I want to approach this subject from a circuits and fields standpoint.
Thesis: I was remembering how, for some reason, schottky diodes do not have a reverse recovery loss, despite having the same current vs. time graph. Where does the energy go?
Well the answer is (or should be), you get it back when the voltage comes back up. A "forward turn-on spike" as it were. Of course, not all of it comes back, it's going to be a lossy capacitor. How much is a good question, though.
So the thought comes, how to measure it?
Now, let me segue a bit onto the *j-rotated complement of electricity, magnetism. 'Magneticians' have this thing called a B-H curve, and the area of the loop equals loss directly (energy density per cycle, to be exact). You can also figure lots of useful things with it, knowing some geometry anyway, like the V-I/time parameters of an inductor or transformer built with the material.
The easy way to measure the B-H curve is to put in an inductor, measure the EMF across it (usually done by adding a secondary, which picks up the EMF = dPhi/dt), and the current through it (magnetization in amp-turns). Since EMF = dPhi/dt, and we want just Phi (which is related to B), we use an integrator on the secondary and we're done. Plot X vs. Y and you get that cute hysteresis loop.
Dividing by the imaginary constant, I will now return to electrics. In Maxwell's Equations, instead of B and H, you get D and E, the displacement and electric field strength. Where H is in A/m, E is V/m. E is easy to measure, apply EMF to a capacitor and divide by the plate seperation (capacitors are even planar quite often, making this a trivial task!). But what the heck is D? B was field density, Wb/m^2, where Wb == V.s. D is charge density, in C/m^2. So that's just the charge over the area of the plates, which we know is Q = V*C = integral I dt, taking account of the plate area of course.
So it should be that, instead of integrating EMF, we integrate current. In the inductor case, we apply some sort of voltage, which causes an EMF = dPhi/dt, making amps rise. In the capacitor case, we apply some sort of current (which causes an "amperomotive force"???), which causes an I = dQ/dt, making volts rise.
For the test circuit, I propose a square wave current source (from a fairly high voltage supply, easily 100V for a "25V" capacitor), a current sense resistor, and for utmost accuracy, a differential amplifier reading the voltage across the capacitor. Send sensed current into an integrator and plot the two outputs. Very simple, of course.
I'm not sure what's equivalent to a secondary winding though. I know you can't make a capacitive transformer. In inductors, series resistance typically limits EMF in terms of terminal voltage (hence the preference for a transformer over an inductor), but the capacitive equivalent would be parallel (leakage) resistance, which is generally negligible (until breakdown). Frequency low enough to avoid ESR is also possible (just as it's possible to minimize parallel eddy current loss by reducing frequency). In inductors, you're sensing V ~ dB/dt; in inductors, you're sensing I ~ dD/dt, which is just the charge accumulating.
Finally, direct applications. If you plop a silicon diode into this circuit, you'll see conduction in one direction, with voltage saturating at Vf. Current is constant, so Q will rise linearly -- power is being delivered in the forward direction. When current reverses, a blob of charge leaks though the junction as it turns off (the amount depending on the the rate of charge I = dQ/dt compared to the recombination time, more or less). For the rest of the reverse-biased cycle, charge varies, but generally goes up (C goes down as V goes up, but Q will always be monotonic, otherwise it wouldn't conserve energy), so it will seem to trace some sort of rising saturation curve. On its way back down, it traces about the same curve, but there will be some offset, due to dielectric loss. When it returns forward, charge goes up again. Through one cycle, direct rectification causes a ratcheting of charge, but if that is subtracted out, the area contained within the recovery and reverse biased loop should correspond to recovery energy plus dielectric loss.
As for the schottky diode, the same general form should hold, except for this: because capacitance is so much larger at low voltages, while recovery is nonexistent, you should see a narrower loop between "reverse recovery" and "forward recovery", suggesting that the charge sucked up during the first few volts is returned during the last few volts. Area under the loop is still energy lost, but it should be distributed more evenly through the loop, rather than a dense wad at reverse recovery.
I suppose I'll conduct tests some time this week, but the labs here don't really have the equipment I'd need -- no high voltage power supply or transistors, and maybe not even schottky diodes as I originally wondered about. Oh well, maybe I'll draw some circuits and put it on the Bench next time I'm home.
Tim