If I didn't miss something, it's the first time that you mentioned different source impedances of the source in both circuits.In both cases (1) simple half wave rectifier circuit and 2)antiparallel diodes circuit) the source is
conjugately matched to the circuit.
I guess, that's the key to understand the results. S-parameters are based on a linearized small signal approximation. A rectifier is however nonlinear and correct analysis must consider this, as harmonic balance does. I don't know, how you did the S-parameter simulation, but if it won't use exactly the same signal level as the final circuit, it introduces an error. I think the only reliable method would be to vary the source impedance until you achieve maximum power delivery for each circuit.The way I calculate the impedance of each circuit is through S-parameter simulation.
I find the impedance of the circuit and then I set the source impedance (power source) with
the corresponding conjugate impedance.
My intuition suggests:The antiparallel circuit DC power is not 2X the DC power of the half wave rectifier circuit as intuition would suggest (at least mine...).
Depending on the source's input power, this varies from 1X up to 1.3X. It never goes higher !!!!
I said, both don't necessarily match, not, they can't match.Why can't you have the the diode+loads circuit getting maximum RF power and at the same time having maximum power delivery to the load resistors?
But I don't understand, why you expect a factor of two in delivered power between both circuits...
So you say that in order to have two times more DC power delivered in the antiparallel circuit, we should have low frequencies where inductive or capacitive effects won't take place?I agree, that your explanation applies for a low frequency voltage source and no complex impedance matching
...that "pulls" the fundamental,...
So you mean that having a small real part (relatively) in the source impedance results in having less DC losses which would mean more DC output power ?The special trick is, that the real load impedance is only a small part of the total impedance. So the losses caused by the DC current specific to the half wave circuit are considerably reduced.
Could you elaborate a little bit more on that?...visualize the effect by watching the relation individual harmonic currents...
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