Question on loading voltage doublers

[Ears]

Am I correct is assuming that as a rule of thumb the use of a voltage doubler halves the current rating available from the source? For example, the doubling a 10V, 2A winding on a transformer will roughly allow a 20V, 1A maximum loading of the doubler.
Are there any hidden considerations that limit the loading of voltage multipliers? (aside from diode current limitations, diode forward voltage drops and capacitor ESR) Thanks.

[Phil_S]

This is a basic application of the Law of Conservation of Energy. 20v * 1A is 20W. 10v * 2A is 20W. This applies better to AC run through a transformer than it does to DC run through a doubler.

The problem is that the real world is not perfect and there is bound to be some loss of energy in the form of heat. Doublers built with caps and resistors tend to have an impact, so don't expect exactly 2x. It will be more or less 2x.

These are cheap to build. Rig one up on an old plank of wood and test it. Here's my test rig. It's presently partially disassembled. I used some old can caps I had around that tested OK on the meter. I don't remember the exact multiple it produced.

[Ears]

This is a basic application of the Law of Conservation of Energy. 20v * 1A is 20W. 10v * 2A is 20W. This applies better to AC run through a transformer than it does to DC run through a doubler.

Thanks Phil. The power relation is what I based my assumption upon. Then I thought to check it in the literature but found none of 'my' texts, nor several web sources, specifically addressed output current and loading of voltage multipliers in any detail. This caused dreaded seeds of uncertainty to arise. Either it's too blindingly obvious or I'd missed something. Happy to know I hadn't.