Question 4 of 6

Question 4 (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.

Approach. Identify the mixing and sampling, find $\beta$ and the loading of the feedback network, compute the open-loop gain $A$ of the loaded basic amplifier, then close the loop with $A_f=A/(1+A\beta)$. Part (b) is a loop-polarity argument.

1. Identify the topology. The feedback voltage $v_f$ at $M_2$'s gate is subtracted from $v_S$ by the differential pair, so the signals are compared as voltages in series (series mixing). The $R_6$–$R_7$ divider is connected directly across the output, so it samples $v_O$ (shunt sampling). The circuit is a series–shunt (voltage-amplifier) feedback amplifier.
2. Feedback factor and loading. $M_2$'s gate draws no current, so $$\beta=\frac{v_f}{v_O}=\frac{R_7}{R_6+R_7}$$ Loading at the input side (output shorted) is $R_6\parallel R_7$ in series with $M_2$'s gate, which carries no current, so it has no effect. Loading at the output side (feedback input port opened) is $R_6+R_7$ across $v_O$, so the basic amplifier sees $$R_L'=R_5\parallel(R_6+R_7)$$
3. Open-loop gain $A$: differential pair. With the feedback removed ($M_2$'s gate at signal ground), $v_S$ divides equally across the two gate-source junctions because the tail is an ideal current source: $i_{d1}=g_m v_S/2$. The load on $M_1$'s drain is $R_1$ in parallel with $Q_3$'s input resistance at the base, which is finite because $\beta_3\neq\infty$ and the emitter is degenerated: $$R_{in3}=r_{\pi3}+(\beta_3+1)R_3,\qquad v_{d1}=-\frac{g_m}{2}\,(R_1\parallel R_{in3})\,v_S$$
4. Open-loop gain $A$: PNP common-emitter stage. A fall in $v_{d1}$ raises $Q_3$'s emitter-base drive, so the base current is $i_{b3}=-v_{d1}/R_{in3}$ and the collector current delivered to the output is $i_o=\beta_3 i_{b3}$. Hence $v_O=i_oR_L'=-\beta_3R_L'\,v_{d1}/R_{in3}$ and $$A=\frac{v_O}{v_S}\bigg|_{\text{open}}=\frac{g_m}{2}\,(R_1\parallel R_{in3})\,\frac{\beta_3R_L'}{r_{\pi3}+(\beta_3+1)R_3}$$ $A$ is positive: there are two inversions ($v_S\to v_{d1}$ and $v_{d1}\to v_O$).
5. Close the loop. $$\boxed{\frac{v_O}{v_S}=A_f=\frac{A}{1+A\beta}}$$ with $A$ and $\beta$ as above. For $A\beta\gg1$ this becomes the precise, transistor-independent gain $$A_f\approx\frac{1}{\beta}=1+\frac{R_6}{R_7}$$ Because the ideal tail makes $i_{d1}=g_m(v_S-v_f)/2$ exactly and $M_2$'s gate draws no current, this feedback result is exact for the stated assumptions: a full nodal solve (six nodes, including $Q_3$'s base current loading $R_1$) gives the same expression. $R_2$ does not appear, since $M_2$'s drain drives nothing.
6. (b) Why $M_1$'s drain. The connection sets the sign of the feedback. Break the loop and raise $M_2$'s gate by $v_f$: the tail current shifts from $M_1$ to $M_2$, so $v_{d1}$ rises by $(g_m/2)(R_1\parallel R_{in3})v_f$ while $v_{d2}$ falls by $(g_m/2)R_2v_f$. The PNP stage inverts, so a rise at $Q_3$'s base lowers $i_o$ and $v_O$, which lowers $v_f$. With $Q_3$ on $M_1$'s drain, the round trip therefore opposes the original change: this is negative feedback, with loop gain $A\beta>0$ in the $A/(1+A\beta)$ form. If $Q_3$ were driven from $M_2$'s drain, the extra inversion would make the round trip reinforce the change. That is positive (regenerative) feedback: $1+A\beta$ could pass through zero, and the circuit would latch against a supply rail or oscillate instead of giving a precise gain. $M_2$'s gate is the inverting input of the whole amplifier only when $Q_3$ hangs on $M_1$'s drain. (Swapping to $M_2$'s drain would work only if $v_S$ and the feedback were also swapped between the two gates.)

As an illustrative magnitude check (values not given in the paper), take $g_m=2\,\text{mA/V}$, $R_1=10\,\text{k}\Omega$, $R_3=1\,\text{k}\Omega$, $r_{\pi3}=2.5\,\text{k}\Omega$, $\beta_3=100$, $R_5=2\,\text{k}\Omega$, $R_6=9\,\text{k}\Omega$, $R_7=1\,\text{k}\Omega$. Then $A\approx14.7$, $A\beta\approx1.47$ and $A_f\approx5.95$, compared with the ideal $1/\beta=10$. The degenerated PNP stage and the loading of $R_1$ by $R_{in3}$ keep $A$ modest, so the closed-loop gain approaches $1/\beta$ only as the loop gain grows.