All tube voltage doubler

[Steven-H]

Gents,
Question - how would you go about making a voltage doubler using tube diodes ? I want to have tube rectification on my next amp project, but it would require about 900v B+ for the power tube, and about 350v B+ for the driver tube (6SN7 into 813). I guess what I am looking for here is a basic primer on voltage multiplier networks so that I can calculate what sort of draw I can pull, etc.

Thanks !
Steve

[jackinnj]

you might be able to use a damper diode -- like the 6BY5 -- although I don't know how much current they can handle.

A voltage multiplier is like a filter -- the value of the capacitors is proportional to the load you need to handle...and the caps have to be able to handle the current...it is far from a trivial excercise...one in which you could easily meet your maker. I keep this around to instill a sense of humility:



might be better off just winding a better transformer -- or getting a mil surplus unit from Fair Radio Sales in Lima OH.

[danFrank]

A voltage doubler would be no problem. I have a couple of Mac MC60's that use Heathkit w-6 power transformers in the voltage doubler configuration. I am using 6CM3 damper diodes, which are probably the beefiest of all the old American damper diodes. They are good for at least 400ma each; they do take a lot of filament current though, 2.5 amps each.
Basically you wire them up " in totem pole" fashion, the cathode of one tube connected to the plate of the other; this is where one of your AC HV leads goes to, the other one goes to the capacitors. Look at photo. B+ is at the bottom of picture.
As for the doubling caps, I use 150uf units with no problems of arcing, stripping, etc. NOS damper diodes are built like tanks. I will probably get flak for using such high capacitance, oh well...
Make sure you don't exceed ratings of the tubes.
If you were planning to use a tripler or quadrupler circuit, all I can say is that I have had no luck with such circuits when drawing any kind of current out of them. I have never had a problem with getting current out of doubler circuits though.
Hope this helps, Daniel

[Steven-H]

Ah, I see. Thanks for the help !
The reason why I wanted to do it this way is because I won't need to get a custom wound transformer with 2 secondaries (900v/450v). I assume that 816's would work as well (obviously on a time delay heater circuit..) considering the provisions needed to take the B+ from the transformer CT ? On the 900v side I will have a draw of 180mA, and on the 450v side I'll have a draw of 20mA. So I was thinking of using 816's in full wave, and a pair of 6x4's or something in full wave for the 450v side (to help drop it down to the 350v I need).

Thanks !
Steve

[Eli Duttman]

Steve,

Is the power trafo you already have equipped with a CT rectifier winding? If that's the case, a Pete Millett devised topology will allow you to extract 2 B+ rails (1 low and 1 high) from the trafo. Both rails are truly full wave rectified.

Hybrid bridge rectify the entire rectifier winding with series connected pairs of UF4007 SS diodes forming the connection to ground and a pair of 6AU4 damper diodes forming the connection from which the high "raw" B+ is taken. Connect both plates of an "ordinary" rectifier, like a 6X4 or 5V4 to the CT of the rectifier winding. The "raw" low B+ is taken from the "ordinary" rectifier's cathode.

A voltage doubler using damper diodes is definitely possible, but keep in mind that each segment in a multiplier is half wave rectified. IMO, anything beyond a doubler should only be done with SS diodes, as the losses with vacuum diodes mount quickly. Look at the last schematic here for a HEAVY DUTY multi-stage topology.

[Steven-H]

Eli,
Yes, the transformer has CT's on all the windings. I will read up on the hybrid rectifiers....

Thanks !
Steve

[Tom Bavis]

With tube rectifiers in a doubler, you may need separate heater windings for each one (even indirectly heated ones), as there will be considerable heater-cathode voltage.

Using a doubler or full-wave bridge on a transformer not designed for it creates extra voltage stress - it would be best to find a hipot tester and make sure the transformer can take the extra voltage... a test at 2-2.5 times the expected peak voltage would make me feel a LOT more confident.

[Steven-H]

Tom,
Thanks - I was concerned about this myself. So, at this point what does everyone feel the best plan of attack would be ? Is it realistically possible to run a voltage divider after the current tap for the 900vdc B+ ? How would the load for the divider network be calculated then, via the plate resistor value ?


Thanks !
Steve

[Original Burnedfingers]

I have a transformer that will give me 313VDC with a SS rectifier. With a SS voltage doubler I get about 615VDC. What I would like to do is make a tube doubler circuit.

I have a 5volt winding that will easily accomodate several 3A rectifiers and planned on using 5R4's since a have a bunch of them.

Any ideas?

[tomchr]

Originally Posted by jackinnj

A voltage multiplier is like a filter -- the value of the capacitors is proportional to the load you need to handle...and the caps have to be able to handle the current...it is far from a trivial excercise...one in which you could easily meet your maker.

The caps need to be able to handle the full load current. Yep. At 180 mA load current, I'd be worried. There's a reason you don't see voltage multipliers in high power circuits. Their most common application are for static voltages where very little current is drawn.

Originally Posted by Eli Duttman

Is the power trafo you already have equipped with a CT rectifier winding? If that's the case, a Pete Millett devised topology will allow you to extract 2 B+ rails (1 low and 1 high) from the trafo. Both rails are truly full wave rectified.

That's not rocket surgery... Use one end of the primary as the 0 V or ground reference. Then the center tap becomes, say, 350 V above ground, and the other end of the primary 700 V above ground. I'm using a 700 V CT (350-0-350) transformer as an example here.
Put a full wave rectifier between the CT and GND. That gives you 1.25*350 V = 438 V rectified. Put a full wave rectifier between the top tap and ground and you'll get 700*1.25 = 875 V rectified. There's your two B+'es.
I use a factor of 1.25 to convert from RMS to peak voltage as for a rectifier, you generally end up somewhere around 1.2~1.3x Vrms on the output rather than the theoretical sqrt(2). The conduction angle of the diodes is non-zero after all...

How you implement the rectifier is up to you. Personally, I'd go with silicon, but a hybrid tube-sand combo is possible as well. Just make sure the rectifiers you choose can handle the current AND the reverse voltage.

I seem to recall Morgan Jones writing about hybrid rectifiers in his book as well... You might be able to pick out a few things there.

~Tom