Circuit for using a PIR Hey guys, I am currently working on an Arduino project where I need to use a PIR(pyroelectric infrared sensor). I am building my own circuit for it from scratch. I am using the circuit example from the datasheet of the sensor I am using (IRA-E700, datasheet: http://www.farnell.com/datasheets/2199772.pdf). I have changed some values, however, to filter out 50Hz signals since i was getting a lot of noise with that frequency. I have almost understood the circuit completely, but there are a couple of things I still haven't understood:

  • What is the purpose of the capacitor C3? I am guessing it's used for some sort of filtering of the noise you get with a PIR. I'm just not sure how such a filter would work?

  • What is the purpose of capacitor C6? I know it has to do with getting the voltage up to 2.5V since the op-amp has 2.5V on its non-inverting input. I'm guessing the capacitor interacts with the op-amp somehow, but I just don't know exactly how.

Thank you in advance for the help guys! I apologize if I left any information out.


2 Answers 2


Without C3 the DC offset of opamp IC1A could be a problem.

C3 blocks DC but conducts AC. This results in that the circuit around opamp IC1A has a DC gain of 1 because for DC the output is more or less directly connected (through R3) to the - input of the opamp.

The AC gain of the circuit around opamp IC1A is roughly R3/R2 (for frequencies where C2 is not involved yet).

The function of C6 is similar even though C6 is in the signal path instead of the in the feedback circuit (like C3). Without C6 the DC output voltage of opamp IC1A would reach and influence the DC operating point of opamp IC1B directly.

This has to be avoided as it can cause problems. Also it is generally a better idea to make circuits operate "on their own" instead of letting their DC operating point be dependent on another circuit.

The DC operating point of opamp IC1B is set by R5 and R6 at half the supply voltage. Again the DC gain of the circuit around opamp IC1B is 1 but the AC gain will be much higher.

To better understand opamp circuits I suggest reading: Opamps for everyone

  • \$\begingroup\$ Ah okay, this makes perfect sense. Thank you very much! \$\endgroup\$
    – Jeppe
    Commented Mar 14, 2018 at 9:46

C3 makes IC1A not to amplify constant DC, only the changes (=the AC component) of the S output of the PIR are amplified

The DC output of IC1A is stopped by C6, only the changes are amplified again in IC1B. IC3 is a comparator which shows both positive and negative peaks at the output of IC1B.


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