# Finding the frequency of a RC network pole given the relative gain at a frequency

In the Texas Instruments application note "Single-Supply, Electret Microphone Pre-Amplifier Reference Design", the following circuit is discussed in section 2.1:

simulate this circuit – Schematic created using CircuitLab

C2 compensates for the parasitic capacitance of U1, however, it also forms a pole with R2 in the amplifier response, as explained in the note. It follows by saying:

The frequency of this pole must be high enough to not affect the microphone transfer function within the audible bandwidth. For this design, a response deviation of -0.1dB at 20kHz is acceptable. The location of the pole can be calculated using the relative gain at 20kHz

Following with equation 7 for a low-pass filter: $$f_p=\frac{f}{\sqrt{\left(\frac{G_0}{G_f}\right)^2-1}}=\frac{\mathrm{20kHz}}{\sqrt{\left(\frac{1}{0.989}\right)^2-1}}=\mathrm{133725kHz}$$

And equation 16 for a high-pass filter (-0.5dB at 20Hz): $$f_c=f\sqrt{\left(\frac{G_0}{G_f}\right)^2-1}=20\sqrt{\left(\frac{1}{0.944}\right)^2-1}=\mathrm{6.986Hz}$$

Where $$\G_0\$$ and $$\G_f\$$ are the gains at low frequency and $$\f\$$ respectively.

They provided no derivation for these equations, as they had for simpler equations earlier in the note. They seems to, and would logically be, related to the RC transfer functions, but I have not been able to derive either thus far.

I was aware of the closed-form equation for finding the -3dB cutoff frequency of an RC low-pass filter as $$\f_c=\frac{1}{2{\pi}RC}\$$, and this is utilized in other literature, and in the note as equation 8 to derive C2, equation 11 to derive C3, equation 14 to find the corner frequency for C6 and R3||R5, and equation 16 for C5.

What were these equations derived from, and if possible, how? If these are inscrutable, is there a different way to find the pole frequency, given the relative gain at a target frequency?

$$Y=\frac{1}{R}+j\omega C ⇒ |Y|=\sqrt{\left(\frac{1}{R}\right)^2+\omega^2 C^2}$$ $$Y_{DC}=\frac{1}{R}$$ $$Y_R=\frac{|Y|}{Y_{DC}}=R\sqrt{\left(\frac{1}{R}\right)^2+\omega^2 C^2}$$ $$f_p=\frac{1}{2\pi RC}$$ $$Y_R=\sqrt{1+R^2\omega^2 C^2}$$ $$Y_R=\sqrt{1+\left(\omega RC\right)^2}$$ $$Y_R=\sqrt{1+\left(2\pi fRC\right)^2}$$ $$Y_R=\sqrt{1+\left(\frac{f}{f_p}\right)^2}$$ $$Y_R^2=1+\left(\frac{f}{f_p}\right)^2$$

$$f_p^2=\frac{f^2}{Y_R^2-1}$$ $$f_p=\frac{f}{\sqrt{Y_R^2-1}}$$

$$\Y_R\$$ is the allowable deviation ratio, i.e. $$\\frac{\mathrm{Value @ 20k} }{\mathrm{Value @ DC}}\$$.

Assuming this is an inverting amp, the gain is the ratio between R1 (not shown) and the effective impedance of R2||C2. Some reciprocals have been handwaved.

All I used was your standard formulae f= 1/RC2pi and a better result in less time than what it would take to read the TI App Note. (and got a similar or better result)

Given:
Drain R=13.7k on Mic,
OA GBW = 20MHz
RO = 50 ohms with 100k shunt. I/O both AC coupled.

I get Max gain = 30 dB (30.68) Rf= 470k , Cf=16 pF RC= 7.520 e-6 f= 21kHz Cin =560 nF fin= 20.7 Hz , If output is loaded with 1M , 1uF will do, otherwise increase Cout.

I don't see any need for academic calculations.

TI has less gain with 275k feedback.