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I want to run a 13V(300mA) valve filament from a 6,3VAC transformer, using a voltage doubler(Greinacher circuit).

I am using PSUD2 to simulate the circuit, Which gives me 2 options for the load, either constant current or resistive. Using either I was able to create a source with ~13V RMS output, with 2,2mF capacitances(Which I could get relatively cheaply as bipolar electrolytic).

I would like to know if this could work, because while nominally the voltage doubler should have ~18V output, because of the strong load, the RMS voltage stays 13V. And this already makes some assumptions: What is the series resistance of bipolar electrolytic capacitors of high value(2.2mF)like these? (I assumed 500mOhm).

I expect reality to be different, so I guess I'll have to measure the voltage and correct with some series resistance or voltage divider, but I'd like to know if its possible at all, and how would I go about creating it?

Also is the heater a resistive or a constant-current source? If neither, which one approximates it better?

What other option do I have if any for using this 6,3VAC tap to power the 13V PCL86 tube(Which sadly doesn't have a center filament tap)?

EDIT: I had went on with the voltage doubler, and I got exactly 13V for the heater without any problems. The simulators results were very precise.

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I agree with @Dave about the modelling. You only need to worry about the resistance at operating temperature. The lower resistance at room temperature might increase the warm-up time, but unless you're worried about that, it should have no significant effect.

An option would be to create an ~8V DC supply and use a boost converter to create a regulated 300mA DC supply. A simple circuit using the LM2577 (or one of the ubiquitous modules) should work fine. You can probably make the circuit smaller and cheaper, and easier on the transformer.

If you do decide to use the capacitor voltage multiplier, make sure that the ripple current rating (RMS ripple current) of the capacitors is observed. Cheap electrolytics may not last long in this service, particularly if they see high temperatures from your tube ('thermionic valve', sorry), on top of the self-heating from the ripple current.

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  • \$\begingroup\$ The boost converter method seems both more complicated and costly(Although more stable), so I will try the voltage doubler and report the results here. \$\endgroup\$
    – akaltar
    Jun 4 '15 at 13:59
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The heater is just a resistor. While its cold resistance may be significantly different from its resistance at operating temperature, this shouldn't be important in terms of your modeling. Just use the value \$\frac{12.6 V}{300 mA} = 42 \Omega\$.

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