Power Electronics: Advanced Conversion Technologies (Solutions, Instructor Solution Manual) Second Edition [2 ed.] 9781138735323, 1138735329

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SOLUTIONS MANUAL FOR

Power Electronics: Advanced Conversion Technologies 2-nd edition

by

Fang Lin Luo Hong Ye

1

SOLUTIONS MANUAL FOR

Power Electronics: Advanced Conversion Technologies 2-nd edition

by

Fang Lin Luo Hong Ye

2

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4

Contents of the Solution Manual for

Power Electronics: Advanced Conversion Technologies 2-nd edition

By Dr. Fang Lin LUO and Dr. Hong YE

Solutions for Chapter 1. Introduction

6

Solutions for Chapter 2. Uncontrolled AC/DC Converters

11

Solutions for Chapter 3. Controlled AC/DC Converters

19

Solutions for Chapter 4. Power Factor Correction Implementing in AC/DC Converters

22

Solutions for Chapter 5. Ordinary DC/DC Converters

26

Solutions for Chapter 6. Voltage-Lift Converters

31

Solutions for Chapter 7. Super-Lift and Ultra-Lift Converters

36

Solutions for Chapter 8. Pulse-Width-Modulated DC/AC Inverters

41

Solutions for Chapter 9. Multi-level Inverters and Soft-Switching DC/AC Inverters

44

Solutions for Chapter 10. Best Switching Angles to Obtain Lowest THD for Multilevel DC/AC Inverters

51

Solutions for Chapter 11. Traditional AC/AC Converters

53

Solutions for Chapter 12. Improved AC/AC Converters

58

Solutions for Chapter 13. AC/DC/AC and DC/AC/DC Converters

61

5

Solutions for Chapter 1: Introduction

1.1:

From Equation (1.1), the impedance Z is

1 1 10  j120  0.01  j C 120 103 10  j 3.77  j 2.65 10  j1.12 

Z  R  j L  j

From Equation (1.2), the phase angle f is

  tan 1

1 C  tan 1 1.12  6.39O 10 R

L 

1.2:

From Equation (1.3), the impedance Z is

Z  R  j L 10  j120  0.01 10  j 3.77 

From Equation (1.4), the phase angle  is

  tan 1

L R

 tan 1

3.77  20.66O 10

6

From Equation (1.5), the time constant  is



L 0.01   1 ms R 10

1.3:

From Equation (1.6), the impedance Z is

1 1  10  j 120 103 C  10  j 2.65 

Z R j

From Equation (1.7), the phase angle f is

   tan 1

1 1   tan 1  14.86O 120  0.01 CR

From Equation (1.8), the time constant  is

τ = RC = 10 × 0.001 = 10 ms

1.4:

The AC supply voltage is

7

v(t )  240 2 sin120 t

From the solution of Question 1.1, we have the impedance of the load. Therefore, by Ohm’s low the circuit current is

i (t ) 

v(t ) 240 2 sin120 t   33.73sin (120 t  6.39O ) A Z 10  j1.12

From equation (1.12), the apparent power S is

S  VI *  240 

33.73 6.39O  5724.216.39O VA 2

From equations (1.15) and (1.16) the real power P and reactive power Q are

P =| S | cos  = 5724.21 cos 6.39 = 5695.4 W

Q =| S | sin  = 5724.21 sin 6.39 = 573.6 VAR

From Equation (1.11) the power factor PF is

PF = cos  = cos 6.39 = 0.994 lagging

1.5:

8

Since the function is a central symmetrical function it is presented in Fourier transform



f (t )  E  bn sin nt n 1

where the Fourier coefficients are

bn 



1





n  1, 3, 5, 7,  

f (t ) sin (nωt )d (ωt )



Substituting the function in f(t)

bn 

1









7 n

8



5 n

8 2 8 f (t ) sin(nt)d (t)  [  sin  d   sin  d ] n 7 n 3 n 8

n n 2 7 n 3n 5n 4 3n  cos  cos  cos [(cos )  (cos )]  (cos ) n n 8 8 8 8 8 8

or

bn 

4 n 3n  cos (cos ) n 8 8

n  1, 3, 5, 7,  

The amplitude of the fundamental harmonic is

b1  V1 / E 

4



(cos

 8

 cos

3 4 )  (0.9239  0.3827)  1.664 8 

From bn we can get the coefficients of the higher-order harmonics

9

b3 

4 3 9 4  cos )  (0.3827  0.9239)   0.2297 (cos 3 8 8 3

b5 

4 5 15 4  cos (cos ) (0.3827  0.9239)  0.1378 5 8 8 5

b7 

4 7 21 4  cos (cos ) (0.9239  0.3827)   0.2377 7 8 8 7

The amplitudes of the harmonics are

V1 = 1.664 E

V3 = 0.2297 E

V5 = 0.1378 E

V7 = 0.2377 E

The HFs are

HF3 = 0.138

HF5 = 0.083

HF7 = 0.143

The THD is



THD 

V

2 n

n2

V1





 HF

2 n

n2

 0.1382  0.0832  0.1432  0.046382  0.2153648

The WTHD is



WTHD 

Vn2  n2 n V1



HFn2 0.1382 0.0832 0.1432     0.010647  0.1031847 3 5 7 n n2 



10

Solutions for Chapter 2: Uncontrolled AC/DC Converters

2.1:

From Equation (2.20)

 = tan-1 (ωL/R) = 78.75O

Check Figure 2.4

β = 272O

2.2:

 = tan-1 (ωL/R) = 72.34O

From Equation (2.20)

β1 = π +  = 252.34O

Let

Step

β

x = Sin (β – )

1

252.34O

0

0.234532

x = 0.234532

2↑

265.907O

+0.234532

0.217508

x = 0.217508

3↓

264.905O

+0.217508

0.218722

x = 0.218722

4↑

264.979O

+0.218722

0.218635

x = 0.218635

5↓

264.977O

+0.218635

0.218641

x = 0.218641

6↓

264.972O

+0.218641

0.218641

x = 0.218641

7

264.972O

y e



R L

sin 

Let x = y

11

So, the extinction angle β = 264.972O with high accuracy.

2.3:

For the parameters given

Z R 2   2 L2 106.9

 = tan–1 (ωL/R) = 0.361rad = 20.7O

ωL/R = 0.377

(a)

From equation (2.25) for current

i = 0.936 sin(ωt – 0.361) + 0.331e–ωt/0.377

β can be found numerically by equating i = 0.

β is found to be 3.5 rad.

(b)

Average output current

3.5

Id   0

1 (0.936sin (t  0.361)  0.331e t /0.377 )d (t ) 2

12

or

Id 

(c)

2V (1  cos  )  0.308 A 2 R

Average output voltage is

Vd 

2V (1  cos  )  30.8V 2

2.4:

a)

From (2.25), the angle α at which D starts to conduct is

  sin 1 m  sin 1

b)

Vc 100  sin 1  40O 2V 110 2

From equation (2.42), γ is

γ = π – 2α =180–80 = 100 deg

c)

From equation (2.43), the average rectified current is

1 I0  2

1400

 110

40

0

2 (sin120 t  110100 2 )d (t) 10.2 A

13

d)

The rms value of the rectified current is

 1 IR    2

e)

1400



400

1/2

 110 2  (sin120 t – 100 ) d (t )  100 2    2

2

 21.2 A

The power delivered by the ac source is

P = RI R2 +VcI0 = 1 × 21.22 + 100 × 10.2 = 1469W

f)

The power factor is

PF 

Power delivered 1469   0.63. VI R 110  21.2

2.5:

Calculation of the angle θ. At ωt = θ, the slopes of the voltage functions are equal to each other as shown in (2.59),

2V sin   (  )/ RC e  RC 1 1   tan   RC Thus     tan 1 ( RC ) 2V cos  

Therefore,

θ = π – tan–1 (100π * 100 * 0.0001) = 180 – 72.34 = 107.66O

14

Calculation of the angle α using Iterative method 1.

Since RC = 0.01 sec, and ωRC = π, we obtain the equation below. At ωt = α, the input voltage is equal to the output voltage,

sin α = (sinθ)e–(2π+α–θ)/ωRC

or

sin α = (sinθ)e–(2π+α–θ)/ωRC = 0.953e–(252.34+α)/π

Using the iterative method 1, define:

x = sin α

y = 0.953e–(252.34+α)/π

Make a table with initial α = 30O:

α (O)

x

y

|x|: y

30

0.5

0.1777

>

20

0.34

0.21

>

12

0.2079

0.2194




12.6

0.21814

0.2187

< 15

12.7

0.21985

0.21858

>

12.63

0.21865

0.21867

≈

We can choose α = 12.63O.

2.6:

Calculation of the angle θ. At ωt = θ, the slopes of the voltage functions are equal to each other.

2V sin  (  )/ RC e  RC 1 1   tan   RC Thus     tan 1 ( RC ) 2V cos  

    tan 1 / (100 *100*0.0001)  180  72.34  107.66O

With comparison with the discharging angle θ in Problem 2.5, it can be seen that both values are same since the R, C and ω have the same value.

Calculation of the angle α using Iterative method 1.

RC = 0.01 sec, and ωRC = π.

We obtain the equation below. At ωt = α, the input voltage is equal to the output voltage,

16

sin α = (sin θ)e–(π + a – θ)/ωRC

sin α = (sin θ)e–(π + a – θ)/ωRC = 0.953e–(72.34 + α)/π

Using the iterative method 1, define:

x = sin α y = 0.953e–(72.34 + a)/π

Make a table with initial α = 45O:

α (O)

x

y

|x| >, =, < y ?

45

0.7071

0.4965

>

35

0.5736

0.5249

>

30

0.5

0.5397




32.2

0.5328

0.5332




32.22

0.53317

0.53311

≈

We can choose α = 32.22O.

The average output voltage Vd is

17

Vd 





    2V    sin(t )d t   0.953e  (t  )/ d t     

2V

 (cos   cos  )  0.953 (1  e   240 2



 (   )/

) 

[(0.846  0.3034)  0.953 (1  0.4597)]

Vd  108.038[(1.1494)  0.953 (0.5403)]  298.94 V

IC = 0

The average capacitor current is zero

The average resistor and diode current is

IR  Id 

Vd 298.49   2.9894 A 100 R

18

Solutions for Chapter 3: Controlled AC/DC Converters

3.1:

From the parameters given,

Vm  120 2  169.7 V Z  R 2   2 L2  12.5 

  tan 1 ( L / R)  0.646 rad ,

 L / R  0.754,    / 3

As α > , the load current is discontinuous.

(a)

Substituting the data into equation (3.17)

i = 13.6 sin (ωt – 0.646) – 21.2e–ωt/0.754

for α ≤ ωt ≤ β.

Solving i(β) = 0 numerically for β. We obtain β = 3.78 rad = 2160.

(b)

The average load current is,

I0 

1





 i d (ωt )  7.05 A



19

(c)

The average output voltage is given by,

v0 

2v



(cos   cos  ) 

2 120



(cos 600  cos 2160 )  70.71V

3.2:

(a)

The firing angle α = 15O, the output voltage and current are continuous. Referring to

the formulae (3.24) (3.25) and (3.26), the output voltage and current are

Vo = 1.17Vin cosα = 1.17 × 240 × cos15O = 271.2 V I0 

(b)

V0 271.2   2.712 A R 100

The firing angle α = 75O, which is greater than π/6 = 30O. The output voltage and

current are discontinuous. Referring to the formulae (3.27) and (3.28), we have the output voltage and current to be

3 sin   1) cos   2 2  0.675  240 (0.224  0.483  1) 120 V

V0  0.675V (

I0 

V0 120  1.2 A R 100

3.3:

(a)

The firing angle α = 20O, the output voltage and current are continuous. Referring to the 20

formula (3.36), we have the output voltage and current to be

VO = 2.34V cos α = 2.34 × 240 cos 20O = 527.73 V

I0 

V0 527.73   5.28 A R 100

(b) The firing angle α = 100O, the output voltage and current are continuous and negative values. Referring to the formula (3.36), we have the output voltage and current to be

VO = 2.34V cos α = 2.34 × 240 cos 100O = –97.5 V

I0 

V0 97.5    0.975 A R 10

21

Solutions for Chapter 4: Power Factor Correction

Implementing in AC/DC Converters

4.1:

Since supply frequency is 60 Hz and switching frequency is 2.4 kHz, there are 20 switching periods in a half supply period (8.33 ms). The voltage transfer gain of the P/O self-lift LuoConverter is

1 Vin 1 k V  V 400  200 2 sin t sin t k  O in   1 400 VO 2

VO 

Duty-cycle k is listed in following Table 4.7.

Table 4.7. Duty ratio k in the 20 chopping-periods in half-cycle (8.33 ms)

ωt (deg)

Input current = 200√2 sin (ωt) (V)

k

9

44.2

0.889

18

87.4

0.781

27

128.4

0.679

36

166.3

0.584

45

200

0.5

54

288.8

0.428

22

63

252

0.37

72

269

0.328

81

279.4

0.302

90

282.8

0.293

99

279.4

0.302

108

269

0.328

117

252

0.37

126

288.8

0.428

135

200

0.5

144

166.3

0.584

153

128.4

0.679

162

87.4

0.781

171

44.2

0.889

180

0

∞

4.2:

Since supply frequency is 60 Hz and switching frequency is 3.6 kHz, there are 30 switching periods in a half supply period (8.33 ms). The voltage transfer gain of the P/O super-lift LuoConverter is

VO  k

2k Vin 1 k VO  2Vin 600  400 2 sin t  VO  Vin 600  200 2 sin t

23

Duty-cycle k is listed in following Table 4.8.

Table 4.8. Duty ratio k in the 20 chopping-periods in half-cycle (10 ms)

ωt (deg)

Input current = 200√2 sin (ωt) (V)

k

6

29.6

0.948

12

58.8

0.891

18

87.4

0.829

24

115

0.763

30

141.4

0.692

36

166.3

0.617

42

189.3

0.539

48

210

0.462

54

228.8

0.383

60

244.9

0.31

66

258.4

0.244

72

269

0.187

78

276.7

0.144

84

281.3

0.117

90

282.8

0.108

96

281.3

0.117

102

276.7

0.144

108

269

0.187

114

258.4

0.244

24

120

244.9

0.31

126

228.8

0.383

132

210

0.462

138

189.3

0.539

144

166.3

0.617

150

141.4

0.692

156

115

0.763

162

87.4

0.829

168

58.8

0.891

174

29.6

0.948

180

0

∞

25

Solutions for Chapter 5: Ordinary DC/DC Converters

5.1.

1 1 V1  20  50 V 1 k 0.4

1) From (5.22), the output voltage is V2 

2) From (5.29), the output voltage ripple is

v2  vC 

kV2 0.6  50  1.5 V fRC 50k  20  20 

3) From (5.28), the inductor L = 10 mH >

Lmin 

k (1  k ) 2 0.6(0.4) 2 R 20  0.0192 mH 2f 2  50k

This boost converter works in CCM.

5.2.

In steady state, the output current is IO. The current of the capacitor C during switching–on is

I C on  I O  The current of the capacitor C during switching–off is 26

I C off 

k I O  1 k

The input current Iin and inductor current IL during switching–off is I in  I L  I L off  I C off  I O 

The power relation is

1 I O  1 k

Pin  Vin I in PO  Ploss  VO I O  I L2 rL

I 1 2 rL Vin I in  VO I O  I L2 rL  VO I O  ( O ) 2 rL  VO I O [1  ( ) ] 1 k 1 k R

i.e. Hence

M

VO I in k k    Vin I [1  ( 1 ) 2 rL ] (1  k )[1  ( 1 ) 2 rL ] (1  k )  rL O 1 k R 1 k R (1  k ) R

Substituting the data into the above Equation, the voltage transfer gain M is, M

i.e.

then

IO 

VO 0.8 0.8 0.8 k      2.4615385  2.46 0.5 0.5 0.325 Vin (1  k )  rL (1  0.8)  0.2  (1  k ) R (1  0.8)20 4

VO MVin 2.46  30    3.69 A and PO  I O2 R  3.692  20  272.322W R R 20

IL 

k 0.8 IO 3.69 IO  I O          and I in  1 k 0.2 1  k 0.2

Finally Pin  Vin I in  30  14.76  442.8W

and



PO 272.322    442.8 Pin

5.3.

1) From (5.45), the output voltage is V2 

1 0.6 V1  20  50 V 1 k 0.4

2) From (5.41), the output voltage ripple is

27

v2  vC 

kV2 0.6  20   0.6 V fRC 50k  20  20 

3) From (5.40), the inductor L = 10 mH > Lmin 

1 k 0.4 R 20  0.2 mH 2f 2  20k

This boost converter works in CCM.

5.4.

Refer to the formulae (5.51a), we obtain

n1 

VO1 6   0.6 kV1 0.5  20

n2 

VO 2 9   0.9 kV1 0.5  20

n3 

VO 3 12  1.2 kV1 0.5  20

The particular winding turn’s number are (with primary winding turn’s N1 = 600)

N2–1 = N1n1 = 600 × 0.6 = 360 N2–2 = N1n2 = 600 × 0.9 = 540 N2–3 = N1n3 = 600 × 1.2 = 720

The first secondary winding has 360 turns, the second secondary winding has 540 turns and

28

the third secondary winding has 720 turns.

5.5.

From the output voltage calculation formula, the turn’s ratio is

n

1 k 0.2 VO  1500  7.5 kVin 0.8  50

The particular turn’s number of the secondary winding is

V2 = N1n = 200 × 7.5 = 1500

The secondary winding has 1500 turns.

5.6.

1) Since the normalized impedance Z N 

R 3000 1   7.5and  2.5, the 2 fL 2  20k  10m 1 k

converter from here we work in DCM. From (5.94) and the output voltage is VO  k (1  k )

R VI  36 V . From here we can see that the output voltage is higher than it in 2 fL

CCM (ME = 1.5).

2) From (5.91), the variation ratio of vO is:

29



k 1 0.6 1   4.9 108 3 3 128 f CCO LO R 128 (20k ) (20  ) 2 10m  3000

30

Solutions for Chapter 6: Voltage Lift Converters

6.1.

1) From (6.47), the output voltage is VO = VI/(1-k) = 20/0.5 = 40 V, i.e. M = 2.

2) From the formulae we can get the ratios:

1 

1 2 1 40   0.1 2 2 2 M fL 2  2 50k 1m

2 

1 40 k R   0.1 2 2 M fLO 2  2 50k 1m



0.5 1 k 1   0.00625 2 fCR 2 50k  20   40

1 



2 1 M 1   0.025 2 fC1 R 2 50k  20   40

k 1 0.5 1 0.5   2 106 3 3 2 128 f LO CCO R 128 (50k ) 1m  (20 ) 40 128  2000

6.2.

The boundary of CCM and DCM of a self-lift negative output Luo-Converter is shown in equation (6.49).

31

The condition is m 

1





M2 22 4    1, i.e. the converter works in DCM. R 1000 5 k 0.5 2 fL 2  50k 1m

Therefore the output voltage is

VO  [1  k 2 (1  k )

R 1000 ]VI  [1  0.52 (1  0.5) ]  20  45 V 2 fL 2  50k 1m

6.3.

Since the enhanced self-lift positive output Luo-Converter has higher voltage transfer gain to be calculated by (6.62), its output voltage is

VO  (1 

1 1 )VI  (1  ) 20  60 V 1 k 1  0.5

6.4.

From (6.123) we obtain the output voltage is VO 

3 3 VI  20 120 V 1 k 1  0.5

The variation ratios:

32





k 1 0.5 1 6.25 104  2 2 16 f CLO 16 (50k )  20  1m

1 

k (1  k ) R 0.5(1  0.5) 300   0.125 2M T fL1 2 6 50k 1m

2 

k (1  k ) R 0.5(1  0.5) 300   0.125 2 M T fL2 2 6 50k 1m





k 3R 0.5 3  300   0.125 M T2 2 fL 62 2  50k 1m

k 1 0.5 1   8.33 104 2 fCR 2 50k  20   300

1 

MT 1 6 1   0.01 2 fC1 R 2 50k  20   300

2 

MT 1 6 1   0.01 2 fC2 R 2 50k  20   300

3 

MT 1 6 1   0.01 2 fC3 R 2 50k  20   300

k 1 0.5 1   2.6 107 3 3 128 f CCO LO R 128 (50k )  (20 ) 2 1m  300

Therefore, the variations are small.

6.5.

From (6.209) the voltage transfer gains of this enhanced double output self-lift DC-DC converters are

33

VO  1 1   M boost1  S   V  (1  k ) 2  (1  0.5) 2  4  in  V 1 M 1  O   1  4 2   boost S  (1  k ) (1  0.5) 2 Vin

Therefore the output voltages are

VO   4Vin  80 V  VO    4Vin   80 V

6.6.

From Equation (6.220), the output voltage is VO 

4k k Vin   80 V . 1 k 1 k

The output voltages corresponding various k are shown in the following table:

Duty cycle k

Output voltage (V)

0.1

8.89

0.2

20

0.3

34.29

0.4

53.33

0.5

80

0.6

120

0.7

186.67

34

0.8

320

0.9

720

35

Solutions for Chapter 7: Super-Lift and Ultra-Lift Converters

7.1.

From formula (7.10), we can get the variation ratio of current iL1,

1 

k (1  k ) 4 R 0.6(1  0.6) 4 100   0.0069 3 3 2(2  k ) fL1 2(2  0.6) 50k 10m

1 

k (1  k ) 2 R 0.6(1  0.6) 2 100   0.00686 2(2  k ) fL1 2 (2  0.6) 50k 10m

From formula (7.8), we can get the output voltage VO  (

From (7.12), its variation ratio is  

2k 2 2  0.6 2 ) Vin  ( ) 20  245 V . 1 k 1  0.6

k 0.6   0.003. 2 RfC4 2 100  50k  20 

7.2. a) The output current is IO. The current of the capacitor C2 during switching–on is

I C 2on  I O  The current of the capacitor C2 during switching–off is

I C 2off 

k I O  1 k

The current of the inductor during switching–off is

I L  I L off  I C 2off  I O 

1 I O  1 k 36

The current of the capacitor C1 during switching–off is

I C1off  I L  The current of the capacitor C1 during switching–on is

The input current is

1 k I L  k I in  k ( I L  I C1on )  (1  k ) I L  (2  k ) I L 

The power relation is

Pin  Vin I in PO  Ploss  VO I O  I L2 rL

I C1on 

I 1 2 rL Vin I in  VO I O  I L2 rL  VO I O  ( O ) 2 rL  VO I O [1  ( ) ] 1 k 1 k R

i.e. Hence

M

VO I in 2k 2k    Vin I [1  ( 1 ) 2 rL ] (1  k )[1  ( 1 ) 2 rL ] (1  k )  rL O 1 k R 1 k R (1  k ) R

b) From above Equation, the voltage transfer gain M is,

M

VO 2k 2  0.6 1.4 1.4      2.8 0.4 0.4 0.5 Vin (1  k )  rL (1  0.6)  0.4  (1  0.6)10 4 (1  k ) R

The output current is,

IO 

VO MVin 2.8  20    5.6 A R R 10

Therefore, the output power is,

PO  I O2 R  5.62 10  313.6W

IO 5.6    1  k 0.4

The inductor current is,

IL 

The input current is,

I in  (2  k ) I L     

Therefore, the input power is,

Pin  Vin I in  20  19.6  392W

Therefore, the efficiency is,



PO 313.6    Pin 392 37

7.3.

From formula (7.27), we can get the variation ratio of current iL1,

1 

k (1  k ) 2 R 0.6(1  0.6) 2 1000   0.04 4(3  k ) fL1 2(3  0.6) 50k 10m

From formula (7.25), we can get the output voltage VO  (

From (7.19), its variation ratio is  

3 k 3  0.6 )Vin  ( )20 120 V . 1 k 1  0.6

k 0.6   0.0003. 2 RfC12 2 1000  50k  20 

7.4.

From formula (7.128), we can get the variation ratio of current iL1, (G3 = 41.875)

1 

k (1  k )3 R 0.6 (1  0.6)3 200   0.00094 2 2 (2  k ) G3 2 fL1 (2  0.6)  41.875 2  50k 10m

From formula (7.127), we can get the output voltage

VO  [(

2k 3 2  0.6 3 )  1]Vin  [( )  1] 20  837.5 V 1 k 1  0.6

From (7.131), its variation ratio is  

k 0.6   0.0015. 2 RfC6 2  200  50k  20  38

7.5.

From formula (7.239), we can get the variation ratio of current iL1,

1 

k (1  k ) 2 R 0.6(1  0.6) 2 400   0.0096 8 fL1 8 50k 10m

From formula (7.236), we can get the output voltage

VO 

2 2 Vin   20 100 V 1 k 1  0.6

From (7.240), its variation ratio is  

k 0.6   0.00075. 2 RfC12 2  400  50k  20 

7.6.

From formula (7.409), we can get the output voltage

VO  [(

j 1 3 5 1 3 )  1]Vin  [( )  1]  20  67480 V 1 k 1  0.6

From (7.407), its variation ratio is  

k 0.6   0.00003. 2 RfC310 2 10000  50k  20 

39

7.7.

From formula (7.436), we can get the value of ξ1,

1 

(1  k) 4 R (1  0.6) 4 10000   1.83  1 2 (2  k ) fL1 2 (2  0.6)  50k 1m

Therefore, this converter works in DCM. From formula (7.443), we can get the voltage transfer gain GDCM

GDCM 

k (1  k ) 2 0.6(1  0.6) 2 10000 ZN   9.6 2 2 50k  1m

The output voltage is .V2 = GDCM × Vin = 9.6 × 20 = 192 V.

40

Solutions for Chapter 8: Pulse-Width-Modulated DC/AC Inverters

8.1.

From Equation (8.3), we have the general rms values

(VO ) h 

Vd (VˆAO ) h 200 (VˆAO ) h (Vˆ )   282.84 AO h V Vd / 2 2 Vd / 2 2 Vd / 2

Checking the data from Table 8.1 we can get the rms values as follows:

Fundamental: (VO)1 = 282.84 × 0.8 = 226.27 V

at 50 Hz

(VO)33 = 282.84 × 0.22 = 62.23 V

at 1650 Hz

(VO)35 = 282.84 × 0.818 = 231.37 V

at 1750 Hz

(VO)37 = 282.84 × 0.22 = 62.23 V

at 1850 Hz

(VO)67 = 282.84 × 0.139 = 39.32 V

at 3350 Hz

(VO)69 = 282.84 × 0.314 = 88.81 V

at 3450 Hz

(VO)71 = 282.84 × 0.314 = 88.81 V

at 3550 Hz

(VO)73 = 282.84 × 0.139 = 39.32 V

at 3650 Hz

etc.

8.2.

Refer to the Figure 8.9, it can be see that the first pulses applied to the switches S1+/S1- and S2+/S2-. Assume that the amplitude of the triangle wave is 1, the amplitude of the sin-wave is

41

0.8. The leading age of the first pulses is at t = 0.

The equation to determine the trailer age of the first pulse to turn-off S1+ is:

0.8sin 100πt = 1600t – 2

Using iterative method, let x = 0.8sin100πt and y = 1600t-2, and initial t0 = 1.6667 ms = 30O

T (ms/degree)

x

y

|x|:y

remarks

1.6667/30O

0.4

0.6667