# How do I find the transfer function of a PID that uses a single op-amp?

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

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.

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: -

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).

• Wow - just love this superposition approach! Oct 26, 2023 at 12:13
• 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. Oct 26, 2023 at 12:16
• Superposition at its finest. Brilliant! Oct 26, 2023 at 12:25
• @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 Oct 26, 2023 at 13:11

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:

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.

• 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. :) Oct 26, 2023 at 13:58
• @periblepsis, oops, I did not see this comment and I added the missing neg. sign. Merci! : ) Jan 28 at 17:33
• Wow. Nice catch after so much time. And... well... I bought your new book, anyway! ;) Jan 28 at 20:40
• Nice catch for you actually : ) Hope you'll like the content, cheers! Jan 28 at 21:54

Simply apply the voltage gain formula of an operational amplifier in inverting configuration.