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Steven-H 2nd July 2007 03:49 PM

Hybrid Rectifier Check Please !
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Can you please double check my schematic here ? For some reason, it just doesn't look right to me in the rectifier bridge stage...

Thanks !

kevinkr 2nd July 2007 04:23 PM

It'll work, but more often the tube rectifiers and silicon are swapped from shown. Your configuration does have the significant advantage that both the 900V and 450V supplies will be delayed by the tube rectifier warm up time.

Depending on the silicon used you might want to consider a couple of diodes in series with snubbing caps and resistors across them - quieter and more robust in the event of a line transient. (Think distant lightning strikes, etc.) Old ARRL handbooks have a lot of information on HV power supply design you might useful.

Steven-H 2nd July 2007 05:44 PM

Thanks for the check. I have no idea why I never thought to check out those old books before - thanks for the mention!

jayme 2nd July 2007 05:58 PM

I don't understand how the 400V B+ works in this circuit.

I thought the secondary center tap is effectively the 0V ground? It seems to me that the bottom B+ has 0V on both legs of the power circuit.

Am I missing something?

Eli Duttman 2nd July 2007 06:53 PM

That circuit is CLEVER. The 900 V. rail is bridge rectified, with the 6AU4s forming the negative half of the bridge and the SS diodes forming the positive half of the bridge. The chokes in the filter keep SS diode switching noise out of the rail. The 400 V. rail is FWCT wired in "reverse", which is perfectly "Kosher". Again, the chokes in the filter ensure SS diode noise stays out of the rail. Snubber cicuitry, as Kevin suggested, is needed only if 1 or more filament windings is present on the trafo.

Keep the fact that PSUs are differential, not inherently single ended, in mind. We make them single ended by arbitrarily grounding a point.

Miles Prower 2nd July 2007 06:57 PM


Originally posted by jayme
Am I missing something?

All four diodes form a full wave bridge, which provides the DC neutral, just as it would if there were no center tap. The center tap and the two 6AU4s form a standard, full wave, rectifier with the polarity reversed, so that the center tap becomes the positive rail with the 6AU4s being in the DC return path. That gives two, positive DC rails with one having approximately twice the voltage of the other. You could also ground the center tap to get a +/- power supply, with a negative and positive rail.

That particular topology has long been used in RF rigs to provide both the high voltage required by the finals, and the lower voltages needed by low level stages and final screen voltage supplies.

ray_moth 2nd July 2007 08:23 PM

One half of the secondary will be more heavily loaded than the other half - but does that matter?

Miles Prower 3rd July 2007 05:25 AM


Not so long as the total secondary VA rating isn't exceeded. Since both rails operate as a full wave, there won't be any DC magnetization, which would be a bad thing, and which would occur if using half wave to develop the low rail.

corne 3rd July 2007 06:27 AM


This circuit can potentially damage the 2x120uF capacitors on the 400V B+.
If the 6AU4s diodes are cold both silicon diodes still form a double phase rectifier with the centre tab as zero.
So after switching on this PSU you'll almost instantaneously have 400V between both B+ leads. The load (imagine a resistor on both 900V to ground and 400V to ground) on both 900V and 400V determines the voltages to ground. If both loads are identical the 900V B+ will be +200V and the 400V B+ will be -200V. This means that the two 120uF capacitors charged in reverse this may cause the capacitors to be damaged.
After warm-up of the 6AU4 diodes the PSU will work as intended.

My suggestion is to replace the silicon diodes by tubes diode and the tube diodes by silicon diodes.
The 400V B+ will switch on instantaneously and the 900V B+ is delayed. If the 400V B+ should be delayed to you should use 4 tube diodes.

Edit :

If there's no load the capacitors on the 900V charge normally but because there's no load the return current must go through the capacitors on the 400V B+ rail. This means the current in these capacitors is reversed. So each time this PSU is switched on the capacitors on the 400V B+ get a reverse current.
This will damage the capacitors.


kevinkr 3rd July 2007 03:17 PM

Something I had missed.. This issue can also be resolved with a single clamp diode with the cathode connected to the transformer center tap and the anode connected to ground. The diode will be reverse biased during normal operation and must withstand 400V - 450V so I'd recommend a 1N4007 or similar. During warm up it will be forward biased and will clamp the +400V output to -0.6V which should be safe with electrolytics - note that this issue is not really a problem with film caps at all. (Solens for example.)

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