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How does one go about finding the transfer function of the following op-amp PID?

single op-amp PID design

I have tried to convert each component of the circuit into its transfer function representations, e.g., making each capacitor be \$\frac{1}{sC}\$. The resulting transfer function is overly complex, though, and not a simple 2nd-order ODE as is expected from what I have studied. From my research, I am supposed to find the constants \$K_i,\$ \$K_p\$ , and \$K_d\$. I don't know how to find them.

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I am supposed to find the constants \$K_i\$, \$K_p\$, and \$K_d\$. I don't know how to find them. Please help.

I would focus on looking at the circuit in three different ways like this: -

enter image description here

And solve each one for the individual constants \$K_i\$, \$K_p\$, and \$K_d\$. No need to evolve an overly complicated and fairly meaningless full transfer function (given the objective).

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  • \$\begingroup\$ Wow - just love this superposition approach! \$\endgroup\$ Oct 26, 2023 at 12:13
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    \$\begingroup\$ Hello Andy, this is a neat way of showing how the various elements combine to form the transfer function. Just for nit-picking a bit, I would say that the differentiation block is a filtered differentiation because you certainly introduce a zero but also a pole, and the two are linked by a ratio \$N\$ in the filtered PID expression that the op ignored in his question I think. \$\endgroup\$ Oct 26, 2023 at 12:16
  • \$\begingroup\$ Superposition at its finest. Brilliant! \$\endgroup\$ Oct 26, 2023 at 12:25
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    \$\begingroup\$ @VerbalKint I actually didn't say it was a differentiator: I was careful to show the components that were valid and left it to the reader to figure out that it wasn't a true differentiator \$\endgroup\$
    – Andy aka
    Oct 26, 2023 at 13:11
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The circuit you've shown in not a PID per se but a filtered PID which implies an extra pole added to break the gain at high frequency and improve noise susceptibility. The formula given by Franc is correct but unusable - no offense meant of course : ) for a so-called design-oriented analysis or D-OA. In other terms, this expression is in a high-entropy form and does not lend itself well to designing the compensator by determining poles and zeroes.

Using the fast analytical circuits techniques or FACTs, leads you to a different solution owing to the fact that two zeroes can immediately be inferred from the circuit: if the impedances made of \$R_D\$ paralleled with \$C_D\$ and the one made of the series connection of \$C_1\$ and \$R_1\$ respectively approaches infinity and becomes a transformed short circuit, then you have a null in the output, indicating the presence of two zeroes immediately factored: \$N(s)=(1+\frac{s}{\omega_{z1}})(1+\frac{s}{\omega_{z2}})\$. The complete low-entropy expression is shown below and features an inverted zero for the most compact notation:

enter image description here

You will also find the coefficients values for the filtered PID and the derivation is shown in my APEC 2012 seminar. Please note that I derived a non-inverting TF while the op-amp-based PID obviously brings a "-" sign in front of the expression.

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  • \$\begingroup\$ If I negate your H0 then I get to the same place. And while it is interesting it also seems almost a necessary approach to this problem. :) \$\endgroup\$ Oct 26, 2023 at 13:58
  • \$\begingroup\$ @periblepsis, oops, I did not see this comment and I added the missing neg. sign. Merci! : ) \$\endgroup\$ Jan 28 at 17:33
  • \$\begingroup\$ Wow. Nice catch after so much time. And... well... I bought your new book, anyway! ;) \$\endgroup\$ Jan 28 at 20:40
  • \$\begingroup\$ Nice catch for you actually : ) Hope you'll like the content, cheers! \$\endgroup\$ Jan 28 at 21:54
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Simply apply the voltage gain formula of an operational amplifier in inverting configuration.

enter image description here

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