HartChapter8solutions.doc

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CHAPTER 8 SOLUTIONS

4/24/10

8-1)

Load:

0,

10 .

Switches:

5 .,

10 0.5 7.07 .

Source :

10 .

(See Example 2-4)

avg rms avg rms m avg rms

I

A

I

A I

I

D

A

I

A



8-2)









/2 max /2 max / / 1/2.4

1

0.1

1

96 ;

0.02 .;

;

19.2 .

1

5

60

5

0.341

19.2 3.94 .

1.66)

)

. 8 1:

( )

(0) 0

( )

1 ( / 2) 19.2 1

6.54 .

T dc dc T t dc o dc o t dc o

V

e

L

V

I

s

T

A

R

e

R

I

A

V

b From Eq

i t

Ae

R

V

i

A

R

V

i t

e

R

i T

e

A

   



    









_

_

 















_

_

















 











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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.

Time 0s 40ms 80ms 120ms 160ms 200ms I(L) -4.0A 0A 4.0A 8.0A (158.333m,3.9485) (8.3333m,6.5486) 8-3) 4.167 max 4.167 min max /.002 ( 1/120)/.002 150 40 1/ 60 ) 7.5 .; 2 ; 4.167 20 20 2 4 Using Eq (8 8), 1 7.5 7.271 . 1 7.271 . Using Eq. (8-5), 7.5 14.8 0 8.33 i 7.5 14.8 dc t o t V L mH T a A ms R R ms e I A e I I A e for t ms e                     _ _               for 8.33ms t 16.7ms      

(3)

b) max ) 7.271 . ) 150 . peak dc c I A d V V V    8-4)

(4)

13.33 max 13.33 min max /.00125 ( 1 125 25 1/ 60 ) 6.25 .; 1.25 ; 13.33 20 20 2 1.25 Using Eq (8 8), 1 6.25 6.25 . 1 6.25 . Using Eq. (8-5), 6.25 12.5 0 8.33 6.25 12.5 dc t o t V L mH T a A ms R R ms e I A e I I A e for t ms i e                     _ _              









/120)/.00125 1/120 2 /.00125 0 2 2 8.33 16.7

) Using the first half-period, 1 6.25 12.5 5.45 . 120 ) 5.25 20 594 . 594 4.75 . 125 t rms rms s dc for ms t ms b I e dt A c P I R W P I A V                



8-5)



 



 









2 2 1 1 1 1 1 1 2 2 , ) 15 2 400 0.01 29.3 8 2 29.3 331 . 4 260 . 4 4 ) ; 2 400 ; ; 2 dc dc dc n n n n n n rms n a Z V I Z V V V V V V V V I b V Z R L I I n Z         _ _               n Vn Zn In,rms 1 331 29.3 8.0 3 110 77 1.02 5 66 127 0.37 2 2 1.02 0.37 0.136 13.6% 8.0 I THD     8-6)

(5)

  



 









2 2 1 1 1 1 1 1 2 2 , ) 2.5 2 120 0.025 31.3 2 2 31.3 88.6 . 4 69.6 . 4 4 ) ; 2 120 ; ; 2 dc dc dc n n n n n n rms n a Z V I Z V V V V V V V V I b V Z R L I I n Z         _ _               n Vn Zn In,rms 1 88.6 31.3 2.0 3 29.5 61.8 0.34 5 17.7 97.5 0.13 2 2 0.34 0.13 0.185 18.5% 2.0 I THD     Using PSpice,

(6)

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 1.200E+02 2.830E+00 1.000E+00 -3.716E+01 0.000E+00 2 2.400E+02 5.377E-06 1.900E-06 -1.203E+02 -4.594E+01 3 3.600E+02 4.778E-01 1.688E-01 -6.658E+01 4.490E+01 4 4.800E+02 3.589E-06 1.268E-06 -1.223E+02 2.629E+01 5 6.000E+02 1.818E-01 6.422E-02 -7.587E+01 1.099E+02 6 7.200E+02 2.858E-06 1.010E-06 -1.162E+02 1.068E+02 7 8.400E+02 9.427E-02 3.331E-02 -8.028E+01 1.798E+02 8 9.600E+02 2.523E-06 8.913E-07 -1.095E+02 1.878E+02 9 1.080E+03 5.743E-02 2.029E-02 -8.292E+01 2.515E+02

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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%.

Time 30ms 35ms 40ms 45ms 50ms I(R) RMS(I(R)) -10A 0A 10A 20A i(t) Peak rms (50.000m,4.7738) (35.134m,8.2603) Time 30ms 35ms 40ms 45ms 50ms I(R) V(A)/10 0 -10 0 10 voltage (100 V) current D3, D4 S3, S4 D1, D2 S1, S2

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8-8)

 

1 1 1 1 0 , 4 ) cos 90 cos cos 55.6 4 4 125 4 cos ; ; ; 2 dc dc dc n n n n n n rms n V a V V V V V I V n Z R jn L I I n Z              _ _         _ _         _ _       n |Vn| Zn In,rms 1 90 12.5 5.08 3 51.6 24.7 1.5 5 4.43 39 0.08 2 2 1.5 0.08 0.29 29% 5.08 I THD     8-9)















 



 

_

_

1 1 0 1 1 1 1, 1 1 1 1 1 1 1 1 4 200 4 ) 255 . 10 2 60 0.035 16.6 255 15.3 . 16.6 15.3 10.9 . 2 ) 30 , 10 2 30 0.035 12.0 15.3 12.0 184 . 4 184

cos cos cos

4 4 200 dc rms dc dc V a V V Z R j L j V I A Z I A b At Hz Z j V I Z V V V V V                                       _ _   _ _       43.7       _ 8-10)

(9)

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.

(10)

From Eq. (8-22), 90 90 _12.86

7

n

   

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,



 



54 114 150 2 2 2 30 66 126 2 360 1 54 30 114 66 150 126 0.730 180 rms m m m rms m m V V dt V dt V dt V V V    _   _      _      _ 



8-13)

(11)

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 5}th_{harmonics and their multiples are eliminated.}

0 0 30 0 30.01 1 54 1 54.01 0 66 0 66.01 1 114 1 114.01 0 126 0 126.01 1 150 1 150.01 0 210 0 210.01 -1 234 -1 234.01 0 246 0 246.01 -1 294 -1 294.01 0 306 0 306.01 -1 330 -1 330 0 360 0

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8-14) a)

 

1



2





3



1 2 3

4

) cos cos cos

48 ; 15 ; 25 ; 55 dc m dc V b V n n n n V V           _   _       

(13)

n 1 3 5 7 9

Vn 149.5 0 -2.79 -3.04 -14.4

 

1

 

2

 

3

cos cos cos

) 0.815 3 i c M        8-15) 1 2 3

To eliminate the third harmonic, cos(3 ) cos(3 ) cos(3 ) 0

Select two of the angles and solve for the third. Examples:       α1 α2 α3 Mi 15 25 55 0.815 20 30 40 0.857 10 30 50 0.831 10 30 70 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.

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8-18)





1 1, 1 0 2 54 2 76.8 . 76.8 0.8 96 32 2 60 .024 32 9.05 rms a dc n V V V V m V Z R jn L jn  jn             From Table 8-3, n Vn/Vdc Vn Zn In=Vn/Zn 1 0.8 76.8 33.3 2.30 mf 17 0.82 78.7 157 0.50 mf - 2 15 0.22 21.1 139 0.151 mf + 2 19 0.22 21.1 175 0.121

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2 2 2 0.50 0.151 0.121 2 2 2 0.23 23% 2.30 2 THD   _   _                 8-19)





1 1, 1 0 2 160 2 226.3 . 226.3 0.9 250 20 2 60 .050 20 18.85 rms a dc n V V V V m V Z R jn L jn  jn             From Table 8-3, n Vn/Vdc Vn Zn In=Vn/Zn 1 0.9 225 27.5 8.18 mf 31 0.71 178 585 0.305 mf - 2 29 0.27 67 547 0.122 mf + 2 33 0.27 67 622 0.108 2 2 2 0.305 0.122 0.108 2 2 2 0.044 4.4% 8.18 2 THD   _   _                

(16)

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

(17)

8-21) Example solution:

 









1 1 1 1 0 0 0.9, 120 2 189 . 0.9 Using Table 8-3, at , 0.71 189 134 . 8%, 0.08 120 2 120 2 13.6 . 12.5 10 2 60 0.020 0.08 13.6 1.09 . 134 123 1.09 123 123 377 0. a m dc a f mf mf mf mf mf f mf f Let m V V V m n m V V for THD I I V I A Z j I A V Z m L I m L                          



020



16.4

Choose odd integer 19 or greater for m ._f  8-22) Example solution:









 





1 1, 1 1 1 1 1 0 1 0 2 100 2 141 . 141 0.9 157 . 0.9 141 4.48 . 30 377 0.025 0.10 0.1 4.48 0.448 . 0.71 157 249 I 0.448 249 26.4 377 0.025

Choose odd integer 29

rms a dc a mf I mf mf mf f mf f V V V V Let m V V m V V I A Z R j L j I THD I A I V Z m L m                             or greater for m ._f

(18)

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

(19)

Bipolar mf = 21:

(20)

Unipolar, mf = 10: 8-24)





1, 1 1 1 1 1, 2 2 500 ) 2 cos cos 3 159 . 3 3 3 3 159 159 6.09 . 26.1 25 377 0.020 4.31 . 2 dc L N rms V a V V V I A Z j I I A          _{ }   _  _  _  _            8-25)

(21)

n,L-N , - ,

Use Eq. (8-42) for V , Z_n  R jn2fL, I_nV_{n L N} /Z_n, and I_{n rms} I_n/ 2. For f = 25 Hz: n VnL-N Zn In In,rms 1 255 11.1 23.0 16.3 5 50.9 25.6 2.0 1.41 7 36.4 34.5 1.06 0.75 11 23.1 52.8 0.44 0.31 13 19.6 62.0 0.32 0.22 2 2 2 2 1.41 0.75 0.31 0.22 0.10 10% 16.3 I THD       2 2 2 2 50.9 36.4 0.22 19.6 0.273 27.3% 255 V THD       For f = 100 Hz, n VnL-N Zn In In,rms 1 255 21.3 11.9 8.43 5 50.9 94.8 0.54 0.38 7 36.4 132 0.27 0.19 11 23.1 208 0.12 0.08 13 19.6 245 0.08 0.06 2 2 2 2 0.38 0.19 0.08 0.06 0.0519 5.19% 8.43 I THD      

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%.

(22)

8-26)





















1 1 1 1 1, 1, 1 1 1 1 30 , 10.7 , 10 2 10.7 151 . 2 2 2 cos cos 0.637 3 3 3 151 237 . 0.637 0.637 60 , 19.5 , 10 2 19.5 276 276 433 . 0.637 dc L N dc L N dc dc At f Hz Z V I Z V V V V V V V At f Hz Z V I Z V V V                   _  _{ }  _ __                  