Gas station without pumps

2019 May 22

Interaction between bias resistor and active high-pass filter

Filed under: Circuits course — gasstationwithoutpumps @ 00:02
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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.

Schematic of bias resistor and active high-pass filter. The input is the current I_in.

The passband gain for the circuit is $R_b\frac{R_f}{R_b + R_i} = (R_b || R_i) \frac{R_f}{R_i}$. 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 $(R_b || R_i) \frac{R_f}{R_b + R_i}$.

There are two ways to get to the correct answer: using Thévenin equivalence and from first principles.

If we replace the current input and $R_b$ with a Thévenin equivalent, whose AC voltage is the AC component of $I R_b$ and whose resistance is $R_b$, then we get a simple active high-pass filter with passband gain $\frac{R_f}{R_i + R_b}$ for a total passband gain of $R_b\frac{R_f}{R_b + R_i}$ and a corner frequency of $\frac{1}{2 \pi (R_i+R_b) C_1}$.

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 $V_{ref}$, and the input node has a voltage, so we get
$V_{input} = V_{dd} - I_b R_b = V_{ref}-I_f \frac{j\omega R_i C_1 + 1}{j \omega C_1}$
which we can rearrange to get
$I_b = \frac{V_{dd} - V_{ref}}{R_b} + I_f \frac{j \omega R_i C_1 + 1}{j\omega R_b C_1}$.
Because $I = I_b + I_f$, we get
$I= \frac{V_{dd} - V_{ref}}{R_b} + I_f \frac{j \omega R_i C_1 + 1}{j\omega R_b C_1} + I_f$
and can solve for $I_f$ to get
$I_f = (I- \frac{V_{dd} - V_{ref}}{R_b}) \frac{j\omega R_b C_1}{1+j\omega(R_b+R_i)C_1}$.

Finally, because $V_{out}-V_{ref} = I_f R_f$, we get
$V_{out}-V_{ref} = R_f (I- \frac{V_{dd} - V_{ref}}{R_b}) \frac{j\omega R_b C_1}{1+j\omega(R_b+R_i)C_1}$.

Our transimpedance gain (including the DC offsets for input current and output voltage) is
$\frac{V_{out}-V_{ref}}{I- \frac{V_{dd} - V_{ref}}{R_b}} = R_f \frac{j\omega R_b C_1}{1+j\omega(R_b+R_i)C_1}$.
At DC, this has the appropriate gain of 0, and for high frequencies (in the passband), the gain is approximately $\frac{R_f R_b}{R_b + R_i}$, as claimed earlier. The corner frequency, where the real and imaginary parts of the denominator match is at $\omega = \frac{1}{(R_b+R_i)C_1}$.