0
\$\begingroup\$

I have to integrate signals very precisely, and Im facing DC drift on inverting integrators using opamps. The usual cure is to add a resistor in parallel with the capacitor, however this produces a different output than the integrator without the resistor, meaning that the initial DC offset of the integrator without a resistor is gone.

To illustrate this consider the following circuit

schematic

simulate this circuit – Schematic created using CircuitLab

Supose that the input waveform to the circuit is a +-1V 1KHz. square signal, the following screenshot displays the input and output signals of the circuit:

enter image description here

In blue theres the input signal to the integrator, a -1 to 1V 1KHz. square signal, and in yellow the output of the integrator, a 3.125V peak to peak triangular waveform spanning from 0 to -3.125V. This can be proven by the following integrals (for the period of the input signal):

$$V_o(t)=-\frac{1}{RC}\left(\int^{0.5ms}_{0}{dt}-\int^{1ms}_{0.5ms}dt\right)$$

Evaluating the first integral yields a line with negative slope from 0 to -3.125V in the [0,0.5ms] interval and the second integral yields a line with positive slope from -3.124V to 0V in the [0.5ms,1ms] interval.

Now suppose I insert a 10M resistor in parallel with the capacitor

schematic

simulate this circuit

Suppose again that the input is a +-1V 1KHz. square signal, the following screenshot displays the input in blue and the output in yellow. Its obvious that the waveform of the output is the same, a triangle signal, however the negative DC offset is lost, so the resulting signal is strictly not the integral because its lacking the DC offset.

enter image description here

My line of thought is that if I add the missing DC offset, then I will get the integral, However, and this is my question: This doesnt seem valid for any waveform or input signal even if its non-periodic?, meaning that just adding a DC offset at the output of the integrator will yield the integral? This seems counter intuitive, for example if the input is a sine wave, the output will be a cosine wave without an offset, adding the offset will not yield the actual integral.

PS Im aware that this is an inverting amplifier and the actual integral has opposite sign.

\$\endgroup\$
  • \$\begingroup\$ For precise integration of signals, over time periods up to 30 minutes, US INS would digitize the accelerometer outputs and "integrate" using digital methods. \$\endgroup\$ – analogsystemsrf May 1 '18 at 3:09
  • \$\begingroup\$ I need to do this with analog methods, the main concern is latency. \$\endgroup\$ – S.s. May 1 '18 at 3:11
  • \$\begingroup\$ What are your error concerns. \$\endgroup\$ – analogsystemsrf May 1 '18 at 3:56
  • 1
    \$\begingroup\$ What exactly are you trying to do? Is there a defined start time and end time for the integral, so you can set the initial condition (and estimate the error over that interval)? \$\endgroup\$ – Spehro Pefhany May 1 '18 at 3:58
1
\$\begingroup\$

You are missing a classic part of integration. You seem to think that $$\int{dt} = t $$ But it doesn't. Instead, $$\int{dt} = t + C $$ For integration, there is always an initial condition which needs to be specified. In your case, the addition of a feedback resistor causes the integrated signal to be symmetric around zero if the input is. That is, if the average input is zero, so will the output.

To get the waveform you're looking for, you need a reset function to zero the integrator. You can do this with an analog switch such as a CD4066 like so

schematic

simulate this circuit – Schematic created using CircuitLab

By applying a reset pulse to the switch and releasing it at just the desired time (the rising edge of your input in this case) you will get the desired waveform. Granted, you may find the synchronization of the reset a major pain in the behind, but there's no help for it.

Also, by adding the feedback resistor you will limit the long-term drift of your integrator. If you watch your first circuit for a long period of time, you'll see the voltage drift one way or the other, and keep on drifting until the integrator output limits either positive or negative. This is due to input bias currents being integrated over the long term. For a low-bias op amp, such as the TL081 (100 pA) the drift will be quite slow. For something really nasty like a 741 (100 nA, typically), the drift will be 1000 times faster.

\$\endgroup\$
  • 1
    \$\begingroup\$ The reset switches are a beast. For integration I worked on, mosfet charge was injected and then removed when the mosfet switch was turned on and off. And it was NOT the same charge both ways. Burr Brown worked to balance their switches so that charge injection was carefully balanced with removal. But it is rare to see. See ACF2101, for example. We eventually found that the package bulk resistance (on the order of 10^11 ohm-cm) itself was way too leaky, too. Had to go to die and wire-bond the circuitry. \$\endgroup\$ – jonk May 1 '18 at 4:22
  • \$\begingroup\$ I forgot about the integration constant, so basically both are the integral of the input but they have different starting conditions? \$\endgroup\$ – S.s. May 1 '18 at 4:34
  • \$\begingroup\$ @jonk - Yeah, charge injection is (for high performance) a problem. In this case, with a 0.1 uF cap, I wouldn't worry too much. \$\endgroup\$ – WhatRoughBeast May 1 '18 at 14:07
  • \$\begingroup\$ @WhatRoughBeast Yeah. We had consistent resolution across devices down to about \$10\:\text{fA}\$. In a custom we did (never went commercial), we got down to \$600\:\text{aA}\$ and were able to see boson "flocking" noise. That required wire bonding of dice, a 3-stage TE cooled stack, and the MCU in dice form itself too, all inside what was slightly smaller than a TO-3 with a sapphire window. Hamamatsu was the hardest nut to crack -- they didn't want to give out detectors in dice. I received a letter from shocked local representation saying I got them. They didn't believe it would happen. \$\endgroup\$ – jonk May 1 '18 at 14:58

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.