Question:
In the circuit shown, the input \(V_i\) is a sinusoidal AC voltage having an RMS value of \(230\,V \pm 20%\). The worst-case peak-inverse voltage seen across any diode is \(\underline{\hspace{2cm}}\) V. (Round off to 2 decimal places.)
In the circuit shown, the input \(V_i\) is a sinusoidal AC voltage having an RMS value of \(230\,V \pm 20%\). The worst-case peak-inverse voltage seen across any diode is \(\underline{\hspace{2cm}}\) V. (Round off to 2 decimal places.)
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In a full-wave bridge rectifier, PIV across each diode equals the maximum peak input voltage, not double the peak as in a center-tapped rectifier.
Updated On: Dec 29, 2025
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Correct Answer: 389
Solution and Explanation
The RMS input is given as \(230V \pm 20%\). Therefore, the maximum RMS voltage is:
\[ V_{rms,max} = 1.2 \times 230 = 276\,V \] The peak voltage corresponding to this RMS value is:
\[ V_{peak} = \sqrt{2} \times 276 = 390.05\,V \] For a full-wave bridge rectifier, the peak-inverse voltage (PIV) across each diode is equal to the maximum peak input voltage:
\[ \text{PIV} = V_{peak} = 390.05\,V \] Thus, the worst-case PIV is:
\[ \boxed{390.05\text{ V}} \]
\[ V_{rms,max} = 1.2 \times 230 = 276\,V \] The peak voltage corresponding to this RMS value is:
\[ V_{peak} = \sqrt{2} \times 276 = 390.05\,V \] For a full-wave bridge rectifier, the peak-inverse voltage (PIV) across each diode is equal to the maximum peak input voltage:
\[ \text{PIV} = V_{peak} = 390.05\,V \] Thus, the worst-case PIV is:
\[ \boxed{390.05\text{ V}} \]
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