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I came across this circuit recently:

enter image description here

Here is my analysis from a glance:

  • Resistor divider R1 and R2 set the DAC point for Vinp. The microphone, Rmic and C1 cause AC perturbations on Vinp.
  • At low frequencies, capacitors act as open circuits. Hence, Vinp is defined entirely by the resistor divider R1 and R2. Capacitor C2 is open circuited, hence R4 serves no purpose. C3 is open circuited and no current flow through R3, so the circuit turns into a unity-gain amplifier
  • At high frequencies, capacitors act a short circuit. The AC signal from Vm superimposes onto Vinp. Capacitor C3 shorts out R3 and capacitor C2 provides a path to ground for R4. The circuit is still in a unity gain configuration because capacitor C3 is a short.

I feel my last point is incorrect. Something tells me R4 path to ground has some other effect.

Am I missing something?

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  • \$\begingroup\$ citmerefwp - Hi, Where did that schematic come from? To comply with the site rule on referencing, details of the original source of copied / adapted material must be provided by you, next to each copied / adapted item. If the original source is online (webpage, PDF, video etc.) please edit the question & add its name & link (URL) (e.g. website name + webpage title + its URL). For offline sources, see the linked site rule. || Always reference to the best of your ability. TY (Please see the tour & help center for the main site rules, as they differ from typical forums.) \$\endgroup\$
    – SamGibson
    Commented May 26 at 22:24
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    \$\begingroup\$ The microphone must be an electret with built-in FET or IC amplifier that needs about 500 uA (I've seen up to 1 mA or so) to operate properly. R1 and R2 are there to place the inputs about halfway between the rails, so probably are the same value. Otherwise, you've boxed things in about right. \$\endgroup\$ Commented May 26 at 22:25

2 Answers 2

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I feel my last point is incorrect. Something tells me R4 path to ground has some other effect.

Am I missing something?

You're on the right path.
C3 becomes a short only at high frequency...at the upper end of audio range, like 20,000 Hz. C3's impedance likely becomes similar to R3 near 20 kHz.

On the other hand, C2 sets the low frequency limit; its reactance becomes equal to R4 down around 20 Hz.

Between 20 Hz and 20,000 Hz, op amp gain is pretty much flat, set by the ratio of R3/R4. (true gain is \$ 1+{{R3}\over{R4}}\$). As periblepsis suggests in comments, you'd likely set R1=R2, so that op amp DC output voltage sits at half-supply. Audio makes it swing above & below this bias point.

I'm assuming reasonably high audio quality. If instead this were for a telephone, you might choose capacitors for frequency limits between 300 Hz to 3000 Hz.


C1 also can affect the low-frequency end. For example, you might also choose to reject audio below 20 Hz here...
\$ C1= {{1}\over{20\times2\pi\times(R_{mic}+ {{R_1\times R_2}\over{R_1+R_2}})}}\$

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  • \$\begingroup\$ very good answer. I understand now. The 1+ R3/R4 flat gain part assumes that the capacitor C3 and C2 impedance is negligible and the resistors R3 and R4 are dominating? The low frequency and high frequency gain are unity. This is basically a bandpass filter? \$\endgroup\$
    – citmerefwp
    Commented May 26 at 22:58
  • \$\begingroup\$ Unsure what you mean by negligible. In the bandpass region, C2's reactance is lower than R4, and C3's reactance is higher than R3. True that R3/R4 dominate in the bandpass region. \$\endgroup\$
    – glen_geek
    Commented May 26 at 23:05
  • \$\begingroup\$ That's what I mean. In the bandpass region, the gain is defined entirely by 1 + R3/R4 even though C2 appears in series with R4. Which means that C2 impedance must be much less than R4 \$\endgroup\$
    – citmerefwp
    Commented May 26 at 23:11
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I feel my last point is incorrect.

You are correct - it is incorrect.

You have taken an entry-level rule of thumb for AC circuit analysis too far, and missed an important concept. The idea that a capacitor is a short circuit at high frequencies is problematic in the real world because the impedance of a cap is inversely proportional to the frequency of the signal going through it. Thus, its impedance is a true short circuit only at infinite Hz. Even this does not take into account the effects of how the capacitor is constructed, which can have a significant effect on its impedance.

At any freq below infinity, the value of the series feedback network (R3 || C3) always is some finite value, the parallel combination of the R3 resistance and the C3 impedance at a specific frequency. Same for the series combination of R4 and C2.

The input impedance is more complex because there are three elements: the Thevenin equivalent resistance of R1 and R2, the impedance of C1 (again, at a specific frequency), and the output impedance of the microphone and Rmic combination.

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