I simulated a miller integrator circuit on multisim online. When I did so I made sure that dc offset in the op amp and the source is zero and also bias and offset currents are zero. But in output to a sine signal (ac) input, the output cosine signal has a negative dc offset. What is causing this offset? Here is the circuit simulated: enter image description here

And here is the input and output waveform: enter image description here

  • \$\begingroup\$ Lack of ground reference? \$\endgroup\$
    – MadHatter
    Commented May 5, 2017 at 19:29
  • 2
    \$\begingroup\$ The answer is staring you in the face ! What are the voltages at t = 0, look at the graph. Both waves start at 0 V because that is the starting point the simulator uses. At this point the capacitor has no charge. Give it an initial condition of 10 mV and the offset will be gone. \$\endgroup\$ Commented May 5, 2017 at 19:32
  • \$\begingroup\$ The choice to start the integration makes a lot of difference. The circuit is working like the math would. \$\int^{t_2}_{t_1} A\cdot sin(x)\:\textrm{d}x=A\cdot\left[cos(t_1)-cos(t_2)\right]\$. If you choose \$t_1=0\$ then the answer is always \$A-A\cdot cos(t_2)\$ and since \$cos(t_2)\$ can't be larger than 1, the integral is always positive or zero. Cool that an electronic circuit can obey the results expected from the math. \$\endgroup\$
    – jonk
    Commented May 5, 2017 at 21:27

1 Answer 1


There's no offset. It works exactly as it should work. Your integrator charges at the positive half cycle of your V1 and discharges back at the negative half cycle. this is the ideal operation.

In practice the integrator drifts due the leakage currents or opamp offset voltage. It can get stucked to max or min output voltage limit or it can gradually forget the starting point and operate as you probably waited. (= zero average due the leak resistance over the capacitor)

In practical integrator circuits there exists a reset switch or an inserted resistance in parallel with C1.

ADDENDUM: Imagine you have a full glass of water. Take a straw and start to soak. The surface of the water sinks. Spit every drop back. The surface returns exactly to its initial position if you do not add anything of your own and do not forget to spit something back. That's a hydromechanical equivalent process for the behaviour of your circuit.


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