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Question 2.SP.29: For the full-wave rectifier circuit of Fig. 2-43(a), let vS ......

For the full-wave rectifier circuit of Fig. 2-43(a), let v_S = 120 \sqrt{2} \sin(120πt)  \text{V},  R_S = 0.001  Ω,  R_L = 5  Ω, and the ideal transformer has a turns ratio of 10:1.    Using SPICE methods and assuming ideal diodes, plot the output voltage v_L and diode currents i_{D1} and i_{D2}.    Compare the results with predicted values based on the solution of Problem 2.28.

2.43
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The netlist code for analysis of the circuit is

Prb2_29.CIR – FW rectifier
vs  1 0 SIN( 0V {sqrt(2)*120V} 60Hz )
Rs 1 2 0.0010ohm
* Ideal transformer, 10:1 ratio
L1  2 0 1H IC=-0.39A
L2 3 0 10mH
L3 0 4 10mH
kall L1 L2 L3 1
D1 3 5 DMOD
D2 4 5 DMOD
RL 5 0 5ohm
.MODEL DMOD D(n=0.0001) ; Ideal diode
.TRAN 1us 16.667ms 0s 1e-6s UIC
.PROBE
.END

Execution of  <Prb2_29.CIR>  and use of the Probe feature of PSpice result in the plots of Fig. 2-44 where the peak values of v_L and i_{D1} have been marked.

Based on the results of Problem 2.28, the predicted peak values of v_L and i_{D1} are given by
i_{D1  \max} = \frac{v_{S  \max}/n}{R_L} = \frac{120 \sqrt{2}/10}{5} = 3.39  \text{A}
v_{L  \max} = \frac{v_{S  \max}}{n} \frac{120 \sqrt{2}}{10} = 16.97  \text{V}

The predicted values and the SPICE results are in agreement.

2.44

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