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If peak voltage for a half wave rectifier circuit is 5V and diode cut in voltage is 0.7, then peak inverse voltage on diode will be?
- a)5V
- b)4.9V
- c)4.3V
- d)6.7V
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If peak voltage for a half wave rectifier circuit is 5V and diode cut ...
PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, If the PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=Vm-Vd=5-0.7=4.3V.
Most Upvoted Answer
If peak voltage for a half wave rectifier circuit is 5V and diode cut ...
Peak inverse voltage (PIV) is the maximum voltage that a diode can withstand when it is reverse biased. In a half-wave rectifier circuit, the diode is reverse biased during the negative half-cycle of the input AC signal.
Given:
Peak voltage = 5V
Diode cut-in voltage = 0.7V
To calculate the PIV, we need to find the maximum reverse voltage across the diode. This occurs when the input signal is at its peak value and the diode is reverse biased.
1. Determine the maximum reverse voltage:
The maximum reverse voltage is equal to the peak voltage minus the diode cut-in voltage.
Maximum reverse voltage = Peak voltage - Diode cut-in voltage
Maximum reverse voltage = 5V - 0.7V
Maximum reverse voltage = 4.3V
2. Therefore, the PIV of the diode is 4.3V.
In this case, the correct answer is option C, 4.3V.
Given:
Peak voltage = 5V
Diode cut-in voltage = 0.7V
To calculate the PIV, we need to find the maximum reverse voltage across the diode. This occurs when the input signal is at its peak value and the diode is reverse biased.
1. Determine the maximum reverse voltage:
The maximum reverse voltage is equal to the peak voltage minus the diode cut-in voltage.
Maximum reverse voltage = Peak voltage - Diode cut-in voltage
Maximum reverse voltage = 5V - 0.7V
Maximum reverse voltage = 4.3V
2. Therefore, the PIV of the diode is 4.3V.
In this case, the correct answer is option C, 4.3V.
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