# What is this RC local feedback configuration?

I'm studying power converter control loops using Christophe Basso's book Designing Control Loops for Linear and Switching Power Supplies.

A very common pattern in the compensator circuits is a resistor in series with a capacitor ($R_2$ and $C_1$ below, please disregard $C_2$) providing local feedback from an op amp's output to it's inverting input: I'm having trouble understanding specifically how this affects the transfer function (like which R's and which C's produce time constants that add a pole or zero) and I've yet to find a place where it's actually spelled out. It seems like one of those things that folks figure is obvious to the reader and never explicitly describe :)

It doesn't match any op amp amplifier circuits I've seen, although there is an augmenting integrator described on p59 of the TI Handbook of Operational Amplifier Applications2 that is quite similar except that the position of R2 and C1 are reversed. From piecing together odds and ends on some application notes and what not, I understand this adds a pole and a zero to the transfer function. But I'd really like to be able to derive that for myself, perhaps helped along by some examples and more description.

Does this configuration have a name that I could search on to learn more? Or is it perhaps easily explained?

• Don't know what your background is, but are you familiar with the s domain ? – efox29 Oct 5 '15 at 8:50
• Yes, Laplace is an old buddy of mine :) – scanny Oct 5 '15 at 11:12

The transfer function of your circuit in standard form (without $C_2$) is

$$G(s) = \frac {1 + sR_2C_1}{sR_1C_1}$$

Visually looking at it, we see that there is one zero and one pole.

The zero being $\omega_z = \frac {1}{R_2C_1}$ and the pole being $\omega_p = \frac {1}{R_1C_1}$

We can also see that as frequency increases, the gain asymptotically approaches $\frac{R_2}{R_1}$ since they become the dominant terms.

This can be shown by taking the limit of G(s). $$G(s) = \frac {1 + sR_2C_1}{sR_1C_1}$$ $$G(s) = \frac{1}{sR_1C_1} + \frac{sR_2C_1}{sR_1C_1}$$ $$\require{cancel}\lim_{s\to \infty} G(s) =\cancel{ \frac{1}{{s}R_1C_1}} + \frac{\cancel{s}R_2\cancel{C_1}}{\cancel{s}R_1\cancel{C_1}}$$

So now you can add a zero or pole by simply adjusting the the Rs and the Cs. This is why standard form is important, is because it makes everything immediately clear.

These types of topologies are used to shape the control loop to make the loop stable (adding phase margin, gain margin).

Google up some of these topologies for more information Type I, Type II and (you guessed it) Type III compensators.

• Thanks for this efox29; +1 on the standard form, I definitely find that a worthwhile extra step. I'm not sure about the gain reducing by $R_2/R_1$ though. At the pole it would decrease by 20dB/decade and at the zero it would increase by 20dB/decade. The only place it would decrease by $R_2/R_1$ would be some arbitrary spot during the transitions, if at all, wouldn't it? – scanny Oct 6 '15 at 20:37
• @scanny added into body to show that gain breaks down to r2/r1 at high frequencies. – efox29 Oct 7 '15 at 14:12
• Ah, ok, now I see it! So this is the asymptotic value of the gain, after the capacitor goes to a short, so to speak, and the pole and zero have had their day, turning it into a plain old inverting amplifier. Got it now, thanks @efox29 :) – scanny Oct 7 '15 at 17:59

It doesn't match any op amp amplifier circuits I've seen

In contrary, it is one of the well-known controller types applied in control systems: Proportional-Integral-Controller (PI). Here is the transfer function:

$$H(s)=-\frac{R_2+\frac{1}{sC_1}}{R}=\frac{R_2}{R}+\frac{1}{sRC_1}$$

Please note that the resistor R is the resistance which is effective for feedback (here: $R=R_1 \parallel R_{\text{lower}}$). More than that, it is important to know that the circuit works NOT as a stand-alone circuit because there is no DC feedback. However, when used as part of an overall negative feedback loop the controller has a stable bias point - provided $V_{\text{ref}}$ is matched to the voltage division produced by $$\frac{R_{\text{lower}}}{R_1+R_{\text{lower}}}$$

• Ah, okay, I think this gives me what I need. Your analysis helped me identify the mistake in my math that was leading to confusing results. One point though, I'm thinking that $R_{lower}$ is shorted by the virtual ground during AC analysis, so $R = R_1$ rather than $R_1 \parallel R_{lower}$. Do you concur? – scanny Oct 5 '15 at 16:18
• Scanny, oh yes - I think you are right. For ac the source Vref presents a short. Thank you. – LvW Oct 5 '15 at 16:57