This question is with regards to the microphone input of a Lectrosonics audio transmitter, in terms of how the amplifier works. They have published that the input is as follows:
A typical 3-wire electret microphone is a Sanken COS-11D (3V-10V phantom power version), as follows:
They specified that the best performing wiring for this type of microphone is as shown in the following diagram:
So the servo bias is set to 4V, and the circuit (minus the limiter, which I'm removing for simplicity) is:
simulate this circuit – Schematic created using CircuitLab
At audio frequencies, the coupling caps look like shorts and the servo bias looks like an open circuit (like a constant current source). After doing this and replacing the JFET with its small signal model (and making \$r_{o} = \infty\$ for simplicity), we are left with:
simulate this circuit
Doing some basic analysis, the gain is \$V_{o}/V_{i} = -R_{2}\cdot g_{m}\$
and the output impedance is \$Z_{o} = R_{1}+R_{2}+R_{3}+R_{1}\cdot R_{2}\cdot g_{m}\$.
So the gain seems rather un-controlled, and the output impedance is large. This however seems to be according to plan because in Lectrosonics faq#061 they say:
At audio frequencies, the servo bias looks like an extremely high
impedance resistor (constant current source) so that none of the
output of the microphone is wasted in a 1k to 4k bias resistor. To
prevent large voltage swings, the input to the first amplifier is a
virtual ground input. This input is very low impedance so that the
current developed by the mic FET is used entirely to drive the virtual
ground input. Since the virtual ground input sees a high impedance
source made of the mic FET's drain and the servo bias, the virtual
ground input has very little loop gain noise. Since the mic's FET is
operating into a virtual ground, there is very little voltage swing on
the FET drain which reduces distortion on the FET compared to a
conventional input.
Why you would want a large source impedance and a low input impedance?
I either made a mistake in my analysis or I am missing an important fundamental in how this circuit works. Analog design experts out there, can you resolve this conundrum?
ADDED:
Alfred's comment and answer prompted me to analyze the output as a current, which makes perfect sense, given the high output impedance and low input impedance.
The circuit then becomes (removing the 300 ohms resistor, which given the high \$Z_{o}\$, makes no difference):
simulate this circuit
This produces a gain of:
$$ I_{o}/V_{i} = g_{m}(R\cdot R_{2}-1)/(R\cdot g_{m}+1) $$
Where \$R = R_{1}||R_{2}\$
If R is sufficiently large (which it is in this particular case), then \$R\cdot g_{m} >> 1\$, and we can cancel out \$g_{m}\$, simplify and get the much nicer:
$$ I_{o}/V_{i} \approx -1/R_{1} $$
Which means that if \$V_{i} = 1mV\$, then \$I_{o}\$ will be around 0.37uA.
The next logical question is if this is a reasonable value to noiselessly amplify.
(Sorry, this question became an analysis.)