For a full-wave bridge rectifier, the supply voltage v(t) = 100 sin ωt. Find the PIV of the diode.
100 V
This solution explains how to determine the Peak Inverse Voltage (PIV) for a diode within a full-wave bridge rectifier circuit, based on the provided AC supply voltage.
The Peak Inverse Voltage (PIV) is the highest reverse voltage that a semiconductor diode can safely withstand when it is not conducting electricity. In rectifier circuits, choosing diodes with an appropriate PIV rating is essential to prevent diode failure (breakdown).
A full-wave bridge rectifier circuit uses four diodes arranged in a bridge configuration. This setup allows it to utilize both the positive and negative halves of the input AC waveform to produce a pulsating DC output. During the negative half-cycle of the input voltage, two specific diodes in the bridge become reverse-biased, and it's across these diodes that the PIV appears.
For a standard full-wave bridge rectifier, the maximum reverse voltage experienced by any diode when it is turned off (reverse-biased) is equal to the peak value of the input AC supply voltage. The PIV requirement for each diode is therefore:
PIV = V_p
Where V_p is the peak voltage of the AC input source.
The AC supply voltage provided is:
v(t) = 100 \sin(\omega t)
From this sinusoidal voltage equation:
100 Volts.Applying the rule for bridge rectifiers, the PIV for each diode is:
PIV = V_p = 100 \text{ V}
The calculated PIV is 100 V. We compare this result with the given options:
| Option Number | Voltage Value | Correctness |
| 1 | 100 V | Correct |
| 2 | 200 V | Incorrect |
| 3 | 50 V | Incorrect |
| 4 | 141 V | Incorrect |
The PIV calculated matches Option 1.
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