Question 4 of 6: Series-Shunt (Voltage-Series) Feedback Amplifier

Question 4: Series-Shunt (Voltage-Series) Feedback Amplifier (15 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: $M_1$/$M_2$ form an NMOS differential pair (tail current $I_{bias}$, ideal) with resistive drain loads $R_1$, $R_2$; $M_1$'s gate is driven by $v_s$, $M_1$'s drain drives $M_3$'s gate (a PMOS common-source second stage whose drain is $v_o$, loaded by $R_3\Vert(R_4+R_5)$); the output is sampled by the $R_4$-$R_5$ divider and fed back to $M_2$'s gate. $g_{m1}=g_{m2}\equiv g_{m,d}$ (matched pair), $g_{m3}$ independent, all $r_o=\infty$.

Find. The closed-loop expression $v_o/v_s$, using feedback-amplifier analysis.

Approach (a). Identify the feedback topology (voltage sampled at the output, series-mixed with $v_s$ at the input via the differential pair), find the open-loop gain $A$ from $v_s-v_f$ to $v_o$ with the feedback network's loading already included, find the feedback factor $\beta=v_f/v_o$, then assemble the standard closed-loop result $A/(1+A\beta)$.

1. Classify the feedback. The output voltage $v_o$ is sensed directly by the $R_4$-$R_5$ divider (voltage/shunt sampling at the output), and the divided voltage is applied to $M_2$'s gate, compared against $v_s$ at $M_1$'s gate through the differential pair (series mixing/differencing at the input, since MOSFET gates draw no current). This is series-shunt (voltage-series) feedback, whose closed-loop form is $A_f=A/(1+A\beta)$.
2. Differential-pair output at $M_1$'s drain. With an ideal tail source, $i_{d1}=-i_{d2}$ regardless of $R_1,R_2$; solving the pair's KCL with $g_{m1}=g_{m2}=g_{m,d}$ and input difference $v_{id}=v_s-v_f$ gives the standard result $i_{d1}=g_{m,d}v_{id}/2$, so at $M_1$'s drain (node driving $M_3$'s gate), $$v_x=-i_{d1}R_1=-\frac{g_{m,d}R_1}{2}(v_s-v_f).$$
3. Second stage, $M_3$ common source. $M_3$'s source sits at $V_{DD}$ (AC ground) and its drain is $v_o$, loaded by $R_3\Vert(R_4+R_5)$ (the feedback network appears as $R_4+R_5$ in series, since $M_2$'s gate draws no current): $$v_o=-g_{m3}\big[R_3\Vert(R_4+R_5)\big]\,v_x.$$
4. Open-loop gain $A$ (from $v_{id}=v_s-v_f$ to $v_o$). Combining the two stages (the two minus signs cancel, giving a non-inverting overall relation, as required for negative feedback to close correctly): $$v_o=A(v_s-v_f),\qquad \boxed{A=\dfrac{g_{m1}R_1\cdot g_{m3}\big[R_3\Vert(R_4+R_5)\big]}{2}}.$$
5. Feedback factor. $M_2$'s gate draws no current, so $R_4$-$R_5$ is an unloaded divider: $$\beta=\frac{v_f}{v_o}=\frac{R_5}{R_4+R_5}.$$
6. Close the loop. $v_o=A(v_s-v_f)=A\,v_s-A\beta v_o\ \Rightarrow\ v_o(1+A\beta)=Av_s$: $$\boxed{\dfrac{v_o}{v_s}=\dfrac{A}{1+A\beta}\ \xrightarrow{A\beta\gg1}\ \dfrac{1}{\beta}=\dfrac{R_4+R_5}{R_5}},$$ the "precise gain" the question asks for — independent of $g_{m1},g_{m3}$ once the loop gain is large, which is exactly why negative feedback is used here.