 × 10−6 ) = 10 × 10−12 C 106

or Q = 10 pC I eq =

b. c.

Q 10 × 10 −12 = TC 10 × 10−6

Q = CV

or

I eq = 1 μ A

so find the time that V0 reaches 99% of its full value.

(

V o = V1 1 − e − t / τ

) where τ = RC

Then 0.99 = 1 − e − t / τ or e − t / τ = 0.01 or t = τ ln (100 )

τ = RC = (10 3 )(10 × 10 −12 ) = 10 −8 s

Then t = 4.61× 10−8 s ______________________________________________________________________________________ 15.22 gain = −10 ⇒

Low frequency f 3dB = 10 × 103 Hz =

Set

f C C2 2π CF

C1 = 10 C2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ f C = 10 f 3dB = 100 kHz

Then C2 2π (10 × 103 ) = = 0.628 100 × 103 CF C1 ,

The largest capacitor is C1 = 30 pF

so let

Then C2 = 3 pF and CF = 4.78 pF ______________________________________________________________________________________ 15.23 a.

Time constant = τ = Req C F where Req =

1 1 = = 2 × 106 Ω 3 fC C1 (100 × 10 )(5 × 10−12 )

Then τ = 2 × 10 6 30 × 10 −12 or τ = 60 μ s

(

b.

)(

vo = −

or Δv o =

1

τ

)

∫ v dt I

(1)(TC ) τ

and TC =

1 fC

So Δv0 =

1 (60 × 10−6 )(100 × 103 )

or Δv0 = 0.167 V

c. or

Now Δv0 = 13 = N (0.167)

N = 78 clock pulses ______________________________________________________________________________________

15.24

(a)

fo =

1

=

1

(

)(

2π 3 RC 2π 3 20 × 10 0.001× 10 − 6 f o = 4.59 kHz, R 2 = 8 R = 160 k Ω

(b) C =

1 2π 3 R f o

=

(

3

1

)(

2π 3 20 × 10 3 25 × 10 3

)

)

C = 184 pF, R 2 = 160 k Ω ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.25 a. ⎛ sRCV ⎞ R ⋅ v0 = ⎜ ⎟ ⋅ v0 R + (1/ sCV ) ⎝ 1 + sRCV ⎠ R ⎛ sRC ⎞ v2 = ⋅v = ⋅v 1 1 ⎜⎝ 1 + sRC ⎟⎠ 1 R+ sC R ⎛ sRC ⎞ v3 = ⋅v = ⋅v 1 2 ⎜⎝ 1 + sRC ⎟⎠ 2 R+ sC R2 v0 = − ⋅ v3 R Then 2 R ⎛ sRC ⎞ ⎛ sRCV ⎞ v0 = − 2 ⎜ ⎟ v0 ⎟ ⎜ R ⎝ 1 + sRC ⎠ ⎝ 1 + sRCV ⎠ Set s = jω v1 =

⎛ ⎞ ⎛ jω RCV ⎞ −ω 2 R 2 C 2 ⎟ ⎜ 2 2 2 ⎟⎜ ⎝ 1 + 2 jω RC − ω R C ⎠ ⎝ 1 + jω RCV ⎠ The real part of the denominator must be zero. 1 − ω 2 R 2 C 2 − 2ω 2 R 2 CCV = 0 so 1 ω0 = R C (C + 2CV ) R2 R

1= −

b. f 0,max =

1 2π (10 ) (10 4

−11

)(10−11 + 2[10−11 ])

f 0,max = 919 kHz f 0,min =

1 2π (10 ) (10 4

−11

)(10−11 + 2[50 × 10−12 ])

f 0,min = 480 kHz

______________________________________________________________________________________ 15.26

(a)

fo =

1

=

(

1

)(

2π 6 RC 2π 6 12 × 10 3 150 × 10 −12 f o = 36.1 kHz, R 2 = 29 R = 348 k Ω

(b) R =

1 2π 6C f o

=

(

1

)(

)

2π 6 0.001× 10 − 6 22 × 10 3

)

R = 2.95 k Ω , R 2 = 29 R = 85.6 k Ω ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.27

v0 − v1 v1 v1 − v2 = + 1 1 R sC sC v or (v0 − v1 ) sC = 1 + (v1 − v2 ) sC R v1 − v2 v2 v2 = + 1 1 R +R sC sC v v ( sC ) or (v1 − v2 ) sC = 2 + 2 R 1 + sRC v0 v2 =− 1 R2 +R sC v v sC =− 0 or 2 1 + sRC R2

(1)

(2)

(3)

so v2 =

−v0 (1 + sRC ) sR2 C

From (2) 1 sC ⎤ ⎡ v1 ( sC ) = v2 ⎢ sC + + R 1 + sRC ⎥⎦ ⎣

or v1 = −

v0 (1 + sRC ) ⎡ 1 1 ⎤ ⋅ ⎢1 + + ⎥ sR2 C ⎣ sRC 1 + sRC ⎦

From (1) 1 ⎡ ⎤ v0 ( sC ) = v1 ⎢ sC + + sC ⎥ − v2 ( sC ) R ⎣ ⎦ Then v0 1 ⎤ ⎡ −v0 (1 + sRC ) ⎤ ⎡1 + sRC 1 ⎤ ⎡ + v0 = ⎢ 2 + ⎥×⎢ ⎥⎢ ⎥ + sR C ⋅ (1 + sRC ) + sRC sR C sRC sRC 1 ⎣ ⎦⎣ ⎦ ⎦ ⎣ 2 2 2 ⎡1 + 2sRC ⎤ ⎡1 + sRC ⎤ ⎡ (1 + sRC ) + sRC ⎤ 1 + sRC −1 = ⎢ ⎢ ⎥ ⎢ ⎥− ⎥ ⎣ sRC ⎦ ⎣ sR2 C ⎦ ⎣ ( sRC )(1 + sRC ) ⎦ sR2 C

−1 =

(1 + 2sRC )(1 + 2 sRC + s 2 R 2 C 2 + sRC ) (1 + sRC )( sRC ) 2 − ( sRC ) 2 ( sR2 C ) ( sRC ) 2 ( sR2 C )

Set s = jω , then − 1 =

(1 + 2 jωRC )(1 + 3 jωRC + ω 2 R 2 C 2 ) − (1 + jωRC )(− ω 2 R 2 C 2 ) (− ω 2 R 2 C 2 )( jωR2 C ) (− ω 2 R 2 C 2 )( jωR2 C )

The real part of the numerator must be zero. 1 − ω 2 R 2 C 2 − 6ω 2 R 2 C 2 + ω 2 R 2 C 2 = 0 6ω 2 R 2 C 2 = 1 so that 1 ω0 = 6 RC Condition for oscillation:

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ −1 = 1=

2 jω RC + 3 jω RC − 2 jω 3 R3 C 3 + jω 3 R 3 C 3 (−ω 3 R 2 C 2 )( jω R2 C )

5 − ω 2 R2 C 2 (ω RC )(ω R2 C )

But 1

ω = ω0 =

6RC

Then 1⎞ ⎛ 2 2 1 ⎜ 5 − ⎟ (6 R C ) 6⎠ ⎝ 6 = 1= ( RC )( R2 C ) RR2 C 2 6 R 2C 2 ⎛ 29 ⎞ ⎜ ⎟ (6 R ) R 6 1= ⎝ ⎠ or 2 = 29 R2 R ______________________________________________________________________________________ 5−

15.28 Let R F 1 = R F 2 = R F 3 ≡ R F ⎛

R ⎞⎛

1

⎞

⎛

R ⎞⎛

1

⎞

υ o1 = ⎜⎜1 + F ⎟⎟⎜ ⎟ ⋅υ o R ⎠⎝ 1 + sRC ⎠ ⎝

υ o 2 = ⎜⎜1 + F ⎟⎟⎜ ⎟ ⋅υ o1 R ⎠⎝ 1 + sRC ⎠ ⎝ υ o3 − υ o 2 υ o3 υ o3 +

R

1 sC

+

R

=0

⎛2 ⎞ υ + sC ⎟ = o 2 R ⎝R ⎠

υ o3 ⎜

⎛

1 ⎞ ⎟ ⋅υ o 2 ⎝ 2 + sRC ⎠

υ o3 = ⎜

RF R R υo = − F R

υo = −

⋅υ o 3

1 ⎞⎛ R F ⎞⎛ 1 ⎞⎛ R F ⎞⎛ 1 ⎞ ⎛ ⎟⎜ ⎟⎜ ⎟ ⋅υ o ⎟⎜1 + ⎟⎜⎜1 + ⎜ sRC R ⎟⎠⎝ 1 + sRC ⎠⎜⎝ R ⎟⎠⎝ 1 + sRC ⎠ 2 + ⎠⎝ ⎝ 2

R ⎛ R ⎞ ⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞ 1 = − F ⎜⎜1 + F ⎟⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ R ⎝ R ⎠ ⎝ 2 + sRC ⎠⎝ 1 + sRC ⎠⎝ 1 + sRC ⎠ Let s = jω 2

1= −

RF R

⎞ ⎞⎛ ⎞⎛ 1 1 1 ⎛ RF ⎞ ⎛ ⎟⎟ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎜⎜1 + ⎟⎟ ⎜⎜ 2 ω 1 ω 1 ω + + R + j RC j RC j RC ⎝ ⎠ ⎝ ⎠ ⎠⎝ ⎠⎝

=−

RF R

⎞⎛ 1 1 ⎛ RF ⎞ ⎛ ⎟⎟⎜⎜ ⎟⎟ ⎜⎜ ⎜⎜1 + R ⎠ ⎝ 2 + jωRC ⎠⎝ 1 + 2 jωRC − ω 2 R 2 C 2 ⎝

=−

RF R

1 ⎛ RF ⎞ ⎛ ⎟⎟ ⎜⎜ ⎜⎜1 + 2 2 2 R 2 4 j ω RC 2 ω R C j ω − + RC − 2ω 2 R 2 C 2 − jω 3 R 3 C 3 ⎠ ⎝ + ⎝

2

2

⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎤ RF ⎛ RF ⎞ ⎡ 1 1+ ⎢ ⎥ R ⎜⎝ R ⎟⎠ ⎣ 2 − 4ω 2 R 2 C 2 + 5 jω RC − jω 3 R 3C 3 ⎦ 2

1= −

Imaginary Term must be zero 5 jω 0 RC − jω 03 R 3 C 3 = 0 5 − jω 02 R 2 C 2 = 0

ω0 =

5 RC

Then ⎡ ⎤ ⎢ ⎥ R ⎛ R ⎞ 1 ⎥ 1 = − F ⎜1 + F ⎟ ⎢ 2 2 R ⎝ R ⎠ ⎢ 4R C − 5 ⎥ ⎢⎣ 2 − R 2 C 2 ⎥⎦ 2

1= −

RF R

2

⎛ RF ⎞ ⎡ 1 ⎤ 1 RF ⎜ 1 + R ⎟ ⎢ 2 − 20 ⎥ = 18 ⋅ R ⎦ ⎝ ⎠ ⎣

⎛ RF ⎞ ⎜1 + R ⎟ ⎝ ⎠

2

2

RF ⎛ RF ⎞ =2 ⎜1 + ⎟ ⇒ R ⎠ R ⎝ ______________________________________________________________________________________ 18 =

RF R

15.29 (a) 1st stage: 1 jωC

⎞⎛ R F ⎞ ⎛ 1 ⎛ R ⎞ ⎟⎜⎜1 + ⋅ ⎜⎜1 + F ⎟⎟ ⋅υ O = ⎜⎜ ⎟ ⋅υ O ⎟ 1 R ⎠ R ⎟⎠ ⎝ 1 + jωRV C ⎠⎝ + RV ⎝ j ωC nd 2 stage: ⎞⎛ R F ⎞ ⎛ 1 ⎟⎟⎜⎜1 + υ O 2 = ⎜⎜ ⎟ ⋅υ O1 R ⎟⎠ ⎝ 1 + jωRC ⎠⎝

υ O1 =

At node of C 3 : υ A −υ O2 υ O2 υ + υ A ( jωC ) + A = 0 ⇒ υ A = R R 2 + jωRC 3rd stage: υ O2 R R υ O = − F ⋅υ A = − F ⋅ R R (2 + jωRC ) Now ⎞⎛ R F ⎞ R ⎞⎛ R F ⎞⎛ ⎛ 1 1 1 ⎟⎜1 + ⎟⎟⎜⎜1 + ⎜⎜ υO = − F ⋅ ⎟ ⋅υ O ⎟⎟⎜⎜ R (2 + jωRC ) ⎝ 1 + jωRC ⎠⎝ R ⎠⎝ 1 + jωRV C ⎟⎠⎜⎝ R ⎟⎠ 2

RF ⎛ RF ⎞ 1 1 1 ⋅ ⋅ ⎟ ⋅ ⎜1 + R ⎜⎝ R ⎟⎠ (2 + jωRC ) (1 + jωRC ) (1 + jωRV C ) From the denominator, we have: (2 + jωRC )(1 + jωRC )(1 + jωRV C ) = 2 + 3 jωRC − ω 2 R 2 C 2 (1 + jωRV C ) 1= −

(

[(2 − ω R C )+ 3 jωRC ](1 + jωR C ) = (2 − ω R C ) + 3 jωRC + jωR C (2 − ω 2

2

)

2

V

2

2

2

V

For oscillation, imaginary part must be zero.

2

)

R 2 C 2 − 3ω 2 RRV C 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

(

)

3ω o RC + ω o RV C 2 − ω o2 R 2 C 2 = 0 3RC + 2 RV C − ω o2 RV R 2 C 3 = 0

So that ω o =

3R + 2 RV RV

1 RC

⎛ RF ⎜⎜1 + R ⎝ Consider the term:

(b) 1 = −

RF R

2

1 ⎞ ⎟⎟ ⋅ 2 2 2 2 2 ⎠ 2 − ω o R C − 3ω o RRV C

(

(

)

2 − ω o2 R 2 C 2 + 3RRV C 2 = 2 − ⎛ 3R + 2 RV = 2 − ⎜⎜ RV ⎝

(

⎞⎛ 3RV ⎟⎜⎜1 + ⎟ R ⎠⎝

1 R C2 2

⎛ 3R + 2 RV ⎜ ⎜ RV ⎝

) ⎞ 2 2 ⎟ R C + 3RRV C 2 ⎟ ⎠

(

⎞ 2 RRV − (3R + 2 RV )(R + 3RV ) ⎟⎟ = RRV ⎠

)

(

2 RRV − 3R 2 + 9 RRV + 2 RRV + 6 RV2 3R 2 + 9 RRV + 6 RV2 =− RRV RRV Then =

RF R

1=

RF R

)

2

RRV ⎛ RF ⎞ ⎜⎜1 + ⎟⎟ ⋅ 2 R ⎠ 3R + 9 RRV + 6 RV2 ⎝ For R = RV 1=

)

(

)

2

R2 ⎛ RF ⎞ ⎟⎟ ⋅ ⎜⎜1 + R ⎠ 18 R 2 ⎝ 2

R ⎛ R ⎞ R Or 18 = F ⎜⎜1 + F ⎟⎟ ⇒ F = 2 R ⎝ R ⎠ R (c) For RV = 15 k Ω , 1 2π 25 × 10 3 0.001× 10 − 6 For RV = 30 k Ω , fo =

(

)(

)

3(25) + 2(15) ⇒ f o = 16.84 kHz 15

1 3(25) + 2(30 ) ⇒ f o = 13.5 kHz −6 30 2π 25 × 10 0.001× 10 So 13.5 ≤ f o ≤ 16.84 kHz ______________________________________________________________________________________ fo =

15.30

(a)

υ O1 − υ O

(

3

)(

+ υ O1 ( jωC ) +

)

υ O1 − υ O 2

R R υ υ 2 ⎛ ⎞ υ O1 ⎜ + jωC ⎟ = O + O 2 R ⎝R ⎠ R

=0

Then (1) υ O1 (2 + jωRC ) = υ O + υ O 2 υ O 2 − υ O1 υ − υ O3 + υ O 2 ( jωC ) + O 2 =0 R R So (2) υ O 2 (2 + jωRC ) = υ O1 + υ O 3 υ O3 − υ O 2 υ + υ O 3 ( j ωC ) + O 3 = 0 R R

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ And (3) υ O 3 (2 + jωRC ) = υ O 2 Also (4) υ O = −

RF ⋅υ O 3 R

υO +υ (2 + jωRC ) O3 υO 2 2 From (2) υ O 3 (2 + jωRC ) = υ O1 + υ O 3 ⇒ υ O 3 (2 + jωRC ) = 2υ O 3 + (2 + jωRC ) υ O υ O 3 (4 + 4 jωRC − ω 2 R 2 C 2 − 2 ) = (2 + jωRC ) υO υ O3 = (2 + jωRC )(2 − ω 2 R 2 C 2 + 4 jωRC ) υO R Using (4) υ O = − F ⋅ R (2 + jωRC )(2 − ω 2 R 2 C 2 + 4 jωRC ) From (1) υ O1 (2 + jωRC ) = υ O + υ O 3 (2 + jωRC ) ⇒ υ O1 =

RF 1 ⋅ 2 2 2 R (2) 2 − ω R C + 8 jωRC + jωRC 2 − ω 2 R 2 C 2 − 4ω 2 R 2 C 2 For oscillation, denominator must be real, so 10 8ω o RC + ω o RC 2 − ω o2 R 2 C 2 = 0 ⇒ ω o = RC R 1 (b) 1 = − F ⋅ R (2) 2 − ω o2 R 2 C 2 − 4ω o2 R 2 C 2 1= −

[ (

)

(

[ (

1= −

(

)

]

)

]

)

R RF R 1 1 1 =− F ⋅ ⋅ =− F ⋅ ( − 56 ) R R 4 − 2ω o2 R 2 C 2 − 4ω o2 R 2 C 2 R ⎡ ⎤ ⎛ 10 ⎞ 2 2 ⎢4 − 6⎜ 2 2 ⎟ R C ⎥ ⎝R C ⎠ ⎦ ⎣

[

]

RF = 56 R 10 10 (c) C = = ⇒ C = 0.00114 μ F 2π f o R 2π 22 × 10 3 20 × 10 3 R F = (56 )R ⇒ R F = 1.12 M Ω ______________________________________________________________________________________

Or

(

15.31

(a)

υ O1 − υ O RV

+ υ O1 ( jωC ) +

)(

υ O1 − υ O 2 R

)

=0

⎛ 1 ⎞ υ υ 1 + + jωC ⎟⎟ = O + O 2 R ⎝ RV R ⎠ RV

υ O1 ⎜⎜

⎛ 1 ⎞ υ υ + jωC ⎟ = O + O 2 ⎜R R ⎟ RV R V ⎝ ⎠

υ O1 ⎜

⎛υ υ ⎞ Then (1) υ O1 (1 + jω (R RV )C ) = (R RV )⎜⎜ O + O 2 ⎟⎟ R ⎠ ⎝ RV From Problem 15.30, (2) υ O 2 (2 + jωRC ) = υ O1 + υ O 3 (3) υ O 3 (2 + jωRC ) = υ O 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ RF ⋅υ O 3 R 2 2 From (2), υ O 3 (2 + jωRC ) = υ O1 + υ O 3 ⇒ υ O1 = υ O 3 (2 + jωRC ) − 1

(4) υ O = −

[

[

]

]

⎤ ⎡υ υ 2 Then (1) υ O 3 (2 + jωRC ) − 1 (1 + jω (R RV )C ) = (R RV )⎢ O + O 3 (2 + jωRC )⎥ R ⎦ ⎣ RV ⎧⎪ ⎫ R RV υ (2 + jωRC )⎪⎬ = (R RV )⋅ O υ O 3 ⎨ (2 + jωRC )2 − 1 (1 + jω (R RV )C ) − R RV ⎪⎩ ⎪⎭ R (R RV ) 1 Then, υO = − F ⋅ ⋅υ O ⋅ R RV ⎧⎪ ⎫ R RV 2 (2 + jωRC )⎪⎬ ⎨ (2 + jωRC ) − 1 (1 + jω (R RV )C ) − R ⎪⎭ ⎩⎪

[

]

[

]

Consider the denominator:

[4 + 4 jωRC − ω (3 − ω

2

2

]

R 2 C 2 − 1 (1 + jω (R RV )C ) −

)

R 2 C 2 + 4 jωRC (1 + jω (R RV )C ) −

R RV

R RV R

R

(2 + jωRC )

(2 + jωRC )

For oscillation, the denominator must be real, so R RV 4 jω o RC + jω o (R RV )C 3 − ω o2 R 2 C 2 − ⋅ jω o RC = 0 R 4 R + (R RV ) 3 − ω o2 R 2 C 2 − R RV = 0

[

[

]

]

4 R + 2(R RV ) = ω (R RV )R 2 C 2 2 o

So that ω o =

⎛ R 4⎜ ⎜R R V ⎝

1 RC

⎞ ⎟+2 ⎟ ⎠

(b) For RV = 15 k Ω , fo =

)

4(25) + 2 ⇒ f o = 22.66 kHz 25 15

1 2π 25 × 10 3 0.001× 10 − 6

4(25) + 2 ⇒ f o = 19.45 kHz 25 30

1 2π 25 × 10 0.001× 10 − 6

(

3

)(

For RV = 30 k Ω , fo =

(

)(

)

So 19.45 ≤ f o ≤ 22.66 kHz ______________________________________________________________________________________ 15.32 a.

We can write ⎛ Zp ⎛ R1 ⎞ vB = ⎜ vA = ⎜ ⎟ v0 ⎜Z +Z s ⎝ R1 + R2 ⎠ and ⎝ p RB 1 Z p = RB = sCB 1 + sRB CB

where

Z s = RA +

and

1 + sRAC A 1 = sC A sC A

Setting vA = vB , we have

⎞ ⎟⎟ v0 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

R1 = R1 + R2

RB 1 + sRB CB 1 + sRAC A RB + 1 + sRB CB sC A

R1 sRB C A = (1) R1 + R2 sRB C A + (1 + sRAC A )(1 + sRB CB ) To find the frequency of oscillation, set s = jω and set the real part of the denominator on the right side of Equation (1) equal to zero. The denominator term is jω RB C A + (1 + jω RA C A )(1 + jω RB CB ) or jω RB C A + 1 + jω RA C A + jω RB CB − ω 2 RA RB C ACB (2) Then from (2), we must have 1 − ω 02 RA RB C ACB = 0 or 1 f0 = 2π RA RB C ACB

b.

To find the condition for sustained oscillation, combine Equations (1) and (2). Then R1 jω RB C A = R1 + R2 jω RB C A + jω RA C A + jω RB CB ) or 1+

R2 R C = 1+ A + B R1 RB C A

Then R2 RA CB = + R1 RB C A ______________________________________________________________________________________ 15.33 a.

We can write ⎛ R1 ⎞ vA = ⎜ ⎟ v0 ⎝ R1 + R2 ⎠ and ⎛ ⎞ R || sL vB = ⎜ ⎟ v0 || + + R sL R sL ⎝ ⎠

Setting vA = vB , we have sRL ⎡ ⎤ ⎢ ⎥ R1 R + sL =⎢ ⎥ ⋅ v0 sRL R1 + R2 ⎢ + R + sL ⎥ ⎢⎣ R + sL ⎥⎦ R1 sRL = R1 + R2 sRL + ( R + sL) 2

(1)

To find the frequency of oscillation, set s = jω and se the real part of the denominator on the right side of Equation (1) equal to zero. The denominator term is:

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ jω RL + ( R + jω L) 2 or (2) jω RL + R 2 + 2 jω RL − ω 2 L2 Then R 2 − ω 02 L2 = 0 or R 1 f0 = ⋅ L 2π b. To find the condition for sustained oscillations, combine Equations (1) and (2). R1 1 jω RL = = R1 + R2 jω RL + 2 jω RL 3 Then 1+

R2 =3 R1

so that R2 =2 R1 ______________________________________________________________________________________ 15.34 1 2πRC 1 1 RC = = = 4.547 × 10 − 6 2π f o 2π 35 × 10 3 Let C = 0.001 μ F, then R = 4.55 k Ω fo =

(

)

R2 =2 R1 ______________________________________________________________________________________

Set

15.35

g m = 2 K n I DQ = 2 (0.7 )(0.8) = 1.497 mA/V C2 = g m R = (1.497 )(2) = 2.993 C1

C 2 ≅ 3(0.02 ) = 0.06 μ F 2π f o =

1 ⎛ CC L⎜⎜ 1 2 ⎝ C1 + C 2

⎞ ⎟⎟ ⎠

⎞ ⎡ (0.02 )(0.06 )× 10 − 6 ⎤ ⎛ 1 ⎟ ⇒ L = 13.8 μ H Then L ⎢ ⎥ = ⎜⎜ 3 ⎟ 0.02 + 0.06 ⎣ ⎦ ⎝ 2π 350 × 10 ⎠ ______________________________________________________________________________________ 2

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.36

Vπ = −V0 V0 ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC2 ⎠

+

V0 V0 − V1 + = g mVπ = − g mV0 RL ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC1 ⎠

⎡ ⎤ 1 V0 ⎢ sC2 + sC1 + + g m ⎥ = V1 ( sC1 ) RL ⎣ ⎦ V1 V0 − V1 + + g mVπ = 0 sL ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC1 ⎠

(1) (2)

⎛ 1 ⎞ V1 ⎜ + sC1 ⎟ = V0 ( sC1 + g m ) ⎝ sL ⎠ V ( sC1 + g m ) V1 = 0 ⎛ 1 ⎞ ⎜ + sC1 ⎟ ⎝ sL ⎠ Then ⎡ ⎤ V ( sC1 )( sC1 + g m ) 1 V0 ⎢ s (C1 + C2 ) + + gm ⎥ = 0 RL ⎛ 1 ⎞ ⎣ ⎦ ⎜ + sC1 ⎟ sL ⎝ ⎠ ⎡ ⎤⎛ 1 1 ⎞ + g m ⎥ ⎜ + sC1 ⎟ = sC1 ( sC1 + g m ) ⎢ s (C1 + C2 ) + R sL ⎝ ⎠ ⎣ ⎦ L g C1 + C2 sC 1 + s 2 C1 (C1 + C2 ) + + 1 + sg m C1 + m = s 2 C12 + sg m C1 L sRL L RL sL

C1 + C2 sC g 1 + s 2 C1C2 + + 1 + m =0 L sRL L RL sL Set s = jω g C1 + C2 jω C1 1 − ω 2 C1C2 + + + m =0 L jω RL L RL jω L Then ω2 =

C1 + C2 C1 + C2 ⇒ ω0 = C1C2 L C1C2 L

and gm ω C1 1 + = ω L ω RL L RL

Then gm (C + C2 )C1 1 + = 1 L RL L C1C2 LRL gm +

1 C1 + C2 = RL C2 RL

g m RL + 1 =

C1 C + 1 or 1 = g m RL C2 C2

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.37 a. V0 V0 V0 + + g mVπ + =0 (1) 1 sL1 R + sL2 sC ⎛ ⎞ ⎜ sL2 ⎟ Vπ = ⎜ (2) ⎟ V0 ⎜ 1 + sL ⎟ ⎜ 2 ⎟ ⎝ sC ⎠ Then g m ( s 2 L2 C ) ⎪⎫ 1 sC ⎪⎧ 1 V0 ⎨ + + + ⎬=0 2 2 ⎩⎪ sL1 R 1 + s L2 C 1 + s L2 C ⎭⎪ 2 2 2 2 ⎪⎧ R(1 + s L2 C ) + ( sL1 )(1 + s L2 C ) s RL1C + g m ( sRL1 )( s L2 C ) ⎪⎫ + ⎨ ⎬=0 ( sRL1 )(1 + s 2 L2 C ) ( sRL1 )(1 + s 2 L2 C ) ⎪⎩ ⎪⎭ Set s = jω . Both real and imaginary parts of the numerator must be zero. R(1 − ω 2 L2 C ) + jω L1 (1 − ω 2 L2 C ) − ω 2 RL1C + ( jω g m RL1 )(−ω 2 L2 C ) = 0 Real part: R(1 − ω 2 L2 C ) − ω 2 RL1C = 0

R = ω 2 RC ( L1 + L2 ) or

ω0 = b.

1 C ( L1 + L2 )

Imaginary part: jω L1 (1 − ω 2 L2C ) − jω gm RL1 (ω 2 L2C ) = 0

(

L1 = ω 2 L1 L2C + gm RL1 ω 2 L2C Now ω 2 =

)

1 ( L1 + L2 )

1=

1 [ L2 C + g m RL2 C ] C ( L1 + L2 )

1=

L2 L (1 + g m R ) ⇒ 1 = (1 + g m R) L1 + L2 L2

or L1 = gm R L2

______________________________________________________________________________________ 15.38

(a) (1) g mVπ +

VC V + C + (VC − Vπ )( jωC ) = 0 R jωL1

(2) Vπ = (VC − Vπ )( jωC )( jωL2 )

(

)

(

Vπ 1 − ω 2 CL 2 = −VC ω 2 CL 2

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Vπ =

(

VC ω 2 L 2 C

)

ω L2 C − 1 2

⎞ ⎛1 1 Then (1) Vπ (g m − jωC ) + VC ⎜⎜ + + jωC ⎟⎟ = 0 ⎠ ⎝ R jωL1

(ω

)

L 2 C (g m − jωC )

⎞ ⎛1 j + VC ⋅ ⎜⎜ − + jωC ⎟⎟ = 0 ω R L 1 ⎠ ⎝ 2 2 ⎡ g m ω L2 C 1 ωC ω L 2 C ⎤ 1 + + j ⎢ωC − − 2 ⎥=0 2 ωL1 ω L 2 C − 1 ⎦ ω L2 C − 1 R ⎣ Set the imaginary part equal to zero. ω 2 L1C − 1 ωC ω 2 L 2 C = 2 ωL1 ω L2 C − 1 VC ⋅

2

ω 2 L2 C − 1

(

(ω

2

)(

(

)

)

) (

)(

L1C − 1 ω 2 L 2 C − 1 = ω 2 L1 C ω 2 L 2 C

)

ω L1 L 2 C − ω L 2 C − ω L1C + 1 = ω L1 L 2 C 2 4

2

2

2

4

ω o2 (L1 + L 2 )C = 1 1 So ω o = (L1 + L2 )C (b) Set the real part equal to zero, g m ω o2 L 2 C 1 + =0 ω o2 L2 C − 1 R g m Rω o2 L2 C = 1 − ω o2 L 2 C

ω o2 L 2 C (g m R + 1) = 1 Then

L 2 C (g m R + 1) (L + L2 )C = 1 ⇒ gm R +1 = 1 (L1 + L2 )C L2 C

We find g m R = (c) C =

1

L1 L2

(2π f o ) (L1 + L2 ) 2

=

1

[2π (750 ×10 )] (100 ×10 ) 3

2

−6

⇒ C = 450 pF

L1 1 1 =1⇒ R = = ⇒ R = 33.3 Ω L2 g m 30 ______________________________________________________________________________________ gmR =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.39 v0 − v1 v1 v1 − vB = + R ⎛ 1 ⎞ R ⎜ ⎟ ⎝ sC ⎠ and vB v −v + B 1 =0 R ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ or

(1)

(2)

1⎞ v ⎛ vB ⎜ sC + ⎟ = 1 ⇒ v1 = vB (1 + sRC ) R ⎝ ⎠ R From (1) 2⎞ v ⎛ v0 ( sC ) = v1 ⎜ sC + ⎟ − B R⎠ R ⎝ or v0 ( sRC ) = vB (1 + sRC )(2 + sRC ) − vB = vB [ (1 + sRC )(2 + sRC ) − 1]

Now ⎛ R ⎞⎡ ⎤ ⎛ R2 ⎞ ⎡ sRC sRC ⎤ T ( s ) = ⎜1 + 2 ⎟ ⎢ = ⎜1 + ⎟ ⎢ ⎥ 2 2 2 ⎥ R sRC sRC R + + − (1 )(2 ) 1 sRC s R C + + − 2 3 1 ⎣ ⎦ ⎦ ⎝ ⎝ 1 ⎠⎣ 1 ⎠ or ⎛ R ⎞⎡ sRC ⎤ T ( s ) = ⎜1 + 2 ⎟ ⎢ 2 2 2 R1 ⎠ ⎣ s R C + 3sRC + 1 ⎥⎦ ⎝ ⎛ R ⎞⎡ ⎤ jω RC T ( jω ) = ⎜1 + 2 ⎟ ⎢ ⎥ 2 2 2 R − R C + j RC ω ω 1 3 ⎦ ⎝ 1 ⎠⎣ Frequency of oscillation: 1 f0 = 2π RC Condition for oscillation: ⎛ R ⎞ ⎡ jω RC ⎤ 1 = ⎜1 + 2 ⎟ ⎢ R1 ⎠ ⎣ 3 jω RC ⎥⎦ ⎝ or R2 =2 R1

______________________________________________________________________________________ 15.40

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ vb − vo vb vb − va + + =0 1 1 R sC sC vb − vo + 2vb ⋅ sC − va ⋅ sC = 0 R (1) Va − Vb Va + =0 1 R sC 1⎞ ⎛ ⎛ 1 + sRC ⎞ Va ⎜ sC + ⎟ = vb ⋅ sC ⇒ vb = va ⎜ ⎟ R⎠ ⎝ ⎝ sRC ⎠ (2) From (1) ⎛1 ⎞ v vb ⎜ + 2 sC ⎟ = o + va ⋅ sC ⎝R ⎠ R Substitute (2) into (1) ⎛ 1 + sRC ⎞ ⎛ 1 + 2 sRC ⎞ vo va ⎜ ⎟⎜ ⎟ = + va ⋅ sC R ⎝ sRC ⎠ ⎝ ⎠ R ⎡ (1 + sRC )(1 + 2 sRC ) ⎤ v − sC ⎥ = o va ⎢ ( sRC ) ⋅ R ⎣ ⎦ R ⎡ (1 + sRC )(1 + 2 sRC ) ⎤ − sRC ⎥ = vo va ⎢ sRC ⎣ ⎦ vo (1 + sRC )(1 + 2 sRC ) − s 2 R 2 C 2 = va sRC va sRC = vo 1 + 3sRC + 2( sRC ) 2 − s 2 R 2 C 2 va sRC = vo 1 + 3sRC + ( sRC ) 2 ⎛ R ⎞ sRC T ( s ) = ⎜1 + 2 ⎟ ⋅ 2 R ⎝ 1 ⎠ 1 + 3sRC + ( sRC ) ⎛ R ⎞⎡ ⎤ jω RC T ( jω ) = ⎜ 1 + 2 ⎟ ⎢ ⎥ 2 2 2 R1 ⎠ ⎣1 − ω R C + 3 jω RC ⎦ ⎝ 2 2 2 So 1 − ω 0 R C = 0 1 fO = 2π RC So

⎛ R 1 = ⎜1 + 2 R1 ⎝ Also

R2 =2 R1

⎞⎛ 1 ⎞ ⎟⎜ ⎟ ⎠⎝ 3 ⎠

So ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.41

v0 − v1 v1 v1 − vB = + sL R R ⎛ sL ⎞ vB = ⎜ ⎟ v1 ⎝ R + sL ⎠ or

(1) (2)

⎛ R + sL ⎞ v1 = ⎜ ⎟ vB ⎝ sL ⎠ Then v0 ⎛ 1 2⎞ v = v1 ⎜ + ⎟ − B sL ⎝ sL R ⎠ R

or v0 ⎛ R + sL ⎞ ⎛ 1 2 ⎞ vB =⎜ ⎟ ⎜ + ⎟ vB − sL ⎝ sL ⎠ ⎝ sL R ⎠ R ⎧⎛ R + sL ⎞⎛ R + 2 sL ⎞ 1 ⎫ = vB ⎨⎜ ⎟⎜ ⎟− ⎬ ⎩⎝ sL ⎠⎝ sRL ⎠ R ⎭

(1)

Then vB =

v0 1 ⋅ sL ⎧ ( R + sL)( R + 2 sL) − ( sL) 2 ⎫ ⋅⎬ ⎨ ( sL)( sRL) ⎩ ⎭

Now ⎛ R ⎞⎛ sRL ⎞ T ( s ) = ⎜1 + 2 ⎟ ⎜ 2 2 2 2 2 ⎟ R1 ⎠ ⎝ R + 3sRL + 2 s L − s L ⎠ ⎝ or ⎛ R ⎞⎛ sRL ⎞ T ( s ) = ⎜1 + 2 ⎟ ⎜ 2 2 2 ⎟ R1 ⎠ ⎝ s L + 3sRL + R ⎠ ⎝ And ⎛ R ⎞⎛ ⎞ jω RL T ( jω ) = ⎜1 + 2 ⎟ ⎜ 2 ⎟ 2 2 R1 ⎠ ⎝ R − ω L + 3 jω RL ⎠ ⎝ R f0 = 2π L Frequency of oscillation: Condition for oscillation: ⎛ R ⎞⎛ 1 ⎞ 1 = ⎜1 + 2 ⎟ ⎜ ⎟ R1 ⎠ ⎝ 3 ⎠ ⎝ or R2 =2 R1

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.42

(

)

R2 f + R 2v R2 ⋅ V REF = − ⋅ V REF R1 R1

υI = −

Set R 2 f = 20 k Ω For υ I = −2 V, set R 2 v = 0 −2 = −

Then

20 (5) ⇒ R1 = 50 k Ω R1

(20 + R2v )

(5) ⇒ R 2v = 20 k Ω 50 ______________________________________________________________________________________ −4 = −

15.43 For υ O = 10 V, υ I = −5 V i = 0.2 mA =

(VTH

10 − (− 5) ⇒ R1 + R 2 = 75 k Ω R1 + R 2

− VTL ) =

0.4 =

R1 (V H − V L ) R2

R1 [10 − (− 10)] ⇒ R2 = 50 R1 R2

R1 + R 2 = R1 + 50 R1 = 75 ⇒ R1 = 1.47 k Ω , R 2 = 73.53 k Ω ______________________________________________________________________________________

15.44 ⎛ R1 (a) VTH − VTL = ⎜⎜ ⎝ R1 + R 2 ⎛ R1 0.2 = ⎜⎜ ⎝ R1 + R 2 Set R1 = 2 k Ω

⎞ ⎟⎟(V H − V L ) ⎠

⎞ ⎛ R1 ⎟⎟[9 − (− 9)] = 18⎜⎜ ⎠ ⎝ R1 + R 2

⎞ ⎟⎟ ⎠

(18)(2) ⇒ R

2 = 178 k Ω 0.2 9 9 (b) i = = ⇒ i = 50 μ A R1 + R 2 2 + 178 ______________________________________________________________________________________

Then 2 + R 2 =

15.45 ⎛ R1 (a) VTH = ⎜⎜ ⎝ R1 + R 2

⎞ ⎛ 2 ⎞ ⎟⎟ ⋅ V H = ⎜ ⎟(10 ) = 0.4 V ⎝ 2 + 48 ⎠ ⎠

⎛ R1 ⎞ ⎛ 2 ⎞ ⎟⎟ ⋅ V L = ⎜ VTL = ⎜⎜ ⎟(− 10 ) = −0.4 V ⎝ 2 + 48 ⎠ ⎝ R1 + R 2 ⎠ (b) For 33.33 ≤ t ≤ 41.67 ms, sin[2π (60 )t ] ⇒ positive half cycle

At υ I = 0 , υ O = +10 V At υ I = +0.4 = 10 sin [2π (60 )t1 ] ⇒ t1 = +0.106 + 33.333 = 33.439 ms So, for 33.333 ≤ t ≤ 33.439 ms, υ O = +10 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 33.439 ≤ t ≤ 41.77 ms, υ O = −10 V 41.77 ≤ t ≤ 50 ms, υ O = +10 V ______________________________________________________________________________________

15.46

a. Upper crossover voltage when v0 = +VP , Now ⎛ R1 ⎞ vB = ⎜ ⎟ (+VP ) ⎝ R1 + R2 ⎠

and ⎛ RA ⎞ ⎛ RB ⎞ vA = ⎜ ⎟ VREF + ⎜ ⎟ VTH ⎝ RA + RB ⎠ ⎝ RA + RB ⎠ v A = vB so that ⎛ R1 ⎞ ⎛ RA ⎞ ⎛ RB ⎞ ⎜ ⎟ VP = ⎜ ⎟ VREF + ⎜ ⎟ VTH + + R R R R 2 ⎠ B ⎠ ⎝ 1 ⎝ A ⎝ RA + RB ⎠ or ⎛ R + RB ⎞ ⎛ R1 ⎞ VTH = ⎜ A ⎟⎜ ⎟ VP ⎝ R1 + R2 ⎠ ⎝ RB ⎠

⎛R ⎞ − ⎜ A ⎟ VREF ⎝ RB ⎠

Lower crossover voltage when v0 = −VP So ⎛ R + RB ⎞ ⎛ R1 ⎞ ⎛ RA ⎞ VTL = − ⎜ A ⎟⎜ ⎟ VP − ⎜ ⎟ VREF R + R R 2 ⎠⎝ B ⎠ ⎝ 1 ⎝ RB ⎠ b. ⎛ 10 + 20 ⎞ ⎛ 5 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ ⎜ ⎟ (10) − ⎜ ⎟ (2) ⎝ 5 + 20 ⎠ ⎝ 20 ⎠ ⎝ 20 ⎠ or VTH = 2 V

and ⎛ 10 + 20 ⎞ ⎛ 5 ⎞ VTL = − ⎜ ⎟ ⎜ ⎟ (10) − 1 ⇒ VTL = −4 V ⎝ 5 + 20 ⎠ ⎝ 20 ⎠ ______________________________________________________________________________________

15.47 a. vB VREF − vB v0 − vB = + R1 R3 R2 ⎛ 1 v 1 1 ⎞ V vB ⎜ + + ⎟ = REF + 0 R3 R2 ⎝ R1 R2 R3 ⎠ VTH = vB

So

when v0 = +VP and VTL = vB when v0 = −VP

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VREF VP + R3 R2 VTH = ⎛ 1 1 1 ⎞ + ⎟ ⎜ + ⎝ R1 R2 R3 ⎠ and VTL =

VREF VP − R3 R2 ⎛ 1 1 1 ⎞ + ⎟ ⎜ + R R R 2 3 ⎠ ⎝ 1

b. VREF

VS =

⎛ 1 1 1 ⎞ + ⎟ R3 ⎜ + ⎝ R1 R2 R3 ⎠ −10 −5 = ⎛ 1 1 1⎞ + ⎟ 10 ⎜ + ⎝ R1 R2 10 ⎠ 1 1 1 1 + = − = 0.10 R1 R2 5 10 ΔVT = VTH − VTL =

0.2 =

2VP R2 ⎛ 1 1 1 ⎞ + ⎟ ⎜ + ⎝ R1 R2 R3 ⎠

2(12) R2 (0.10 + 0.10)

R = 600 kΩ So 2 Then 1 1 + = 0.10 R1 R2

1 1 + = 0.10 ⇒ R1 = 10.17 kΩ R1 600

c. VTH = −5 + 0.1 = −4.9 VTL = −5 − 0.1 = −5.1 ______________________________________________________________________________________ 15.48

a.

If the saturated output voltage is where

then the circuit behaves as a comparator

v0 < 6.2 V.

If the saturated output voltage is no control. b.

VP < 6.2 V,

VP > 6.2 V,

the output will flip to either +VP or −VP and the input has

Same as part (a) except the curve at vI ≈ 0 will have a finite slope.

Circuit works as a comparator as long as v01 < 8.7 V and v02 > −3.7 V. Otherwise the input has no control. ______________________________________________________________________________________ c.

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.49 ⎛ R2 (a) When υ O = 0 , υ + = V S = ⎜⎜ ⎝ R1 + R 2 When υ O = V H , υ + = VTH

⎞ ⎟⎟ ⋅ V REF ⎠

V REF − VTH VTH − V H = R1 R2 ⎛ 1 ⎛ R + R2 V REF V H 1 ⎞ ⎟⎟ = VTH ⎜⎜ 1 + = VTH ⎜⎜ + R1 R2 R R 2 ⎠ ⎝ 1 ⎝ R1 R 2 ⎛ R2 VTH = V REF ⎜⎜ ⎝ R1 + R 2

⎞ ⎛ R1 ⎟⎟ + V H ⎜⎜ ⎠ ⎝ R1 + R 2

⎞ ⎟⎟ ⎠

⎞ ⎛ R1 ⎟⎟ = V S + V H ⎜⎜ ⎠ ⎝ R1 + R 2

⎞ ⎟⎟ ⎠

⎛ R1 ⎞ ⎟⎟ VTL = V S + V L ⎜⎜ ⎝ R1 + R 2 ⎠ (b) V S = −1.75 V, R1 = 4 k Ω ⎛ 4 VTH = −1.5 = −1.75 + (12 )⎜⎜ ⎝ 4 + R2

⎞ ⎟⎟ ⇒ R 2 = 188 k Ω ⎠

⎛ 188 ⎞ − 1.75 = ⎜ ⎟ ⋅ V REF ⇒ V REF = −1.787 V ⎝ 188 + 4 ⎠ 12 − (− 1.787 ) 13.787 (c) (i) For υ O = 12 V, i = = ⇒ i = 71.8 μ A R1 + R 2 4 + 188 12 − 1.787 ⇒ i = 53.2 μ A 192 ______________________________________________________________________________________

(ii) For υ O = −12 V, i =

15.50

a. Switching point when v0 = 0. Now ⎛ R2 v+ = VREF = ⎜ ⎝ R1 + R2

⎞ ⎟ vI ⎠ where vI = VS .

Then ⎛ R + R2 VS = ⎜ 1 ⎝ R2

⎞ ⎛ R1 ⎞ ⎟ VREF = ⎜ 1 + ⎟ VREF R 2 ⎠ ⎠ ⎝ v1

Now upper crossover voltage for VTH − VREF VREF − VL = R1 R2

occurs when v0 = VL and v+ = VREF . Then

⎛ R1 R ⎞ ⋅ VL + VREF ⎜ 1 + 1 ⎟ R2 R ⎝ 2 ⎠ R1 or VTH = VS − ⋅ VL R2 Lower crossover voltage for vI occurs when v0 = VH and vI = VREF . Then or VTH = −

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ VH − VREF VREF − VTL = R2 R1 ⎛ R1 R ⎞ ⋅ VH + VREF ⎜ 1 + 1 ⎟ R2 R ⎝ 2 ⎠ R1 or VTL = VS − ⋅ VH R2 or VTL = −

b.

VTH = −1

For and R1 = 0.833 kΩ so that Now ⎛ R ⎞ VS = ⎜ 1 + 1 ⎟ VREF R ⎝ 2 ⎠

VTL = −2, VS = −1.5 V.

VTL = VS −

Then

R1 R ⋅ VH ⇒ −2 = −1.5 − 1 (12) R2 20

⎛ 0.833 ⎞ −1.5 = ⎜ 1 + ⎟ VREF 20 ⎠ ⎝ which gives VREF = −1.44 V

______________________________________________________________________________________ 15.51 ⎛ R1 ⎞ 25 ⎞ ⎟ ⋅ Vγ = ⎛⎜ (a) VTH = ⎜⎜ ⎟(0.7 ) = 0.175 V ⎟ ⎝ 25 + 75 ⎠ ⎝ R1 + R3 ⎠ ⎛ R1 ⎞ ⎟(− Vγ ) = −0.175 V VTL = ⎜⎜ ⎟ ⎝ R1 + R3 ⎠ (c) (i) υ I = 2 V, υ O = −0.7 V I D1 = 0

⎛ − 0.7 − (− 10) ⎞ I R 2 = −⎜ ⎟ = −0.465 mA 20 ⎝ ⎠

⎛ − 0.175 − (− 0.7 ) ⎞ I R 3 = +⎜ ⎟ ⇒ I R3 = 7 μ A 75 ⎝ ⎠ I R 3 + I R 2 + I D 2 = 0 ⇒ I D 2 = − I R 3 − I R 2 = −0.007 − (− 0.465) = 0.458 mA

(ii) υ I = −2 V, υ O = +0.7 V I D2 = 0 ⎛ 10 − 0.7 ⎞ I R2 = ⎜ ⎟ = 0.465 mA ⎝ 20 ⎠

⎛ 0.175 − (0.7 ) ⎞ I R3 = ⎜ ⎟ ⇒ I R 3 = −7 μ A 75 ⎝ ⎠ I D1 = I R 2 + I R 3 = 0.465 − 0.007 = 0.458 mA ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.52 (a) Vγ + V Z = 0.7 + 5.6 = 6.3 V

VTH 0 − (− 6.3) R = ⇒ VTH = 1 (6.3) R1 R2 R2 VTL = − VTH − VTL = 0.6 =

Then

R1 (6.3) R2

R1 [6.3 − (− 6.3)] = (12.6) R1 R2 R2

R2 = 21 , Set R1 = 4 k Ω , then R 2 = 84 k Ω R1

(b) Maximum current in R 2 , iR2 =

6.3 6.3 = = 0.075 mA R2 84

10 − 6.3 = 0.8 + 0.075 = 0.875 mA R 10 − 6.3 R= = 4.23 k Ω 0.875 ______________________________________________________________________________________ iR =

15.53 a. v0 = VREF + 2Vγ 5 = VREF + 2(0.7)

or VREF = 3.6 V

b. ⎛ R1 ⎞ VTH = ⎜ ⎟ (VREF + 2Vγ ) ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ 0.5 = ⎜ ⎟ (5) ⎝ R1 + R2 ⎠ or 1 +

R2 R = 10 ⇒ 2 = 9 R1 R1

For example, let R2 = 90 kΩ and R1 = 10 kΩ c.

For

For vI = 10 V, and v0 is in its low state. D1 is on and D2 is off.

v1 − (v1 + 0.7) VREF − v1 v1 − v0 + = 100 1 1 v1 = −0.7,

then 10 − 0 3.6 − (−0.7) −0.7 − v0 + = 100 1 1

or v0 = −5.1 V

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.54

For

v0 = High = (VREF + 2Vγ ).

Then switching point is when.

⎛ R1 ⎞ vI = vB = ⎜ ⎟ v0 ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ or VTH = ⎜ ⎟ (VREF + 2Vγ ) ⎝ R1 + R2 ⎠

Lower switching point is when ⎛ R1 v1 = vB = ⎜ ⎝ R1 + R2

⎞ ⎟ v0 ⎠ and v0 = −(VREF + 2Vγ )

so ⎛ R1 ⎞ VTL = − ⎜ ⎟ (VREF + 2Vγ ) ⎝ R1 + R2 ⎠ ______________________________________________________________________________________

15.55 By symmetry, inverting terminal switches about zero.

Now, for v0 low, upper diode is on. VREF − v1 = v1 − v0 v0 = 2v1 − VREF where v1 = −Vγ

so v0 = −(VREF + 2Vγ )

Similarly, in the high state v0 = (VREF + 2Vγ ) Switching occurs when non-inverting terminal is zero. So for v0 low.

VTH − 0 0 − ⎡⎣ − (VREF + 2Vγ ) ⎤⎦ = R1 R2

R1 ⋅ (VREF + 2Vγ ) R2 By symmetry R VTL = − 1 ⋅ (VREF + 2vγ ) R2 ______________________________________________________________________________________ or VTH =

15.56 ⎛ R1 (a) υ + = ⎜⎜ ⎝ R1 + R 2

⎞ ⎛ 10 ⎞ ⎟⎟ ⋅υ O = ⎜ ⎟(5) = 1.667 V ⎝ 10 + 20 ⎠ ⎠ ⎛

⎛ t t ⎞ ⎟⎟ = 5 − 6.667 exp⎜⎜ − ⎝ τX ⎠ ⎝ τX

υ X = 5 + (− 1.667 − 5) exp⎜⎜ −

⎞ ⎟⎟ , for 0 < t < t1 ⎠

⎛ ⎛ t − t1 ⎞ ⎞ ⎛ (t − t1 ) ⎞ ⎟⎟ ⎟ = −5 + 6.667 exp⎜⎜ − ⎟⎟ , for t1 < t < T ⎟ τ τ X X ⎝ ⎝ ⎠ ⎠ ⎠ ⎝

υ X = −5 + (1.667 − (− 5)) exp⎜⎜ − ⎜⎜

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ t (b) 1.667 = 5 − 6.667 exp⎜⎜ − 1 ⎝ τX ⎛ t 6.667 exp⎜⎜ − 1 ⎝ τX

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ = 3.333 ⎠

⎛ t ⎞ exp⎜⎜ + 1 ⎟⎟ = 2 ⇒ t1 = τ X ln (2) ⎝ τX ⎠ By symmetry, T − t1 = τ X ln (2 )

Then T = 2τ X ln (2 )

(

)(

)

f =

1 1 = , τ X = R X C X = 40 × 10 3 0.02 × 10 −6 = 8 × 10 −4 s T 2τ X ln (2)

f =

2(8 × 10 ) ln (2 )

1

−4

= 902 Hz

Duty cycle = 50% ______________________________________________________________________________________ 15.57 ⎛ R1 (a) υ + (+ ) = ⎜⎜ ⎝ R1 + R 2 ⎛

R1 ⎝ R1 + R 2

υ + (− ) = ⎜⎜

⎞ ⎛ 10 ⎞ ⎟⎟(5) = ⎜ ⎟(5) = 1.667 V ⎝ 10 + 20 ⎠ ⎠ ⎞ ⎛ 10 ⎞ ⎟⎟(− 10 ) = ⎜ ⎟(− 10 ) = −3.333 V ⎝ 10 + 20 ⎠ ⎠ ⎛

υ X = 5 + (− 3.333 − 5) exp⎜⎜ −

⎛

t ⎞

t ⎞

⎟⎟ = 5 − 8.333 exp⎜⎜ − ⎟⎟ , ⎝ τX ⎠ ⎝ τX ⎠ ⎛

υ X = −10 + (1.667 − (− 10 )) exp⎜⎜ − ⎝

⎛ t (b) 1.667 = 5 − 8.333 exp⎜⎜ − 1 ⎝ τX

for 0 < t < t1

(t − t1 ) ⎞⎟ τX

⎛ (t − t1 ) ⎞ ⎟⎟ , for t1 < t < T ⎟ = −10 + 11.667 exp⎜⎜ − τ X ⎠ ⎝ ⎠

⎞ ⎟⎟ ⎠

⎛ −t ⎞ ⎛ t 8.333 exp⎜⎜ 1 ⎟⎟ = 3.333 ⇒ exp⎜⎜ + 1 ⎝τX ⎠ ⎝ τX t1 = τ X ln (2.5)

⎞ ⎟⎟ = 2.5 ⎠

⎛ (T − t1 ) ⎞ ⎟ Also − 3.333 = −10 + 11.667 exp⎜⎜ − τ X ⎟⎠ ⎝

⎛ (T − t1 ) ⎞ ⎛ (T − t1 ) ⎞ ⎟⎟ = 6.667 ⇒ exp⎜⎜ + ⎟ = 1.75 11.667 exp⎜⎜ − τ τ X ⎟⎠ X ⎝ ⎠ ⎝ T − t1 = τ X ln (1.75)

Now T = τ X [ln(2.5) + ln (1.75)] = τ X ln[(2.5)(1.75)]

1 , τ X = R X C X = 8 × 10 −4 s T 1 f = = 847 Hz 8 × 10 − 4 ln[(2.5)(1.75)] ______________________________________________________________________________________ f =

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.58 f =

1 2.2 RX C X

1 1 = 2.2 f (2.2)(12 × 103 ) RX C X = 3, 788 × 10−5 RX C X =

RX = 56 K

For example, Let C X = 680 pF 1 Within 2 of 1% of design specification.

______________________________________________________________________________________ 15.59

( )(

)

t1 = 1.1R X C X = (1.1) 10 4 0.1× 10 −6 ⇒ t1 = 1.1 ms

(

0 < t < t1 , υ Y = 10 1 − e

−t / τ Y

)

τ Y = RY C Y = (2 × 10 )(0.02 × 10 −6 ) 3

= 4 × 10 −5 s

Now

t1

τY

= 2.75

⇒ CY

completely charges during each cycle. ______________________________________________________________________________________ 15.60 a.

Switching voltage ⎛ R1 + R3 ⎞ ⎛ 10 + 10 ⎞ vX = ⎜ ⎟ ⋅ VP = ⎜ ⎟ (±10) R + R + R ⎝ 10 + 10 + 10 ⎠ 3 2 ⎠ ⎝ 1

So vX = ±6.667 V Using Equation (15.83(a)) 2 ⎛ 2 ⎞ v X = V P + ⎜ − V P − V P ⎟e −t1 / τ X = V P 3 ⎝ 3 ⎠ 5 2 Then 1 − ⋅ e −t1 / τ X = 3 3 1 5 − t1 / τ X = ⋅e or t1 = τ X ln (5) 3 3 T 1 1 t1 = = = ⇒ t1 = 0.001 s 2 2 f 2(500 )

So b.

(

10 −3 = τ X ln (5) ⇒ τ X = 6.21× 10 −4 = R X 0.01× 10 −6 RX = 62.1 kΩ

Switching voltage ⎛ ⎞ R1 vX = ⎜ ⎟ (±VP ) ⎝ R1 + R3 + R2 ⎠

10 1 ⎛ ⎞ =⎜ ⎟ (±VP ) = ⋅ (±VP ) 10 + 10 + 10 3 ⎝ ⎠ Using Equation (15.83(a))

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 ⎞ ⎛ 1 v X = V P + ⎜ − V P − V P ⎟e −t1 / τ X = V P 3 ⎝ 3 ⎠ 4 1 Then 1 − e −t1 / τ X = 3 3 2 4 −t1 / τ X = ⋅e 3 3

(

)

t1 = τ X ln (2 ) = 6.21× 10 −4 ln (2 ) = 4.30 × 10 −4 s −4

T = 2t1 = 8.6 × 10 s 1 ⇒ f = 1.16 kHz T ______________________________________________________________________________________ f =

15.61

(a) β =

R1 20 = = 0.625 R1 + R 2 20 + 12

T = 250 × 10 − 6 = τ X

0.7 ⎤ ⎡ ⎢ 1 + 10 ⎥ ln ⎢ ⎥ = τ X (1.0485) ⎢1 − 0.625 ⎥ ⎣⎢ ⎦⎥

So τ X = 2.384 × 10 −4 = R X C X Set C X = 0.01 μ F, then R X = 23.84 k Ω (b) For t < 0 , υ X = 0.7 V So υ Y must be < υ X ⇒ υ I < 0 ⎛ (T ′ − T ) ⎞ ⎟ (c) υ X = Vγ = 10 + [− β (10) − 10] exp⎜⎜ − τ X ⎟⎠ ⎝ ⎛ (T ′ − T ) ⎞ ⎛ (T ′ − T ) ⎞ ⎟ = 9.3 ⎟⎟ ⇒ 16.26 exp⎜⎜ − 0.7 = 10 − 16.25 exp⎜⎜ − τX ⎠ τ X ⎟⎠ ⎝ ⎝ ⎛ (T ′ − T ) ⎞ ⎟ = 1.747 exp⎜⎜ + τ X ⎟⎠ ⎝ T ′ − T = τ X ln (1.747 ) = 2.384 × 10 −4 ln (1.747 ) T ′ − T = 133 μ s ______________________________________________________________________________________

15.62

(a) T = τ X

⎡ Vγ ⎢1 + VP ln ⎢ ⎢ 1− β ⎢ ⎣⎢

⎤ ⎥ ⎥ , V = 0.7 V, V = 5 V, β = 0.5 γ P ⎥ ⎥ ⎦⎥

τ X = R X C X = (20 × 10 3 )(1.2 × 10 −6 ) = 2.4 × 10 −2 s

(

T = 2.4 × 10

−2

)

⎡ 0.7 ⎤ ⎢1 + 5 ⎥ ln ⎢ ⎥ ⇒ T = 19.78 ms ⎢ 1 − 0.5 ⎥ ⎣⎢ ⎦⎥

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) T ′ − T ≅ 0.4τ X = 9.6 ms ______________________________________________________________________________________ 15.63 a.

From Equation (15.95)

T = 1.1 RC For T = 60 s = 1.1 RC then RC = 54.55 s

For example, let C = 50 μ F and R = 1.09 MΩ b. Recovery time: capacitor is discharged by current through the discharge transistor. 5 − 0.7 IB ≅ = 0.043 mA + V = 5 V, 100 If then If β = 100, I C = 4.3 mA

VC =

1 Ic I C dt = ⋅ t ∫ C C

2 + ⋅ V = 3.33 V Capacitor has charged to 3

t=

So that

VC ⋅ C (3.33)(50 × 10−6 ) = 4.3 × 10−3 IC ⋅

So recovery time t ≈ 38.7 ms ______________________________________________________________________________________ 15.64

T = 1.1 RC 5 × 10−6 = 1.1 RC −6 so RC = 4.545 ×10 s

For example, let C = 100 pF

and R = 45.5 kΩ From Problem (15.63), recovery time t≅

VC ⋅ C (3.33)(100 × 10−12 ) = IC 4.3 × 10−3

or t = 77.4 ns ______________________________________________________________________________________

15.65

Duty cycle = 60% = 0.60 =

R A + RB × 100% R A + 2R B

25 + R B ⇒ R B = 50 k Ω 25 + 2 R B

f = 80 × 10 3 =

1

(0.693)(25 + 2(50))×10 3 C

⇒ C = 144.3 pF

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.66 f =

1 (0.693)( RA + 2 RB )C

RA = R1 = 10 kΩ, RB = R2 + xR3 So 10 kΩ ≤ RB ≤ 110 kΩ 1 = 627 kHz (0.693)(10 + 2(110)) × 103 × (0.01× 10−6 ) 1 = = 4.81 kHz (0.693)(10 + 2(10)) × 103 × (0.01× 10−6 )

f min = f max

So 627 Hz ≤ f ≤ 4.81 kHz R + RB Duty cycle = A × 100% RA + 2 RB Now 10 + 10 × 100% = 66.7% 10 + 2(10) and 10 + 110 × 100% = 52.2% 10 + 2(110) So 52.2 ≤ Duty cycle ≤ 66.7% ______________________________________________________________________________________ 15.67

1 kΩ ≤ RA ≤ 51 kΩ 1 kΩ ≤ RB ≤ 51 kΩ 1 = 1.40 Hz (0.693)(1 + 2(51)) × 103 × (0.01× 10−6 ) 1 f max = = 2.72 kHz (0.693)(51 + 2(1)) × 103 × (0.01× 10 −6 ) or 1.40 kHz ≤ f ≤ 2.72 kHz f min =

Duty cycle =

RA + RB × 100% RA + 2 RB

1 + 51 × 100% = 50.5% 1 + 2(51)

or 51 + 1 × 100% = 98.1% 51 + 2(1) or 50.5% ≤ Duty cycle ≤ 98.1% ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.68 a. V + − 3VEB R1 A + R1B = 0.7

IE3 = IE 4 =

Assume VEB

IE3 = IE 4 =

22 − 3(0.7) = 0.398 mA 25 + 25

Now ⎛ 20 ⎞ I C 3 = I C 4 = I C 5 = I C 6 = ⎜ ⎟ (0.398) ⎝ 21 ⎠ I C 3 = I C 4 = I C 5 = I C 6 = 0.379 mA 0.398 ⎛ 20 ⎞ ⎜ ⎟ ⇒ I C1 = I C 2 = 0.018 mA 21 ⎝ 21 ⎠ I D = 0.398 mA, current in D1 and D2 b. ⎛I ⎞ ⎛ 0.398 × 10−3 ⎞ VBB = 2VD = 2VT ln ⎜ D ⎟ = 2(0.026) ln ⎜ ⎟ −13 ⎝ 10 ⎠ ⎝ IS ⎠ I C1 = I C 2 =

or VBB = 1.149 V = VBE 7 + VEB 8 Now IC 7 ≈ IC 4 + IC 9 + I E8 I C 4 = 0.379 mA ⎛ 20 ⎞ I B9 = IC 8 = ⎜ ⎟ I E 8 ⎝ 21 ⎠

So ⎛I ⎞ I E 8 = 1.05 I B 9 = 1.05 ⎜ C 9 ⎟ ⎝ 100 ⎠ ⎛ 100 ⎞ ⎛ 21 ⎞ IC 7 = IC 4 + ⎜ ⎟ I E 8 + I E 8 = I C 4 + (96.24) ⎜ ⎟ I C 8 1.05 ⎝ ⎠ ⎝ 20 ⎠

So I C 7 = 0.379 mA + 101I C 8 and ⎛I ⎞ ⎛I ⎞ VBE 7 = VT ln ⎜ C 7 ⎟ ; VEB 8 = VT ln ⎜ C 8 ⎟ ⎝ IS ⎠ ⎝ IS ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then ⎡ ⎛I ⎞ ⎛ I ⎞⎤ 1.149 = 0.026 ⎢ ln ⎜ C 7 ⎟ + ln ⎜ C 8 ⎟ ⎥ ⎝ I S ⎠⎦ ⎣ ⎝ IS ⎠ ⎡ I (0.379 × 10−3 ) + 101I C 8 ⎤ 44.19 = ln ⎢ C 8 ⎥ (10−13 ) 2 ⎣ ⎦ (10−13 ) 2 exp (44.19) = 101I C2 8 + 3.79 × 10−4 I C 8 1.554 × 10−7 = 101I C2 8 + 3.79 × 10 −4 I C 8 (3.79 × 10 −4 ) 2 + 4(101)(1.554 × 10 −7 ) −3.79 × 10−4 ± 2(101) 2(101) I C 8 = 37.4 μ A

IC 8 =

I C 7 = 0.379 + 101(0.0374) ⇒ I C 7 = 4.16 mA ⎛ 21 ⎞ I C 9 = 4.16 − 0.379 − 0.0374 ⎜ ⎟ ⎝ 20 ⎠ I C 9 = 3.74 mA P = (0.398 + 0.398 + 4.16)(22) ⇒ P = 109 mW c. ______________________________________________________________________________________

15.69 a.

b.

From Figure 15.44, 3.7 W to the load V + ≈ 19 V P=

c. or

1 VP2 2 RL

VP = 2 RL P = 2(10)(3.7) ⇒ VP = 8.6 V

______________________________________________________________________________________ 15.70 ⎛ R ⎞ R R (a) ⎜⎜1 + 2 ⎟⎟ = 12 ⇒ 2 = 11 , and 4 = 12 R1 ⎠ R1 R3 ⎝ 2 V (b) PL = L ⇒ V L = 2 R L PL = 2(12)(15) = 18.97 V 2R L

υ O1

max

IL =

= υ O2

max

= 9.49 V

V L 18.97 = = 1.58 A RL 12

⎛ I ⎞⎛ V ⎞ ⎛ 0.8 ⎞⎛ 24 ⎞ ⎟⎜ ⎟ = 9.6 W (c) (i) PL = ⎜⎜ L ⎟⎟⎜⎜ L ⎟⎟ = ⎜⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ V 24 = 30 Ω (ii) R L = L = I L 0.8 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.71

a. Then

v01 = iR2 + vI where

i=

vI R1

⎛ R ⎞ v01 = vI ⎜1 + 2 ⎟ R1 ⎠ ⎝

Now ⎛R ⎞ v02 = −iR3 = −vI ⎜ 3 ⎟ ⎝ R1 ⎠

So ⎛ R ⎞ ⎡ ⎛ R ⎞⎤ vL = v01 − v02 = vI ⎜ 1 + 2 ⎟ − ⎢ −vI ⎜ 3 ⎟ ⎥ R1 ⎠ ⎣ ⎝ ⎝ R1 ⎠ ⎦ v R R Av = L = 1 + 2 + 3 vI R1 R1 Av = 10 ⇒

b.

Want

R2 R3 + =9 R1 R1

⎛ R2 ⎞ R3 ⎜1 + ⎟ = R1 ⎠ R1 Also want ⎝ R2 ⎛ R2 ⎞ + ⎜1 + ⎟ = 9 R1 ⎝ R1 ⎠

R2 =4 R1

so R = 200 kΩ For R1 = 50 kΩ, 2 and Then

R3 =5 R1

c. or

R = 250 kΩ so 3 1 VP2 P= 2 RL

VP = 2 RL P = 2(20)(10) = 20 V

So peak values of output voltages are v01 = v02 = 10 V

20 =1 A 20 ______________________________________________________________________________________ Peak load current =

15.72 ⎛ R ⎞ (a) υ O1 = ⎜⎜1 + 2 ⎟⎟ ⋅υ I R1 ⎠ ⎝ R4 R ⎛ R ⎞ ⋅υ O1 = − 4 ⎜⎜1 + 2 ⎟⎟ ⋅υ I R3 R3 ⎝ R1 ⎠ υ L = υ O1 − υ O 2

υO2 = −

Aυ =

υ L ⎛ R 2 ⎞⎛ R 4 ⎞ ⎟ ⎟⎜1 + = ⎜1 + υ I ⎜⎝ R1 ⎟⎠⎜⎝ R3 ⎟⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ R ⎞ R (b) ⎜⎜1 + 2 ⎟⎟ = 12 ⇒ 2 = 11 , Set R 2 = 120 k Ω , then R1 = 10.9 k Ω R1 ⎠ R1 ⎝ R4 = 1 , Set R3 = R 4 = 120 k Ω R3

(c) (i) PL =

V L2 (16)2 = 5.12 W = 2 R L 2(25)

V L 16 = = 0.64 A R L 25 ______________________________________________________________________________________

(ii) I L =

15.73 (a) From Problem 15.72 R R υ O 2 = − 4 ⋅υ O1 , Set 4 = 1 R3 R3 ⎛ R ⎞⎛ R ⎞ Aυ = ⎜⎜1 + 2 ⎟⎟⎜⎜1 + 4 ⎟⎟ R1 ⎠⎝ R3 ⎠ ⎝ ⎛ R ⎞ R 25 = ⎜⎜1 + 2 ⎟⎟(2 ) ⇒ 2 = 11.5 R1 ⎠ R1 ⎝ Set R 2 = R3 = R 4 = 100 k Ω , then R1 = 8.69 k Ω ⎛ V ⎞⎛ I ⎞ ⎛ 24 ⎞⎛ 1.2 ⎞ ⎟ = 14.4 W ⎟⎜ (b) (i) PL = ⎜⎜ L ⎟⎟⎜⎜ L ⎟⎟ = ⎜⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠⎝ 2 ⎠ V 24 = 20 Ω (ii) R L = L = I L 1.2 V L2 (24)2 = 7.2 W = 2 R L′ 2(40 ) ______________________________________________________________________________________

(c) PL =

15.74 Line regulation =

ΔV0 ΔV +

Now ΔI =

ΔV + R1

and ΔVZ = rZ ⋅ ΔI and ΔV0 = 10ΔVZ

So ΔV0 = 10 ⋅ rZ ⋅

ΔV + R1

So ΔV0 10(15) = ⇒ 1.61% ΔV + 9300 ______________________________________________________________________________________

Line regulation =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.75

(a) R of =

ΔVO ΔI O

=

8 ⇒ R of = 4 m Ω 2

(b) ΔVO = Rof ⋅ ΔI O = (10 )(1.2 ) ⇒ ΔVO = 12 mV ______________________________________________________________________________________ 15.76

For V0 = 8 V V + (min) = V0 + I 0 (max) R11 + VBE11 + VBE10 + VEB 5

This assumes VBC 5 = 0. Then V + (min) = 8 + (0.1)(1.9) + 0.6 + 0.6 + 0.6 V + (min) = 9.99 V ______________________________________________________________________________________ 15.77 a. IC 3 = IC 5 =

VZ − 3VBE (npn) R1 + R2 + R3

6.3 − 3(0.6) = 0.571 mA 0.576 + 3.4 + 3.9 1 ⎛ 0.6 ⎞ = ⎜ ⎟ = 0.106 mA 2 ⎝ 2.84 ⎠

IC 3 = IC 5 = IC 8

Neglecting current in Q9 , total collector current and emitter current in Q5 is 0.571 + 0.106 = 0.677

Now I Z 2 R4 + VEB 4 = VEB 5 ⎛I ⎞ VEB 4 = VT ln ⎜ Z 2 ⎟ ⎝ I5 ⎠ ⎛I ⎞ VEB 5 = VT ln ⎜ C 5 ⎟ ⎝ 2I S ⎠ ⎛ I ⎞ I Z 2 R4 = VT ln ⎜ C 5 ⎟ ⎝ 2I Z 2 ⎠ Then ⎛ 0.677 ⎞ 0.026 R4 = ⋅ ln ⎜ ⎟ 0.25 ⎝ 2(0.25) ⎠

or R4 = 31.5 Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ b.

From Example 15.16, VB 7 = 3.43 V . Then ⎛ R13 ⎞ ⎜ ⎟ V0 = VB 8 = VB 7 ⎝ R12 + R13 ⎠ or

⎛ 2.23 ⎞ ⎜ ⎟ (12) = 3.43 ⎝ 2.23 + R12 ⎠ 3.43(2.23 + R12 ) = (2.23)(12) which yields R12 = 5.57 kΩ

______________________________________________________________________________________ 15.78 regulation =

Line Now

ΔV0 ΔV +

ΔVB 7 = ΔI C 3 ⋅ R1

⎛ R13 ⎞ ⎜ ⎟ (ΔV0 ) = ΔVB 7 = ΔI C 3 R1 R + R13 ⎠ and ⎝ 12 ΔI C 3 =

and ΔI Z =

and Then

ΔVZ ΔI Z ⋅ rZ = R1 + R2 + R3 R1 + R2 + R3 ΔV + r0

r0 =

where

VA IZ

⎛ 0.015 ⎞ (0.4288)(ΔV0 ) = ΔI C 3 (3.9) = (3.9)ΔI Z ⎜ ⎟ ⎝ 7.876 ⎠ 50 r0 = = 87.6 kΩ 0.571 Then ⎛ ΔV + ⎞ (0.4288)(ΔV0 ) = (0.00743) ⎜ ⎟ ⎝ 87.6 ⎠

So ΔV0 = 0.0198% ΔV + ______________________________________________________________________________________

15.79

(a) R1 + rz =

V + 25 = = 2.0833 k Ω IZ 12

R1 = 2.0833 − 0.012 = 2.0713 k Ω

(b) V + = 5.6 + (0.012 )(12 ) = 5.744 V For x = 0 , ⎛ R3 + R 4 ⎞ 1+ 2 +1⎞ ⎟ ⋅ VO ⇒ VO = ⎛⎜ 5.744 = ⎜⎜ ⎟(5.744) = 7.659 V ⎟ ⎝ 2 +1 ⎠ ⎝ R 2 + R3 + R 4 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For x = 1 , ⎛ ⎞ R4 1+ 2 +1⎞ ⎟ ⋅ VO ⇒ VO = ⎛⎜ 5.744 = ⎜⎜ ⎟(5.744) = 22.976 V ⎟ R + R + R ⎝ 1 ⎠ 3 4 ⎠ ⎝ 2 So 7.659 ≤ VO ≤ 22.976 V

(c) For x = 1 , V1 =

1 VO 4 1 4

υ d = V REF − V1 = V REF − VO ⎛I VO = AOLυ d − V BE = AOLυ d − VT ln⎜⎜ O ⎝ IS

⎞ ⎟ ⎟ ⎠

⎛I ⎞ 1 ⎛ ⎞ VO = AOL ⎜V REF − VO ⎟ − VT ln⎜⎜ O ⎟⎟ 4 ⎠ ⎝ ⎝ IS ⎠ ⎛I ⎞ ⎛ 1 ⎞ VO ⎜1 + AOL ⎟ = AOLV REF − VT ln⎜⎜ O ⎟⎟ ⎝ 4 ⎠ ⎝ IS ⎠ ⎛I ⎞ AOLV REF − VT ln⎜⎜ O ⎟⎟ ⎝ IS ⎠ VO = 1 1 + AOL 4 V (NL ) − VO (FL ) Load regulation = O VO (NL )

AOLV REF − VT ln[I O (NL ) I S ] ⎡ AOLV REF − VT ln[I O (FL ) I S ]⎤ −⎢ ⎥ 1 + (1 4 )AOL 1 + (1 4 )AOL ⎣ ⎦ = AOLV REF − VT ln[I O (NL ) I S ] 1 + (1 4)AOL

⎡ ⎛ I (FL ) ⎞ ⎛ I (NL ) ⎞⎤ ⎛ I (FL ) ⎞ ⎟ − ln⎜ O ⎟⎥ VT ⎢ln⎜⎜ O ⎟ VT ln⎜⎜ O ⎟ ⎜ I ⎟ ⎟ I S S ⎠ ⎝ ⎠⎦⎥ ⎣⎢ ⎝ ⎝ I O (NL ) ⎠ = = ⎛ I (NL ) ⎞ ⎛ I (NL ) ⎞ ⎟ ⎟ AOLV REF − VT ln⎜⎜ O AOLV REF − VT ln⎜⎜ O ⎟ ⎟ ⎝ IS ⎠ ⎝ IS ⎠ Let I O (FL ) = 5 A, I O (NL ) = 1% = 0.05 A ⎛ I (NL ) ⎞ ⎟ ≅ 0.7 V Let VT ln⎜⎜ O ⎟ ⎝ IS ⎠

(0.026) ln⎛⎜

5 ⎞ ⎟ 0.11973 ⎝ 0.05 ⎠ = Load regulation = 5 × 10 3 (5.744) − 0.7 2.8719 × 10 4

(

)

Load regulation = 4.169 × 10 −4 % ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 15.80

(a) I E =

V Z 6.8 = = 1.36 mA R2 5

⎛ β I O = ⎜⎜ ⎝ 1+ β For V BC = 0 ,

⎞ ⎛ 80 ⎞ ⎟⎟ ⋅ I E = ⎜ ⎟(1.36 ) = 1.343 mA ⎝ 81 ⎠ ⎠ VO = 20 − V Z − V EB = 20 − 6.8 − 0.6 = 12.6 V V 12.6 R L (max ) = O = = 9.38 k Ω I O 1.343

So 0 ≤ R L ≤ 9.38 k Ω ⎛ R1 (b) V + = ⎜⎜ ⎝ R1 + rZ

⎛ 10 4 ⎞ + ⎞ + ⎟ ⎟⎟ V − V ZO = ⎜ 4 ⎜ 10 + 20 ⎟ V − 6.8 ⎠ ⎝ ⎠

(

)

(

)

⎛ 10 4 ⎞ ⎟(13.2) = 13.17365 V For V + = 20 V, V + = ⎜⎜ 4 ⎟ ⎝ 10 + 20 ⎠ ⎛ 80 ⎞⎛ 20 − 13.17365 ⎞ I O = ⎜ ⎟⎜ ⎟ = 1.3484 mA 5 ⎝ 81 ⎠⎝ ⎠ ⎛ 10 4 ⎞ ⎟(16 − 6.8) = 9.181637 V For V + = 16 V, V + = ⎜⎜ 4 ⎟ ⎝ 10 + 20 ⎠ ⎛ 80 ⎞⎛ 16 − 9.181637 ⎞ I O = ⎜ ⎟⎜ ⎟ = 1.3468 mA 5 ⎝ 81 ⎠⎝ ⎠

So 1.3468 ≤ I O ≤ 1.3484 mA ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 16 16.1 ⎛ k ′ ⎞⎛ W ⎞ (a) υ O = V DD − ⎜⎜ n ⎟⎟⎜ ⎟ R D 2(υ I − VTN )υ O − υ O2 ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.1 ⎞⎛ W ⎞ 2 0.1 = 3.3 − ⎜ ⎟⎜ ⎟(40) 2(3.3 − 0.5)(0.1) − (0.1) 2 L ⎝ ⎠⎝ ⎠

[

]

[

]

⎛W ⎞ ⎛W ⎞ 0.1 = 3.3 − ⎜ ⎟(1.1) ⇒ ⎜ ⎟ = 2.91 ⎝L⎠ ⎝L⎠ ⎛ k ′ ⎞⎛ W ⎞ (b) ⎜⎜ n ⎟⎟⎜ ⎟ R DVO2t + VO t − V DD = 0 ⎝ 2 ⎠⎝ L ⎠ ⎛ 0.1 ⎞ 2 ⎜ ⎟(2.91)(40 )VO t + VO t − 3.3 = 0 ⎝ 2 ⎠ 5.82VO2t + VO t − 3.3 = 0 ⇒ VO t = 0.672 V

VO t = V I t − VTN ⇒ V I t = 1.172 V 3.3 − 0.1 ⇒ i D , max = 80 μ A 40 = (80 )(3.3) = 264 μ W

(c) i D , max = PD , max

______________________________________________________________________________________ 16.2 (a) (i) K n R DVO2t + VO t − V DD = 0

(0.05)(100)VO2t + VO t − 3.3 = 0 ⇒ VO t

[

= 0.7185 V

⇒ V I t = 1.219 V

(ii) υ O = 3.3 − (0.05)(100 ) 2(3.3 − 0.5)υ O − υ O2

]

We find 5υ − 29υ O + 3.3 = 0 ⇒ υ O = 0.116 V 2 O

(b) (i) (0.05)(30)VO2t + VO t − 3.3 = 0 ⇒ VO t = 1.187 V

[

⇒ V I t = 1.687 V

(ii) υ O = 3.3 − (0.05)(30 ) 2(3.3 − 0.5)υ O − υ O2

]

Or 1.5υ − 9.4υ O + 3.3 = 0 ⇒ υ O = 0.373 V 2 O

(c) (i) (0.05)(5)VO2t + VO t − 3.3 = 0 ⇒ VO t = 2.147 V

[

⇒ V I t = 2.647 V

(ii) υ O = 3.3 − (0.05)(5) 2(3.3 − 0.5)υ O − υ O2

]

Or 0.25υ − 2.4υ O + 3.3 = 0 ⇒ υ O = 1.663 V ______________________________________________________________________________________ 2 O

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.3 (a) P = IV 0.25 = I (3.3) ⇒ I = 75.76 μ A 3.3 − 0.15 = 41.6 k Ω 0.07576 ⎛ k ′ ⎞⎛ W ⎞ 2 I = ⎜⎜ n ⎟⎟⎜ ⎟ 2(VGS − VTN )V DS − V DS ⎝ 2 ⎠⎝ L ⎠

R=

[

]

[

]

⎛W ⎞ ⎛ 100 ⎞⎛ W ⎞ 2 75.76 = ⎜ ⎟⎜ ⎟ 2(3.3 − 0.5)(0.15) − (0.15) ⇒ ⎜ ⎟ = 1.85 2 L ⎝L⎠ ⎠⎝ ⎠ ⎝ V − V DS (sat ) ⎛ k ′ ⎞⎛ W ⎞ 2 (b) I D = ⎜⎜ n ⎟⎟⎜ ⎟(VGS − VTN ) = DD R ⎝ 2 ⎠⎝ L ⎠ 3.3 − V DS (sat ) ⎛ 0.1 ⎞ 2 ⎜ ⎟(1.85)V DS (sat ) = 41.6 ⎝ 2 ⎠

2 (sat ) + V DS (sat ) − 3.3 = 0 ⇒ V DS (sat ) = 0.805 V Or 3.848V DS V DS (sat ) = VGS − VTN 0.805 = VGS − 0.5 ⇒ VGS = 1.305 V Then 0.5 ≤ VGS ≤ 1.305 V ______________________________________________________________________________________

16.4 (a) From Equation (16.21) (V DD − υ O − VTNL )2 K D (W L ) D = = (W L )L 2(υ I − VTND )υ O − υ O2 KL =

(W L )D (W L )L

(1.8 − 0.08 − 0.4)2 2 2(1.4 − 0.4)(0.08) − (0.08)

=

1.7424 0.1536

= 11.34

PD , max = i D , max ⋅ V DD

0.3 = i D , max (1.8) ⇒ i D , max = 0.1667 mA

[

⎛ 0.1 ⎞⎛ W ⎞ 2 i D , max = 0.1667 = ⎜ ⎟⎜ ⎟ 2(1.4 − 0.4 )(0.08) − (0.08) ⎝ 2 ⎠⎝ L ⎠ D

]

⎛W ⎞ ⎛W ⎞ Which yields ⎜ ⎟ = 21.7 and ⎜ ⎟ = 1.91 L ⎝ ⎠D ⎝ L ⎠L (b) V I t =

(

1.8 − 0.4 + (0.4 ) 1 + 11.34

) = 0.7206 V

1 + 11.34 0.4 ≤ V I ≤ 0.7206 V ______________________________________________________________________________________

16.5

(a)

[

]

KD 2 2 2(3 − 0.5)(0.25) − (0.25) = [3 − 0.25 − 0.5] KL KD K (1.1875) = (5.0625) ⇒ D = 4.26 KL KL

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

[

]

KD 2 2 2(2.5 − 0.5)(0.25) − (0.25) = [3 − 0.25 − 0.5] KL KD K (0.9375) = (5.0625) ⇒ D = 5.4 KL KL

⎛ 0.1 ⎞ 2 2 (c) i D = K L (VGSL − VTN ) = ⎜ ⎟(1)(3 − 0.25 − 0.5) = 0.253 mA ⎝ 2 ⎠ P = i D ⋅ V DD = (0.253)(3) = 0.759 mW ______________________________________________________________________________________ 16.6

(V DD − υ O − VTNL ) KD = K L 2(υ I − VTND )υ O − υ O2 2

(a)

υ I = Logic 1 = V DD − VTNL = 3 − 0.5 = 2.5 V

K D (W L ) D (3 − 0.1 − 0.5)2 5.76 = = = (W L )L 2(2.5 − 0.5)(0.1) − (0.1)2 0.39 KL (W L )D = 14.77 (W L )L PD , max = i D , max ⋅ V DD

0.4 = i D , max (3) ⇒ i D , max = 0.1333 mA

[

⎛ 0.1 ⎞⎛ W ⎞ 2 i D , max = 0.1333 = ⎜ ⎟⎜ ⎟ 2(2.5 − 0.5)(0.1) − (0.1) ⎝ 2 ⎠⎝ L ⎠ D

]

⎛W ⎞ ⎛W ⎞ Which yields ⎜ ⎟ = 6.84 and ⎜ ⎟ = 0.463 L ⎝ ⎠D ⎝ L ⎠L

(b) V I t =

(

3 − 0.5 + (0.5) 1 + 14.77

1 + 14.77 VO t = 1.016 − 0.5 = 0.516 V

) = 1.016 V

______________________________________________________________________________________ 16.7 We have KD 2 ⎡ 2 ( vI − VTND ) vO − vO2 ⎤⎦ = (VDD − vO − VTNL ) KL ⎣

(W / L )D ⎡ 2 2 2 (V − V − V )( 0.08VDD ) − ( 0.08VDD ) ⎤ = (VDD − 0.08VDD − VTN ) ⎦ (W / L )L ⎣ DD TN TN (W / L )D ⎡ 2 2 2 ⎤⎦ = ⎣⎡( 0.92 − 0.2 ) VDD ⎦⎤ = 0.5184VDD 2 (VDD − 2 ( 0.2 )VDD ) ( 0.08VDD ) − 0.0064VDD ⎣ (W / L )L (W / L )D (W / L ) D = 5.4 [0.096] = 0.5184 ⇒ (W / L )L (W / L )L ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.8

VOH = VB − VTN = Logic 1

So

VB = 4 V ⇒ VOH = 3 V

(a)

VB = 5 V ⇒ VOH = 4 V

(b)

VB = 6 V ⇒ VOH = 5 V

(c)

VB = 7 V ⇒ VOH = 5 V ,since VDS = 0

(d)

For vI = VOH K D ⎡⎣ 2 ( vI − VT ) vO − vO2 ⎤⎦ = K L [VB − vO − VT ]

2

Then (a)

(1) ⎡⎣ 2 ( 3 − 1)VOL − VOL2 ⎤⎦ = ( 0.4 ) [ 4 − VOL − 1]

(b)

2

(c) (d)

⇒ VOL = 0.657 V

2

(1) ⎡⎣ 2 ( 4 − 1)VOL − VOL2 ⎤⎦ = ( 0.4 ) [5 − VOL − 1]

⇒ VOL = 0.791 V

2

(1) ⎡⎣ 2 ( 5 − 1)VOL − VOL2 ⎤⎦ = ( 0.4 ) [6 − VOL − 1]

⇒ VOL = 0.935 V

Load in non-sat region iDD = iOL 2 2 ⎤⎦ = ( 0.4 ) ⎡ 2 ( 7 − VOL − 1)( 5 − VOL ) − ( 5 − VOL ) ⎤ (1) ⎡⎣ 2 ( 5 − 1)VOL − VOL ⎣ ⎦ 2 2 8VOL − VOL = ( 0.4 ) ⎡⎣ 2 ( 6 − VOL )( 5 − VOL ) − ( 25 − 10VOL + VOL ) ⎤⎦

2 = ( 0.4 ) ⎡⎣ 2 ( 30 − 11VOL + VOL ) − 25 + 10VOL − VOL2 ⎤⎦ 2 2 ⎤⎦ = ( 0.4 ) ⎡⎣60 − 22VOL + 2VOL − 25 + 10VOL − VOL

2 2 8VOL − VOL = 14 − 4.8VOL + 0.4VOL 2 1.4VOL − 12.8VOL + 14 = 0

VOL =

12.8 ± 163.84 − 4 (1.4 )(14 ) 2 (1.4 )

VOL = 1.27V

For load

VDS ( sat ) = 7 − 1.27 − 1 = 4.73V VDS = 5 − 1.27 = 3.73 non-sat

______________________________________________________________________________________ 16.9

(a)

(

)

KD V I t − VTND = −VTNL KL

(

)

500 V I t − 0.5 = −(− 0.8) ⇒ V I t = 0.8578 V 100 For Driver: VO t = V I t − VTND = 0.8578 − 0.5 = 0.3578 V

For Load: (b)

[

VO t = V DD + VTNL = 3.3 + (− 0.8) = 2.5 V

]

KD 2 2(υ I − VTND )υ O − υ O2 = (− VTNL ) KL

[

]

⎛ 500 ⎞ 2 2 ⎟ 2(3.3 − 0.5)υ O − υ O = [− (− 0.8)] ⎜ ⎝ 100 ⎠

We find 5υ O2 − 28υ O + 0.64 = 0 ⇒ υ O = 0.0230 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) i D , max = K L (− VTNL ) = 100[− (− 0.8)] = 64 μ A 2

2

PD , max = i D , max ⋅ V DD = (64)(3.3) = 211 μ W

______________________________________________________________________________________ 16.10 2 2 ⎛ 500 ⎞ ⎡ ⎜ ⎟ 2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤⎦ = ( −VTNL ) ⎝ 50 ⎠ ⎣

So

( −VTNL )

2

= 4.9 ⇒ VTNL = −2.21 V

______________________________________________________________________________________ 16.11 (a) PD , max = i D , max ⋅ V DD

80 = i D , max (1.8) ⇒ i D , max = 44.44 μ A

⎛W ⎞ ⎛ 100 ⎞⎛ W ⎞ 2 i D , max = 44.44 = ⎜ ⎟⎜ ⎟ [− (− 0.6)] ⇒ ⎜ ⎟ = 2.47 ⎝ L ⎠L ⎝ 2 ⎠⎝ L ⎠ L KD 2 2 2(1.8 − 0.3)(0.06 ) − (0.06 ) = [− (− 0.6 )] KL (W L )D KD K (0.1764) = (0.36) ⇒ D = 2.04 = (W L )L KL KL

[

]

⎛W ⎞ Then ⎜ ⎟ = 5.04 ⎝ L ⎠D

(

)

KD V I t − VTND = −VTNL KL

(b)

(

)

5.04 V I t − 0.3 = [− (− 0.6 )] ⇒ V I t = 0.720 V 2.47 For Driver: VO t = 0.720 − 0.3 = 0.420 V

For Load:

VO t = 1.8 − 0.6 = 1.2 V

(c) PD , max = 80 μ W

[

]

⎛ 5.04 ⎞ 2 2 2⎜ ⎟ 2(1.8 − 0.3)υ O − υ O = [− (− 0.6)] 2 . 47 ⎠ ⎝

We find 4.08υ O2 − 12.24υ O + 0.36 = 0 ⇒ υ O = 0.0297 V ______________________________________________________________________________________ 16.12 a.

From Equation (16.27(b)): 2 ⎛W ⎞ ⎡ ⎛W ⎞ 2 ⎜ ⎟ ⎣ 2 ( 2.5 − 0.5 )( 0.05 ) − ( 0.05 ) ⎤⎦ = ⎜ ⎟ [− ( −1)] ⎝ L ⎠D ⎝ L ⎠L ⎛W ⎞ ⎜ ⎟ =1 ⎝ L ⎠L

Then

⎛W ⎞ ⎜ ⎟ = 5.06 ⎝ L ⎠D

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

b. or

2 ⎛ 80 ⎞ iD = ⎜ ⎟ (1) ⎡⎣ − ( −1) ⎤⎦ ⎝ 2⎠

iD = 40 μ A P = iD ⋅ VDD = ( 40 )( 2.5 ) ⇒ P = 100 μ W

______________________________________________________________________________________ 16.13

a.

vI = 0.5 V ⇒ iD = 0 ⇒ P = 0

i.

vI = 5 V, From Equation (16.11), v0 = 5 − ( 0.1)( 20 ) ⎡⎣ 2 ( 5 − 1.5 ) v0 − v02 ⎤⎦

ii.

2v02 − 15v0 + 5 = 0 v0 =

15 ±

(15 ) − 4 ( 2 )( 5) ⇒ v0 = 0.35 V 2 ( 2) 2

5 − 0.35 = 0.2325 mA 20 P = iD ⋅ VDD = ( 0.2325 )( 5 ) ⇒ P = 1.16 mW

iD =

b. ii.

vI = 0.25 V ⇒ iD = 0 ⇒ P = 0

i.

vI = 4.3 V, From Equation (16.21), 2 100 ⎡⎣ 2 ( 4.3 − 0.7 ) v0 − v02 ⎤⎦ = 10 [5 − v0 − 0.7 ] 10 ⎡⎣7.2v0 − v02 ⎤⎦ = 18.49 − 8.6v0 + v02

Then 11v02 − 80.6v0 + 18.49 = 0 v0 =

80.6 ±

(80.6 ) − 4 (11)(18.49 ) ⇒ v0 = 0.237 V 2 (11) 2

Then iD = 10 [5 − 0.237 − 0.7 ] = 165 μ A 2

P = iD ⋅ VDD = (165 )( 5 ) ⇒ P = 825 μ W

c.

vI = 0.03 V ⇒ iD = 0 ⇒ P = 0

i.

vI = 5 V iD = K L ( −VTNL ) = (10 ) ⎣⎡ − ( −2 ) ⎦⎤ = 40 μ A 2

2

P = iD ⋅ VDD = ( 40 )( 5 ) ⇒ P = 200 μ W ii. ______________________________________________________________________________________

16.14 (a) υ I = V DD − VTNL = 5 − 0.5 = 4.5 V

[

]

K D 2(υ I − VTND )υ O1 − υ O2 1 = K L [V DD − υ O1 − VTNL ]

[

]

2

10 2(4.5 − 0.5)υ O1 − υ O2 1 = (1)[5 − υ O1 − 0.5]

2

We find 11υ O21 − 89υ O1 + 20.25 = 0 ⇒ υ O1 = 0.234 V υ O 2 = V DD − VTNL = 4.5 V (b) υ I = 0.234 V, ⇒ υ O1 = 4.5 V From part (a), υ O 2 = 0.234 V ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.15

(a)

[

]

KD 2 2(υ I − VTND )υ O − υ O2 = (− VTNL ) KL

[

]

⎛4⎞ 2 2 ⎜ ⎟ 2(5 − 0.8)υ O1 − υ O1 = [− (− 1.2 )] 1 ⎝ ⎠

We find 4υ O2 1 − 33.6υ O1 + 1.44 = 0 ⇒ υ O1 = 0.0431 V ⇒ υ O2 = 5 V (b) For υ I = 0.0431 V, υ O1 = 5 V From part (a), υ O 2 = 0.0431 V ______________________________________________________________________________________ 16.16 (a) υ O = V DD − VTNLO = 2.5 − 0.5 = 2.0 V

[

( 2φ + υ − 2φ )] = 2.5 − [0.5 + 0.25( 0.7 + υ − 0.7 )]

(b) υ O = V DD − VTNLO + γ

υO

fp

O

fp

O

υ O − 2.209 = −0.25 0.7 + υ O υ O2 − 4.418υ O + 4.88 = 0.0625(0.7 + υ O ) υ O2 − 4.4805υ O + 4.836 = 0 ⇒ υ O = 1.81 V ______________________________________________________________________________________ 16.17

(a)

KD (υ I − VTND ) = (− VTNL ) KL

Or υ I = [− (VTNL )] (b) VTNL = VTNLO + γ

(

KL 20 + VTND = [− (− 0.6)] + 0.4 = 0.6683 V KD 100

) 0.7 ) = −0.460 V

2φ fp + υ O − 2φ fp

(

= −0.6 + 0.25 0.7 + 1.25 −

20 + 0.4 = 0.6057 V 100 ______________________________________________________________________________________

υ I = [− (− 0.460)]

6.18

(a)

[

]

KD 2 2(υ I − VTND )υ O − υ O2 = (− VTNL ) KL

[

]

KD 2 2 2(1.8 − 0.4 )(0.1) − (0.1) = [− (− 0.6 )] KL KD K (0.27 ) = 0.36 ⇒ D = 1.33 KL KL

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

[ 2(1.333)[2(1.8 − 0.4)υ

⎛K (b) 2⎜⎜ D ⎝ KL

]

⎞ ⎟⎟ 2(υ I − VTND )υ O − υ O2 = [− VTNL ]2 ⎠ O

]

− υ O2 = [− (− 0.6 )]

2

We find 2.667υ O2 − 7.467υ O + 0.36 = 0 ⇒ υ O = 49.1 mV ⎛ k ′ ⎞⎛ W ⎞ ⎛ 0.1 ⎞ 2 2 (c) i D , max = ⎜⎜ n ⎟⎟⎜ ⎟ (− VTNL ) = ⎜ ⎟(1)[− (− 0.6)] ⇒ i D , max = 18 μ A ⎝ 2 ⎠ ⎝ 2 ⎠⎝ L ⎠ L P = i D , max ⋅ V DD = (18)(1.8) = 32.4 μ W

______________________________________________________________________________________ 6.19 (a) One input high, KD K 2 2 2(3 − 0.5)(0.1) − (0.1) = [− (− 1)] ⇒ D = 2.04 KL KL

[

]

(b) P = i D ⋅ V DD 0.1 = i D (3) ⇒ i D = 33.33 μ A ⎛ k′ i D = ⎜⎜ n ⎝ 2

⎞⎛ W ⎞ 2 ⎟⎟⎜ ⎟ [− VTNL ] ⎠⎝ L ⎠ L

⎛W ⎞ ⎛ 100 ⎞⎛ W ⎞ 33.33 = ⎜ ⎟⎜ ⎟ (1) ⇒ ⎜ ⎟ = 0.667 ⎝ L ⎠L ⎝ 2 ⎠⎝ L ⎠ L

⎛W ⎞ Then ⎜ ⎟ = (2.04)(0.667 ) = 1.36 ⎝ L ⎠D

[

]

(c) 3(2.04 ) 2(3 − 0.5)υ O − υ O2 = [− (− 1)]

2

6.12υ O2 − 30.6υ O + 1 = 0 ⇒ υ O = 0.0329 V ______________________________________________________________________________________

16.20

(a)

[

]

KD 2 2 2(2.5 − 0.4)(0.05) − (0.05) = [− (− 0.6 )] KL KD K (0.2075) = 0.36 ⇒ D = 1.735 KL KL

(b) P = i D , max ⋅ V DD

50 = i D , max (2.5) ⇒ i D , max = 20 μ A ⎛ 100 ⎞⎛ W ⎞ 2 i D , max = 20 = ⎜ ⎟⎜ ⎟ [− (− 0.6 )] L 2 ⎠⎝ ⎠ L ⎝

⎛W ⎞ ⎛W ⎞ We find ⎜ ⎟ = 1.11 and ⎜ ⎟ = 1.93 ⎝ L ⎠L ⎝ L ⎠D

[

]

(c) (i) 2(1.735) 2(2.5 − 0.4)υ O − υ O2 = 0.36

3.47υ − 14.574υ O + 0.36 = 0 ⇒ υ O = 24.9 mV 2 O

[

]

(ii) 3(1.735) 2(2.5 − 0.4 )υ O − υ O2 = 0.36

5.205υ − 21.861υ O + 0.36 = 0 ⇒ υ O = 16.5 mV 2 O

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

[

]

(iii) 4(1.735) 2(2.5 − 0.4)υ O − υ O2 = 0.36 6.94υ O2 − 29.148υ O + 0.36 = 0 ⇒ υ O = 12.4 mV ______________________________________________________________________________________

16.21 a. P = iD ⋅ VDD

250 = iD ( 5 ) ⇒ iD = 50 μΑ ⎛ k' ⎞⎛W ⎞ 2 iD = ⎜ n ⎟ ⎜ ⎟ [ −VTNL1 ] ⎝ 2 ⎠ ⎝ L ⎠ ML1 2 ⎛ 60 ⎞ ⎛ W ⎞ 50 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ − ( −2 ) ⎤⎦ ⎝ 2 ⎠ ⎝ L ⎠ ML1

So that

⎛W ⎞ ⎜ ⎟ = 0.417 ⎝ L ⎠ ML1

KD 2 ⎡ 2 ( vI − VTND ) vO − vO2 ⎤⎦ = [ −VTNL ] KL ⎣ 2 KD ⎡ 2 2 ( 5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤ = ⎡⎣ − ( −2 ) ⎤⎦ ⎦ KL ⎣

or

KD ⎛W ⎞ = 3.23 ⇒ ⎜ ⎟ = 1.35 KL ⎝ L ⎠ MD1

b. For vX = vY = 0 ⇒ v01 = 5 and v03 = 4.2 Then K D 2 ⎣⎡ 2 ( vO1 − VTND ) vO 2 − vO2 2 ⎤⎦ + K D 3 ⎡⎣ 2 ( vO 3 − VTND ) vO 2 − vO2 2 ⎤⎦ = K L 2 [ −VTNL 2 ]

2

K D 2 ∝ 8, K D 3 ∝ 8, K L 2 ∝ 1 2 2 ⎤⎦ + 8 ⎡⎣ 2 ( 4.2 − 0.8 ) v02 − v02 ⎤⎦ = (1) ⎡⎣ − ( −2 ) ⎤⎦ 8 ⎡⎣ 2 ( 5 − 0.8 ) v02 − v02

2

2 2 + 54.4v02 − 8v02 =4 67.2v02 − 8v02

Then 16v02 − 121.6v0 + 4 = 0 v02 =

121.6 ±

(121.6 ) − 4 (16 )( 4 ) 2 (16 ) 2

v = 0.0330 V

So 02 ______________________________________________________________________________________ 16.22

⎛ 100 ⎞ 2 2 2 (a) i D , max = ⎜ ⎟(1)[V DD − υ O − VTN ] = 50[3.3 − υ O − 0.4] = 50[2.9 − υ O ] ⎝ 2 ⎠ The output is small, so neglect υ O2 . Then i D , max ≅ 50[8.41 − 5.8υ O ] (Eq. 1) Also,

[

⎛ 100 ⎞ 2 i D , max = ⎜ ⎟(12 ) 2(υ GSX − VTN )υ DSX − υ DSX ⎝ 2 ⎠

[

]

]

2 = 600 2(2.9 − 0.4)υ DSX − υ DSX ≅ 3000υ DSX

(Eq. 2)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ And

[

]

⎛ 100 ⎞ 2 i D , max = ⎜ ⎟(12 ) 2(υ GSY − VTN )υ DSY − υ DSY ⎝ 2 ⎠ We have υ GSY = υ Y − υ DSX ; υ DSY = υ O − υ DSX

[

i D , max = 600 2(2.9 − υ DSX − 0.4 )(υ O − υ DSX ) − (υ O − υ DSX ) ≅ 600[2(2.5 − υ DSX )(υ O − υ DSX )]

Now Also

2

]

(Eq.3)

3000υ DSX = 50[8.41 − 5.8υ O ] υ DSX = 0.016667[8.41 − 5.8υ O ] (Eq. 4) 3000υ DSX = 600[2(2.5 − υ DSX )(υ O − υ DSX )] ≅ 1200[2.5υ O − 2.5υ DSX ] 2.5υ DSX = 2.5υ O − 2.5υ DSX

Or υ DSX ≅ 0.5υ O Then from Eq. 4, 0.5υ O = 0.140 − 0.09667υ O ⇒ υ O ≅ 0.2346 V (b) υ DSX ≅ 0.5υ O ≅ 0.1173 V υ GSX = 2.9 V υ GSY ≅ 2.9 − 0.1173 ≅ 2.783 V υ DSY = υ O − υ DSX ≅ 0.1173 V ______________________________________________________________________________________ 16.23 a.

We can write 2 2 ⎤⎦ = K y ⎡⎣ 2 ( vY − vDSX − VTNY ) vDSY − vDSY ⎤⎦ = K L [ −VTNL ] K x ⎡⎣ 2 ( v X − VTNX ) vDSX − vDSX

2

2 From the first and third terms, (neglect vDSX ),

4 ⎡⎣ 2 ( 5 − 0.8 ) vDSX ⎤⎦ = (1) ⎡⎣ − ( −1.5 ) ⎤⎦

or

2

vDSX = 0.067 V

2 From the second and third terms, (neglect vDSY ),

4 ⎡⎣ 2 ( 5 − 0.067 − 0.8 ) vDSY ⎤⎦ = (1) ⎡⎣ − ( −1.5 ) ⎤⎦ v

or DSY Now and

2

= 0.068 V

vGSX = 5, vGSY = 5 − 0.067 ⇒ vGSY = 4.933 V v0 = vDSX + vDSY ⇒ v0 = 0.135 V

Since v0 is close to ground potential, the body-effect has little effect on the results. ______________________________________________________________________________________ 16.24

(a)

[

]

1 ⎛ KD ⎜ 4 ⎜⎝ K L

⎞ ⎟⎟ 2(υ I − VTND )υ O − υ O2 = (− VTNL )2 ⎠

1 ⎛ KD ⎜ 4 ⎜⎝ K L

⎞ ⎟⎟ 2(3.3 − 0.4 )(0.1) − (0.1)2 = [− (− 0.6)]2 ⎠

⎛ KD ⎜⎜ ⎝ KL

[

⎞ ⎛K ⎟⎟(0.1425) = 0.36 ⇒ ⎜⎜ D ⎠ ⎝ KL

]

⎞ ⎟⎟ = 2.53 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) P = i D , max ⋅ V DD

100 = i D , max (3.3) ⇒ i D , max = 30.3 μ A ⎛ 100 ⎞⎛ W ⎞ 2 i D , max = 30.3 = ⎜ ⎟⎜ ⎟ [− (− 0.6 )] ⎝ 2 ⎠⎝ L ⎠ L

⎛W ⎞ ⎛W ⎞ Which yields ⎜ ⎟ = 1.68 and ⎜ ⎟ = 4.26 ⎝ L ⎠L ⎝ L ⎠D ______________________________________________________________________________________ 16.25 Y = [A OR (B AND C)] AND D ______________________________________________________________________________________

16.26 Considering a truth table, we find A B Y 0 0 0 0 1 1 1 0 1 1 1 0 which shows that the circuit performs the exclusive-OR function. ______________________________________________________________________________________ 16.27 ( A + B)(C + D )

______________________________________________________________________________________ 16.28

(a)

Carry-out = A • ( B + C ) + B • C

(b)

For vO1 = Low = 0.2 V 2 KD ⎡ 2 2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = ⎡⎣ − ( −1.5 ) ⎤⎦ ⇒ ⎦ KL ⎣

⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = 1, ⎜ ⎟ = 1.37 then ⎝ L ⎠ D For ⎝ L ⎠ L ⎛W ⎞ M 6 : ⎜ ⎟ = 1.37 ⎝ L ⎠6

So, for

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ To achieve the required composite conduction parameter, ⎛W ⎞ M 1 − M 5 : ⎜ ⎟ = 2.74 ⎝ L ⎠1−5

For ______________________________________________________________________________________ 16.29 (a)

(b)

(W L ) A

[

]

2(2.5 − 0.4)(0.05) − (0.05) = [− (− 0.6 )] 1 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ (0.2075) = 0.36 ⇒ ⎜ ⎟ = 1.735 ⎝ L ⎠A ⎝ L ⎠A 2

2

⎛W ⎞ ⎛W ⎞ ⇒⎜ ⎟ = 2⎜ ⎟ = 3.47 ⎝ L ⎠ B ,C , D ⎝ L ⎠A ______________________________________________________________________________________

16.30 Design Problem ______________________________________________________________________________________ 16.31 V DD 2.5 = = 1.25 V 2 2 = 1.25 − (− 0.4 ) = 1.65 V

(a) By symmetry, V I t = VOP t

VON t = 1.25 − 0.4 = 0.85 V

(c) For υ I = 1.1 V, NMOS in saturation, PMOS in nonsaturation

[

K n [υ I − VTN ] = K p 2(V DD − υ I + VTP )(V DD − υ O ) − (V DD − υ O ) 2

(1.1 − 0.4) = 2(2.5 − 1.1 − 0.4)(V DD − υ O ) − (V DD − υ O ) (V DD − υ O )2 − 2(V DD − υ O ) + 0.49 = 0 ⇒ (V DD − υ O ) = 0.2859 2

2

Or υ O = 2.5 − 0.2859 = 2.214 V For υ I = 1.4 V, NMOS in nonsaturation, PMOS in saturation

[

2

]

K n 2(υ I − VTN )υ O − υ O2 = K p (V DD − υ I + VTP )

2(1.4 − 0.4 )υ O − υ O2 = (2.5 − 1.4 − 0.4 )

2

υ O2 − 2υ O + 0.49 = 0 ⇒ υ O = 0.286 V

2

V

]

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 16.32 ⎛ 100 ⎞ ⎛ 40 ⎞ 2 2 (a) K n = ⎜ ⎟(2 ) = 100 μ A/V ; K p = ⎜ ⎟(5) = 100 μ A/V ⎝ 2 ⎠ ⎝ 2 ⎠ (i) By symmetry V 3.3 V I t = DD = = 1.65 V 2 2 ⇒ VOP t = 1.65 − (− 0.4) = 2.05 V

⇒ VON t = 1.65 − 0.4 = 1.25 V

(iii) For υ O = 0.25 V; NMOS in nonsaturation, PMOS in saturation 2(υ I − VTN )υ O − υ O2 = (V DD − υ I + VTP )

2

2(υ I − 0.4 )(0.25) − (0.25) = (3.3 − υ I − 0.4 ) 2

2

0.5υ I − 0.2 − 0.0625 = 8.41 − 5.8υ I + υ I2

υ I2 − 6.3υ I + 8.6725 = 0 ⇒ υ I = 2.03 V For υ O = 3.05 V; NMOS in saturation, PMOS in nonsaturation

(υ I − VTN )2 = 2(V DD − υ I + VTP )(V DD − υ O ) − (V DD − υ O )2 (υ I − 0.4)2 = 2(3.3 − υ I − 0.4)(3.3 − 3.05) − (3.3 − 3.05)2 υ I2 − 0.8υ I + 0.16 = 0.5(2.9 − υ I ) − 0.0625

υ I2 − 0.3υ I − 1.2275 = 0 ⇒ υ I = 1.27 V ⎛ 100 ⎞ ⎛ 40 ⎞ 2 2 (b) K n = ⎜ ⎟(4) = 200 μ A/V ; K p = ⎜ ⎟(5) = 100 μ A/V 2 2 ⎝ ⎝ ⎠ ⎠ 3.3 − 0.4 +

(i) V I t = ⇒ VOP t

200 (0.4) 100

= 1.436 V 200 1+ 100 = 1.436 + 0.4 = 1.836 V

⇒ VON t = 1.436 − 0.4 = 1.036 V

(iii) For υ O = 0.25 V; NMOS in nonsaturation, PMOS in saturation

[

]

200 2(υ I − 0.4)(0.25) − (0.25) = 100(3.3 − υ I − 0.4 ) 2

2

2(0.5υ I − 0.2 − 0.0625) = 8.41 − 5.8υ I + υ I2

υ I2 − 6.8υ I + 8.935 = 0 ⇒ υ I = 1.78 V For υ O = 3.05 V ; NMOS in saturation, PMOS in nonsaturation

[

200(υ I − 0.4) = 100 2(3.3 − υ I − 0.4 )(3.3 − 3.05) − (3.3 − 3.05) 2

(

)

2 υ I2 − 0.8υ I + 0.16 = 0.5(2.9 − υ I ) − 0.0625

2

]

2υ − 1.1υ I − 1.0675 = 0 ⇒ υ I = 1.06 V ______________________________________________________________________________________ 2 I

16.33 ⎛ 100 ⎞ ⎛ 40 ⎞ 2 2 (a) K n = ⎜ ⎟(4) = 200 μ A/V ; K p = ⎜ ⎟(12 ) = 240 μ A/V 2 2 ⎝ ⎠ ⎝ ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.3 − 0.4 +

200 (0.4) 240

= 1.707 V 200 1+ 240 ⇒ VOP t = 1.707 + 0.4 = 2.107 V

(i) V I t =

⇒ VON t = 1.707 − 0.4 = 1.307 V

(ii) For υ O = 3.1 V ; NMOS in saturation, PMOS in nonsaturation

[

200(υ I − 0.4) = 240 2(3.3 − υ I − 0.4 )(3.3 − 3.1) − (3.3 − 3.1) 2

2

]

υ I2 − 0.8υ I + 0.16 = 1.2[0.4(2.9 − υ I ) − 0.04]

υ I2 − 0.32υ I − 1.184 = 0 ⇒ υ I = 1.26 V (iii) For υ O = 0.2 V ; NMOS in nonsaturaton, PMOS in saturation

[

]

200 2(υ I − 0.4 )(0.2) − (0.2) = 240(3.3 − υ I − 0.4 ) 2

(

0.4υ I − 0.2 = 1.2 8.41 − 5.8υ I + υ I2

2

)

1.2υ I2 − 7.36υ I + 10.292 = 0 ⇒ υ I = 2.157 V ⎛ 100 ⎞ ⎛ 40 ⎞ 2 2 (b) K n = ⎜ ⎟(6 ) = 300 μ A/V ; K p = ⎜ ⎟(4 ) = 80 μ A/V 2 2 ⎝ ⎠ ⎝ ⎠ 3.3 − 0.4 +

300 (0.4) 80

= 1.25 V 300 1+ 80 (ii) For υ O = 3.1 V; NMOS in saturation, PMOS in nonsaturation

(i) V I t =

[

300(υ I − 0.4) = 80 2(3.3 − υ I − 0.4)(3.3 − 3.1) − (3.3 − 3.1) 2

(

)

3.75 υ − 0.8υ I + 0.16 = 0.4(2.9 − υ I ) − 0.04 2 I

2

]

3.75υ I2 − 2.6υ I − 0.52 = 0 ⇒ υ I = 0.855 V

(iii) For υ O = 0.2 V; NMOS in nonsaturation, PMOS in saturation

[

]

300 2(υ I − 0.4)(0.2) − (0.2) = 80(3.3 − υ I − 0.4 ) 2

2

3.75(0.4υ I − 0.2 ) = 8.41 − 5.8υ I + υ I2

υ I2 − 7.3υ I + 9.16 = 0 ⇒ υ I = 1.61 V ______________________________________________________________________________________ 16.34

a.

For N1

vO1 = 0.6 < VTN ⇒ vO 2 = 5 V

in nonsaturation and P1 in saturation. From Equation (16.43),

⎡ 2 ( vI − 0.8 )( 0.6 ) − ( 0.6 )2 ⎤ = [5 − vI − 0.8]2 ⎣ ⎦ 1.2vI − 1.32 = 17.64 − 8.4vI + vI2

or vI2 − 9.6vI + 18.96 = 0 vI =

or

9.6 ±

( 9.6 )

2

− 4 (1)(18.96 ) 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ vI = 2.78 V V0 Nt ≤ v02 ≤ V0 Pt

b.

From symmetry, VIt = 2.5 V V0 Pt = 2.5 + 0.8 = 3.3 V and V0 Nt = 2.5 − 0.8 = 1.7 V 1.7 ≤ v ≤ 3.3 V

02 So ______________________________________________________________________________________

16.35

V0 Nt ≤ v01 ≤ V0 Pt

a.

By symmetry, VIt = 2.5 V and So b.

V0 Pt = 2.5 + 0.8 = 3.3 V V0 Nt = 2.5 − 0.8 = 1.7 V

1.7 ≤ v01 ≤ 3.3 V

For

vO 2 = 0.6 < VTN ⇒ vO 3 = 5 V

N 2 in nonsaturation and P2 in saturation. From Equation (16.43), ⎡ 2 ( vI 2 − 0.8 )( 0.6 ) − ( 0.6 )2 ⎤ = [5 − vI 2 − 0.8]2 ⎣ ⎦ 1.2vI 2 − 1.32 = 17.64 − 8.4vI 2 + vI22

or So

vI22 − 9.6vI 2 + 18.96 = 0 vI 2 = v01 = 2.78 V

For v01 = 2.78, both N1 and P1 in saturation. Then vI = 2.5 V

______________________________________________________________________________________ 16.36 2.5 = 1.25 V 2 2 For 0.4 ≤ υ I ≤ 1.25 V; i D = 120(υ I − 0.4) μ A

(a) V I t =

i D , peak = 120(1.25 − 0.4) = 86.7 μ A 2

For 1.25 ≤ υ I ≤ 2.1 V; i D = 120(2.5 − υ I − 0.4) μ A 2

⎛ 1.8 ⎞ (b) V I t = ⎜ ⎟ = 0.9 V ⎝ 2 ⎠

For 0.4 ≤ υ I ≤ 0.9 V; i D = 120(υ I − 0.4 ) μ A 2

i D , peak = 120(0.9 − 0.4 ) = 30 μ A 2

For 0.9 ≤ υ I ≤ 1.4 V; i D = 120(1.8 − υ I − 0.4 ) μ A ______________________________________________________________________________________ 2

16.37

⎛ 80 ⎞ ⎛ 40 ⎞ (a) K n = ⎜ ⎟(2 ) = 80 μ A/V 2 , K p = ⎜ ⎟(4 ) = 80 μ A/V 2 ⎝ 2 ⎠ ⎝ 2 ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.8 VI t = = 0.9 V 2 2 i D , peak = 80(0.9 − 0.35) = 24.2 μ A ⎛ 80 ⎞ ⎛ 40 ⎞ (b) K n = ⎜ ⎟(2 ) = 80 μ A/V 2 , K p = ⎜ ⎟(6) = 120 μ A/V 2 2 ⎝ ⎠ ⎝ 2 ⎠ 1.8 − 0.35 + VI t = 1+

80 (0.35) 120 80 120

= 0.9556 V

i D , peak = 80(0.9556 − 0.35) = 29.34 μ A 2

⎛ 80 ⎞ ⎛ 40 ⎞ (c) K n = ⎜ ⎟(4 ) = 160 μ A/V 2 , K p = ⎜ ⎟(4 ) = 80 μ A/V 2 ⎝ 2 ⎠ ⎝ 2 ⎠ 1.8 − 0.35 + VI t =

160 (0.35) 80

160 1+ 80

= 0.8057 V

i D , peak = 160(0.8057 − 0.35) = 33.23 μ A 2

______________________________________________________________________________________ 16.38

⎛ 100 ⎞ ⎛ 40 ⎞ 2 2 (a) K n = ⎜ ⎟(3) = 150 μ A/V , K p = ⎜ ⎟(7.5) = 150 μ A/V ⎝ 2 ⎠ ⎝ 2 ⎠ 3.3 VI t = = 1.65 V 2 2 i D , peak = 150(1.65 − 0.4 ) = 234 μ A ⎛ 100 ⎞ ⎛ 40 ⎞ 2 2 (b) K n = ⎜ ⎟(4) = 200 μ A/V , K p = ⎜ ⎟(4 ) = 80 μ A/V ⎝ 2 ⎠ ⎝ 2 ⎠ 3.3 − 0.4 + VI t =

200 (0.4) 80

200 1+ 80

= 1.369 V

i D , peak = 200(1.369 − 0.4 ) = 188 μ A 2

⎛ 100 ⎞ ⎛ 40 ⎞ 2 2 (c) K n = ⎜ ⎟(3) = 150 μ A/V , K p = ⎜ ⎟(12 ) = 240 μ A/V ⎝ 2 ⎠ ⎝ 2 ⎠ 3.3 − 0.4 + VI t =

150 (0.4) 240

150 1+ 240

= 1.796 V

i D , peak = 150(1.796 − 0.4 ) = 292 μ A 2

______________________________________________________________________________________ 16.39

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

( )( ) = (10 )(0.2 × 10 )(1.8)

2 (a) P = fC LV DD = 10 7 0.2 × 10 −12 (2.5) ⇒ P = 12.5 μ W 2

−12 2 7 ⇒ P = 6.48 μ W (b) P = fC LV DD ______________________________________________________________________________________ 2

16.40 2 (a) P = 4 × 10 6 150 × 10 6 0.12 × 10 −12 (1.8) ⇒ P = 233 W

(

)(

(

)( ) )(300 ×10 )(0.12 ×10 )V

6 −12 2 (b) P = 233.28 = 4 × 10 DD ⇒ V DD = 1.27 V ______________________________________________________________________________________ 6

16.41

(a) (b)

P=

3 = 3 × 10−7 W 107

2 ⇒ CL = P = fCLVDD

CL =

(i) CL =

(ii) CL =

3 × 10−7

( 5 ×10 ) ( 5) 6

⇒ CL = 0.0024 pF

2

3 × 10−7

2

⇒ CL = 0.00551pF

2

⇒ CL = 0.0267 pF

( 5 ×10 ) ( 3.3) 6

3 × 10−7

( 5 ×10 ) (1.5) 6

P 2 fVDD

(iii) ______________________________________________________________________________________ 16.42

(a) (b)

P=

10 = 2 × 10−6 W 5 × 106

CL = CL =

(i) CL =

(ii) CL =

P 2 fVDD 2 × 10−6

(8 ×10 ) ( 5) 6

2

⇒ CL = 0.01 pF

2 × 10−6

2

⇒ CL = 0.023 pF

2

⇒ CL = 0.111 pF

(8 ×10 ) ( 3.3) 6

2 × 10−6

(8 ×10 ) (1.5) 6

(iii) _____________________________________________________________________________ 16.43 (a) For υ I ≅ V DD , NMOS in nonsaturation

[

2 i D = K n 2(υ I − VTN )υ DS − υ DS

]

di D 1 = ≅ K n [2(υ I − VTN )] So rds dυ DS

and υ DS very small

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 1 Or rds ≅ = ⎛ k n′ ⎞⎛ W ⎞ ⎛W ⎞ ⎜⎜ ⎟⎟⎜ ⎟ [2(V DD − VTN )] k n′ ⎜ ⎟ (V DD − VTN ) 2 L ⎝ L ⎠n ⎝ ⎠⎝ ⎠ n For υ I ≅ 0 , PMOS in nonsaturation

[

2 i D = K p 2(V DD − υ I + VTP )υ SD − υ SD

]

and υ SD very small

⎛ k ′p ⎞⎛ W ⎞ di D 1 ⎛W ⎞ = = ⎜⎜ ⎟⎟⎜ ⎟ [2(V DD − υ I + VTP )] ≅ k ′p ⎜ ⎟ (V DD + VTP ) rsd dυ SD ⎝ 2 ⎠⎝ L ⎠ p ⎝ L ⎠p 1 Or rsd = ⎛W ⎞ k ′p ⎜ ⎟ (V DD + VTP ) ⎝ L ⎠p

So

⎛W ⎞ ⎛W ⎞ (b) Let ⎜ ⎟ = 2 , ⎜ ⎟ = 4 , and V DD = 5 V ⎝ L ⎠n ⎝ L ⎠p For NMOS: 1 ⇒ r = 1.11 k Ω rds = (0.1)(2)(5 − 0.5) ds υ 0.5 i d = ds = = 0.45 mA rds 1.11 For PMOS: 1 ⇒ r = 1.39 k Ω rsd = (0.04)(4)(5 − 0.5) sd υ 0.5 = 0.36 mA i d = sd = rsd 1.39 ______________________________________________________________________________________

16.44 3 (V DD + VTP − VTN ) = 0.5 + 3 (3.3 − 0.5 − 0.5) 8 8 = 1.3625 V

(a) V IL = VTN + ⇒ V IL

V IH = VTN + ⇒ V IH

5 (V DD + VTP − VTN ) = 0.5 + 5 (3.3 − 0.5 − 0.5) 8 8 = 1.9375 V

1 [2(1.3625) + 3.3 − 0.5 + 0.5] = 3.0125 V 2 1 VOLU = [2(1.9375) − 3.3 − 0.5 + 0.5] = 0.2875 V 2 Then NM L = 1.3625 − 0.2875 = 1.075 V

(b) VOHU =

NM H = 3.0125 − 1.9375 = 1.075 V ______________________________________________________________________________________

16.45

(a)

K n 100 = =2 Kp 50

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V IL = 0.35 +

(2.5 − 0.35 − 0.35) ⎡2 ⎢ (2 − 1) ⎣⎢

⎤ 2 − 1⎥ 2 + 3 ⎦⎥

⇒ V IL = 0.8268 V

V IH = 0.35 +

(2.5 − 0.35 − 0.35) ⎡⎢ 2(2) − 1⎤⎥ (2 − 1) ⎣⎢ 3(2 ) + 1 ⎦⎥

⇒ V IH = 1.2713 V 1 [(1 + 2)(0.8268) + 2.5 − (2)(0.35) + 0.35] 2 ⇒ VOHU = 2.3152 V

(b) VOHU =

VOLU =

(1.2713)(1 + 2) − 2.5 − (2)(0.35) + 0.35 2(2)

⇒ VOLU = 0.2410 V Then NM L = 0.8268 − 0.2410 = 0.5858 V NM H = 2.3152 − 1.2713 = 1.0439 V ______________________________________________________________________________________

16.46

⎛ 100 ⎞ ⎛ 40 ⎞ 2 2 (a) K n = ⎜ ⎟(2 ) = 100 μ A/V , K p = ⎜ ⎟(5) = 100 μ A/V ⎝ 2 ⎠ ⎝ 2 ⎠ 3 V IL = 0.4 + (3.3 − 0.4 − 0.4) = 1.3375 V 8 5 V IH = 0.4 + (3.3 − 0.4 − 0.4 ) = 1.9625 V 8 1 VOHU = [2(1.3375) + 3.3 − 0.4 + 0.4] = 2.9875 V 2 1 VOLU = [2(1.9625) − 3.3 − 0.4 + 0.4] = 0.3125 V 2 Then NM L = 1.3375 − 0.3125 = 1.025 V NM H = 2.9875 − 1.9625 = 1.025 V ⎛ 100 ⎞ ⎛ 40 ⎞ 2 2 (b) K n = ⎜ ⎟(4) = 200 μ A/V , K p = ⎜ ⎟(12) = 240 μ A/V ⎝ 2 ⎠ ⎝ 2 ⎠ K n 200 = = 0.8333 K p 240 V IL = 0.4 +

(3.3 − 0.4 − 0.4) ⎡2 (0.8333 − 1) ⎢⎢⎣

V IH = 0.4 +

(3.3 − 0.4 − 0.4) ⎡⎢ (0.8333 − 1) ⎢⎣

0.8333 ⎤ − 1⎥ = 1.4127 V 3.8333 ⎥⎦ ⎤ 2(0.8333) − 1⎥ = 2.0370 V 3(0.8333) + 1 ⎥⎦

1 [(1 + 0.8333)(1.4127 ) + 3.3 − (0.8333)(0.4) + 0.4] 2 ⇒ VOHU = 2.9783 V

Now VOHU =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.0370(1 + 0.8333) − 3.3 − (0.8333)(0.4) + 0.4 VOLU = 2(0.8333) ⇒ VOLU = 0.3007 V Then NM L = 1.4127 − 0.3007 = 1.1112 V NM H = 2.9783 − 2.0370 = 0.9413 V ______________________________________________________________________________________ 16.47

v A = vB = 5 V

a.

N1

and N 2 on, so vDS 1 ≈ vDS 2 ≈ 0 V

P1 and P2 off So we have a P3 − N 3 CMOS inverter. By symmetry, vC = 2.5 V (Transition Point).

For vA = vB = vC ≡ vI

b. Want

K n ,eff = K p , eff

kn′ ⎛ W ⎞ k ′ ⎛ 3W ⎞ ⋅⎜ ⎟ = P ⋅⎜ ⎟ 2 ⎝ 3L ⎠ n 2 ⎝ L ⎠ P kn′ = 2k P′ ,

With

Or c.

then

2 1 ⎛W ⎞ 1 ⎛W ⎞ ⋅ ⋅⎜ ⎟ = ⋅3⋅⎜ ⎟ 2 3 ⎝ L ⎠n 2 ⎝ L ⎠ P 9 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⋅⎜ ⎟ ⎝ L ⎠n 2 ⎝ L ⎠ P

We have ⎛ k ′ ⎞ ⎛ W ⎞ ⎛ 2k ′p ⎞ ⎛ 9 ⎞⎛ W ⎞ Kn = ⎜ n ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠ ⎝ L ⎠n ⎝ 2 ⎠ ⎝ 2 ⎠⎝ L ⎠ p ⎛ k ′p ⎞ ⎛ W ⎞ Kp = ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ L ⎠p

Then from Equation (16.41) 5 + ( −0.8 ) + VIt = 1+

Kn ⋅ ( 0.8 ) Kp Kn Kp

Now Kn ⎛9⎞ = ( 2) ⎜ ⎟ = 9 Kp ⎝ 2⎠

Then VIt =

5 + ( −0.8 ) + 3 ( 0.8 ) 1+ 3

⇒ VIt = 1.65 V

______________________________________________________________________________________ 16.48 By definition, NMOS is on if gate voltage is 5 V and is off if gate voltage is 0 V. N1 N2 N3 N4 N5 State

v0

1

0

off

on

off

on

on

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2 off off on on off 0 3 on on off off on 5 4 on on off on on 0 v OR vY ) ⊗ ( v X AND vZ ) Logic function ( X

Y ) Z ) with ( X Exclusive OR of ( X ______________________________________________________________________________________

v OR v

v AND v

16.49

NMOS in Parallel 4-PMOS in series

(b)

CL

⎛W ⎞ ⇒⎜ ⎟ =2 ⎝ L ⎠n ⎛W ⎞ ⇒ ⎜ ⎟ = 4 ( 4 ) = 16 ⎝ L ⎠p

doubles ⇒ current must double to maintain switching speed.

⎛W ⎞ ⇒⎜ ⎟ =4 ⎝ L ⎠n ⎛W ⎞ ⎜ ⎟ = 32 ⎝ L ⎠p

______________________________________________________________________________________ 16.50 ⎛W ⎞ ⎜ ⎟ = 4 ( 2) = 8 4-NMOS in series ⎝ L ⎠n ⎛W ⎞ ⎜ ⎟ =4 L 4-PMOS in parallel ⎝ ⎠ p ⎛W ⎜ ⎝L ⎛W ⎜ ⎝L

⎞ ⎟ = 16 ⎠n ⎞ ⎟ =8 ⎠p

(b) ______________________________________________________________________________________ 16.51 ⎛W ⎞ ⇒⎜ ⎟ =2 (a) NMOS in parallel ⎝ L ⎠ n ⎛W ⎞ ⇒ ⎜ ⎟ = 3 ( 4 ) = 12 3-PMOS in series ⎝ L ⎠ P

⎛W ⎜ ⎝L ⎛W ⎜ ⎝L

⎞ ⎟ =4 ⎠n ⎞ ⎟ = 24 ⎠p

(b) ______________________________________________________________________________________ 16.52

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛W ⎞ ⎜ ⎟ = 3( 2) = 6 (a) 3-NMOS in series ⎝ L ⎠ n ⎛W ⎞ ⎜ ⎟ =4 L 3-PMOS in parallel ⎝ ⎠ p

⎛W ⎜ ⎝L ⎛W ⎜ ⎝L

⎞ ⎟ = 12 ⎠n ⎞ ⎟ =8 ⎠p

(b) ______________________________________________________________________________________

16.53 (a) Y = ABC + DE

(b)

(c)

(W L ) An, Bn,Cn = 6 , (W L )Dn, En All (W L ) p = 8

=4

______________________________________________________________________________________ 16.54

(a) (b)

Y = A( BD + CE )

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) NMOS: 3 transistors in series for pull down mode. ⎛W ⎞ ⎜ ⎟ = 2 ( 3)( 2 ) = 12 For twice the speed: ⎝ L ⎠ n ⎛W ⎞ ⎜ ⎟ = 2 ( 4) = 8 L PMOS: ⎝ ⎠ P , A ⎛W ⎞ = 2 ( 2 )( 4 ) = 16 ⎜ ⎟ ⎝ L ⎠ P , B ,C , D , E

______________________________________________________________________________________ 16.55

(a) (b)

(c)

Y = A + BC + DE

⎛W ⎞ ⎜ ⎟ =2 L NMOS: ⎝ ⎠ n , A

⎛W ⎞ =4 ⎜ ⎟ ⎝ L ⎠ n , B ,C , D , E

PMOS: 3 transistors in series for the pull-up mode ⎛W ⎞ ⎜ ⎟ = 3 ( 4 ) = 12 ⎝ L ⎠p

______________________________________________________________________________________ 16.56 (a) Y = A[B + CD ]

(b) (c) (W L ) An, Bn = 4 , (W L )Cn , Dn = 8

(W L ) Ap

= 4 , (W L ) Bp ,Cp , Dp = 8

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.57 (a) For Y = AB C + A B C + A BC

(

)(

)(

We have Y = AB C + A B C + A BC = A + B + C A + B + C A + B + C

)

(b) All (W L )n = 3 All (W L ) p = 6 ______________________________________________________________________________________ 16.58 (a)

(b)

(W L ) An, Bn,Cn = 2 , (W L )Dn, En (W L ) Ap, Dp , Ep = 6 , (W L )Bp,Cp

=1 = 12

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.59 (a) Y = C ( A + B ) (b) All (W L )n = 4 (W L )Cp = 4 , (W L ) Ap, Bp = 8

______________________________________________________________________________________ 16.60 (a) All (W L )n = 2 All (W L ) p = 20

(b) All (W L )n = 10 All (W L ) p = 4

______________________________________________________________________________________ 16.61 By definition: NMOS off if gate voltage = 0 NMOS on if gate voltage = 5 V PMOS off if gate voltage = 5 V PMOS on if gate voltage = 0

State 1 2 3 4 5 6

N1

P1

NA

NB

NC

v01

N2

P2

v02

off on off on off on

on off on off on off

off on off off off off

off off off off off on

off off off on off on

5 5 5 5 5 0

on on on on on off

off off off off off on

0 0 0 0 0 5

Logic function is

v02 = ( v A OR vB ) AND vC

______________________________________________________________________________________ 16.62 State

v01

v02

v03

1 2 3 4 5 6

5 0 5 5 5 0

5 0 5 0 5 5

0 5 0 5 0 0

Logic function:

v03 = ( v X OR vZ ) AND vY

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.63

______________________________________________________________________________________ 16.64

______________________________________________________________________________________ 16.65

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.66

2 I = −C

dVC dt

So 1 ( 2I ) ⋅ t C ΔVC = −0.5 V

ΔVC = −

For

−0.5 = −

2 ( 2 x 10−12 ) ⋅ t 25 x 10−15

⇒ t = 3.125 ms

______________________________________________________________________________________ 16.67 (a) (i) υ O = 0 (ii) υ O = φ − VTN = 3.3 − 0.4 = 2.9 V (iii) υ O = 2.5 V (b) (i) υ O = 0 (ii) υ O = φ − VTN = 1.8 − 0.4 = 1.4 V (iii) υ O = φ − VTN = 1.8 − 0.4 = 1.4 V ______________________________________________________________________________________

16.68 (a) (i) υ O = 0 (ii) υ O = φ − VTN = 2.5 − 0.5 = 2 V (iii) υ O = 1.8 V (b) (i) υ O = 0 (ii) υ O = φ − VTN = 2 − 0.5 = 1.5 V (iii) υ O = φ − VTN = 2 − 0.5 = 1.5 V ______________________________________________________________________________________ 16.69 (a) υ O1 = 2.5 − 0.4 = 2.1 V υ O 2 = 2.5 V (b) υ I′1 = 2.5 − 0.4 = 2.1 V

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (W L )1 2 2 2(2.1 − 0.4 )(0.1) − (0.1) = (2.5 − 0.1 − 0.4 ) (W L )2

[

]

(W L )1 [0.33] = 4

⎛W ⎞ ⇒ ⎜ ⎟ = 12.1 ⎝ L ⎠1

υ I′ 3 = 2.1 V

(W L )3 2 2 [ 2(2.1 − 0.4 )(0.1) − (0.1) ] = [− (− 0.6)] (W L )4

⎛W ⎞ ⇒ ⎜ ⎟ = 1.09 ⎝ L ⎠3 ______________________________________________________________________________________

(W L )3 [0.33] = 0.36

16.70 A B Y 0 0 1 0 1 0 1 0 0 ⇒ indeterminate 1 1 0.1 Without the top transistor, the circuit performs the exclusive-NOR function. ______________________________________________________________________________________ 16.71 A 0 0 1 1

A

1 1 0 0

B 0 1 0 1

B

1 0 1 0

Y 0 1 1 1

Z 1 0 0 0

Y = A + AB = A + B Z = Y or Z = AB

______________________________________________________________________________________ 16.72 (a)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)

______________________________________________________________________________________ 16.73 (a) (i) Y = 0 (ii) Y = 2.5 V (iii) Y = 0 (iv) Y = 2.5 V (b) (i) Y = 0 (ii) Y = 0 (iii) Y = 2.5 V (iv) Y = 2.5 V (c) For φ = 1 , φ = 0 ; then Y = B

For φ = 0 , φ = 1 ; then Y = A A multiplexer ______________________________________________________________________________________ 16.74 Y = AC + BC

______________________________________________________________________________________ 16.75 (a) (i) Y = 0 (ii) Y = 2.5 V (iii) Y = 2.5 V (iv) Y = 0

(b) Y = AB + A B = A ⊗ B ______________________________________________________________________________________ 16.76 A 0 1 0 1

B 0 0 1 1

Y 0 1 1 0

Exclusive-OR function. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.77 This circuit is referred to as a two-phase ratioed circuit. The same width-to-length ratios between the driver and load transistors must be maintained as discussed previously with the enhancement load inverter. When φ1 is high, v01 becomes the complement of vI . When φ2 goes high, then v0 becomes the complement of v01 or is the same as vI . The circuit is a shift register. ______________________________________________________________________________________

16.78

Want Q to be the transition point of M 5 − M 6 . From Equation (16.26(b)),

(

)

K5 V I t − VTN 5 = −VTN 6 K6

(

)

100 V I t − 0.4 = [− (− 0.6)]⇒ V I t = Q = 0.7795 V 40 This is region where M 1 and M 3 are biased in the saturation region.

(

)

K1 V I t − VTN 1 = −VTN 3 K3

(

)

150 V I t − 0.4 = [− (− 0.6 )] ⇒ V I t = S = 0.7098 V 40 This analysis neglected the effect of M 2 starting to conduct. ______________________________________________________________________________________ 16.79 vIt =

3.3 + ( −0.4 ) + 0.5

vI = 1.5 V

1+1

= 1.7 V

NMOS Sat; PMOS Non Sat

( vI − 0.5)

2 = ⎡ 2 ( 3.3 − vI − 0.4 )( 3.3 − vo1 ) − ( 3.3 − vo1 ) ⎤ ⇒ vo1 = 2.88 V ⎣ ⎦ vI = 1.6 V vo1 = 2.693 V vI = 1.7 V vo1 = variable (switching region) vI = 1.8 V NMOS Non Sat; PMOS Sat 2

( 3.3 − VI − 0.4 )

2

= ⎡⎣ 2 ( vI − 0.5 ) vo1 − vo21 ⎤⎦ ⇒ vo1 = 0.607 V

Now vI = 1.5 V, vo1 = 2.88 V ⇒ vo ≈ 0V vI = 1.6 V, vo1 = 2.693 V

NMOS Non Sat; PMOS Sat

( 3.3 − vo1 − 0.4 )

2

= ⎡⎣ 2 ( vo1 − 0.5 ) vo − vo2 ⎤⎦

vo = 0.00979 V vI = 1.7 V, v o1 =

Switching Mode ⇒ v0 = Switching Mode.

vI = 1.8 V, vo1 = 0.607 V

( v01 − 0.5)

2

NMOS Sat; PMOS Non Sat

2 = ⎡ 2 ( 3.3 − v01 − 0.4 )( 3.3 − v0 ) − ( 3.3 − v0 ) ⎤ ⇒ v0 = 3.298 V ⎣ ⎦

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.80 For R = φ = VDD and S = 0 ⇒ Q = 0, Q = 1 For S = φ = VDD and R = 0 ⇒ Q = 1, Q = 1 The signal φ is a clock signal. For φ = 0, The output signals will remain in their previous state. ______________________________________________________________________________________

16.81

a.

Positive edge triggered flip-flop when CLK = 1, output of first inverter is D and then Q = D = D .

b.

For example, put a CMOS transmission gate between the output and the gate of M 1 driven by a

CLK pulse. ______________________________________________________________________________________

16.82

For J = 1, K = 0, and CLK = 1; this makes Q = 1 and Q = 0 . For J = 0, K = 1, and CLK = 1 , and if Q = 1, then the circuit is driven so that Q = 0 and Q = 1. If initially, Q = 0, then the circuit is driven so that there is no change and Q = 0 and Q = 1. J = 1, K = 1, and CLK = 1, and if Q = 1, then the circuit is driven so that Q = 0. If initially, Q = 0 , then the circuit is driven so that Q = 1. So if J = K = 1, the output changes state. ______________________________________________________________________________________ 16.83

For J = vX = 1, K = vY = 0, and CLK = vZ = 1, then v0 = 0. For J = vX = 0, K = vY = 1, and CLK = vZ = 1, then v0 = 1.

Now consider J = K = CLK = 1. With vX = vZ = 1, the output is always v0 = 0, So the output does not change state when J = K = CLK = 1. This is not actually a J − K flip-flop. ______________________________________________________________________________________ 16.84 (a) 256K ⇒ 262,144 cells ⇒ 512× 512 Each decoder ⇒ 9 inputs (b) (i) Row 52, address = 000110011 (ii) Row 129, address = 010000000 (iii) Row 241, address = 011110000 (c) (i) Column 24, address = 000010111 (ii) Column 165, address = 010100100 (iii) Column 203, address = 011001010 ______________________________________________________________________________________ 16.85

(a)

1-Megabit memory ⇒ = 1, 048,576 ⇒ 1024 × 1024 = 10

Number of input row and column decodes lines necessary (b) 250K × 4 bits ⇒ 262,144 × 4 bits ⇒ 512 × 512 For 512 lines ⇒ 9 row and column decoder lines necessary. ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 16.86

32 rows × 16 columns Each column contains 8 bits ______________________________________________________________________________________

16.87 Assume the address line is initially uncharged, then dVC 1 I VC = ∫ Idt = ⋅ t C C dt or −12 V ⋅ C ( 2.7 ) ( 5.8 × 10 ) t= C = ⇒ 250 × 10−6 I Then t = 6.26 × 10−8 s ⇒ 62.6 ns I =C

______________________________________________________________________________________ 16.88

(a)

V DD − Q ⎛ k n′ ⎞⎛ W ⎞ = ⎜⎜ ⎟⎟⎜ ⎟ 2(VGSD − VTN )Q − Q 2 R ⎝ 2 ⎠⎝ L ⎠ D 2.5 − 0.02 ⎛ 80 ⎞⎛ W ⎞ 2 = ⎜ ⎟⎜ ⎟ 2(2.5 − 0.4 )(0.02) − (0.02) 1 ⎝ 2 ⎠⎝ L ⎠ D

[

]

[

]

⎛W ⎞ ⎛W ⎞ 0.062 = ⎜ ⎟ (0.0836 ) ⇒ ⎜ ⎟ = 0.74 ⎝ L ⎠D ⎝ L ⎠D (b) 16K ⇒ 16,384 cells V 1.2 i D ≅ DD = = 1.2 μ A 2 1 P = (1.2 )(1.2 )(16,384 ) ⇒ P = 23.6 mW ______________________________________________________________________________________

16.89 16 K ⇒ 16,384 cells 200 ⇒ 12.2 μW 16,384 V P 12.2 2.5 iD = = = 4.88 μ A ≅ DD = ⇒ R = 0.512 M Ω 2.5 VDD R R PT = 200 mW ⇒ Power per cell =

If we want vO = 0.1 V for a logic 0, then ⎛ k ′ ⎞⎛ W ⎞ iD = ⎜ n ⎟ ⎜ ⎟ ⎡⎣ 2 (VDD − VTN ) vO − vO2 ⎤⎦ ⎝ 2 ⎠⎝ L ⎠ 2 ⎛ 35 ⎞⎛ W ⎞ 4.88 = ⎜ ⎟⎜ ⎟ ⎡ 2 ( 2.5 − 0.7 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ ⎜ ⎟ = 0.797 L So ⎝ ⎠

______________________________________________________________________________________ 16.90 D = V DD = 2.5 V

Assume M P 3 in saturation; M NA , M N 1 in nonsaturation I DP 3 = I DNA = I DN 1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ k ′p ⎞⎛ W ⎞ ⎛ 35 ⎞ 2 2 I DP 3 = ⎜⎜ ⎟⎟⎜ ⎟ (V SG 3 + VTP ) = ⎜ ⎟(1)(2.5 − 0.4 ) = 77.175 μ A L 2 2 ⎝ ⎠ ⎝ ⎠⎝ ⎠ 3 I DP 3 = I DN 1

[

⎛ 80 ⎞ 77.175 = ⎜ ⎟(2) 2(2.5 − 0.4 )Q − Q 2 ⎝ 2 ⎠

]

We find Q 2 − 4.2Q + 0.9647 = 0 ⇒ Q = 0.244 V I DP 3 = I DNA

[

⎛ 80 ⎞ 2 77.175 = ⎜ ⎟(1) 2(2.5 − 0.244 − 0.4 )(D − 0.244) − (D − 0.244) ⎝ 2 ⎠

[

1.9294 = 3.712(D − 0.244 ) − (D − 0.244 )

2

]

]

= 3.712 D − 0.9057 − D + 0.488 D − 0.05954 2

We find D − 4.2 D + 2.895 = 0 ⇒ D = 0.869 V 2nd approximation, M P 3 in nonsaturation Assume V DS = V DD − D ≅ 2.5 − 0.869 = 1.631 V 2

[

]

⎛ 35 ⎞ 2 I DP 3 = ⎜ ⎟(1) 2(2.5 − 0.4)(1.631) − (1.631) = 73.33 μ A ⎝ 2 ⎠ I DP 3 = I DN 1

[

⎛ 80 ⎞ 73.33 = ⎜ ⎟(2) 2(2.5 − 0.4)Q − Q 2 ⎝ 2 ⎠

]

Q 2 − 4.2Q + 0.9166 = 0 ⇒ Q = 0.231 V I DP 3 = I DNA

[

⎛ 80 ⎞ 2 73.33 = ⎜ ⎟(1) 2(2.5 − 0.231 − 0.4)(D − 0.231) − (D − 0.231) ⎝ 2 ⎠

[

(

1.833 = 3.738(D − 0.231) − D 2 − 0.462 D + 0.05336

]

)]

We find D − 4.2 D + 2.75 = 0 ⇒ D = 0.812 V ______________________________________________________________________________________ 2

16.91

Approximation, M N 2 cutoff I DP 2 = I DB , assume D = 0

Both M P 2 and M B in nonsaturation

[

(

) (

⎛ 35 ⎞ ⎜ ⎟(4) 2(2.5 − 0.4 ) 2.5 − Q − 2.5 − Q ⎝ 2 ⎠

)] 2

[

[ (

⎛ 80 ⎞ = ⎜ ⎟(1) 2(2.5 − 0.4 )Q − Q 2 ⎝ 2 ⎠

) (

)]

[

]

70 4.2 2.5 − Q − 6.25 − 5Q + Q 2 = 40 4.2Q − Q 2

]

We find 0.75Q + 2.8Q − 7.4375 = 0 ⇒ Q = 1.794 V Approximation, M P1 cutoff; assume D = 2.5 V I DNA = I DN 1 Both M A and M N 1 in nonsaturation 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 16 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

[

⎛ 80 ⎞ 2 ⎜ ⎟(1) 2(2.5 − Q − 0.4 )(2.5 − Q ) − (2.5 − Q ) ⎝ 2 ⎠

]

[

⎛ 80 ⎞ = ⎜ ⎟(2 ) 2(2.5 − 0.4 )Q − Q 2 ⎝ 2 ⎠

(

2(2.1 − Q )(2.5 − Q ) − (2.5 − Q ) = 2 4.2Q − Q 2 2

(

2 5.25 − 2.1Q − 2.5Q + Q

2

]

)

) − (6.25 − 5Q + Q ) = 8.4Q − 2Q 2

2

We find 3Q 2 − 12.6Q + 4.25 = 0 ⇒ Q = 0.370 V ______________________________________________________________________________________ 16.92

For Logic 1, v1:

( 5)( 0.05) + ( 4 )(1) = (1 + 0.05) v1 ⇒ v1 = 4.0476 V v2 : (5)(0.025) + (4)(1) = (1 + 1.025)v2 ⇒ v2 = 4.0244 V

For Logic 0, v1:

(0)(0.05) + (4)(1) = (1 + 0.05)v1 ⇒ v1 = 3.8095 V v2 : (0)(0.025) + (4)(1) = (1 + 0.025)v2 ⇒ v2 = 3.9024 V

______________________________________________________________________________________ 16.93

Design Problem ______________________________________________________________________________________ 16.94

Design Problem ______________________________________________________________________________________ 16.95 Design Problem ______________________________________________________________________________________ 16.96

(a) Quantization error Or LSB ≤ 0.10 V For a (b)

6-bit word , LSB = 1 − LSB =

=

1 LSB ≤ 1% ≤ 0.05 V 2

5 = 0.078125 V 64

5 = 0.078125 V 64

3.5424 × 64 = 45.34 ⇒ n = 45 5 (c) Digital Output = 101101

45 × 5 = 3.515625 64

Δ = 3.5424 − 3.515625 = 0.026775