Circuit design of a Low-pass filter borrowed from Wikipedia Article on the subject:

Circuit design of a Low-pass filter borrowed from Wikipedia Article on the subject.

Circuit design of a High-pass filter borrowed from Wikipedia Article on the subject:

Circuit design of a High-pass filter borrowed from Wikipedia Article on the subject.

I'm thinking of implementing an electronic filter Circuit that will pass certain frequencies and attenuate others. It is for the purpose of adding an effect to an audio signal. I notice that such a circuit uses resistors to function. My question is, will there be a reduction in amplitude to the signal after it passes through the Circuit, and is this strictly desirable?

If not, is there a way to actively compensate for this loss with some sort of amplifier that operates proportionally to the resistance in the Circuit? Is this perhaps what "First order active Low/High-pass filter Circuits" do, using op amps?

Additionally, if I were to implement the resistor as a potentiometer, what should I research in order to actively 'measure' the current resistance in the resistor and amplify proportionally?

  • \$\begingroup\$ Start with a basic op-amp like the TL082 of LM358. Get a power supply and breadboard to get going. Even a single 9 V battery will work if you use single supply circuits. \$\endgroup\$
    – skvery
    Mar 3, 2017 at 17:50

1 Answer 1


Those passive filters have a voltage gain of 1 in the passband as long as there is no load (i.e. a resistance/impedance from \$V_{\text{out}}\$ to ground). Consider the low pass filter, for example: at low frequencies the capacitor behaves as an open circuit and, since \$V_{\text{out}}\$ is also an open circuit, there is no current through the resistor and therefore no voltage drop across it. In other words, \$V_{\text{in}} = V_{\text{out}}\$, which is a gain of 1. There is some attenuation in the passband near the corner frequency because no filter -- passive or active -- is perfect at the corner frequency. In the low pass filter's stopband the gain is 0 since the capacitor acts as a short circuit.

However, in practice you are unlikely to see an open circuit at \$V_{\text{out}}\$. There will be some finite load resistance or impedance, and this does reduce the gain of the low pass filter. If you have a load resistor equal to the filter resistor, for example, you have a gain of 1/2 (the two resistances form a voltage divider that divides the input in half).

Here's a passive low pass filter for which you can simulate the frequency response and adjust the load resistor to see what happens:


simulate this circuit – Schematic created using CircuitLab

The easy way to fix this passive filter so that it has a gain of 1 even with a load is to add an op amp buffer to the output of the passive filter (which makes the overall filter an active one):

enter image description here

Source: http://www.electronics-tutorials.ws/filter/filter_5.html

This works because the input of the op amp (which is connected to the output of the passive filter) has a very high impedance -- it's nearly an open circuit. Thus you have a gain of 1 (in the passband) at the output of the passive filter, and the buffer has a gain of 1 as well, so the overall gain is 1 in the passband. You could also use a non-inverting op amp amplifier instead of a buffer to provide gain >1, if necessary/desired.

The passive filter may be acceptable if your load resistance is known (and constant) and you can accept a gain less than 1. In that case, you would choose the filter resistor so that it gives an acceptable gain in the passband even with the load resistor.

  • \$\begingroup\$ Definition: a passive filter followed by an amplifier is not generally called an active filter, in my experience. An active filter uses amplifiers along with (usually) resistors and capacitors to get responses that could not be achieved with resistors and capacitors alone. Flat pass bands, sharp roll-offs, and very narrow band pass responses can be obtained. \$\endgroup\$
    – user128351
    Mar 3, 2017 at 20:54
  • \$\begingroup\$ @stretch The filter with the op amp buffer that I've shown is a passive filter with an active element, so the overall circuit is an active one. Yes, there are more complex and better active filters possible but what I've shown solves the OP's question about the gain. \$\endgroup\$
    – Null
    Mar 3, 2017 at 21:01

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