In article ,
Richard Clark wrote:

We have an audio Z match. We have a diode Z match. There is no sense
that if this is a high-Z or a low-Z for either/both/neither.

Richard-

We have a diode detector, which is nonlinear. If you allow a filter
capacitor on its output, there is current flow only on alternate RF
voltage peaks, in order to charge up the capacitor for the amount lost
to the audio load during the cycle. Its effect on circuit Q may be a
fraction of the total, also consisting of coil Q and input circuit
matching.

Therefore you want the diode connected to the high end of the coil, not
transformed to a lower impedance point on the coil.

Load impedance will affect audio level, so should be kept high for that
reason. However there may be some value of load that produces the
greatest audio output voltage. If too high, the filter capacitor would
charge to the peak of the modulation waveform and stay there.

Worst case is a low impedance audio load and no filter capacitor. Half
of the RF waveform is essentially cut off, leaving a peak-to-peak value
of one half the open circuit value. Since energy is extracted from the
circuit during the conduction cycle, the effect on Q should be
approximately the same as if a resistor having the same peak-to-peak
effect was connected across the circuit, disregarding the RMS value of a
half-sine waveform.

Fred
K4DII

On Wed, 17 Nov 2010 14:57:22 -0500, Fred McKenzie
wrote:

In article ,
Richard Clark wrote:

We have an audio Z match. We have a diode Z match. There is no sense
that if this is a high-Z or a low-Z for either/both/neither.

Hi Fred,

The subtext of that observation was the lack of numbers.
Quantification will aid immensely and give perspective to the direness
of the problem.

We have a diode detector, which is nonlinear. If you allow a filter
capacitor on its output, there is current flow only on alternate RF
voltage peaks, in order to charge up the capacitor for the amount lost
to the audio load during the cycle. Its effect on circuit Q may be a
fraction of the total, also consisting of coil Q and input circuit
matching.

This is dangerously close to decoupling the problem by looking at
minutia. Q is the ratio of power in to power consumed.

Therefore you want the diode connected to the high end of the coil, not
transformed to a lower impedance point on the coil.

Your detector placement serves the consumption side and consumption -
listening to program content - is the whole point of detection.

Preserving Q does not serve that goal.

Load impedance will affect audio level, so should be kept high for that
reason.

Audio level "should" follow power applied, be it low voltage high
current (a speaker); or high voltage low current (a piezo). Preserving
power levels while changing its form is the purpose transduction.

However there may be some value of load that produces the
greatest audio output voltage. If too high, the filter capacitor would
charge to the peak of the modulation waveform and stay there.

What you describe is a clamp, not a filter; but point is well taken.
What you have is the wrong balance of charge time to discharge time.
In other words, you have selected the wrong detector for the chosen
transducer, or vice-versa. Choose both to complement each other.

Consider, what would be lost if you clipped out that pesky cap?

73's
Richard Clark, KB7QHC