HartChapter8solutions.doc

CHAPTER 8 SOLUTIONS 4/24/10

8-1)

Load: I avg  0, I rms  10 A. Switches: I avg  5 A., I rms  I m D  10 0.5  7.07 A. Source : I avg  I rms  10 A.

(See Example 2-4)

8-2)

I max

V  dc R

 1  e T /2 L 0.1 1 V 96 ;    0.02s.; T  ; dc   19.2 A.  T /2 R 5 60 R 5  1 e   0.341   3.94 A.  1.66)

I max  19.2 

b) From Eq. 8  1: io (t )  io (0)  0  A  

Vdc  Ae  t / R

Vdc R

Vdc 1  e  t /   R i (T / 2)  19.2  1  e 1/2.4   6.54 A. io (t ) 

c) PSpice results are consistent with parts (a) and (b). The current waveform reaches steady state after approximately 100 ms, corresponding to 5 time constants. 8.0A (8.3333m,6.5486)

(158.333m,3.9485) 4.0A

0A

-4.0A 0s

40ms

80ms

120ms

160ms

I(L) Time

8-3) Vdc 150 L 40 mH   7.5 A.;     2 ms; R 20 R 20 

a)

T 1 / 60   4.167 2 4 ms

Using Eq (8  8),  1  e 4.167  7.271 A. 4.167  1 e 

I max  7.5 

I min   I max  7.271 A. Using Eq. (8-5), for 0  t  8.33 ms

 7.5  14.8e  (t 1/120)/.002

for 8.33 ms  t  16.7 ms

io  

 7.5  14.8e t /.002 

200ms

b)

c ) I peak  7.271 A. d ) Vmax  Vdc  150 V .

8-4)

Vdc 125 L 25 mH   6.25 A.;     1.25 ms; R 20 R 20 

a)

T 1 / 60   13.33 2 1.25 ms

Using Eq (8  8),  1  e 13.33  6.25 A. 13.33  1 e 

I max  6.25 

I min   I max  6.25 A. Using Eq. (8-5),  6.25  12.5e t /.00125 

for 0  t  8.33 ms

 6.25  12.5e  (t 1/120)/.00125

for 8.33 ms  t  16.7 ms

io  

b) Using the first half-period, 1  120

I rms

1/120

  6.25  12.5e

 t /.00125

0



2

dt  5.45 A.

2 c) P  I rms R   5.25  20  594 W . 2

Is 

P 594   4.75 A. Vdc 125

8-5) Z1  152   2  400   0.01

a)



2

 29.3 



V1  I1Z1  8 2  29.3   331 V . V1 

4Vdc 

b)

 Vdc  Vn 

4Vdc ; n n 1 3 5

THDI 

8-6)

 V1  260 V . 4 Z n  R 2   2 400 L  Vn 331 110 66

1.022  0.37 2  0.136  13.6% 8.0

2

; Zn 29.3 77 127

In 

Vn Zn

; I n ,rms  In,rms 8.0 1.02 0.37

In 2

a ) Z1  2.52   2  120   0.025



2

 31.3 



V1  I1Z1  2 2  31.3  88.6 V . V1 

4Vdc 

b) Vn 

 Vdc  4Vdc ; n

 V1  69.6 V . 4 Z n  R 2   2 120 L 

n 1 3 5

Vn 88.6 29.5 17.7 THDI 

Using PSpice,

2

;

In 

Vn Zn

Zn 31.3 61.8 97.5

0.342  0.132  0.185  18.5% 2.0

; I n ,rms 

In,rms 2.0 0.34 0.13

In 2

FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L) DC COMPONENT = -3.668708E-06 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 2 3 4 5 6 7 8 9

1.200E+02 2.400E+02 3.600E+02 4.800E+02 6.000E+02 7.200E+02 8.400E+02 9.600E+02 1.080E+03

2.830E+00 5.377E-06 4.778E-01 3.589E-06 1.818E-01 2.858E-06 9.427E-02 2.523E-06 5.743E-02

1.000E+00 -3.716E+01 0.000E+00 1.900E-06 -1.203E+02 -4.594E+01 1.688E-01 -6.658E+01 4.490E+01 1.268E-06 -1.223E+02 2.629E+01 6.422E-02 -7.587E+01 1.099E+02 1.010E-06 -1.162E+02 1.068E+02 3.331E-02 -8.028E+01 1.798E+02 8.913E-07 -1.095E+02 1.878E+02 2.029E-02 -8.292E+01 2.515E+02

TOTAL HARMONIC DISTORTION = 1.847695E+01 PERCENT

8-7) Using a restricted time interval of 33.33 ms to 50 ms to analyze steady-state current, the peak value is 8.26 A and the rms value is 4.77 A. The THD from the output file is 32%. 20A

10A

Peak (35.134m,8.2603)

rms (50.000m,4.7738)

i(t) 0A

-10A 30ms I(R)

35ms RMS(I(R))

40ms

45ms

50ms

Time

voltage (100 V)

10

current

S1, S2

D3, D4

0 D1, D2 S3, S4

-10 30ms I(R)

35ms V(A)/10 0

40ms Time

45ms

50ms

8-8)  4Vdc  cos      

a ) V1  

 90   V1  1  55.6    cos    4Vdc  4  125 

  cos 1 

 4Vdc Vn    cos  n   n  n 1 3 5

;

Z n  R  jn0 L ; I n  |Vn| 90 51.6 4.43 THDI 

8-9) V1 

a)

Vn Zn

; I n ,rms 

Zn 12.5 24.7 39

In,rms 5.08 1.5 0.08

1.52  0.082  0.29  29% 5.08

4Vdc 4  200    255 V .  

Z1  R  j0 L  10  j 2 60  0.035   16.6  V1 255   15.3 A. Z1 16.6

I1 

I1,rms  b)

15.3 2

 10.9 A.

At 30 Hz ,

Z1  10  j 2 30  0.035   12.0  V1  I1Z1   15.3  12.0   184 V .  4Vdc  cos      

V1  

8-10)

 184   V1  1  43.7    cos   4 200    4Vdc  

   cos 1 

In 2

α = 30° Using the FFT function in Probe shows that voltages at frequencies at multiples of n = 3 are absent.

b) α = 15° Using the FFT function in Probe shows that voltages at frequencies at multiples of n = 5 are absent.

8-11)

From Eq. (8-22),



90 90   12.86 n 7

Using the FFT function in Probe, the n = 7 harmonic is absent.

8-12) Letting T = 360 seconds and taking advantage of half-wave symmetry, Vrms 

2   360 

Vrms  Vm

8-13)

54



30

114

Vm2 dt 



66



150

Vm2 dt 

V

2 m dt

126

 

1   54  30    114  66    150  126   0.730Vm 180 

The VPWL_FILE source is convenient for this simulation. A period of 360 seconds is used, making each second equal to one degree. A transient simulation with a run time of 360 second and a maximum step size of 1m gives good results. The FFT of the Probe output confirms that the 3rd and 5th harmonics and their multiples are eliminated. 0 30 30.01 54 54.01 66 66.01 114 114.01 126 126.01 150 150.01 210 210.01 234 234.01 246 246.01 294 294.01 306 306.01 330 330 360

0 0 1 1 0 0 1 1 0 0 1 1 0 0 -1 -1 0 0 -1 -1 0 0 -1 -1 0 0

8-14) a)

b)

Vm 

4Vdc  cos  n1   cos  n 2   cos  n 3  n 

Vdc  48 V ; 1  15 ;  2  25 ;  3  55

n Vn c) M i 

1 149.5

3 0

cos  1   cos   2   cos   3  3

5 -2.79

7 -3.04

9 -14.4

 0.815

8-15) To eliminate the third harmonic, cos(31 )  cos(3 2 )  cos(3 3 )  0 Select two of the angles and solve for the third. Examples: α1 15 20 10 10

α2 25 30 30 30

α3 55 40 50 70

Mi 0.815 0.857 0.831 0.731

8-16) This inverter is designed to eliminate harmonics n = 5, 7, 11, and 13. The normalized coefficients through n = 17 are n Vn/Vdc 1 4.4593 3 -0.8137 5 0.0057 ≈ 0 7 -0.0077 ≈ 0 9 -0.3810 11 0.0043 ≈ 0 13 -0.0078 ≈ 0 15 -0.0370 17 0.1725 The coefficients are not exactly zero for those harmonics because of rounding of the angle values.

8-17)

8-18) V1  V1,rms 2  54 2  76.8 V . ma 

V1 76.8   0.8 Vdc 96

Z n  R  jn0 L  32  jn 2 60  .024   32  jn9.05 From Table 8-3,

mf mf - 2 mf + 2

n 1 17 15 19

Vn/Vdc 0.8 0.82 0.22 0.22

Vn 76.8 78.7 21.1 21.1

Zn 33.3 157 139 175

In=Vn/Zn 2.30 0.50 0.151 0.121

2

2

0.121  0.50  0.151          2  2    2 THD  2.30 2

2

 0.23  23%

8-19) V1  V1, rms 2  160 2  226.3 V . ma 

V1 226.3   0.9 Vdc 250

Z n  R  jn0 L  20  jn 2 60  .050   20  jn18.85 From Table 8-3,

mf mf - 2 mf + 2

2

n 1 31 29 33

Vn/Vdc 0.9 0.71 0.27 0.27

2

0.108  0.305  0.122          2  2    2 THD  8.18 2

Vn 225 178 67 67

Zn 27.5 585 547 622

2

 0.044  4.4%

In=Vn/Zn 8.18 0.305 0.122 0.108

8-20) The circuit “Inverter Bipolar PWM Function” is suitable to verify the design results. The parameters are modified to match the problem values.

Transient Analysis and Fourier Analysis are establish in the Simulation Setup menu:

The output file contains the THD of the load current, verifying that the THD is less than 10%. TOTAL HARMONIC DISTORTION = 9.387011E+00 PERCENT

8-21) Example solution: Let ma  0.9, Vdc 

Vm 120 2   189 V . ma 0.9

Using Table 8-3, at n  m f , Vmf  0.71 189   134 V . for THD  8%, I mf  0.08I1 I1 

V1 120 2 120 2    13.6 A. Z1 10  j 2 60  0.020  12.5

I mf  0.08  13.6   1.09 A. Z mf  mf 

Vmf I mf



134  123   m f 0 L 1.09

123 123   16.4 0 L 377  0.020 

Choose odd integer 19 or greater for m f .

8-22) Example solution: V1  V1,rms 2  100 2  141 V . Let ma  0.9  Vdc  I1 

V1 141   157 V . ma 0.9

V1 V1 141    4.48 A. Z1 R  j0 L 30  j 377  0.025 

THDI  Z mf  mf 

I mf I1

Vmf I mf



 0.10



0.71 157  0.448

I mf  0.1 4.48   0.448 A.

 249   m f 0 L

249  26.4 377  0.025 

Choose odd integer 29 or greater for m f .

8-23) Use the bipolar PWM function circuit of Fig. 8-23a, and use the unipolar PWM function circuit of Fig.8-26 with mf = 10. Use ma = 0.8 for V1 = 120 V from the 150-V dc source.

The THD for bipolar, mf = 21, is 10.2 %, for bipolar mf = 41 is 5.2%, and for unipolar mf = 10 is 5.9%.

Bipolar mf = 21:

Bipolar mf = 41:

Unipolar, mf = 10:

8-24) a ) V1, L  N  I1 

V1 159 159    6.09 A. Z1 25  j 377  0.020  26.1

I1,rms 

8-25)

2Vdc     2     500  2  cos    cos     3  159 V . 3   3  3    3

I1 2

 4.31 A.

Use Eq. (8-42) for Vn,L-N , Z n  R  jn2 fL , I n  Vn, L- N / Z n , and I n, rms  I n / 2. For f = 25 Hz: n 1 5 7 11 13 THDI 

VnL-N 255 50.9 36.4 23.1 19.6

Zn 11.1 25.6 34.5 52.8 62.0

In 23.0 2.0 1.06 0.44 0.32

In,rms 16.3 1.41 0.75 0.31 0.22

1.412  0.752  0.312  0.222  0.10  10% 16.3 THDV 

For f = 100 Hz, n 1 5 7 11 13

THDI 

50.92  36.42  0.222  19.62  0.273  27.3% 255

VnL-N 255 50.9 36.4 23.1 19.6

Zn 21.3 94.8 132 208 245

In 11.9 0.54 0.27 0.12 0.08

In,rms 8.43 0.38 0.19 0.08 0.06

0.382  0.192  0.082  0.062  0.0519  5.19% 8.43

The THD for current is reduced from 10% to 5.19% as f is increased from 25 Hz to 100 Hz. The THD of the line-to-neutral voltage remains at 27.3%. These results can also be determined from a PSpice simulation for the six-step inverter.

8-26)





At f  30 Hz , Z1  10.7 , V1  I1Z1  10 2  10.7   151 V . V1, L  N  Vdc 

2Vdc     2    2  cos    cos    Vdc  0.637  3   3  3 

V1, L  N 0.637



151  237 V . 0.637





At f  60 Hz , Z1  19.5 , V1  I1Z1  10 2  19.5   276 V Vdc 

276  433 V . 0.637