# Coupling capacitor location wrt input impedance

Discussion in 'General Electronics Chat' started by hrs, Dec 7, 2015.

1. ### hrs Thread Starter Member

Jun 13, 2014
98
8
Hi,

Suppose I have a coupling capacitor in front of a buffer op-amp, how do I find the impedance required for calculating the -3dB frequency? Is the impedance of interest defined only by the stuff after the capacitor or also the stuff in front of it?

To keep it simple I will assume the buffer impedance to be a resistance, Rbuffer. Per attached image I might think that the impedance of the upper configuration is R1 + (R2 * Rbuffer / (R2 + Rbuffer)) and the impedance of the lower configuration might be just Rbuffer.

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2. ### dl324 Distinguished Member

Mar 30, 2015
3,561
680
3dB down when $\small X_c=\frac{1}{{\omega}fC}$ is half of the impedance.

Last edited: Dec 7, 2015
3. ### AnalogKid Distinguished Member

Aug 1, 2013
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1,332
In the upper schematic, you have a highpass filter with cutoff frequency f. The output of the filter is the C1-R1 node.
f = 1 / (2 pi (R1+R2) C).

Because the buffer amp has such a high input impedance, it basically is picking off the output of the filter partway down an attenuator.
Vbufferinput = (Vfilter (the result of the highpass filter at the C1-R1 node) x R2) / (R1+R2)

ak

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4. ### crutschow Expert

Mar 14, 2008
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3,506
The resistance to use for the frequency calculation is the sum of all the equivalent resistance values in series with the capacitor on both sides of the capacitor (makes no difference which side).

Incidentally your lower circuit won't work because there's no DC path for the op amp input bias current for the (+) input.
The op amp output will go to a saturated value (either plus or minus depending upon the polarity of the bias current) and stay there.

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5. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,019
1,123
To be able to find -3dB frequency for the circuit with only one capacitor we need to find a equivalent resistance seen from capacitors terminals. But we first need to short all the voltage sources.

Fc ≈ 0.16/(C*Req)

So for the upper case we have:

Req = R1 + R2||Rbuffer

And for the lower case

Req = Rbuffer + R1||R2

And this circuit has a small error. There is no DC path for input bias current. In real life this circuit might work with TL07x for sometime. But such connection is fundamentally incorrect.

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