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In the below circuit, do Rbias and R2 both affect the cut-off frequency? Or is it simply 1/2*pi*R2*C2

enter image description here

zud
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1 Answers1

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They both do.

The electret microphone with its JFET buffer behaves as a current source, with a very high output impedance. In terms of the AC analysis, this is in parallel with Rbias, since the upper end of Rbias is at AC ground. The Thevenin equivalent of this is a voltage source in series with Rbias.

This means that the current flowing toward the opamp is determined by the Thevenin source voltage divided by the series impedance of Rbias+C2+R2. The cutoff frequency is

$$\frac{1}{2\pi(R_{bias}+R_2)C_2}$$

Dave Tweed
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  • I did a little quick hand analysis where I modeled the input current from the JFET as a current source and considered the output quantity of interest the current that flows into the inverting op-amp -- just thought I'd share it since it fleshes out the math a bit more. [![Analysis][1]][1] [1]: https://i.stack.imgur.com/B0V9F.jpg – rusty_old_jfet May 30 '17 at 21:08
  • Also, see [here](https://electronics.stackexchange.com/a/306315/11683) for some thoughts about better ways to bias a microphone. – Dave Tweed May 30 '17 at 22:17