Question 2 of 5: Series–Shunt Feedback Amplifier — Feedback Factor, Design Ratio and Loop Gain

Question 2: Series–Shunt Feedback Amplifier — Feedback Factor, Design Ratio and Loop Gain (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-stage amplifier: $M_1$ is a common-source stage loaded by $R_{D1}$ whose drain drives the gate of $M_2$; $M_2$ is a common-source stage loaded by $R_{D2}$ whose drain is the output $v_O$. A resistive ladder returns from $v_O$ through $R_2$ to the source of $M_1$, and $R_1$ runs from that source node to ground. The signal $v_S$ (on a $V_{GG}$ pedestal) drives the gate of $M_1$. For part (d), $g_{m1}=g_{m2}=4$ mA/V and $R_{D1}=R_{D2}=10$ kΩ; $r_{o1}$ and $r_{o2}$ are to be neglected throughout.

Find. The feedback network and its $\beta$, the ideal closed-loop gain, the ratio $R_2/R_1$ (and $R_2$) for an ideal gain of 10 V/V, an expression for the loop gain $A\beta$, and the numerical values of $A$, $A\beta$ and $A_f$.

Approach. Recognise the topology as voltage-series (series–shunt), lift the $R_1$–$R_2$ ladder out as a two-port to read off $\beta$ and the loading it imposes, then evaluate the loaded open-loop gain stage by stage and close the loop with $A_f = A/(1+A\beta)$.

1. Part (a) — identify the topology and the feedback network. The ladder $R_2$–$R_1$ connects across the output node and returns to the source of $M_1$, not to its gate. Because it bridges the output node it samples the output voltage (shunt sampling), and because the returned voltage appears in series with the gate–source loop of $M_1$ it subtracts from $v_S$ in series (series mixing). That is voltage-series, i.e. series–shunt feedback: the ideal voltage amplifier, whose input resistance is raised and output resistance lowered by the loop. The $\beta$ circuit is therefore just the two resistors.
2. Write the feedback factor. The ladder is driven by $v_O$ and, with the gate current of a MOSFET zero, is unloaded at its own output, so it is a plain voltage divider: $$\beta=\frac{v_f}{v_O}=\frac{R_1}{R_1+R_2}$$ with $v_f$ the voltage fed back to the source of $M_1$. Note that $\beta$ here is dimensionless because both the sampled and the mixed quantities are voltages — a property of this topology alone.
3. State the ideal or upper-bound closed-loop gain. When the loop gain is large, $A_f=A/(1+A\beta)\to 1/\beta$, so $$A_f\big|_{\text{ideal}}=\frac{1}{\beta}=1+\frac{R_2}{R_1}$$ This is an upper bound and not an equality: a real, finite $A$ always leaves $A_f$ a little below $1/\beta$. It is also why the closed-loop gain is set by a resistor ratio, which matches and tracks on an integrated circuit far better than any device parameter.
4. Part (b) — solve the design ratio. Setting the ideal gain to 10 V/V, $$1+\frac{R_2}{R_1}=10\quad\Longrightarrow\quad \boxed{\frac{R_2}{R_1}=9}$$ and with the stated $R_1=1$ kΩ this gives $\boxed{R_2 = 9\ \text{k}\Omega}$. Both values are carried into part (d).
5. Part (c) — load the amplifier with the feedback network before writing $A$. The $\beta$ network is not weightless: in the two-port method its loading is applied to the basic amplifier before the loop is closed. Looking into the ladder from the source of $M_1$ with the output shorted gives $R_1\parallel R_2$ in series with that source; looking in from the output node with the input port open (series mixing draws no current) gives $R_1+R_2$ shunting the drain of $M_2$. Hence the loaded open-loop gain $$A=\underbrace{\frac{-g_{m1}R_{D1}}{1+g_{m1}(R_1\parallel R_2)}}_{\text{stage 1, degenerated}}\times\underbrace{\Big[-g_{m2}\big(R_{D2}\parallel(R_1+R_2)\big)\Big]} _{\text{stage 2, loaded}}$$
6. Write the loop gain. Multiplying by the feedback factor gives the expression the question asks for: $$A\beta=\frac{g_{m1}g_{m2}\,R_{D1}\big(R_{D2}\parallel(R_1+R_2)\big)} {1+g_{m1}(R_1\parallel R_2)}\cdot\frac{R_1}{R_1+R_2}$$ Two inversions make $A$ positive, which is what a stable negative-feedback loop requires with this (subtracting) mixing arrangement — a sign check worth doing before substituting numbers.
7. Part (d) — evaluate the loading terms. With $R_1=1$ kΩ and $R_2=9$ kΩ, $$\begin{aligned}R_1\parallel R_2&=\frac{(1)(9)}{1+9}=0.9\ \text{k}\Omega\\ R_1+R_2&=10\ \text{k}\Omega\\ R_{D2}\parallel(R_1+R_2)&=10\parallel10=5\ \text{k}\Omega\end{aligned}$$ The degeneration term is the one that does the damage: $g_{m1}(R_1\parallel R_2) =(4\times10^{-3})(900)=3.6$, so the first stage keeps only $1/4.6$ of its undegenerated gain.
8. Evaluate the two stage gains and the open-loop gain. Substituting, $$\begin{aligned}A_1&=\frac{-(4\times10^{-3})(10\times10^{3})}{1+3.6} =\frac{-40}{4.6}=-8.696\\ A_2&=-(4\times10^{-3})(5\times10^{3})=-20.0\end{aligned}$$ so that $$\boxed{A=A_1A_2=(-8.696)(-20.0)=173.9\ \text{V/V}}$$ For comparison, ignoring the feedback network’s loading altogether would give $g_{m1}R_{D1}\cdot g_{m2}R_{D2}=1600$ — nine times too large, and the single most common error on this question.
9. Close the loop. With $\beta=1/10=0.1$, $$A\beta=(173.9)(0.1)=\boxed{17.39},\qquad A_f=\frac{A}{1+A\beta}=\frac{173.9}{18.39}=\boxed{9.456\ \text{V/V}}$$ The amount of feedback is $1+A\beta=18.39$, i.e. 25.3 dB of desensitivity.
10. Check the result against the design target and against an exact solve. The realised 9.456 V/V sits 5.4 per cent below the ideal 10 V/V, which is exactly what a loop gain of only 17.4 predicts — the fractional shortfall is $1/(1+A\beta)=5.4$ per cent, so the two numbers confirm one another. Solving the three-node small-signal circuit directly (source of $M_1$, drain of $M_1$, output), with no feedback formalism at all, returns $v_O/v_S=9.458$, confirming the two-port bookkeeping to four figures.