SOLUTION MANUAL to Introduction to Modern Power Electronics
by Andrzej M. Trzynadlowski John Wiley & Sons, Inc.
Chapter 1 (a) From From Fig. Fig. 1.18 1.18(a (a), ), V o,dc P1.3. (a) o,dc = 0.32 pu. From Eq. (1.44), V o, dc =
1 π
[1 + cos ( 90 o )] )] = 0.318 pu pu
(b) (b) From From Fig. Fig. 1.18 1.18(a (a), ), V o,1,p 2 = 0.35 pu o,1,p = 0.5 pu. Thus, V o,1 o,1 = 0.5/ % P1.4. (a) (a) From From Fig. Fig. 1.18 1.18(b (b), ), V o,1,p o,1,p = 0.6 pu V o = 1
(b)
V o,1 o,1
1 2π
d h(V) h(V)
π
1
2
2
+
sin ( π )]
= 0.5 pu pu
= 0.6/%2 = 0.424 pu V o, h =
(c)
[ π -
0. 5 2 - 0. 424 2 = 0.265 pu pu
= 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 2 π for T and ωt for t ). ). The output voltage waveform, v o(ωt ) has both the odd and half-wave symmetries, so that there are no even harmonics. Consequently, th the peak value, V o,k,p o,k,p, of the k harmonic can be found as
2
Chapter 1 (a) From From Fig. Fig. 1.18 1.18(a (a), ), V o,dc P1.3. (a) o,dc = 0.32 pu. From Eq. (1.44), V o, dc =
1 π
[1 + cos ( 90 o )] )] = 0.318 pu pu
(b) (b) From From Fig. Fig. 1.18 1.18(a (a), ), V o,1,p 2 = 0.35 pu o,1,p = 0.5 pu. Thus, V o,1 o,1 = 0.5/ % P1.4. (a) (a) From From Fig. Fig. 1.18 1.18(b (b), ), V o,1,p o,1,p = 0.6 pu V o = 1
(b)
V o,1 o,1
1 2π
d h(V) h(V)
π
1
2
2
+
sin ( π )]
= 0.5 pu pu
= 0.6/%2 = 0.424 pu V o, h =
(c)
[ π -
0. 5 2 - 0. 424 2 = 0.265 pu pu
= 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 2 π for T and ωt for t ). ). The output voltage waveform, v o(ωt ) has both the odd and half-wave symmetries, so that there are no even harmonics. Consequently, th the peak value, V o,k,p o,k,p, of the k harmonic can be found as
2
V o k, p, = V o k, s, =
2 π
π
∫ v ( ω t )s i n( k ω t ) d ω t k = 1 , 3 , 5 o
0
which yields V o,k, p =
4 V i
k = 1,3,5,...
k π
It can be seen that the amplitude, V o,1,p o,1,p, of the fundamental voltage is as much as 27% higher than the dc input voltage, V i. P1.6. (a) V o,dc =
115
2
π
[1 + cos ( 30 o )] )] = 96.6 V
(b) V o = 115
2
1
[ π -
2π
π
1
6
2
+
sin ( 60
o
)] = 113.3 V
Eq. (1.45) for a generic ac voltage controller was used. It is also valid for the gene generi ric c rect rectif ifie ier, r, sinc since e the the rms rms valu value e does does not depe depend nd on the the sign sign of a function.
V o, ac =
(c)
k r(V) r(V)
113. 32 - 96. 6 2 = 59.2 V
= 59.2/96.6 = 0.613
P1.7. f 1 = 120 Hz
3
P1.8. This is a rather tedious problem. The waveform of output voltage, v o(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 V o, k, p =
For instance, with
αf =
2
2
V o, k, s + V o, k, c
90o, the peak value of fundamental voltage, V o,1, p =
0. 5 2 + (-0.3183
V o,1,p,
is
2
) = 0.5927
P1.9. The waveform of output voltage of the generic rectifier has no halfwave 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. t ON + t OFF = t ON/D = 0.2/0.4 = 0.5 ms
4
f sw
= 1/( t ON +
t OFF)
= 1/0.5 = 2 kHz
P1.13. t ON + t OFF = 1/1250 = 0.8 ms t ON
= D(t ON +
t OFF)
t OFF
= 0.8 - 0.24 = 0.56 ms
= 0.3
Η 0.8
= 0.24 ms
P1.14. Since M = D = t ON/(t ON + t OFF), then t ON/t OFF = M/(1
-
M)
P1.15. t ON + t OFF = 1/(50 t ON
= 0.8/0.2 = 4
Η 60)
= 1/3000 s = 1/3 ms
6 f sw
= 3 kHz
= 0.3/3 = 0.1 ms
P1.16. (a) no. of pulses = f sw/f = 1500/60 = 25 pulses per cycle
(b)
D1,2
= 0.333,
D5
(c) pulse width =
= 0.667
t ON
=
D1,2(t ON
+
t OFF)
=
D1,2/f sw =
ms P1.17. (a) S1 & S2
6
ON
S3 & S4
6
OFF
S5
6
OFF
(b) S1 & S2
6
OFF
S3 & S4
6
OFF
S5
6
ON
(c) S1 & S2
6
OFF
S3 & S4
6
ON
S5
6
OFF
P1.18. Δ = 360o/8 = 45 o αn
= (n - 0.5)
Dn
= 0.75sin(αn)
γ n
=
o
Η 45
o
Dn Η 45
5
0.333/1500 = 0.222
n
αn
Dn
γ n
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
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
M
io( t1 )=io( t0 )+ [ V _i p , s ( ω it _0 ) n- Rio( t0 )- E ∆ T L and io ( t 2 ) = i o ( t 1 )[1 -
R L
(1 - M) ∆T] -
E (1 - M) ∆T L
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(t 1)
= 0.971 io(t 0) + 19.65 |sin(377 t 0)| - 5.21
io(t 2)
= 0.971 io(t 1) - 5.21
and
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
6
i o (0) = I o =
V o - E R
where V o =
2
V i, p M = 108 V
π
Thus, io(0) = 36 A. Computed values of the current are tabulated below. t (s)
io
t (s)
(A)
io
(A)
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: t (s)
io
t (s)
(A)
io
(A)
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
7
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
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 (1.75), yields io(t 1)
= 0.4io(t 0) + 9.76sin(314 t 0)
io(t 2)
= 0.8io(t 1)
ΔT
= 1/500 s, to Eqs. (1.73) and
and
The initial current, io(0), can be found using Eq. (1.70), which gives -5.92 A. Computed values of the current are tabulated below. t (s)
io
(A)
t (s)
io
(A)
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
8
io(0)
=
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 value equal to the initial value: t (s)
io
(A)
t (s)
io
(A)
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).
9
Chapter 4 P4.1.
V o,dc =
Io,dc
3 3 2π
230 2 _ = 155.3 V 3
= 155.3/10 = 15.53 A
P4.2. The output voltage of the rectifier varies between V LN,p and V LN,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 V LL,p and %3V LL,p/2, that is, between 650.5 V and 563.4 V. Therefore, with E = 480 V, the conduction is continuous.
10
P4.4. k E = 480/(460 %2) = 0.74 ν
= tan-1(377
Η 0.05/2.1)
= 83.6 o
The diagram in Fig. 4.10 indicates continuous conduction. Thus, V o Io
= (3/π)
Η 460%2
= 621 V
= (621 - 480)/2.1 = 67 A
P4.5. k E = 600/(460 %2) = 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. k E = 260/( %2 ν
Η 230)
= tan-1(377
= 0.8
Η 0.0007/0.15)
= 60.4 o
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, o
V o
= (3/π)
Η %2 Η 230 Η cos(25
) = 281.5 V
Io
= (281.5 - 260)/0.15 = 143.4 A
and
P4.11. k E = 260/( %2 ν
Η 230)
= tan-1(377
= 0.8
Η 0.0007/0.15)
= 60.4 o
11
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 t -π /4
ω
o io ( ω t) = 1069[ sin( ω t - 0.4 ) - 1.62+ 0.918e 1.76 ]
Computing the current waveform allows to pinpoint the extinction angle, 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, V o,dc, can now be found from Eq. (4.47) as The average current, Io,dc, is Io,dc
= 263 V.
= (263 - 260)/0.15 = 20 A
P4.12. V LL,p = %2 ν
V o,dc
Η 230
= tan-1(377
= 325 V
Η 0.0007/0.15)
= 60.4 o
Substituting ν and α f in condition (4.45) gives k E < -0.52 (this value can also be read from the diagram in Fig. 4.22). Consequently, E < -0.52 Η 325 V, that is, E < -169 V. P4.13. From Eq. (4.49), F p =
(3/π)cos(60o) = 0.477
P4.14. V LL,p = %2
Η 460
= 650.5 V
With an ideal ac source, V o,dc = (3/π) Η 650.5 Η cos(40o) = 476 V With a source with inductance, V o,dc = 0.9 Η 476 = 428.4 V, that is, the dc voltage is reduced by ΔV o,dc = 47.6 V. From Eq. (4.64), X sIo,dc = (π/3)ΔV o,dc = (π/3) substituted in Eq. (4.63), yields μ
Η 47.6
= 49.8 V. This, when
= |cos-1[cos(40o) - 2 Η 49.8/650.5] - 40o| = 12.2o
12
P4.15. n
αn
Dn
γn
α1,n
α2,n
1
7.5o
0.487
7.30o
3.85 o
11.15o
2
22.5o
0.635
9.52o
17.74o
27.26o
3
37.5o
0.739
11.08 o
31.96o
43.04o
4
52.5o
0.793
11.90 o
46.55o
58.45o
n
αn
Dn
γn
α1,n
α2,n
1
7.5o
0.705
10.58 o
2.21 o
12.79o
2
22.5o
0.792
11.88 o
16.56o
28.44o
3
37.5o
0.795
11.92 o
31.54o
43.46o
0.774
11.61 o
46.70o
58.31o
P4.16.
4 52.5o P4.17. From Eq. (4.93), t ON( min ) = 0.05
2π 24_2 π _50
sin
(
π
24
) = 5.4_ 10 -6 s = 5.4 µ s
and from Eq. (4.95), t OFF(
min
)
=
π 2π [1 - cos ( )] = 7.1_ 10 -6 s = 7.1 µ s 24_2 π _50 24
P4.18. αf1 = 60o, αf2 = 120o V o =
3
460
π
2 cos ( 60 o ) = 311 V
At ωt = 15o, v o,1
=
v CB
v o,2 = v CA
= 460%2cos(15o + 30o) = 460 V = -460%2sin(15o) = -168.4 V
13
Δv o
= 460 - 168.4 = 291.6 V
P4.19. V LL,pk(max) = 1200/1.4 = 857 V V LL(max)
= 857/ %2 = 606 V
V o(max) = Io(max) = Rmin
(3/π)
Η 857
= 818 V
100/1.2 = 83 A
= 818/83 = 9.9
Ω
P4.20. 3
V LL, pk(rat)
π
V LL, pk(rat) =
cos ( 30
400 3
cos ( 30 π
V rat ∃ n
o
1.2 Η 1.1 Η 483.7 = 638.5 kV
= 638.5/5 = 128
14
) = 400
= 483.7 kV o
)
Chapter 5
P5.2. 377_0.005 _ = tan -1 ( ) = 43. 3o 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.
M
V o
30o
1.0
120 V
90o
0.8
96 V
150o
0.2
24 V
αf
15
P5.3. From Fig. 5.3, for ν = 0 and M = 0.75, αf = 85o P5.4. The output voltage is the line-to-neutral voltage, hence V o = M 460/ %3. The magnitude control ratio can be found from Fig. 5.12.
M
V o
20o
0.98
260 V
90o
0.55
146 V
130o
0.12
32 V
αf
P5.5. V o = %0.65
Η 115
= 93 V
P5.6. t ON + t OFF = 1/(60
Η 20)
= 8.33
t ON
= 0.7
t OFF
= 833 - 583 = 250 μs
Η 833
Η
-4
Η 10
s = 833 μs
= 583 μs
P5.7. Similarly to the input voltage, the output voltage, v o, is the line-to-line voltage. V o
=
V o,1
%0.4 Η 230
= 0.4
Η 230
= 145 V
= 92 V
P5.8. From Eq. (5.38), V o,LN ,1 =
Hence, V o,LL,1 = 351
Η %3
3
_0.8_460
π
= 608 V
16
= 351 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. Io(rat)
V rat ∃ %2 Η (1
+ 0.4)
Η 460
= 10000/( %3
Η 460)
= 12.6 A
Irat ∃ %2/π Η (1
+ 0.2)
Η 12.6
= 911 V
= 6.8 A
P5.11. Based on results of P5.10, Irat ∃ 6.8/%3 = 3.9 A
P5.12. t ON(min) = 2π/(30 t OFF(min) =
2π/(30
Η 377) Η 0.04 Η 377) Η (1
= 22.2
-6
Η 10
- 0.96) = 22.2
s = 22.2 μs -6
Η 10
s = 22.2 μs
Chapter 6 P6.1. (1) First quadrant (M > 0, see answer to the next Problem)
(2)
V o
(3)
Io
= 0.7
Η 240
= 168 V
= 168/10 = 16.8 A
P6.2. No. There is no EMF in the load, which therefore cannot supply any power.
P6.3. t ON + t OFF = 1/1200 = 8.33
(a) (b)
t ON
= 0.7
t OFF
Η 833
-4
Η 10
s = 833 μs
= 583 μs
= 833 - 583 = 250 μs
(c) From Eq. (6.20), f sw(pu) = 0.02/10
Η 1200
17
= 2.4 pu
From Eq. (6.18), Io,ac(pu) = 0.7(1 - 0.7)/(2 %3 Η 2.4) = 0.025 pu From Eq. (6.19), Io,ac = 240/10 (d)
Io
Η 0.025
= 0.6 A
= 168/10 = 16.8 A
k r(I)
= 0.6/16.8 = 0.036
P6.4. (a) f sw(pu) = 0.02/10
(b)
Io,ac(pu)
Η 1200
= 2.4 pu
= 0.7(1 - 0.7)/(2 %3 Η 2.4) = 0.025 pu
P6.5. (a) Fourth quadrant (see Table 6.1 or Fig. 6.21)
(b)
V o
(c)
Io
= -0.7
Η 150
= -105 V
= (-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)
(c) (d)
t ON
+ t OFF = 1/900 = 1.11
t ON
= 0.3
t OFF
τ
s = 1.11 ms
= 0.33 ms
= 1.11 - 0.33 = 0.78 ms
= 1.7
f sw(pu)
-3
Η 10
/0.5 = 0.0034 s
= 0.0034
Io,ac(pu) Io,ac
Η 1.11
-3
Η 10
= 0.7
Η 900
Η (1
= 0.0198
= 3.06 pu
- 0.7)/(2%3
Η 150/0.5
Η 3.06)
= 0.0198 pu
= 5.94 A
P6.7. (a) First quadrant, since V o > 0, while V o > E implies Io > 0
(b)
Io
= (125 - 100)/0.5 = 50 A
18
(c)
M
= 125/150 = 0.833
(d)
D1
= 0.833
P6.8. V o = E + RIo
(a) First quadrant, since D1
= (210 + 0.1
E
> 0 and
Η 150)/300
Io
= 0.75
(b) Fourth quadrant, since E < 0 and D4
= (-210 + 0.1
P6.9. (a) V o = 0.5 Io
Η 300
>0
Η 150)/300
Io
>0
+ 1 = 0.65
= 150 V
= (150 - 100)/0.1 = 500 A
(b) The
M
= -0.8 value is erroneous. Change it to -0.6.
V o = -0.6 Η 300 = -180 V Io = (-180 + 200)/0.1 = 200
A
P6.10. From Eq. (6.18), with M = 0.5 (which results in the highest ripple), 0.02 >
0.5(1 - 0.5) 2 3 f sw(pu)
which is satisfied if f sw(pu) > 3.6 τ
= 0.0005/0.1 = 0.005 s
f sw
> 3.6/0.005 = 720 Hz
P6.11. First quadrant: M
= 0.6 ... 1
E
= 240 V
D1
E/V i =
= 0.6 ... 1
19
240/400 = 0.6
Second quadrant: M
E
= 0 ... 0.6
D2 E
Third quadrant: M
E
0.6
E/V i =
-0.6
= 0.6 ... 1
= -240 V
D4
= -0.6 ... 0
E/V i =
= 0.4 ... 1
= -240 V
D3
= -1 ... -0.6
Fourth quadrant: M
= 240 V
E/V i =
-0.6
= 0.4 ... 1
P6.12. V o,p = 12/(1 - 0.9) = 120 V t ON
+ t OFF = 1/2000 = 5
t OFF
= (1 - 0.9)
Η 500
-4
Η 10
s = 500 μs
= 50 μs
P6.13. V rat > 12 V Irat >
150/12 = 12.5 A
P6.14. V rat > V i,p = %2 Irat > Io,rat =
= 325 V
3000/300 = 10 A
P6.15. V rat > V i,p = %2 Irat > Io,rat =
Η 230
Η 460
= 651 V
10000/600 = 17 A
P6.16. V rat > V o,p = 6/(1 - 0.9) = 60 V ID(rat) > Io,dc(rat) =
240/6 = 40 A
(the average output voltage of 6 V determine the rated average chopper)
is used to current of the IS(rat) >
0.9
Η 40/(1
- 0.9) = 360 A
20
Chapter 7 P7.1. Simple square-wave mode: V o,1 = 0.9
Η 310
Optimal square-wave mode: V o,1 = 0.828
= 279 V
Η 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 π
ck =
4 π
2
4
π
∫ s i (nk t ) d t = k [ c o( sk ) - c o( sk 2 ) ] ω
ω
π
α d
α d
for k = 1, 3, 5, .... In the simple square-wave mode,
21
αd
= 0, and in the optimal
squre-wave mode,
αd
= 0.405 rad. MODE:
simple squarewave
optimal squarewave
k
ck
ck
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 360 o/30 = 12o wide. α13
= (13 - 1/2)
F (m,α13)
γ13
o
Η 12
= 150o
= 0.6sin(150 o) = 0.3
= 0.3
o
Η 12
= 3.6o
α1,13
= 150o - 3.6o/2 = 148.2o
α2,13
= 150o + 3.6o/2 = 151.8o
P7.4. V LL,1,p = 2/%3 V LL,1
Η 620
= 716 V
= 716/ %2 = 506 V
V LN,1 =
506/%3 = 292 V
(peak line-to-line) (rms line-to-line) (rms line-to-neutral)
P7.5. 5 - 1 - 3 - 2 - 6 - 4 - ...
22
P7.6. The switching intervals are 15o wide. α4
= (4 - 0.5)
Phase A:
o
Η 15
0
= 52.5o # α4
< π/3
F (m, α4) a
= (1 + 0.785)/2 = 0.8925
γA,4
Phase B:
= 2 Η 0.9 Η cos(52.5o - 60o) - 1 = 0.785
= 0.8925
o
Η 15
= 13.3875 o
α1A,4
= 52.5o + 13.3875 o/2 = 59.19375 o
α2A,4
= 52.5o - 13.3875o/2 = 45.80625 o
4π/3 #
α4
F (m, α4 b
- 120o < 5π/3
- 120o) = -1
= (1 - 1)/2 = 0
no switchings in this phase Phase C:
2π/3 #
α4
F (m, α4-
- 240o < π
240o) = 2
o
Η 0.9 Η sin(52.5
- 240o - 30o) - 1 =
0.096 c
= (1 + 0.096)/2 = 0.548
γC,4
= 0.548
o
Η 15
= 8.22o
α1C,4
= 52.5o + 8.22o/2 = 56.61o
α2C,4
= 52.5o - 8.22o/2 = 48.39 o
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.
23
P7.7. The switching intervals are 10o wide and 292 μs long. m
= 400%2/620 = 0.912
α19 =
(19 - 0.5)
o
Η 10
= 185o
6
Sextant 4
6 α
= 5o
X = 3 = 011 2, Y = 1 = 001 2, Z1 = 0002 = 0 State sequence: X - Y - Z1 = 3 - 1 - 0 d X
=
d 3
= 0.912sin(60 o - 5o) = 0.747
d Y
=
d 1
= 0.912sin(5o) = 0.079
d Z = d 0
= 1 - 0.747 - 0.079 = 0.174
t 3 =
0.747
Η 292
= 218 μs
t 1 =
0.079
Η 292
= 23 μs
t 0 =
0.174
Η 292
= 51 μs
P7.8. The switching intervals are 10o wide and 292 μs long. m
= 400%2/620 = 0.912
α19 =
(19 - 0.5)
o
Η 10
= 185o
6
Sextant 4
X = 3, Y = 1, Z = 0 State sequence: X - Y - Z = 3 - 1 - 0 d X
=
d 3
= 0.912sin(60 o - 5o) = 0.747
d Y
=
d 1
= 0.912sin(5o) = 0.079
d Z = d 0
= 1 - 0.747 - 0.079 = 0.174
t 3 =
0.747
Η 292
= 218 μs
t 1 =
0.079
Η 292
= 23 μs
24
6 α
= 5o
t 0 =
0.174
Η 292
= 51 μs
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 highefficiency sequence 0 - 1 - 3. P7.9. In the angle domain, the 19 th switching interval extends from 180o to 190o. State 3 occupies the first 0.747 of this interval, from 180o to 180o + 0.747 Η 10o = 187.47 o. State 1 occupies the next 0.079 of the interval, from 187.47 o to 187.47o + 0.079 Η 10o = 188.26 o. 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, 18 th, interval was State 7. Consequently, prior to the beginning of the switching interval in question, all switching variables are 1 (State 7). At 180 o, switching variable a changes from 1 to 0 (State 3), at 187.47 o, variable b changes from 1 to 0 (State 1), and at 188.26o, variable c follows suit (State 0). At 190 o, variable c changes from 0 to 1 (State 1). Thus:
Switch SA: Switch SA': Switch SB: Switch SB': Switch SC: Switch SC':
turns turns turns turns turns turns
off at 180o on at 180o off at 187.47o on at 187.47o off at 188.26o and turns on at 190o on at 188.26o and turns off at 190o
P7.10. In the angle domain, the 19 th switching interval extends from 180o to 190o. State 3 occupies the first 0.747 of this interval, from 180o to 180o + 0.747 Η 10o = 187.47 o. State 1 occupies the next 0.079 of the interval, from 187.47 o to 187.47o + 0.079 Η 10o = 188.26 o. 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, 18 th, 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 180 o since the first state in the 19 th switching interval is State 3 again. At 187.47 o, variable b changes from 1 to 0 (State 1), and at 188.26 o, 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 switched Switch SA': is not switched Switch SB: turns off at 187.47o
25
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. P7.11. Switching variable a changes from 0 to 1 at 0o, 14.80o, 89.07o, 92.07o, 170.52 o, 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 = (2%3/π) IL,1
Η 200
= 220.5 A
= 220.5/%2 = 155.9 A
V LN,1
= 2 Η 155.9 = 311.8 V
(the
voltage its contribution negligible) V LL,1
= 311.8
Η %3
load
inductance produces only spikes during switchings, and to the output voltage is
= 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 177.76 o (OFF). Switching pattern for switch SA' is shifted by 180o, that is, 182.24 o (ON), 185.6o (OFF), ..., 357.76o (OFF).
P7.14. V i = (3/π) V LL,p
Η %2 Η 1.2
= 1.065
Η 1.62
= 1.62 kV = 1.73 kV
V LL
= 1.73/%2 = 1.22 kV
V LN
= 1.22/%3 = 0.7 kV
P7.15. 2010 = 2023
6 a
= 2, b = 0,
c
=2
26
P7.16. See Fig. 7.55. Shifting the v AB waveform rightwards by 15 o produces the even and half-wave symmetry. The rms value of the line-to-line voltage can be calculated as
1
V L L=
2 π
5
π
3
∫
[ 1 d ω t + 2
∫
1
0
π
12
2 1 5 1 0 5. d ω t ]= [ π +0 . 2 5 ( π - π ) ]= 0 . 7 6 π 4 12 4 2
π
4
The peak value of the fundamental of this voltage is 1
V LL,1, p =
4 π
5
π
4
π
12
∫
[ 1 cos( ω t)d ω t +
∫ 0.5 cos(
1
0
t)d ω t
ω
π
4
=
4
1
5
1
[ sin( π ) - sin(0)+ 0.5 sin( π ) - 0.5 sin( π )] = 1. 4 12 4 π
and the rms value, V LL,1 = 1.065/%2 = 0.7531. Hence, the harmonic content is and the total harmonic distortion of the line-to-line voltage is d h,LL= 0.1274/0.7531 = 0.169.V LL, h = 0. 7638 2 - 0. 7531 2 = 0.1274 Similarly, to acquire the even and half-wave symmetries, the should be shifted leftwards by 15o. Then
v AN
waveform
and, using the same approach as before, the rms value, V LN, of the line-toneutral voltage is calculated as 0.441, i.e., equal to V LL/%3. The rms value, V LN,1, of the fundamental line-to-neutral voltage is 0.4348. Thus, 2
2
0. 441 - 7.4, 0. 4348 V LN ,in h =Example the = dc0.074 link voltage is given by P7.17. Since, as calculated
27
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 vo = 200 - 243.8 cos (64550t + 0.609) V 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 Η 248 = 322.4 V. Thus, in comparison with the unclamped dc link, the required voltage rating of the inverter switches is reduced by 27%. P7.18. According to Eqs. (7.12) and (4.89), duty ratio of switch SA in the n th switching interval is given by
1 2 1 = = _1 + m [ sin ( ) + sin(3α n )] _ α n D A,n an 2 6 3 where α n = (n
1 2π - ) 2 18
Thus, the on-time of the switch is t ON =
D A, n 18_120
and the off-time is t OFF =
1 - D A, n 18_120
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 t ON and t OFF for m = 1 and individual switching intervals are listed in the following table. n
DA,n
t ON (μs)
t OFF (μs)
1
0.6484
300.2
162.8
28
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
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
t ON
and t OFF 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, V rat ∃
1.4 Η %2
Η 460
= 911 V
The average output voltage of the rectifier is (3/ π) Η %2 Η 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 Η 439) = 78.9 A. Consequently, the rated current, IS(rat), of the inverter switches must satisfy the condition
29
IS(rat) ∃ (%2/π) Η 1.25 Η 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, V rat ∃
1.4 Η %2
Η 2400/2
= 2376 V
The average output voltage of the rectifier is (3/π) Η %2 Η 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 Η 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 Η 2441) = 47.3 A. Consequently, the rated current, IS(rat), of the inverter switches must satisfy the condition IS(rat) ∃ (%2/π) Η 1.25 Η 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
30
Chapter 8 P8.1. D = 32/(32 + 8) = 0.8 V i = V o/D f sw
= 1/(32 + 8) = 0.025 MHz = 25 kHz
P8.2. V o = 0.6 Io
= 12/0.8 = 15 V
Η 50
= 30 V
= 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),
31
L = 2
30
1- 0 _ = 6.67_ 10-4 H 3 2_3 15_ 10
and from Eq. (8.9),
∆V o = 30
1 - 0.6 2
-4
8_6.67_ 10 _50_ 10 -6 _(15_ 10 3 )
= 0.2 V
P8.3. D = 32/(32 + 8) = 0.8 V i = V o Η (1 f sw
- D) = 12
Η (1
- 0.8) = 2.4 V
= 1/(32 + 8) = 0.025 MHz = 25 kHz
P8.4. D = 32/(32 + 8) = 0.8 V i = V o Η (1 f sw
- 1/D) = 12
Η (1
- 1/0.8) = -3 V
= 1/(32 + 8) = 0.025 MHz = 25 kHz
P8.5. Assume f sw = 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, can be specified from Eq. (8.9) as
C ≥
1- 0 -3
3
2
8_ 10 _(25_ 10 )
D
= 0, it
_100 = 2_ 10-5 F = 20 µ F
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, V o = 12/(1 - 0.5) = 24 V.
Assuming
f sw
= 25 kHz and
R
= 200 Ω, the required capacitance, for the worst
32
case of D = 1, can be found from Eq. (8.18) as 1
C ≥
200_25_ 10
_100 = 2_ 10 -5 F = 20 µ F
3
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, f sw = 25 kHz and R = 200 Ω, the required capacitance can be found from Eq. (8.18) as 1
C ≥
P8.8. V o = 1
Η 0.4 Η 15
200_25_ 10
_100 = 2_ 10 -5 F = 20 µ F
3
=6V
For the worst case of D = 0, the required inductance can be found from Eq. (8.11) as
L ≥
6 3
2_25_ 10 _2
(1 - 0) = 6_ 10 -5 = 60 µ H
Then, from Eq. (8.9),
∆V o =
1 - 0.4 2
-5
8_6_ 10 _160_ 10 -6 _(25_ 10 3 )
_6 = 0.075 V = 75 mV
P8.9. Io = 100/25 = 4 A
From Eq. (8.55), D =
100 V o = = 0.769 + 0.5_60 + 100 k N V i V o
and from Eq. (8.56),
33
2
(1 - 0.769 ) _100
f sw >
2_0. 5 2 _4_0.13_ 10 -3
= 20563 Hz ≈ 20.5 kHz
Then, from Eq. (8.18),
∆V o =
0.769 25_0.2_ 10 -3 _20.5_ 10 3
_100 = 0.75 V
P8.10. From Eq. (8.55), V o = 1
0.35
100 = 53.8 V 1 - 0.35 53.8 = 2.69 A I o = 20
For the worst case of D = 0, the inductances can be determined from Eqs. (8.57) and (8.50) as 2
L1 = 2
(1 - 0 ) _53.8 2
2_ 1 _2.69_30_ 10
3
= 6.67_ 10 - 4 H
and 2
L 2 = 2
(1 - 0 ) _53.8 2_2.69_30_
10
3
= 6.67_ 10 - 4 H
Then, from Eq. (8.9),
∆V o =
1 - 0.35 2
-4
8_6.67_ 10 _25_ 10 -6 _(30_ 10 3 )
P8.11. From Eq. (8.58), D =
80 2_2_50
and from Eq. (8.59),
34
= 0.4
_53.8 = 0.29 V
∆V o =
1 - 2_0.4 2
32_0.1_ 10 -3 _20_ 10 -6 _(20_ 10 3 )
_80 = 0.63 V
P8.12. From Eq. (8.60), for D = Dmax= 0.5, V i = 100(2
Η 0.5)
P8.13. From Eq. (8.58), for D = Dmax= 0.5, V i = 100(2
Η 2 Η 0.5)
P8.17. From Eq. (8.63),
from Eq. (8.62), Z o =
0.1_ 10 -3 = 20 0.25_ 10 -6
Ω
from Eq. (8.65), 35000
k f =
31831
= 1.1
and from Eq. (8.66), r =
15 20
= 0.75
P8.18. From Eq. (8.63),
f o =
1 - 6
2π 60_ 10 _120_ 10 -9
= 59314 Hz
from Eq. (8.62), Z o =
60_ 10 -6 120_ 10 - 9
from Eq. (8.65),
35
= 22.36
Ω
= 100 V = 50 V
k f =
24000
= 0.405
59314
from Eq. (8.66), 20
r =
22.36
= 0.8944
and from Eq. (8.75), 1
K v =
2
π
2 1 + [ 8_0.8944
Thus,
V o
P8.19.
= 0.1657 R
Η 120
(0.405 -
= 19.88 V and
= 502/500 = 5
Io
1 0.405
= 0.1657 ) ]
2
= 19.88/20 = 0.994 A
Ω
From Eq. (8.63),
f o =
1 2π 12_ 10 -6 _3.3_ 10 -6
= 25291 Hz
from Eq. (8.62), Z o =
12_ 10 -6 3.3_ 10 -6
= 1.91
from Eq. (8.65), k f =
30000 = 1.186 25291
from Eq. (8.66), r =
5 1.91
= 2.62
and from Eq. (8.81),
36
Ω
