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English Pages 63 Year 2017
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 5724.216.39O 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 nt 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(nt)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
n2
V1
HF
2 n
n2
0.1382 0.0832 0.1432 0.046382 0.2153648
The WTHD is
WTHD
Vn2 n2 n V1
HFn2 0.1382 0.0832 0.1432 0.010647 0.1031847 3 5 7 n n2
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
2k 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 20 272.322W 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 14.76 442.8W
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 4.9 108 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 2 106 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 6.25 104 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 8.33 104 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 2.6 107 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
2k 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 2on I O The current of the capacitor C2 during switching–off is
I C 2off
k I O 1 k
The current of the inductor during switching–off is
I L I L off I C 2off I O
1 I O 1 k 36
The current of the capacitor C1 during switching–off is
I C1off 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 C1on ) (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 C1on
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 2k 2k 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 2k 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 10 313.6W
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 392W
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 [(
2k 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