Question

An asymmetrical periodic pulse train \( v_{\text{in}} \) of 10 V amplitude with on-time \( T_{\text{on}} = 1 \, \text{ms} \) and off-time \( T_{\text{off}} = 1 \, \mu \text{s} \) is applied to the circuit shown in the figure. The diode \( D_1 \) is ideal. \[ \text{The difference between the maximum voltage and minimum voltage of the output waveform \( v_o \) (in integer) is V.} \] 

Hint:

In a circuit with a diode and capacitor, the voltage across the capacitor depends on the charging and discharging times, which are governed by the RC time constant.

Answer 1

Correct Answer: 10

The diode \( D_1 \) conducts during the positive half-cycle of the input pulse, charging the capacitor to the maximum voltage of \( 10 \, \text{V} \). During the off-time \( T_{\text{off}} \), the diode is reverse-biased and no current flows, so the capacitor discharges through the resistor. The time constant \( \tau \) of the RC circuit is given by: \[ \tau = R \cdot C = 500 \, \text{k}\Omega \cdot 20 \, \text{nF} = 10 \, \text{ms} \] Thus, the capacitor charges to \( 10 \, \text{V} \) and discharges according to the time constant. The difference between the maximum and minimum voltage across the capacitor is \( 10 \, \text{V} \). Thus, the difference between the maximum and minimum voltage of the output waveform is \( 10 \, \text{V} \).