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I have seen the standard high pass filter with the -3dB cutoff frequency = \$ 1/2\pi RC\$.


simulate this circuit – Schematic created using CircuitLab

However I have seen this particular circuit which I'm seeing here and there, especially on the inputs and outputs of ADCs DACs etc. Is the resistor there to limit the input/output current? Would the -3dB cutoff frequency change at all? If so, what frequency would the new one be at?


simulate this circuit

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The corner frequency is not 2 pi RC, it's ONE OVER 2 pi RC! – Jon Watte Feb 28 '14 at 17:20
up vote 5 down vote accepted

Spehro gave you the answer, now let me tell you why you could (should?) have known that by looking at the circuit.

Two resistors in series are indistinguishable from one resistor with the sum of the two resistances. (I hope you knew this?)

Hence when we take Vout in the second circuit from the R1/C1 junction, we have exactly the same circuit as the first (but with R1' = R1 + R2).

But instead we take the output from the R1/R2 junction. These two resistors form a pure resistive voltage divider, for which frequency is totally irrelevant. So the fact that we take the output from R1/R2 instead of from C1/R1 can't influence the frequency response, except from the constant factor R2/(R1+R2).

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Excellent point. Also remember a capacitor is open for DC and short for high (OK, infinite) frequency so you can see that the circuit is high pass due to the placement of the cap. – John D Feb 27 '14 at 22:14
The corner frequency depends on both resistor values. See my answer. I think we're on the same page but it isn't clear in your last sentence. – Alfred Centauri Feb 28 '14 at 0:03

First, that's a high-pass filter.

Secondly, if you do not load Vout (on the second circuit), then I think you can see that the series resistance R1 + R2 is equivalent to R1 (on the first circuit), so you can find the -3dB frequency.

The output voltage at high frequencies (second circuit) will be \$ V_{OUT} = V_{IN} \cdot \$\$R_2 \over R_1 + R2\$, and the voltage at the cutoff frequency will be -3dB relative to that.

To see immediately that it's a high-pass filter, just remember that capacitors act as open circuits at low frequencies (and inductors act as open circuits at high frequencies).

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Sorry, I meant to write high pass! Corrected now – midnightBlue Feb 28 '14 at 10:52

What is the purpose of this simple 1 capacitor 2 resistor [high] pass filter?

The extra resistor gives you an extra degree of freedom.

For the standard RC high-pass filter in the first schematic, the transfer function is

$$H(j\omega) = \frac{j\omega R_1C}{1 + j\omega R_1C} $$

So, the asymptotic high frequency gain is 1 and the corner frequency is \$f_c=\frac{1}{2\pi R_1 C}\$.

But what if you want something different than 1 for the high frequency gain? Add another resistor.

It's straightforward to show that, for the second schematic, the transfer function is

$$H(j \omega) = \frac{R_2}{R_S}\frac{j \omega R_SC}{1 + j \omega R_SC}$$


$$R_S = R_1 + R_2$$

So, you still have the high-pass filter but now, the asymptotic high frequency gain is \$\frac{R_2}{R_S}\$ and the corner frequency is \$f_c=\frac{1}{2\pi R_S C}\$.

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+1 because I love mathematical explanations. You can't argue with a clear proof :) – Brian Onn Feb 28 '14 at 5:25

This is a filter with a resistor divider. It's NOT a lowpass filter, it's a highpass filter, often used to block DC. The divider is presumably used to scale the input to an appropriate range for whatever is connected to Vout. Assuming the load on Vout is infinite impedance, the corner frequency should be -

$$ f = \frac{1}{2 \pi (R1 + R2) C}$$

At frequencies much higher than the cutoff, the transfer function approaches -


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The circuit in question is also high-pass filter, just like the typical cap + resistor one you show. But the added resistor provides level shifting. R1 and R2 create a voltage divider, without affecting the frequency response that C1 + Rt high pass filter creates.

Simply put, it is a high-pass filter + level shifter.

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While the other answers were great, I just felt they were very technical sounding. A simple explanation was warranted imho. – Passerby Mar 1 '14 at 2:06

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