Capacitive wireless power transfer question

No power is transferred. You'd have to have something in your circuit to absorb power, and your description mentions none.

Jeroen Belleman

following set up.

nce at this frequency is about

There's no power being transferred "across" the capacitor.

You are just charging it up and discharging it. Ideal capacitance is non-di ssipative so there's no energy going anywhere. FR4 isn't a particularly per fect dielectric so there may be some power dissipation in the dielectric, b ut that power isn't going anywhere.

You could detect the dissipation in the FR4 by measuring the temperature o f the copper on the grounded side of the capacitor, which might be percepti bly warmer when the oscillator was running. An interchangable thermistor mi ght do it.

charge pump? ----+---+ C2 ^ |R |C2

---||---+ | | ___ ^ | | ___ | | | ___ ___ ___

(input signal from the left) measure voltage on c2 for various R(a poti) if you look up charge pumps you'll see the efficiency will drop with higher load currents, because charging the capcitor from a Voltage source will burn half the energy in the source resistance.

The higher the load current, the more voltage difference on c1 to compensate each cycle.

This is a perfect explanation for why radio transmitters never work. Because nearly all of them have a DC blocking cap somewhere in the power amplifier chain, and it fails to transfer any power.

Thanks!

Although, it does make me wonder why people persist in building their transmitters that way...

Wait -- with one plate grounded, how can power be transferred? It can only be burned up in various losses.

Snicker >:-} ...Jim Thompson

Whups -- sorry, didn't notice the OP's comment about one end grounded.

But -- power transfer across capacitances is a real thing.

Doesn't the capacitor contribute the "imaginary part?" ;-)

Thanks. Maybe I am not reading the diagram correctly, but you are suggesting a series connection of C2-R-C1, with C1 grounded ? Am I correct ?

I think there are rectifiers in there, somewhere.

minimal ascii.....

RL

Sorry. I was under the assumption you were building some kind of charge pump thingy.... Ignore the post.

if you ground the other end, of course there won't be any dissipation in the capacitor itself, but in the series resistance. (source R and capacitor losses).

That is, if that's what you want to know.