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
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
|