22-Elec-B5 Advanced Electronics · December 2018

Question 3 of 5: Miller Compensation of a Two-Pole Op Amp

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 3: Miller Compensation of a Two-Pole Op Amp (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A two-node model of a transconductance stage: an input node of resistance $R_1$ shunted by $C_1$, an output node of resistance $R_L$ shunted by $C_L$, and a compensation capacitor $C_C$ bridging the two. The two open-loop pole frequencies and the 100 dB d.c. gain read off the supplied Bode plot fix the two node resistances.

Find. A value of $C_C$ that makes the amplifier stable in unity-gain feedback, the reasoning behind it, the two new pole frequencies, and the resulting phase margin.

Approach. Recover the node resistances from the printed poles, write the exact two-pole denominator with $C_C$ present, choose $C_C$ from an explicit stability criterion rather than from the dominant pole alone, and then read the new poles and the phase margin off that same denominator.

1. Recover the node resistances. Each uncompensated pole is the $RC$ product of its own node, so $$\begin{aligned}R_1 &= \frac{1}{2\pi f_{P1} C_1} = \frac{1}{2\pi (10^5)(100\ \text{pF})} = 15.92\ \text{k}\Omega \\ R_L &= \frac{1}{2\pi f_{P2} C_L} = \frac{1}{2\pi (10^6)(5\ \text{pF})} = 31.83\ \text{k}\Omega.\end{aligned}$$ The output stage alone therefore contributes $g_m R_L = 1273$, or 62.1 dB, of the 100 dB total — a reminder that $A_0$ must be read from the plot and not computed from $g_m R_L$.
2. Show why compensation is needed at all. With $A_0 = 10^{100/20} = 10^5$ and both poles far below crossover, the loop crosses unity on the −40 dB/decade segment at $$f_{t,\text{unc}} = \sqrt{A_0 f_{P1} f_{P2}} = \sqrt{(10^5)(10^5)(10^6)} = 100\ \text{MHz},$$ where the phase lag is $180^\circ - \arctan(f_t/f_{P1}) - \arctan(f_t/f_{P2}) = 0.6^\circ$ of margin. In unity-gain feedback the amplifier is on the edge of oscillation, which is exactly what the supplied plot shows for the low-$1/\beta$ (50 dB) case.
3. Write the exact denominator with $C_C$ in place. Two node equations with the bridging capacitor give $V_o/I_s$ a denominator $D(s) = 1 + b_1 s + b_2 s^2$ with $$\begin{aligned}b_1 &= C_1R_1 + C_LR_L + C_C\left(g_m R_1 R_L + R_1 + R_L\right) \\ b_2 &= R_1R_L\left[C_1C_L + C_C(C_1+C_L)\right].\end{aligned}$$ Because the roots are real and widely separated, the new poles are $f'_{P1} = 1/(2\pi b_1)$ and $f'_{P2} = b_1/(2\pi b_2)$. The term $g_m R_1 R_L C_C$ dominates $b_1$: that is Miller multiplication pulling the first pole down, while $b_2$ grows only linearly in $C_C$, which is what pushes the second pole up. The pair of movements is what "pole splitting" means.
4. State the stability criterion before choosing a value. This is a design question with one free parameter, so the criterion must be declared. Placing the unity-gain crossover at half the second pole, $$f_t = A_0 f'_{P1} = \tfrac{1}{2} f'_{P2},$$ buys a phase margin near 63° — a well-damped, non-peaking closed-loop response. (Setting $f_t = f'_{P2}$ instead would leave only 45° and a visibly peaked step response.) Substituting the two expressions above reduces the criterion to the single equation $b_1^{\,2} = 2A_0 b_2$, one quadratic in $C_C$, whose positive root is 29.7 pF.
5. Choose a standard value and evaluate the new poles. Rounding up to the nearest convenient value, take $$\boxed{C_C = 30\ \text{pF}}$$ which puts $b_1 = 611\ \mu\text{s}$ and $b_2 = 1.85 \times 10^{-12}\ \text{s}^2$, so $$\boxed{\begin{aligned}f'_{P1} &= 260\ \text{Hz} \\ f'_{P2} &= 52.6\ \text{MHz}\end{aligned}}$$ The first pole has moved down by a factor of 384 and the second up by a factor of 53 from their uncompensated positions.
6. Find the crossover and the phase margin. With a single dominant pole the gain falls at 20 dB/decade to $f_t = A_0 f'_{P1} = 10^5 \times 260 = 26.0$ MHz, and $$\text{PM} = 180^\circ - \arctan\!\frac{f_t}{f'_{P1}} - \arctan\!\frac{f_t}{f'_{P2}} = 180^\circ - 90.0^\circ - 26.3^\circ = \boxed{63.7^\circ}$$ Comfortably stable, and by construction: the second pole sits one octave above crossover, so it can only remove about 26°.
7. Charge the right-half-plane zero. The feed-forward path through $C_C$ puts a zero at $f_z = g_m/(2\pi C_C) = 212$ MHz. It adds gain but subtracts phase, so the honest margin is $$\text{PM}_z = 63.7^\circ - \arctan\!\frac{f_t}{f_z} = 63.7^\circ - 7.0^\circ = 56.7^\circ,$$ still well damped. The zero sits eight times above crossover here only because $g_m$ is large; in a low-power design it would be the limiting factor and would need a nulling resistor in series with $C_C$.
8. Check the result against the physical ceiling, and against the naive design. No amount of $C_C$ can push the second pole past $g_m/[2\pi(C_1+C_L)] = 60.6$ MHz, and 52.6 MHz respects that. More instructively, a designer who assumes the second pole stays at 1 MHz and lowers the crossover to 0.5 MHz by Miller multiplication alone would specify $$C_C = \frac{1}{2\pi f'_{P1}R_1(1+g_mR_L)} - \frac{C_1}{1+g_mR_L} \approx 1570\ \text{pF},$$ 52 times larger, for a closed-loop bandwidth of 0.5 MHz instead of 26 MHz. Recognising that the second pole moves is the whole point of the technique.