Why 2 diodes in series for full bridge rectifier

[es1]

A friend of mine builds tube audio amplifiers for a hobby. Being an electrical engineer, who works exclusively with micro-electronics, I thought I might mess around in spice a bit to better understand some of the standard circuit topologies they use.

My friend is not an engineer so he doesn't really know the history or justification of many of the components so they typical answer to "why is this here?" is "because everybody does it that way."

Here is an example circuit very similar to what he is currently building:
http://www.electra-print.com/300bdrd.php

The second schematic on this page shows how to make a full bridge rectifier

One thing that looked odd to me was the two diodes in series on the rectifier.
I understand what the filtering caps and resistors in parallel to the diodes are for, but why put two diodes in series? Maybe it's a safety thing? I've seen redundancies like this in some battery circuits when there is concern of shorts. I also notice the 5V supply does not have them.

 

[es1]

One other quick question: Are the 390K resistors in parallel with the 220uF capacitors there to make sure the caps discharge in a safe way and in a reasonable amount of time when the circuit is turned off? I suspect they are there to dominate the internal leakage of the cap as the leakage is an uncontrolled parameter.

 

[Bob S]

Are you referring to the full-wave center-tap rectifier circuit? The high voltage PRV (peak reverse voltage) exceeds the value of one diode, so two are used.

Bob S

 

[es1]

Is that after derating? I looked at this datasheet for the 1N4007 and it lists the PRV at 1KV. I though this would be enough for a ~700V supply. I suppose it is not unreasonable for it to get ~60C and it if leaked 1uA we would have to derate by ~40% which would require two.

http://www.fairchildsemi.com/ds/1N%2F1N4001.pdf

But definitely it makes sense. Thanks.

 

[Bob S]

The required PRV in that circuit is 2 x √2 x 570 volts.

Bob S

 

[vk6kro]

This shows the moment when the diodes are under most stress. When the transformer voltage reaches its negative peak ( of √2 times 570 V) and the capacitor input filter is charged up to the peak value ( of √2 times 570 V minus diode drops) then the two voltages combine across the diodes to give a value of about 1610 volts. Which is exactly what Bob S said.

[PLAIN]http://dl.dropbox.com/u/4222062/full%20wave%20rectifier%20PIV.PNG

Yes, the 390 K resistors are there to discharge the capacitors after the power is turned off.
They also help to distribute the voltage equally between the capacitors which are in series.

 

[es1]

Thanks Guys, Somehow I totally forgot about the negative swing. It's pretty obvious now that I have it in spice. I should have run it before asking.