bridge-rectifierrectifier
I know that in a full wave rectifier the PIV of the diodes is 2*VP, whilst in a full wave bridge rectifier we have just Vp.
It is also said that the bridge configuration is more suitable in applications that require higher voltages due to the PIV parameter. I can't understand the reason.
If the PIV specifies the maximum reverse voltage that a diode can support without going at breakdown, wouldn't it be better to have a higher PIV with a higher input voltage? What am I missing?
2*Vsecondary/Pi - Vk or as you put it 0.636*Vac - Vk.
Consider the circuit diagram of a center-tapped full wave rectifier where D1 is forward biased and D2 is reverse biased.
The maximum reverse voltage appearing across will be 2*Vp. Where Vp is the amplitude of input signal.
So the diode that is used in a center-tapped full wave rectifier should have a PIV of atleast twice the peak voltage of input sine wave. Otherwise diode breakdown will happen and current will flow through the reverse biased diode. And the circuit is not a rectifier anymore.
Similarly, PIV for a full wave bridge rectifier will be Vp.
Best Answer
I think you may be getting confused by the terminology.
The rectifier PIV requirement is lower in a bridge rectifier, therefor it's better to use one for high voltages, if you had a peak voltage of 400 V you would need 800 V rectifiers for a full wave but only 400 V rectifiers for a bridge.
On the other hand a full wave with center tapped transformer is better at low voltages because it eliminates one diode drop. The difference between a 0.7 V drop and the 1.4 V drop of a bridge doesn't matter much at 100 V, but if you're making a high current 2 V supply that extra 0.7 V would make a big difference.