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Unformatted text preview: AV [Nu i WWW“; EE610 HW#13 2010 1) Using the voltage waveform shown in Figure 13.3—2, use a Fourier Series to derive the funda—
mental component of the phase voltage V“. 2) Given a single—phase Eel—bridge inverter. The output voltage is shown. The output load is induc—
tive with L 2 l mH, f; : 1000112, iom.(0) : ——2 A. Plot to“, from 0 to Zn. Label the devices conducting over the entire waveform. 3) Given a 3—phase inverter operating under sinemti'iangle modulation. Vdc 2 200 V. The inverter is sourcing a wyeconnneeted 3—phase resistive load with mom! 3 1 £2. The phase—a duty cycle under is d0 = 50 (303(96 — (b and C have the same amplitude but diplaced by 120 degrees). a) Plot the fundamental component of the phase—a voltage you would expect to see out of the
inverter on the following plot. \ t
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{it “:1. {1‘3 . is {it t e. * bm'g‘é’“ 4) For the circuit shown below, the switching signal and a—phase current are shown on Page 2.
Label the piots which correspond to the switches T1,Dl,T4,D4 'clc Inverter Circuit i l
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at ' "‘3 " ‘7 _ .__..____.________._h. _____r_.___...MM.V WWiW %>\ a“? Ex“ 4“ "[10. 0(3E1ﬂi'lf3 {I Iii'iHltlfi U. OBI5‘20? [.1. 0951310 I]. (“Kiln :3 H. H3001 / 5. Given 3—phase 6pulse inverter circuit shown in Problem 4. a) Assume the load is purely resistive. Thephase resistances are not equal. i’hasea has a 2 Ohm
resistor, phase I) and 0 each have 4 Ohm resistors. The dc link voltage is 50 volts. A 180
degree switching strategy is employed. Piot the phase current, transistor 1 and diode l cur—
rents over a single switching cycle. b) Given the same inverter with a balanced load. Using sine—triangle PWM, the phase—a duty
cycle waveform has a value a", = dcos(BC) e 519003696) . Phaseb and c duty cycles are similar (with appropriate phase dis— placement). Derive the the resulting fundamental component of the phasea voltage assuming dc, < l . What is the bound on d to ensure this condition? Based upon your result, is it better to introduce a 3rd or 9th harmonic? 6) KW Chapter 13, #13, #14, #16. I," “‘2 1, ‘ TLTZ an, n M,
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'16 “SAABLW (4.958“, D “mu«Mr—»;....4.._;u,....._...kh...._...,¢ A..L..,_..u...v.;,‘...‘_._ .. ... ... _ . _ , . 6&3 12/2/05 8:31 AM D:\Documents\steve\courses\ee610\induc3.m 1 of 1 % Steady State Induction Motor Analysis % Parameters of Induction Machine P = 4; we : 2*pi*60; rs 0.435; LS .754/377; Lm = 26.13/377; Lr .754/377; rr .816; Vdc = 280; d1 linspace(0,1,20); d 2 linspace(l,5.20); fd = 0.5*sqrt(l~(1./d).AZ)+1/4.*d.*(pim2*acos(l./d)}:
vs z [vdc*d1/2/sqrt(2) 2*vdc/pi*fd/Sqrt{2)3;
do = [d1 d1; %Handy Number 1:: = Lr + Lm; H ii 1% wr = 2*1710/60*2*pi; %Determine operating characteristics for i = l:length(do) s = (we — wr)/we; vasfi).= vs(i); zin = rs + j*we*LS + l.0/(l.0/(j*we*Lm) + 1.0/(rr/s + j*we*Lr)};
ias(i) = vas(i)/zin; %vas(i) a zin*ias(i); iar(i) = ~ia5(i)*j*we*Lm/{rr/s + j*we*lrr); iasmag(i} abs(ias(i)}; iarmag(i) = abs(iar(i)); teti} 2 3*P/2*we*Lm/377*real(j*conj(ias(i)}*iar(i)};
pin(i)=3.0*real{conj{ia5(i)}*va5(i}); poutii) : te(i)*wr*2.0/P: eff(i) = 100*pout(i)/pin(i); l
pf(i) : abs(real(conj(ias(i))*vas(i)))/(abs(ias(i))*abs(vas(i))): I
rpm(i} = wr*2/P*60/(2*pi}; i il end figure(1) plottdo,te) tit1e('torque Versus duty cycle')
xlabel('duty') ylabel('Nm') grid : 91...."4 .. ,..i. Kama; ..a..;.«;;.;_ .; Nm 14 torque Versus duty cycle I 0.5 1 1.5 2 2.5 3 3.5 4 4.5
duty ...
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 Fall '09
 ANDRISANI

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