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Can I implement the I action in an analog PID controller as follows?:

schematic

simulate this circuit – Schematic created using CircuitLab

And am I right when I say that the voltage follower is used to prevent distortion in the input impedance of the second opamp?

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Your basic idea is right, but not the implementation. Vin going to the wrong opamp input. Vin should drive the + input. The - input should be simply the output voltage fed back.

Otherwise, yes, this overall circuit will result in the negative integral of the input voltage. The first opamp buffers Vin so that it doesn't have to be a low impedance source.

Note that the initial condition is not specified. This is the voltage on C when this circuit is first started up. If this is inside the feedback path, and some glitching at startup is acceptable, then this may be OK.

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  • \$\begingroup\$ What do you mean by: 'Note that the initial condition is not specified'? \$\endgroup\$
    – klopr
    Feb 1 '18 at 17:54
  • \$\begingroup\$ Am I right about the fact about the impedance? \$\endgroup\$
    – klopr
    Feb 1 '18 at 17:57
  • \$\begingroup\$ @klopr - Assuming the input is zero, when you turn on the circuit and it finds a stable operating point, charge may well (almost certainly will) get injected into the cap. The result is that immediately after turn-on the output will almost certainly not be zero. How far from zero it gets will depend on the op amp and the capacitor size. This is usually handled by placing a switch across the cap which shorts it out, and the switch automatically opens up shortly after power is applied, letting the integrator work properly. \$\endgroup\$ Feb 1 '18 at 21:54
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And am I right when I say that the voltage follower is used to prevent distortion in the input impedance of the second opamp?

No, it's just a buffer and not needed if the source impedance of the driving signal (the signal at the input) is much lower than R.

Can I implement the I action in an analog PID controller as follows?

You can approximate it but remember that op-amps are not ideal and an op-amp won't have infinite gain at DC (as a true integrator has). There may also be input bias currents in the low nano amp range for an op-amp and these can self charge the capacitor so you need to decide whether these currents can be ignored in your application.

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  • \$\begingroup\$ For the buffer, I do not now if the impedance of the driving signal is always much lager than \$\text{R}\$, because it is a part in a PID controller. So is it a good idea to put the buffer into there? \$\endgroup\$
    – klopr
    Feb 1 '18 at 18:19
  • \$\begingroup\$ There is no generic PID circuit that I have to reference your question to so I cannot answer. The source impedance needs to much smaller and not much larger although I do like lager very much lol. \$\endgroup\$
    – Andy aka
    Feb 1 '18 at 19:18

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