AdvancedA-004-001-003
In a full-wave centre-tap power supply, regardless of load conditions, the peak inverse voltage (PIV) will be _____ times the RMS voltage:
D
Answer
Measurements and troubleshooting
Type
A
0.707
B
1.4
C
0.636
D
2.8
Answer Notes
In a full-wave centre-tap rectifier, each diode must withstand the full peak voltage of the entire secondary winding across it when it is reverse-biased, not just one half. The peak voltage of an AC sine wave is approximately 1.414 times its RMS voltage.
Because the non-conducting diode is effectively connected across the entire secondary winding (both halves), the Peak Inverse Voltage (PIV) it experiences is twice the peak voltage of one half of the winding. Therefore, the PIV is 2 x 1.414 = 2.828 times the RMS voltage of one half of the secondary.
Rounding this gives a factor of 2.8. Distractors like 1.4 represent the standard peak-to-RMS ratio, but fail to account for the doubled voltage stress placed across the reverse-biased diode in a centre-tapped configuration.
Previous · A-004-001-002
In a half-wave power supply with a capacitor input filter and a load drawing little or no current, the peak inverse voltage (PIV) across the diode can reach _____ times the RMS voltage.
Next · A-004-001-004
A full-wave bridge rectifier circuit makes use of both halves of the AC cycle, but unlike the full-wave centre-tap rectifier circuit it does not require: