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Razavi gives the answer of why the compensation capacitor should not be placed between the gate and drain of M3 transistor.

Below are the quotations of his question and answer:

Question:

Explain why, in the circuit of Fig. 10.76, the compensation capacitor should not be placed between the gate and the drain of M2 or M3.

Answer:

Putting Cc "across" M3 only affects the dominant pole and we cannot take advantage of pole-splitting to widen the bandwidth.

If I remember correctly, shouldn't pole-splitting narrow the bandwidth instead of widening it, as the following answer suggests?

Regarding Miller compensation

Answer

Here is the picture of Fig.10.76

Fig.10.76

After breaking the loop, we will eventually arrive at the following equivalent circuit. Let's call it Figure A.

Figure A

After certain calculation, we will get the pole at node A,B and Y.

$$\omega_A=\frac{1}{C_Ar_{o1}}=1.24\times10^9 \frac{rad}{s}\tag{1}$$ $$\omega_B=\frac{1}{C_Br_{o3}}=1.30\times10^7 \frac{rad}{s}\tag{2}$$ $$\omega_Y=\frac{1}{C_Yr_{o2}}=1.89\times10^9 \frac{rad}{s}\tag{3}$$

Thus,without compensation, \$\omega_B\$ should be the dominant pole. \$\omega_A\$ should be the 2nd pole, and \$\omega_Y\$ should be the 3rd pole.

Here is his entire explanation:

For Problem 10.14, Cc should not be placed "across" M2 or M3 because of the location of the poles, since the dominant pole was at node B, the second pole at node A, and the third pole at node Y, we need to split the first two poles by placing Cc across M1.

Putting Cc "across" M2 only splits the 2nd and 3rd pole keeping the dominant pole unchanged, It moves he 2nd pole toward the dominant pole and 3rd pole away, that cannot give a 60 degree P.M.

Putting Cc "across" M3 only affects the dominant pole and we cannot take advantage of pole-splitting to widen the bandwidth.

Reference: Razavi, Design of Analog CMOS Integrated Circuits

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    \$\begingroup\$ You seem to be misreading the linked answer; the response is in regards to Miller compensation, or compensation in general, where we choose an op-amp of additional bandwidth, then reduce it down to complement the plant's response. The pole-splitting question, it seems, was overlooked and not addressed. At least, that's my reading of it. \$\endgroup\$ Commented Feb 27 at 19:52

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I cannot see how pole-splitting could ever extend the bandwidth. The whole idea of the technique is to alter the sum of the poles, but not its product (otherwise we'd be increasing the order of the system) such that both dominant pole and the 2nd pole become further away from each other to obtain more phase margin (2nd pole must be sufficiently below the 0dB line).

Since the dominant pole is shifted to a lower frequency, that means the UGF, must be, inevitably, reduced as well.

So I'd say that statement is wrong. It's also not clear to me that Razavi wrote or verified those solutions you have found. Beware.

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