I have to build an analog PID controller (with op-amps), I've made a schematic:

enter image description here

My questions:

  • Can I build my PID controller this way?
  • My PID tranferfunction (in the s-domain) looks like:


But what in my schematic represents the variables from my transferfunction?

  • 1
    \$\begingroup\$ I would think you need some resistors on the outputs of those first three op-amps, otherwise they will fight each other. Also, they are effectively tied to ground through the inverting amplifier, so some series resistance is definitely in order. \$\endgroup\$
    – uint128_t
    Feb 2, 2016 at 16:48
  • \$\begingroup\$ Ok, I'll fix that! What about the tranferfunction question? \$\endgroup\$ Feb 2, 2016 at 17:21
  • \$\begingroup\$ Not sure, but once you have a working circuit, you can easily derive the transfer function. \$\endgroup\$
    – uint128_t
    Feb 2, 2016 at 17:24

1 Answer 1


You need 3 more resistors of value R3 to make this add the components equally. This is a textbook parallel PID controller. Assuming R1 = R2**

\$\tau_i\$ is R3*C1

\$\tau_d\$ is R4*C2

While you can certainly build this, it has a number of problems. The derivative will cause noise problems. Usually a filter has to be placed in front of the differentiator if you really need the derivative function (it's often not necessary and causes more problems than not).

You also have to consider saturation in each of the three terms as well as the output. As part of that consideration of saturation, if the integrator continues to integrate while the output is saturated you'll have integrator windup, which will usually cause a lot of over/undershoot, especially at start-up.

** If they are not equal, or if the summing resistors are different from R3 then it will effectively scale each of the three terms. You can probably see that that's (ideally) the same as scaling the time constants. In practice, you cannot use an integration capacitor 1/100 the size and use a large summing resistor because the op-amp will saturate before it can apply enough integral action to force the output over the whole range.

  • \$\begingroup\$ I've found a right schematic online, but what are in this schematic my varibles from my transferfunction? \$\endgroup\$ Feb 2, 2016 at 17:37
  • \$\begingroup\$ There are some scaling factors, which you can work out, as I mentioned above. The transfer function is modified by the resistor R4 which limits the derivative gain, to deal with the problem I mentioned. \$\endgroup\$ Feb 2, 2016 at 17:39
  • \$\begingroup\$ I dont now which resistors I've to choose with the same value \$\endgroup\$ Feb 2, 2016 at 17:42
  • 2
    \$\begingroup\$ You deleted your original circuit, which means that the part numbers I've given above make no sense to anyone reading it. \$\endgroup\$ Feb 2, 2016 at 19:05

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