Amie Wrote:.... If a floating detector is used, and it only has its own leakage as a load then the actual signal output voltage would be as high as it could possibly be.
That's (debatably) true, though there are complications!
If the leakage resistance is super-high, then the detector output load capacitor will really charge right up to the signal peak, with virtually no load. And as the AF signal increases to the modulation peak, the detector output will follow the modulation.
But as the AF signal cycle passes its peak and heads downwards, with the AM signal heading towards the modulation trough, the detector load capacitance will stay charged because the leakage resistance is so high. If the AF is at a low frequency, it might discharge so that the detector output can follow the modulation envelope. But if the AF is at a high frequency, it very likely won't, so not only will the AF output drop, it'll also be distorted as the rising portion of the AF cycle will be correct but the falling bit will just be a segment of an exponential discharge curve. Which could well be the distortion you mentioned.
So under these conditions, reducing the load will actually give an INCREASE in detector AF signal output voltage.
