SOLUTION-Introduction to Modern Power Electronics

SOLUTION MANUAL

to

Introduction to Modern Power

Electronics

by

Andrzej M. Trzynadlowski

John Wiley & Sons, Inc.

2

Chapter 1

P1.3. (a) From Fig. 1.18(a), Vo,dc = 0.32 pu. From Eq. (1.44),

(b) From Fig. 1.18(a), Vo,1,p = 0.5 pu. Thus, Vo,1 = 0.5/2 = 0.35 pu

P1.4. (a) From Fig. 1.18(b), Vo,1,p = 0.6 pu

(b) Vo,1 = 0.6/2 = 0.424 pu

(c) dh(V) = 0.265/0.424 = 0.625

P1.5. As in the all subsequent problems involving spectral analysis, formulas provided in Appendix B are used, in the angle-domain version (substitute 2π for T and ωt for t). The output voltage waveform, vo(ωt) has both the odd and half-wave symmetries, so that there are no even harmonics. Consequently, the peak value, Vo,k,p, of the kth harmonic can be found as

which yields

3

It can be seen that the amplitude, Vo,1,p, of the fundamental voltage is as much as 27% higher than the dc input voltage, Vi.

P1.6. (a)

(b)

Eq. (1.45) for a generic ac voltage controller was used. It is also valid for the generic rectifier, since the rms value does not depend on the sign of a function.

(c) kr(V) = 59.2/96.6 = 0.613

P1.7. f1 = 120 Hz

P1.8. This is a rather tedious problem. The waveform of output voltage, vo(ωt), given by

has only the half-wave symmetry. Therefore, only odd harmonics are present in the spectrum, and the amplitudes of odd harmonics must be computed as

4

For instance, with αf = 90o, the peak value of fundamental voltage, Vo,1,p, is

P1.9. The waveform of output voltage of the generic rectifier has no half-wave symmetry, so that all harmonics are present, including the dc component ("zero harmonic"). Analogous waveform in the generic ac voltage controller has the half-wave symmetry, and only odd harmonics appear in the spectrum of output voltage. The dc component and even harmonics are absent.

P1.10. S1 & S2: D1,2 = 70/100 = 0.7

S3 & S4: D3,4 = 0

S5: D5 = 1 - 0.7 = 0.3

S5 is switching complementarily with S1 & S2.

P1.11. S1 & S2: D1,2 = 0

S3 & S4: D3,4 = 35/100 = 0.35

S5: D5 = 1 - 0.35 = 0.65

S5 is switching complementarily with S3 & S4.

P1.12. tON + tOFF = tON/D = 0.2/0.4 = 0.5 ms

fsw = 1/(tON + tOFF) = 1/0.5 = 2 kHz

P1.13. tON + tOFF = 1/1250 = 0.8 ms

tON = D(tON + tOFF) = 0.3 0.8 = 0.24 ms

tOFF = 0.8 - 0.24 = 0.56 ms

5

P1.14. Since M = D = tON/(tON + tOFF), then

tON/tOFF = M/(1 - M) = 0.8/0.2 = 4

P1.15. tON + tOFF = 1/(50 60) = 1/3000 s = 1/3 ms fsw = 3 kHz

tON = 0.3/3 = 0.1 ms

P1.16. (a) no. of pulses = fsw/f = 1500/60 = 25 pulses per cycle

(b) D1,2 = 0.333, D5 = 0.667

(c) pulse width = tON = D1,2(tON + tOFF) = D1,2/fsw = 0.333/1500 = 0.222 ms

P1.17. (a) S1 & S2 ON S3 & S4 OFF S5 OFF

(b) S1 & S2 OFF S3 & S4 OFF S5 ON

(c) S1 & S2 OFF S3 & S4 ON S5 OFF

P1.18. Δ = 360o/8 = 45o

αn = (n - 0.5) 45o

Dn = 0.75sin(αn)

γn = Dn 45o

n αn Dn γn

1 22.5o 0.287 12.9o

2 67.5o 0.693 31.2o

3 112.5o 0.693 31.2o

4 157.5o 0.287 12.9o

6

5 202.5o 0.287 12.9o

6 247.5o 0.693 31.2o

7 292.5o 0.693 31.2o

8 337.5o 0.287 12.9o

P1.21. Eqs. (1.73) and (1.75) can be modified to

and

where ΔT = 1/720 s. Substituting the data (if you have the first print of the book, correct the inductance, L, to 12 mH), yields

io(t1) = 0.971io(t0) + 19.65 |sin(377t0)| - 5.21

and

io(t2) = 0.971io(t1) - 5.21

Since 720/60 = 12, there are twelve switching intervals per cycle. The initial value, io(0), of the current can be assumed to equal the average ouput current, Io,that is

where

Thus, io(0) = 36 A. Computed values of the current are tabulated below.

t (s) io (A) t (s) io (A)

7

0.0000 36.0 0.0090 29.1

0.0007 29.7 0.0097 23.0

0.0014 23.7 0.0104 27.0

0.0021 27.6 0.0111 21.0

0.0028 21.6 0.0118 32.2

0.0035 32.8 0.0125 26.0

0.0042 26.6 0.0132 39.7

0.0049 40.3 0.0139 33.4

0.0056 33.9 0.0146 44.2

0.0063 44.7 0.0153 37.7

0.0069 38.2 0.0160 41.2

0.0083 35.3 0.0167 34.8

Correction of the initial current to 33.8 A yields the final value of the current equal to the initial value:


0.0000 33.8 0.0090 27.6

0.0007 27.6 0.0097 21.6

0.0014 21.6 0.0104 25.6

0.0021 25.6 0.0111 19.6

0.0028 19.6 0.0118 30.8

0.0035 30.9 0.0125 24.7

0.0042 24.8 0.0132 38.5

0.0049 38.5 0.0139 32.1

0.0056 32.2 0.0146 43.0

0.0063 43.0 0.0153 36.6

0.0069 36.6 0.0160 40.1

0.0083 33.8 0.0167 33.8

8

The substantial ripple of the current results from the low switching frequency. In practice, it would be an order of magnitude higher. Note that once the current equations are found, a simple computer program allows easy repeating of the calculations with various values of io(0).

P1.22. Substituting the data, including ΔT = 1/500 s, to Eqs. (1.73) and (1.75), yields

io(t1) = 0.4io(t0) + 9.76sin(314t0)

and

io(t2) = 0.8io(t1)

The initial current, io(0), can be found using Eq. (1.70), which gives io(0) = -5.92 A. Computed values of the current are tabulated below.


0.0000 -5.92 0.0115 3.14

0.0015 -2.37 0.0120 2.51

0.0020 -1.89 0.0135 -4.73

0.0035 4.98 0.0140 -3.79

0.0040 3.98 0.0155 -10.80

0.0055 10.88 0.0160 -8.64

0.0060 8.70 0.0175 -12.74

0.0075 12.76 0.0180 -10.19

0.0080 10.21 0.0195 -9.81

0.0095 9.82 0.0200 -7.85

0.0100 7.86

The poor convergence is due to the low load inductance and low switching frequency. The initial current must be corrected to -7.87 A to yield the final

9

value equal to the initial value:


0.0000 -7.87 0.0115 3.16

0.0015 -3.15 0.0120 2.53

0.0020 -2.52 0.0135 -4.71

0.0035 4.73 0.0140 -3.77

0.0040 3.78 0.0155 -10.78

0.0055 10.79 0.0160 -8.63

0.0060 8.63 0.0175 -12.74

0.0075 12.74 0.0180 -10.19

0.0080 10.19 0.0195 -9.84

0.0095 9.82 0.0200 -7.87

0.0100 7.86

Note that once the current equations are found, a simple computer program allows easy repeating of the calculations with various values of io(0).

10

Chapter 4

P4.1.

Io,dc = 155.3/10 = 15.53 A

P4.2. The output voltage of the rectifier varies between VLN,p and VLN,p/2 (see Fig. 4.3), that is, between 187.8 V and 93.9 V. Therefore, with E = 180 V, the conduction is discontinuous.

P4.3. The output voltage of the rectifier varies between VLL,p and 3VLL,p/2, that is, between 650.5 V and 563.4 V. Therefore, with E = 480 V, the conduction is continuous.

P4.4. kE = 480/(4602) = 0.74

= tan-1(377 0.05/2.1) = 83.6o

The diagram in Fig. 4.10 indicates continuous conduction. Thus,

11

Vo = (3/π) 4602 = 621 V

Io = (621 - 480)/2.1 = 67 A

P4.5. kE = 600/(4602) = 0.922

αc = sin-1(0.922) - π/3 = 7.2o

αe = 60o - 7.2o = 52.8o

β = 52.8o - 7.2o = 45.6o

P4.8. As seen from the diagram in Fig. 4.21, it is the firing angle of 30o that is always feasible. This is because the SCRs are fired at the instant when the corresponding line-to-line voltage is peaking.

P4.10. kE = 260/(2 230) = 0.8

= tan-1(377 0.0007/0.15) = 60.4o

According to Fig. 4.21, the firing angle of 25o is feasible, and according to Fig. 4.22, the conduction is continuous with this angle. Thus,

Vo = (3/π) 2 230 cos(25o) = 281.5 Vand

Io = (281.5 - 260)/0.15 = 143.4 A

P4.11. kE = 260/(2 230) = 0.8

= tan-1(377 0.0007/0.15) = 60.4o

According to Fig. 4.21, the firing angle of 45o is feasible, and according to Fig. 4.22, the conduction is discontinuous with this angle. Thus, the extinction angle must be found from the current waveform given by Eq. (4.46). Substituting the data, yields

Computing the current waveform allows to pinpoint the extinction angle,

12

αe, at which the current reaches zero. It occurs at ωt of about 82.5o, which gives the conduction angle, β, as

β = 82.5o - 45o = 37.5o

The average voltage, Vo,dc, can now be found from Eq. (4.47) as Vo,dc = 263 V. The average current, Io,dc, is

Io,dc = (263 - 260)/0.15 = 20 A

P4.12. VLL,p = 2 230 = 325 V

= tan-1(377 0.0007/0.15) = 60.4o

Substituting and αf in condition (4.45) gives kE < -0.52 (this value can also be read from the diagram in Fig. 4.22). Consequently, E < -0.52 325 V, that is, E < -169 V.

P4.13. From Eq. (4.49),

Fp = (3/π)cos(60o) = 0.477

P4.14. VLL,p = 2 460 = 650.5 V

With an ideal ac source, Vo,dc = (3/π) 650.5 cos(40o) = 476 VWith a source with inductance, Vo,dc = 0.9 476 = 428.4 V, that is, the dc voltage is reduced by ΔVo,dc = 47.6 V.

From Eq. (4.64), XsIo,dc = (π/3)ΔVo,dc = (π/3) 47.6 = 49.8 V. This, when substituted in Eq. (4.63), yields

μ = |cos-1[cos(40o) - 2 49.8/650.5] - 40o| = 12.2o

P4.15.

n αn Dn γn α1,n α2,n

1 7.5o 0.487 7.30o 3.85o 11.15o

13

2 22.5o 0.635 9.52o 17.74o 27.26o

3 37.5o 0.739 11.08o 31.96o 43.04o

4 52.5o 0.793 11.90o 46.55o 58.45o

P4.16.

n αn Dn γn α1,n α2,n

1 7.5o 0.705 10.58o 2.21o 12.79o

2 22.5o 0.792 11.88o 16.56o 28.44o

3 37.5o 0.795 11.92o 31.54o 43.46o

4 52.5o 0.774 11.61o 46.70o 58.31o

P4.17. From Eq. (4.93),

and from Eq. (4.95),

P4.18. αf1 = 60o, αf2 = 120o

At ωt = 15o,

vo,1 = vCB = 4602cos(15o + 30o) = 460 V

vo,2 = vCA = -4602sin(15o) = -168.4 V

Δvo = 460 - 168.4 = 291.6 V

P4.19. VLL,pk(max) = 1200/1.4 = 857 V

VLL(max) = 857/2 = 606 V

14

Vo(max) = (3/π) 857 = 818 V

Io(max) = 100/1.2 = 83 A

Rmin = 818/83 = 9.9 Ω

P4.20.

Vrat 1.2 1.1 483.7 = 638.5 kV

n = 638.5/5 = 128

15

Chapter 5

P5.2.

Values of the magnitude control ratio, M, for individual firing angles, αf, can now be estimated from Fig. 5.3 as approximately equidistant from the characteristics for the purely resistive and purely inductive loads.

αf M Vo

30o 1.0 120 V

90o 0.8 96 V

150o 0.2 24 V

P5.3. From Fig. 5.3, for = 0 and M = 0.75, αf = 85o

P5.4. The output voltage is the line-to-neutral voltage, hence Vo = M 460/3. The magnitude control ratio can be found from Fig. 5.12.

16

αf M Vo

20o 0.98 260 V

90o 0.55 146 V

130o 0.12 32 V

P5.5. Vo = 0.65 115 = 93 V

P5.6. tON + tOFF = 1/(60 20) = 8.33 10-4 s = 833 μs

tON = 0.7 833 = 583 μs

tOFF = 833 - 583 = 250 μs

P5.7. Similarly to the input voltage, the output voltage, vo, is the line-to-line voltage.

Vo = 0.4 230 = 145 V

Vo,1 = 0.4 230 = 92 V

P5.8. From Eq. (5.38),

Hence, Vo,LL,1 = 351 3 = 608 V

P5.9. Permitted states of the matrix converter: 73, 74, 76, 81, 82, 84, 97, 98, 100, 137, 138, 140, 145, 146, 148, 161, 162, 164, 265, 266, 268, 273, 274, 276, 289, 290, 292

P5.10. Vrat 2 (1 + 0.4) 460 = 911 V

17

Io(rat) = 10000/(3 460) = 12.6 A

Irat 2/π (1 + 0.2) 12.6 = 6.8 A

P5.11. Based on results of P5.10, Irat 6.8/3 = 3.9 A

P5.12. tON(min) = 2π/(30 377) 0.04 = 22.2 10-6 s = 22.2 μs

tOFF(min) = 2π/(30 377) (1 - 0.96) = 22.2 10-6 s = 22.2 μs

Chapter 6

P6.1. (1) First quadrant (M > 0, see answer to the next Problem)

(2) Vo = 0.7 240 = 168 V

(3) Io = 168/10 = 16.8 A

P6.2. No. There is no EMF in the load, which therefore cannot supply any power.

P6.3. tON + tOFF = 1/1200 = 8.33 10-4 s = 833 μs

(a) tON = 0.7 833 = 583 μs

(b) tOFF = 833 - 583 = 250 μs

(c) From Eq. (6.20), fsw(pu) = 0.02/10 1200 = 2.4 pu

From Eq. (6.18), Io,ac(pu) = 0.7(1 - 0.7)/(23 2.4) = 0.025 pu

From Eq. (6.19), Io,ac = 240/10 0.025 = 0.6 A

(d) Io = 168/10 = 16.8 A

18

kr(I) = 0.6/16.8 = 0.036

P6.4. (a) fsw(pu) = 0.02/10 1200 = 2.4 pu

(b) Io,ac(pu) = 0.7(1 - 0.7)/(23 2.4) = 0.025 pu

P6.5. (a) Fourth quadrant (see Table 6.1 or Fig. 6.21)

(b) Vo = -0.7 150 = -105 V

(c) Io = (-105 + 120)/0.5 = 30 A

P6.6. (a) The chopper operates in the fourth quadrant (see P.6.5). Thus,D4 = 1 - 0.7 = 0.3

(b) tON + tOFF = 1/900 = 1.11 10-3 s = 1.11 ms

tON = 0.3 1.11 = 0.33 ms

(c) tOFF = 1.11 - 0.33 = 0.78 ms

(d) τ = 1.7 10-3/0.5 = 0.0034 s

fsw(pu) = 0.0034 900 = 3.06 pu

Io,ac(pu) = 0.7 (1 - 0.7)/(23 3.06) = 0.0198 pu

Io,ac = 0.0198 150/0.5 = 5.94 A

P6.7. (a) First quadrant, since Vo > 0, while Vo > E implies Io > 0

(b) Io = (125 - 100)/0.5 = 50 A

(c) M = 125/150 = 0.833

(d) D1 = 0.833

P6.8. Vo = E + RIo

19

(a) First quadrant, since E > 0 and Io > 0

D1 = (210 + 0.1 150)/300 = 0.75

(b) Fourth quadrant, since E < 0 and Io > 0

D4 = (-210 + 0.1 150)/300 + 1 = 0.65

P6.9. (a) Vo = 0.5 300 = 150 V

Io = (150 - 100)/0.1 = 500 A

(b) The M = -0.8 value is erroneous. Change it to -0.6.

Vo = -0.6 300 = -180 VIo = (-180 + 200)/0.1 = 200 A

P6.10. From Eq. (6.18), with M = 0.5 (which results in the highest ripple),

which is satisfied if fsw(pu) > 3.6

τ = 0.0005/0.1 = 0.005 s

fsw > 3.6/0.005 = 720 Hz

P6.11. First quadrant: E = 240 V E/Vi = 240/400 = 0.6

M = 0.6 ... 1 D1 = 0.6 ... 1

Second quadrant: E = 240 V E/Vi = 0.6

M = 0 ... 0.6 D2 = 0.4 ... 1

Third quadrant: E = -240 V E/Vi = -0.6

20

M = -1 ... -0.6 D3 = 0.6 ... 1

Fourth quadrant: E = -240 V E/Vi = -0.6

M = -0.6 ... 0 D4 = 0.4 ... 1

P6.12. Vo,p = 12/(1 - 0.9) = 120 V

tON + tOFF = 1/2000 = 5 10-4 s = 500 μs

tOFF = (1 - 0.9) 500 = 50 μs

P6.13. Vrat > 12 V

Irat > 150/12 = 12.5 A

P6.14. Vrat > Vi,p = 2 230 = 325 V

Irat > Io,rat = 3000/300 = 10 A

P6.15. Vrat > Vi,p = 2 460 = 651 V

Irat > Io,rat = 10000/600 = 17 A

P6.16. Vrat > Vo,p = 6/(1 - 0.9) = 60 V

ID(rat) > Io,dc(rat) = 240/6 = 40 A (the average output voltage of 6 V is used to determine the rated average current of the chopper)

IS(rat) > 0.9 40/(1 - 0.9) = 360 A

21

Chapter 7

P7.1. Simple square-wave mode: Vo,1 = 0.9 310 = 279 V

Optimal square-wave mode: Vo,1 = 0.828 310 = 257 V (see pp. 276-277)

P7.2. In both modes of the inverter, the output voltage waveform has both the odd and half-wave symmetry. Therefore, Eq. (B.23) can be used, which, in the angle-domain version yields

for k = 1, 3, 5, .... In the simple square-wave mode, αd = 0, and in the optimal squre-wave mode, αd = 0.405 rad.

MODE: simple square-wave

optimal square-wave

k ck ck

22

1 1.273 1.170

3 0.424 0.148

5 0.255 0.112

7 0.182 0.173

Note that the low-order voltage harmonics in the optimal square-wave mode are significantly reduced in comparison with those in the simple square-wave mode.

P7.3. The switching intervals are 360o/30 = 12o wide.

α13 = (13 - 1/2) 12o = 150o

F(m,α13) = 0.6sin(150o) = 0.3

γ13 = 0.3 12o = 3.6o

α1,13 = 150o - 3.6o/2 = 148.2o

α2,13 = 150o + 3.6o/2 = 151.8o

P7.4. VLL,1,p = 2/3 620 = 716 V (peak line-to-line)

VLL,1 = 716/2 = 506 V (rms line-to-line)

VLN,1 = 506/3 = 292 V (rms line-to-neutral)

P7.5. 5 - 1 - 3 - 2 - 6 - 4 - ...

P7.6. The switching intervals are 15o wide.

α4 = (4 - 0.5) 15o = 52.5o

Phase A: 0 α4 < π/3

F(m, α4) = 2 0.9 cos(52.5o - 60o) - 1 = 0.785

23

a = (1 + 0.785)/2 = 0.8925

γA,4 = 0.8925 15o = 13.3875o

α1A,4 = 52.5o + 13.3875o/2 = 59.19375o

α2A,4 = 52.5o - 13.3875o/2 = 45.80625o

Phase B: 4π/3 α4 - 120o < 5π/3

F(m, α4 - 120o) = -1

b = (1 - 1)/2 = 0

no switchings in this phase

Phase C: 2π/3 α4 - 240o < π

F(m, α4- 240o) = 2 0.9 sin(52.5o - 240o - 30o) - 1 = 0.096

c = (1 + 0.096)/2 = 0.548

γC,4 = 0.548 15o = 8.22o

α1C,4 = 52.5o + 8.22o/2 = 56.61o

α2C,4 = 52.5o - 8.22o/2 = 48.39o

The first angle indicates the turn-on instant of the upper (common-anode) switch in a given phase (leg) of the inverter, the second angle indicates the turn-off instant of this switch. Vice-versa, the other (common-cathode) switch turns off at the first angle and turns on at the second angle.

P7.7. The switching intervals are 10o wide and 292 μs long.

m = 4002/620 = 0.912

α19 = (19 - 0.5) 10o = 185o Sextant 4 α = 5o

X = 3 = 0112, Y = 1 = 0012, Z1 = 0002 = 0

24

State sequence: X - Y - Z1 = 3 - 1 - 0

dX = d3 = 0.912sin(60o - 5o) = 0.747

dY = d1 = 0.912sin(5o) = 0.079

dZ = d0 = 1 - 0.747 - 0.079 = 0.174

t3 = 0.747 292 = 218 μs

t1 = 0.079 292 = 23 μs

t0 = 0.174 292 = 51 μs

P7.8. The switching intervals are 10o wide and 292 μs long.

m = 4002/620 = 0.912

α19 = (19 - 0.5) 10o = 185o Sextant 4 α = 5o

X = 3, Y = 1, Z = 0

State sequence: X - Y - Z = 3 - 1 - 0

dX = d3 = 0.912sin(60o - 5o) = 0.747

dY = d1 = 0.912sin(5o) = 0.079

dZ = d0 = 1 - 0.747 - 0.079 = 0.174

t3 = 0.747 292 = 218 μs

t1 = 0.079 292 = 23 μs

t0 = 0.174 292 = 51μs

States and their durations are the same for both versions of the space-vector PWM technique. In Problems 7.7 and 7.8, the same state sequence, 3 - 1 - 0, applies to the switching interval in question. However, in the next switching interval, the high-quality state sequence would be 1 - 3 - 7, and the high-efficiency sequence 0 - 1 - 3.

25

P7.9. In the angle domain, the 19th switching interval extends from 180o to 190o. State 3 occupies the first 0.747 of this interval, from 180o to 180o + 0.747 10o = 187.47o. State 1 occupies the next 0.079 of the interval, from 187.47o to 187.47o + 0.079 10o = 188.26o. State 0 occupies the remaining part of the interval, from 188.26o to 190o. In the next, 20th, switching interval, the first state is State 1. The last state in the previous, 18th, interval was State 7. Consequently, prior to the beginning of the switching interval in question, all switching variables are 1 (State 7). At 180o, switching variable a changes from 1 to 0 (State 3), at 187.47o, variable b changes from 1 to 0 (State 1), and at 188.26o, variable c follows suit (State 0). At 190o, variable c changes from 0 to 1 (State 1). Thus:

Switch SA: turns off at 180o

Switch SA': turns on at 180o

Switch SB: turns off at 187.47o

Switch SB': turns on at 187.47o

Switch SC: turns off at 188.26o and turns on at 190o

Switch SC': turns on at 188.26o and turns off at 190o

P7.10. In the angle domain, the 19th switching interval extends from 180o to 190o. State 3 occupies the first 0.747 of this interval, from 180o to 180o + 0.747 10o = 187.47o. State 1 occupies the next 0.079 of the interval, from 187.47o to 187.47o + 0.079 10o = 188.26o. State 0 occupies the remaining part of the interval, from 188.26o to 190o. In the next, 20th, switching interval, the first state is State 0 again. The last state in the previous, 18th, interval was State 3. Consequently, prior to the beginning of the switching interval in question, a = 0, b = 1, and c = 1 (State 3). The switching variables do not change at 180o since the first state in the 19th switching interval is State 3 again. At 187.47o, variable b changes from 1 to 0 (State 1), and at 188.26o, variable c follows suit (State 0). At 190o, the 20th switching interval begins, and the inverter remains in State 0. Thus:

Switch SA: is not switchedSwitch SA': is not switchedSwitch SB: turns off at 187.47o

Switch SB': turns on at 187.47o

Switch SC: turns off at 188.26o

Switch SC': turns on at 188.26o

Comparison of results of this problem with those of Problem 7.9 well illustrates the reduction in the number of switchings when the high-quality state sequence is replaced with the high-efficiency sequence.

26

P7.11. Switching variable a changes from 0 to 1 at 0o, 14.80o, 89.07o, 92.07o, 170.52o, 189.48o, 267.93o, 270.93o, and 345.2o. It changes from 1 to 0 at 9.48o, 87.93o, 90.93o, 165.2o, 180o, 194.8o, 269.07o, 272.07o, and 350.52o.

P7.12. IL,1,p = (23/π) 200 = 220.5 A

IL,1 = 220.5/2 = 155.9 A

VLN,1 = 2 155.9 = 311.8 V (the load inductance produces only voltage spikes during switchings, and its contribution to the output voltage is negligible)

VLL,1 = 311.8 3 = 540.1 V

P7.13. Switching pattern for switch SA is: 2.24o (ON), 5.6o (OFF), 21.26o, 30o, 38.74o, 54.4o, 57.76o, 122.24o, 125.6o, 141.26o, 150o, 158.74o, 174.4o (ON), and 177.76o (OFF). Switching pattern for switch SA' is shifted by 180o, that is, 182.24o (ON), 185.6o (OFF), ..., 357.76o (OFF).

P7.14. Vi = (3/π) 2 1.2 = 1.62 kV

VLL,p = 1.065 1.62 = 1.73 kV

VLL = 1.73/2 = 1.22 kV

VLN = 1.22/3 = 0.7 kV

P7.15. 2010 = 2023 a = 2, b = 0, c = 2

P7.16. See Fig. 7.55. Shifting the vAB waveform rightwards by 15o produces the even and half-wave symmetry. The rms value of the line-to-line voltage can be calculated as

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The peak value of the fundamental of this voltage is

and the rms value, VLL,1 = 1.065/2 = 0.7531. Hence, the harmonic content is

and the total harmonic distortion of the line-to-line voltage is dh,LL= 0.1274/0.7531 = 0.169.

Similarly, to acquire the even and half-wave symmetries, the vAN waveform should be shifted leftwards by 15o. Then

and, using the same approach as before, the rms value, VLN, of the line-to-neutral voltage is calculated as 0.441, i.e., equal to VLL/3. The rms value, VLN,1, of the fundamental line-to-neutral voltage is 0.4348. Thus,

P7.17. Since, as calculated in Example 7.4, the dc link voltage is given by

then the peak value of this voltage is 200 + 243.8 = 443.8 V, and the inverter switches must be able to withstand at least this voltage (in contrast, if a regular inverter supplied from the same dc voltage of 200 V was considered, the switches would have to withstand only the 200 V).

As a result of the clamping, the voltage gain of the inverter is reduced by 24% (see the conclusion of Example 7.4). To compensate for this drop, the dc input voltage must be raised by the same 24%, to 248 V. Since voltage pulses are clipped to 1.3 of that voltage, their amplitude is 1.3 248 = 322.4 V. Thus, in comparison with the unclamped dc link, the required voltage rating of the

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inverter switches is reduced by 27%.

P7.18. According to Eqs. (7.12) and (4.89), duty ratio of switch SA in the nth

switching interval is given by

where

Thus, the on-time of the switch is

and the off-time is

The extremal values of DA,n occur when m = 1. Indeed observing the waveform of third-harmonic modulating function in Fig. 4.41, it can be seen that, at certain values of ωt, it reaches the extremal values of -1 and +1, corresponding to DA,n = 0 and DA,n = 1, respectively. The values of tON and tOFF

for m = 1 and individual switching intervals are listed in the following table.

n DA,n tON (μs) tOFF (μs)

1 0.6484 300.2 162.8

2 0.8849 409.7 53.3

3 0.9904 458.5 4.4

4 0.9944 460.4 2.6

5 0.9811 454.2 8.7

6 0.9944 460.4 2.6

7 0.9904 458.5 4.4

8 0.8849 409.7 53.3

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9 0.6484 300.2 162.8

10 0.3516 162.8 300.2

11 0.1151 53.3 409.7

12 0.0096 4.4 458.5

13 0.0056 2.6 460.4

14 0.0189 8.7 454.2

15 0.0056 2.6 460.4

16 0.0096 4.4 458.5

17 0.1151 53.3 409.7

18 0.3516 162.8 300.2

It can be seen that the shortest tON and tOFF times are both 2.6 μs.

P7.19. The peak value of the voltage supplying the diode rectifier constitutes the highest voltage in the system. Thus,

Vrat 1.4 2 460 = 911 V

The average output voltage of the rectifier is (3/π) 2 460 = 621 V, and it is equal to the maximum available value of the peak fundamental line-to-line voltage of the inverter. Therefore, the rms value of this voltage is 621/2 = 439 V, and it is the rated voltage of the inverter. The rated current of the inverter is 60000/(3 439) = 78.9 A. Consequently, the rated current, IS(rat), of the inverter switches must satisfy the condition

IS(rat) (2/π) 1.25 78.9 = 45 A

while the rated current, ID(rat), of the inverter diodes can be a half of that value, that is, ID(rat) 45/2 = 23 A

P7.20. The peak value of the voltage supplying the diode rectifier constitutes the highest voltage in the system. Thus, disregarding the possibility of faulty operation of the inverter,

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Vrat 1.4 2 2400/2 = 2376 V

The average output voltage of the rectifier is (3/π) 2 2400 = 3241 V and, assuming the square-wave operation, the voltage gain of the inverter is 1.065. Hence, the maximum available value of the peak fundamental line-to-line voltage of the inverter is 1.065 3241 = 3452 V, and the rms value of this voltage is 3452/2 = 2441 V, and it is the rated voltage of the inverter. The rated current of the inverter is 200000/(3 2441) = 47.3 A. Consequently, the rated current, IS(rat), of the inverter switches must satisfy the condition

IS(rat) (2/π) 1.25 47.3 = 27 A

while the rated current, ID(rat), of the inverter diodes can be a quarter of that value, that is, ID(rat) 27/4 = 7 A

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Chapter 8

P8.1. D = 32/(32 + 8) = 0.8

Vi = Vo/D = 12/0.8 = 15 V

fsw = 1/(32 + 8) = 0.025 MHz = 25 kHz

P8.2. Vo = 0.6 50 = 30 V

Io = 30/10 = 3 A

It can reasonably be assumed that the minimum value of inductance L for continuous conduction is to be determined for the worst case, that is, the duty ratio, D, equal zero. Then, from Eq. (8.11),

and from Eq. (8.9),

P8.3. D = 32/(32 + 8) = 0.8

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Vi = Vo (1 - D) = 12 (1 - 0.8) = 2.4 V

fsw = 1/(32 + 8) = 0.025 MHz = 25 kHz P8.4. D = 32/(32 + 8) = 0.8

Vi = Vo (1 - 1/D) = 12 (1 - 1/0.8) = -3 V

fsw = 1/(32 + 8) = 0.025 MHz = 25 kHz

P8.5. Assume fsw = 25 kHz as in the preceding Problems

Assume L = 1mH

D = 6/12 = 0.5

Assuming that the capacitor is selected for the worst case, that is, D = 0, it can be specified from Eq. (8.9) as

P8.6. Conversion of a 12-V voltage into a 6-V is, of course, impossible in a boost converter. Here, the solution could end. However, for comparison with the buck converter in the preceding Problem, let us assume the same duty ratio, D, of 0.5. Then, Vo = 12/(1 - 0.5) = 24 V.

Assuming fsw = 25 kHz and R = 200 Ω, the required capacitance, for the worst case of D = 1, can be found from Eq. (8.18) as

It can be seen that, in contrast with the buck converter, the ripple of the output voltage depends on the load, here assumed as 1 kΩ.

P8.7. Assuming, as in the preceding example, fsw = 25 kHz and R = 200 Ω, the required capacitance can be found from Eq. (8.18) as

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P8.8. Vo = 1 0.4 15 = 6 V

For the worst case of D = 0, the required inductance can be found from Eq. (8.11) as

Then, from Eq. (8.9),

P8.9. Io = 100/25 = 4 A

From Eq. (8.55),



P8.10. From Eq. (8.55),

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For the worst case of D = 0, the inductances can be determined from Eqs. (8.57) and (8.50) as

and


P8.11. From Eq. (8.58),


P8.12. From Eq. (8.60), for D = Dmax= 0.5, Vi = 100(2 0.5) = 100 V

P8.13. From Eq. (8.58), for D = Dmax= 0.5, Vi = 100(2 2 0.5) = 50 V

P8.17. From Eq. (8.63),

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from Eq. (8.62),

from Eq. (8.65),


P8.18. From Eq. (8.63),

from Eq. (8.62),

from Eq. (8.65),

from Eq. (8.66),


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Thus, Vo = 0.1657 120 = 19.88 V and Io = 19.88/20 = 0.994 A

P8.19. R = 502/500 = 5 Ω

From Eq. (8.63),

from Eq. (8.62),

from Eq. (8.65),

from Eq. (8.66),


Thus, Vi = 50/0.74 = 67.6 V.

P8.20. For r = 4, Eq. (8.82) gives

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and, as seen in Fig. 8.35, the maximum voltage gain, Kv, occurs with the frequency ratio, kf, somewhere between 1.25 and 1.5. A computer can be used to search this range. It turns out that Kv reaches the maximum value of 2.844 when kf = 1.41. Since

then Lr = (1.41/36π)2 = 1.55 10-4 H = 0.155 mH.

SOLUTION-Introduction to Modern Power Electronics

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