In the circuit shown, the input V_i is a sinusoidal AC voltage having an rms value of 230 V ± 20%. The worst case peak-inverse voltage seen across any diode is _______ V. (Round off to 2 decimal places)

The correct answer is 390.27 V (or 390.32 V depending on the approximation of $\sqrt{2}$).

1. Analyzing the Circuit Diagram:
The given circuit diagram shows a full-wave bridge rectifier connected to a filter capacitor $C$ and a load resistor $R$:

In a bridge rectifier with a capacitor filter, the peak inverse voltage (PIV) across any diode under steady-state conditions is equal to the peak value of the input sinusoidal AC voltage, $V_m$.

2. Determining the Input Voltage Parameters:
The input voltage $V_i$ is a sinusoidal AC voltage with an RMS value given by:
$V_{rms}=230\text{V}\pm 20\%$

To find the worst-case (maximum) peak-inverse voltage, we must use the maximum possible RMS input voltage, which occurs at the upper limit of the tolerance range (+20%):
$V_{rms,max}=230\times (1+0.20)=276\text{V}$

3. Calculating the Worst-Case Peak Voltage:
The peak voltage $V_m$ for a sinusoidal wave is related to its RMS value by:
$V_m=\sqrt{2}\times V_{rms,max}$

Depending on the precision used for the value of $\sqrt{2}$:

• If we use $\sqrt{2}\approx 1.414$:
$V_m\approx 1.414\times 276=390.264\text{V}$
Rounding to two decimal places gives 390.27 V (or 390.266 V).

• If we use a more precise value $\sqrt{2}\approx 1.41421356$:
$V_m\approx 1.41421356\times 276\approx 390.32\text{V}$
Rounding to two decimal places gives 390.32 V.