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Transcript
ECE 321 - Homework #5 H-Bridge, DC to AC, AC to DC Conversion. Due October 6th 1) H-Bridge: Determine the currents Ib1, Ib2, Ic1, Ic2 for 1a) A = 0V, B = 0V +10V 9.3mA T1 is ON 1k T2 is OFF 9.3V 0V T1 48mA ON IB1 = 9.3mA, IC1 = 48mA IB2 = 0mA 100 IC2 = 0mA Y 9.8V +5V 1k 0V T2 OFF 0mA 1b) A = 10V, B = 10V +10V T1 is OFF 1k T2 is ON IB1 = 0mA OFF T1 10V 100 IC1 = 0mA Y IB2 = 9.3mA IC2 = 48mA 0.2V +5V 1k 10V 48mA T2 ON 9.3mA +10V 1c) A = 10V, B = 0V 1k T1 is OFF T1 10V OFF 100 T2 is OFF IB1 = 0mA IC1 = 0mA IB2 = 0mA IC2 = 0mA Y 5V +5V 1k 0V T2 OFF 2) DC to AC. With two circuits from above, you can apply +10V / -10V to the 100 Ohm resistor. What is the "optimal" value of A: the time all transtors are off? Include your definition of "optimal" Include calculations for finding this "optimal" time Define "optimal" to be highest percentage of energy in the 1st harmonic Compuations: Guess the turn on time 'a' Compute the 1st harmonic Compute the efficiency Example: a = 1/12. The efficiency should be 91% t = [0:0.0001:1]'; -->a = 1/12 0.0833333 -->f = 24*(t>a) .* (t<0.5-a) - 24*(t>0.5+a).*(t<1-a); -->a1 = sum(f .* sin(2*%pi*t)) / sum( (sin(2*%pi*t)).^2 ) 26.462187 -->n = mean( (a1*sin(2*%pi*t)) .^ 2) / mean( f .^ 2 ) 0.9118716 Check. Now Now sweep from (0, 0.25) t = [0:0.0001:1]'; for i=1:249 a = i / 1000; f = 24*(t>a) .* (t<0.5-a) - 24*(t>0.5+a).*(t<1-a); a1 = sum(f .* sin(2*%pi*t)) / sum( (sin(2*%pi*t)).^2 ); n(i) = mean( (a1*sin(2*%pi*t)) .^ 2) / mean( f .^ 2 ); end The best you can do is 92.2% efficient -->max(n) 0.9225980 Efficiency vs. Turn-On Time as a percentage of one cycle. The green line shows a=1/12 "Optimal" waveform to approximate a sine wave. Turn on time is a = 0.065 with eff = 92.2% 3-5) AC to DC (SCR) The following AC to DC converter uses two diodes and two SCR's 23.3V V1 = max(A,B) V1 SCR 0 + A: 24sin(wt) RL Q 180 Vout 100 - B: -24sin(wt) 0 180 V2 = min(A,B) V2 -23.3V 3) Assume the firing angle (Q) is zero degrees (i.e. the SCR's behave like normal diodes). Plot the voltage Vout(t) -->t = [0:0.001:1]'; -->y = 46.4 * sin(%pi*t); -->VL = y; -->Vrms = sqrt( mean( VL .^ 2 ) ) 32.793362 -->plot(t,VL,t,Vrms) -->xlabel('Time'); -->ylabel('Volts'); Voltage at VL for one cycle (blue) and it's RMS value (green) 4) Assume the firing angle (Q) is 90 degrees (i.e. the SCR turns on when the sine-wave is at it's peak). Plot the voltage Vout (note: if the current is zero, Vout = zero) -->VL = y .* (t > 0.5); -->Vrms = sqrt( mean( VL .^ 2 ) ) Vrms = 23.165209 -->plot(t,VL,t,Vrms) Voltage at VL for one cycle (blue) and it's RMS value (red) 5) Plot Vout vs the firing angle from 0 degrees to 180 degrees -->for i=1:1000 --> Q = i/1000; --> VL = y .* (t > Q); --> Vrms(i) = sqrt( mean( VL .^ 2 ) ); --> end -->size(Vrms) ans = 1000. 1. -->size(t) ans = 1001. 1. -->Q = [1:1000]' / 1000; -->plot(Q*180,Vrms) -->plot(Q*180,Vrms) -->xlabel('Firing Angle (degrees)'); -->ylabel('VL (rms)'); RMS Output Voltage vs. Firing Angle Sidelight - typical problem for ECE 437: Power Electronics Suppose the load has a large inductance (i.e. is a DC motor). R = 10 Ohms L = 100mH Also assume that only the voltage across the resistor matters (i.e. the AC component just produces heat. The DC terms is what spins the motor and produces work.) L = 0.1H V1 = max(A,B) SCR 46.6V + A: 24sin(wt) Q RL Q Vout 10 B: -24sin(wt) - V2 = min(A,B) Determine the voltage across the resistor for a firing angle of 30 degrees If the inductance is large, the current will be constant plus some ripple. This means a pair of diodes is always on This means that once a diode is turned on, it stays on until its counterpart turns on -->t = [0:0.001:1]'; -->Q = 30/180; -->Vin = 46.6*sin(%pi*(t+Q)); -->c0 = mean(Vin) 25.666239 -->plot(t,Vin,t,c0) time The input voltage vs time (blue) and its average value (green) for a firing angle of 30 degrees with a large inductive load. Since the inductance is large, one diode pair must be on at all times. This results in the input voltage going negative towards the end of a cycle. The input signal is the blue waveform. Since this is a periodic signal, it has A DC term A 1st harmonic, and Other higher harmonics Taking the first two terms: The DC term is -->c0 = mean(Vin) 25.666239 The 1st harmonic is: -->a1 = sum(Vin .* sin(2*%pi*t) ) / sum( sin(2*%pi*t) .^ 2) 19.777605 -->b1 = sum(Vin .* cos(2*%pi*t) ) / sum( cos(2*%pi*t) .^ 2) - 17.093806 meaning V in ≈ 25.66 + 19.77 sin (2πt) − 17.09 cos (2πt) + ... Now, compute the current to the load at DC, and at 120 Hz (1st harmonic assuming a 60Hz input through a full wave rectifier) Using phasors at harmonic n (n = 0, 1, 2, 3, ...) V in = a + jb R ⎞ V L = ⎛⎝ R+jωL ⎠ V in At DC (n=0) -->Vin = c0 25.666239 -->w = 0; -->VL0 = (R / (r + j*w*L)) * Vin 25.666239 At 120Hz (n=1) -->Vin = b1 - j*a1 - 17.093806 - 19.777605i -->VL1 = (R / (r + j*w*L)) * Vin - 2.8732331 + 1.886062i meaning for a 60Hz input (120Hz 1st harmonic): VL ≈ 25.66 − 2.87 cos (240πt) − 1.88 sin (240πt) -->VL = VL0 + real(VL1)*cos(2*%pi*t) - imag(VL1)*sin(2*%pi*t); -->plot(t,Vin,t,VL) -->xlabel('Time'); -->ylabel('Voltage'); Input voltage (blue) and voltage across the 10 Ohm load (green) with a large inductive load. Note that the voltage has a DC offset - meaning the current never goes to zero (a diode is always on) The DC term determines the current to the resistive load. The 120Hz term (1st harmonic) determines the AC component (i.e. the ripple - approximately. There are other harmonics we're ignoring, but they'll be small) Repeating for a firing angle from 0 to 90 degrees: DC voltage at the load (blue - Volts) along with its AC component (green - Vp) for firing angles from 0 to 90 degrees - assuming a large inductive load Note that the assumption that the current is always on (meaning a diode is always on) falls apart near a firing angle of 90 degrees. If the ripple is more than the DC level, the voltage (meaning current) tries to go negative. The diodes won't allow this - meaning our assumption is wrong - meaning the computations will be off. Code: t R L w j = = = = = [0:0.001:1]'; 10; 0.1; 2*%pi*120; sqrt(-1); Vdc = []; Vac = []; for i=1:500 Q = i/1000; Vin = 46.6*sin(%pi*(t + Q)); Vin0 = mean(Vin); VL0 = Vin0; a1 = sum(Vin .* sin(2*%pi*t) ) / sum( sin(2*%pi*t) .^ 2); b1 = sum(Vin .* cos(2*%pi*t) ) / sum( cos(2*%pi*t) .^ 2); Vin1 = b1 - j*a1; VL1 = Vin1 * R / (R + j*w*L); Vdc(i) = VL0; Vac(i) = abs(VL1); end plot(t,Vdc,t,Vac) dc voltage to the resistive load peak voltage for the ripple