I want to calculate the integral of a voltage pulse. I have designed this circuit to do the integration. I am using a non inverting integrator:


The op_amp_2 is the voltage pulse. Comp_1 is used to discharge the cap once the pulse has ended. So that the integral of each pulse can be found.

This is how the output looks like:


I wonder why the output falls down as the pulse ends. Is this because of the bias current of Op-AMP? I am specifically using a low bias (~pA) current opamp.

How can I avoid this?

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    \$\begingroup\$ What do you think R8 is doing? \$\endgroup\$ – Harry Svensson Nov 27 '17 at 23:26
  • \$\begingroup\$ Because your circuit ain't an integrator? \$\endgroup\$ – Marko Buršič Nov 27 '17 at 23:30
  • \$\begingroup\$ R8 is used to set the opamp output DC operating point. \$\endgroup\$ – Ash Nov 27 '17 at 23:31
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    \$\begingroup\$ An integrator has no DC bias, no DC operating point. Where did you get the concept of this? \$\endgroup\$ – Marko Buršič Nov 27 '17 at 23:34
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    \$\begingroup\$ R8 discharges C2 over time, so this circuit is a high pass, not an integrator. \$\endgroup\$ – Janka Nov 27 '17 at 23:34

Another thing that might be worth mentioning is that it's not an actual integrator that you got there. It may mimic one and fool you.

Here's a schematic of how an actual integrator would look like. Also, look at the size of the "R8" I'm using. It's 100 kΩ here, not 1 kΩ. 1 kΩ is way too small.

enter image description here

As you can see, I'm simply integrating a square wave, the actual integrator shows a triangle wave, as it should. While the bottom "integrator" which you are using show a triangle wave on top of the square wave input.

Here's the link in case you want to simulate it and see for yourself.

To make this answer appear less as a comment. The reason for why your output voltage is drooping is because of R8. Your small capacitor (only 1 nF) is discharging through a very small resistor (R8).

And to answer the title "Integration Output Doesn't make sense", well that's because it's not proper integration as I proved above with the circuit simulation.

  • \$\begingroup\$ The circuit you showed as my integrator is not correct. You don't have the R and C filter on the non inverting pin matching with the R and C of the inverting pin. \$\endgroup\$ – Ash Dec 7 '17 at 20:18
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    \$\begingroup\$ @Ash If that's what you want to believe. Believe that. \$\endgroup\$ – Harry Svensson Dec 7 '17 at 20:59

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