[Amps] PIV requirement for identical, individual diodes

[Mike Tubby]

On 23/04/2012 16:10, Ron Youvan wrote:
>     Gary K4FMX Schafer wrote:
>
>>> Just draw a diode bridge circuit and you can easily figure out how much
>>> voltage is across each diode. Consider that at any given time one side
>>> of the transformer is at ground and the other side is connected to the
>>> filter capacitor. Then look at what diodes are reversed biased.
>> But the filter capacitor is ALWAYS connected across the transformer with a
>> full wave bridge. Any spike from the power line will always be shunted by
>> the large filter capacitor. No way for a large spike to reverse bias any of
>> the diodes.
>     How about when the capacitor is charged to say, +1,600 Volts and the transformer's outingput is
> -1,600 Volts and then an inductive kick causes the transformer outingput to jump to 2,000 Volts or more?


Consider a 1000V AC transformer secondary and a simple half-wave 
rectifier (single diode string) and a capacitor.

One one half-cycle the capacitor charges to around 1414V (when the diode 
forward conducts) then on the other half-cycle the transformer output 
appears 'negative' reversing the voltage - at the highest point there is 
2828V across the diode(s).  Hence the PIV needs to be >3000V.

Its for this reason that when I built HT supplies for 3-500, 8877, etc. 
and use something like a 2400VAC @ 1.4A CCS transformer I put 10 x 
UF5408 (3A 1000V PIV) diodes in each leg of the bridge.  The UF5408 is 
the ultra-fast recovery version of the 1N5408 - I use them because:

     * 3A (continuous) 1000V PIV
     * low forward drop
     * very low leakage current
     * don't need caps and resistors across them
     * they run cool

I always build a full bridge with 40 or 48 diodes from the same batch 
and then tape a few 'spares' inside the PSU.  I have never needed to use 
one of the spares in over 15 years!


Mike G8TIC