In grading the preamplifier lab, I made a mistake when correcting a number of student papers. Students who had used a bias resistor rather than a transimpedance amplifier to convert the microphone’s current output to voltage had not taken into consideration the interaction between the bias resistor and the input impedance of the next stage, which was usually an active high-pass filter. In grading, I overcorrected the student work, changing both the i-to-v gain and the first-stage gain, when the correct action would have been to change either one, leaving the other alone.
The passband gain for the circuit is . The first version corrects the gain of the filter, while the second version corrects the gain of the current-to-voltage conversion. In my grading, I mistakenly applied the correction twice getting
.
There are two ways to get to the correct answer: using Thévenin equivalence and from first principles.
If we replace the current input and with a Thévenin equivalent, whose AC voltage is the AC component of
and whose resistance is
, then we get a simple active high-pass filter with passband gain
for a total passband gain of
and a corner frequency of
.
For those who don’t quite trust themselves to do Thévenin equivalence, we can use first principles to reason about the various currents in the schematic. The negative-feedback loop holds the op amp’s negative input to , and the input node has a voltage, so we get
which we can rearrange to get
.
Because , we get
and can solve for to get
.
Finally, because , we get
.
Our transimpedance gain (including the DC offsets for input current and output voltage) is
.
At DC, this has the appropriate gain of 0, and for high frequencies (in the passband), the gain is approximately , as claimed earlier. The corner frequency, where the real and imaginary parts of the denominator match is at
.