|Above Von the function is linear below it is square law.|
|Another problem with the modeling is its response to temperature.|
This graph being linear you can see the effect of the square law detector. So the analysis of a crystal set will show we see a gain in the detector circuit. Mr Tongue explains this and give some analysis of the impedance matching effects. Here is where the problem comes to light. The crystal set functions without any amplification other than what the circuit can provide. A high Q tank can boost the signal and the square law detector can boost the signal. If we had a signal strong enough to operate in the linear region it would be a good thing I don't think we would want to attenuate the signal to force square law detection. How can this be? When teaching introductory level classes we use something like this:
Using an ideal diode this is true and with large signal detection it is still close enough. But with small signal detection we are operating below the threshold and the detector is not switching it is acting as a non linear device. In other words rather than clip the negative half cycle it reduces its amplitude. The effect is rather than passing half the signal amplitude it is passing half plus a suppressed negative half cycle. So the filter capacitor charges to a voltage above half the input and the DC bias it produces is what the output rides on.
Take another look at the center signal and add negative pulses a quarter the amplitude of the positive pulses and shift the zero reference down.
If the input peaks at 8mv and we chop it to get 4mv with the ideal diode then add back 1mv because the diodes doesn't switch cleanly we get a 25% boost in signal.