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In Reply to: huh? posted by zarniwoop on May 05, 2002 at 17:36:46:
No, NOT PEAK! Peak to Peak! Think about it! 1000 V or greater diodes is necessary.
Follow Ups:
I thought about it... and I still stand by my logic. With a FW bridge the diodes only ever see 1.4 x RMS VAC, the peak voltage. ( <450V in my case)You are right for the case of a half-wave or full-wave center tapped rectifier though. In those cases the diodes would need a PIV rating of 2.8 x VAC.
PIV seen by the diodes is the capacitor voltage plus a few diode drops...
G.
Crank it up...
I agree with you. I overstated the need for a higher diode voltage, but normally I would still keep the voltage much higher than absolutely necessary to keep line transients from triggering the diodes into reverse conduction.
I prefer to be safe rather than logical, but then they must have put those 1KV diodes in the original power supply for a reason. My understanding is that it will be twice the DC output which is across the cap. After all, the cap charges up and then the AC goes equally in the opposite direction. I don't have your circuit in front of me, but it is not a good idea to replace components with lower voltage types in any electronic circuit.
What if you double up on the diodes by putting two in series? I understand that this doubles the voltage handling capacity, is that right?Retsel
On further consideration, I understand Zarniwoop's position. However, I would STILL use a diode that has at least the voltage rating of the devices that I am replacing. Hi AC input voltage, line transients, etc can still cause failure sooner or later. Series devices would work, more or less, but IF they are not matched for leakage, one device can leak more than the other and you again get a dangerous situation. The best way is to use Hexfreds or soft recovery rectifiers from Motorola or Harris that work as well as Schottky diodes at this voltage. Of course, Schottky diodes work better with voltages below 15V or so.
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