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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