0
\$\begingroup\$

I'm working through a solved example problem on compensation of a two-stage op amp. For extra context, this example problem is originally from an exam of a now-concluded course offering, so it is meant to be solve-able without simulation, numerical methods, etc. The link has all the relevant figures/details including bode plots but I've included a picture of the topology and given component values below.

The example problem (labeled problem 2.i in the attached images) presents a set of values for the amplifier including the Miller compensation capacitor \$C_C\$. The question asks for the minimum value of \$C_C\$ that gives a 45 degree phase margin in unity gain feedback. The given answer is 0.03 pF, or anything in the range of 0.01 pF to 0.1 pF. This is a smaller range than what I can come up with, reasoning below.

First approach:

One expression for the non-dominant pole location of this amplifier topology comes from the dominant-pole approximation. Note that although as seen in the attached plots the uncompensated pole locations are at the same frequency (1 Grad/s), for the given value of \$C_C\$ they are actually reasonably far apart, so this approximation seems justified. For a transfer function

$$H(s)=\frac{A(s)}{1+b_1 s + b_2 s^2}$$

where \$p_1, p_2\$ are not very close together, we have that

$$p_1 \approx\frac{-1}{b_1}=\frac{-1}{R_{o1}(C_C+C_1)+R_{o2}(C_C+C_2)+G_{m2}R_{o2}R_{o1}C_C}\approx\frac{-1}{G_{m2}R_{o2}R_{o1}C_C}$$

where the last approximation holds if the Miller effect dominates and the compensation capacitor is relatively large. We have for the nondominant pole

$$p_2\approx\frac{-b_1}{b_2}=\frac{1}{p_1 b_2}\approx\frac{-G_{m2}R_{o2}R_{o1}C_C}{R_{o1}R_{o2}(C_1C_2+C_C C_1 +C_C C_2)}=\frac{-G_{m2}C_C}{C_1 C_2 +C_C(C_1+C_2)}$$

For a 45 degree phase margin, the nondominant pole sits on top of the unity-gain crossover frequency, assuming that the dominant pole has contributed all 90 degrees of its phase shift. So the condition is

$$\omega_u=|p_2|=\frac{G_{m2}C_C}{C_1 C_2 +C_C(C_1+C_2)}$$

As \$C_C\$ decreases, \$p_1\$ increases, \$\omega_u\$ increases, and \$p_2\$ decreases. An approximate result for \$\omega_u\$ is \$\omega_u=|A_0||p_1|=\frac{G_{m1}}{C_C}\$. For the given values, this is \$10^{10}\$ rad/s, which matches with the plots. Plugging in and solving, we get \$C_C=0.24 pF\$, which is far outside the answer's given range (0.01 to 0.1 pF).

Second approach:

Assume a simpler relationship for \$p_2\$, so instead assume that \$p_2\$ is proportional to \$C_C\$, so we have \$p_2=p_{2,0}\frac{C_C}{C_{C,0}}\$, where the extra 0 subscript indicates the nominal values (\$p_{2,0}=10^{11}\$ rad/s, \$C_{C,0} = 1\$ pF). \$\omega_u\$ was already assumed inversely proportional to \$C_C\$. This leads to

$$C_C=\sqrt{\frac{\omega_{u,0}}{p_{2,0}}C_{C,0}^2}\approx 0.32 \text{ pF}$$

which is even further out of the previous range.

Third approach:

Assume that exactly one of \$\omega_u\$ or \$p_2\$ are constant with respect to \$C_C\$ and the other one is inversely or directly proportional to \$C_C\$, respectively, as in the second approach. This approach leads to

$$C_C=\frac{\omega_{u,0}}{p_{2,0}}C_{C,0}=\text{0.1 pF}$$

This is the top of the provided reference range of 0.01 to 0.1 pF. The reference solution uses a non-inclusive less-than so it seems that 0.1 pF may or may not have been considered close enough, but at least this is progress. However, I'm not sure how to derive anything closer to 0.01 pF or the "main" given answer of 0.03 pF.

Fourth approach:

This approach doesn't actually work at all but I'm including for completeness, as it's the only other thing I can think of besides running a simulation. We could assume that the unity-gain frequency is fixed at its nominal value and instead assume the more accurate expression for the nondominant pole, i.e.

$$p_2=\frac{-G_{m2}C_C}{C_1 C_2 +C_C(C_1+C_2)}$$

However this approach fails as solving yields \$C_C=-0.5 \text{ pF}\$ (this isn't because of the minus sign in the above expression; when plugging in and solving I'm actually using the magnitude of \$p_2\$ since \$p_2\$ is a pole of the s-domain transfer function).

Are there any other approaches I'm missing for trying to derive the given result?

component values amplifier

\$\endgroup\$

1 Answer 1

0
\$\begingroup\$

Yes. Build the thing with more than enough compensation to make it definately stable and then gradually reduce the compensation until you get 45 degrees phase margin.

\$\endgroup\$
5
  • \$\begingroup\$ Thanks for the response. I'm a bit confused what you mean since I think what I do in the different approaches I go through are solving for this minimum compensation? Additionally maybe I should have noted that this example question was originally given in an exam setting so simulation/other numerical solutions wouldn't have been possible. \$\endgroup\$
    – Halleff
    Oct 19, 2023 at 22:10
  • \$\begingroup\$ That's fine if you're limited to a mathamatical approach but it's sometimes good to remember, whilst we are getting very involved in theoretical calculations, that a practical real-world approach can be more effective at yielding realistic results. \$\endgroup\$
    – user350400
    Oct 20, 2023 at 0:32
  • \$\begingroup\$ That's fair. If you were going to attempt a hand calculation though, can you speak to which of the approaches I describe are most accurate / are there any errors? I think the most accurate one should be the first. \$\endgroup\$
    – Halleff
    Oct 20, 2023 at 0:43
  • \$\begingroup\$ I don't think that I could really give a helpful opinion on that. How much faith do you have in the given solution (0.01 - 0.1 pF) actually being correct? \$\endgroup\$
    – user350400
    Oct 20, 2023 at 3:23
  • 1
    \$\begingroup\$ Not terribly confident. I decided to check with Matlab since I might as well and I find that the minimum possible capacitance is about 0.175 pF: i.imgur.com/71JdhMt.png (x-axis is CC in pF, y-axis is degrees of phase margin). I have no idea how one would get something in the given range so it seems like most likely an error to me. \$\endgroup\$
    – Halleff
    Oct 20, 2023 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.