[Amps] PIV requirement for identical, individual diodes

[Gary Schafer]

> -----Original Message-----
> From: amps-bounces at contesting.com [mailto:amps-bounces at contesting.com]
> On Behalf Of Mike Tubby
> Sent: Monday, April 23, 2012 11:50 AM
> To: ka4inm at gmail.com
> Cc: amps; Ron Youvan
> Subject: Re: [Amps] PIV requirement for identical, individual diodes
> 
> 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.
> 
> 
> Mike G8TIC

That is true for a half wave rectifier but it is not true for a full wave
bridge. 
In a full wave bridge each leg of the bridge will never see more than the
peak voltage of the transformer.

73
Gary  K4FMX