# 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.

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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.

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