0
\$\begingroup\$

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:

Schematic

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:

Waveform

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?

\$\endgroup\$
7
  • 3
    \$\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
  • 3
    \$\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
  • 2
    \$\begingroup\$ R8 discharges C2 over time, so this circuit is a high pass, not an integrator. \$\endgroup\$ – Janka Nov 27 '17 at 23:34
0
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\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
  • 1
    \$\begingroup\$ @Ash If that's what you want to believe. Believe that. \$\endgroup\$ – Harry Svensson Dec 7 '17 at 20:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.