1
QUESTION 2(E).
-57.45e-30V
-57.45e-30V
V+
D1
D4
Dbreak
R1
I
0W
100
Dbreak
W
438.5e-21V
V1
VOFF = 0
VAMPL = 170
FREQ = 60
AC = 0
V+
0V
438.5e-21V
V-
D3
D2
Dbreak
L1
0W
10mH
Dbreak
0V
0V
V-
0
Figure 2.0: Circuit Diagram for a Full-Wave Rectifier with RL Load.
200
150
100
50
-0
-50
-100
-150
-200
0s
V(V1:+,V1:-)
5ms
V(R1:2,L1:1)
10ms
-I(R1)
AVG(W(R1))
15ms
AVG(-I(R1))*10
20ms
25ms
30ms
35ms
40ms
45ms
Time
Figure 2.1: Average Current Io waveform of a Full-wave Rectifier.
50ms
2
From the PSpice simulation shown below, the average current is given to be AVG(-I(R1)) *10 =
10.635A. If divided by 10, the Io will be 1.063A. This PSpice result is really close to my analytical
result which is given to 1.08A, since PSpice result is usually peak, it is expected that that there
would be a slight difference in my result.
Figure 2.2: Average Current and the Power absorbed by the Resistor of a Full-wave Rectifier.
The figure above shows the power across the resistor which is almost the same with my analytical
result. This result is pretty close.
Pr from the PSpice = 140watts
Analytical result is given to be = 141watts
200
150
100
50
-0
-50
-100
-150
-200
0s
V(D1:2,0)
5ms
AVG(V(R1:2,0))
-I(R1)
10ms
W(L1)
V(D1:1,D4:1)
15ms
V(D1:1,D4:1)
20ms
25ms
Time
Figure 2.3: Average Voltage Vo of a Full-wave Rectifier.
30ms
35ms
40ms
45ms
3
Figure 2.4: Average Voltage Vo of a Full-wave Rectifier.
The PSpice result is really close to the analytical result for Vo. Vo is given as 108Volts while
PSpice simulated result is given as 105volts
Question 3 (E).
L1
R1
25.94e-21W
72.00V
72.00V
50mH
5
D1
Dbreak
W
I
D4
Dbreak
36.00V
V1
VOFF = 0
VAMPL = 338
FREQ = 60
AC = 0
V+
72.00V
V+
36.00V
VI
D3
72
D2
V2
-5.186nW
I
W
Dbreak
Dbreak
0V
V-
0
Figure 3.0: Circuit Diagram for a Full-Wave Rectifier with RL Load.
4
2.8K
2.4K
2.0K
1.6K
1.2K
0.8K
0.4K
0
-0.4K
0s
V(V1:+,V1:-)
5ms
V(R1:2,0)
I(R1)
10ms
AVG(W(R1))
15ms
AVG(W(V2))
-I(V1)
20ms
I(V2)
25ms
30ms
35ms
40ms
45ms
50ms
Time
Figure 3.2:
Power across the DC circuit and the Power absorbed by the Resistor Waveform of a
Full-wave Rectifier with RL Load.
Figure 3.3:
Power across the DC circuit and the Power absorbed by the Resistor Waveform of a
Full-wave Rectifier with RL Load.
In this simulation, power across the DC voltage source is found to be 1596watts, while the power across
the resistor PR is found to be 2726watts
5
6
Question 5 (c).
L1
0W
-90.68e-30V
10mH
V+
-90.68e-30V
D1
D4
Dbreak
R1
15
Dbreak
-685.2e-21V
V1
VOFF = 0
VAMPL = 170
FREQ = 60
AC = 0
V+
C1
10000u
0V
-685.2e-21V
V-
D3
D2
Dbreak
Dbreak
0V
0V
V-
0
Figure 5.0: Circuit Diagram for a Full-Wave Rectifier with LC Filter.
200V
150V
100V
50V
-0V
-50V
-100V
-150V
-200V
200ms
V(D1:1,V1:-)
205ms
AVG(V(R1:2,0))
210ms
215ms
220ms
225ms
230ms
235ms
240ms
245ms
250ms
Time
Figure 5.1: Waveform of an Average Voltage Vo for a Full-Wave Rectifier with an LC Filter on
5ohms Resistor.
7
Figure 5.1: Average Voltage Vo for a Full-Wave Rectifier with an LC Filter on 5ohms Resistor.
200V
150V
100V
50V
-0V
-50V
-100V
-150V
-200V
200ms
V(D1:1,V1:-)
205ms
AVG(V(R1:2,0))
210ms
215ms
220ms
225ms
230ms
235ms
240ms
245ms
250ms
Time
Figure 5.1: Waveform of an Average Voltage Vo for a Full-Wave Rectifier with an LC Filter on
15ohms Resistor.
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Figure 5.1: Average Voltage Vo for a Full-Wave Rectifier with an LC Filter on 15ohms Resistor.