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

Question 2: Series–Shunt Feedback Amplifier — Feedback Factor, Loop Gain and Closed-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 MOSFET amplifier in which the signal enters the gate of $M_1$, the drain of $M_1$ drives the gate of $M_2$, and the output at the drain of $M_2$ is returned through $R_2$ to the source of $M_1$, where $R_1$ completes the path to ground:

Find. the identity of the feedback network and expressions for $\beta$, for the ideal closed-loop gain and for the loop gain $A\beta$; the divider ratio that sets an ideal gain of 10 V/V; and the numerical values of $A\beta$, $A$ and $A_f$ for the stated device data.

Approach. Identify the sampling and mixing topology, read $\beta$ off the feedback divider, load the basic amplifier with the two-port terminations of that divider to build the $A$ circuit, then combine $A$ and $\beta$ in the standard feedback expressions.

1. Part (a) — identify the topology and the $\beta$ network. The output voltage $v_O$ is sampled by connecting $R_2$ directly across the output node (shunt sampling), and the sampled signal is returned as a voltage in series with the gate–source loop of $M_1$ by lifting its source above ground (series mixing). The feedback network is therefore nothing more than the resistive divider formed by $R_2$ and $R_1$, tapped at the source of $M_1$, and it delivers$$\beta = \frac{v_f}{v_O} = \frac{R_1}{R_1+R_2}$$Because the loop is voltage-series it stabilises a voltage gain, raises the input resistance and lowers the output resistance.
2. State the ideal closed-loop gain. When the loop gain is large, $A_f = A/(1+A\beta) \to 1/\beta$, so the upper bound the topology can approach is$$A_f\big|_{\text{ideal}} = \frac{1}{\beta} = \boxed{1 + \frac{R_2}{R_1}}$$which is the familiar non-inverting-amplifier form. It depends only on a resistor ratio, which is precisely the benefit the feedback is bought for.
3. Part (b) — choose the divider ratio. Setting the ideal gain to 10 V/V,$$1 + \frac{R_2}{R_1} = 10 \;\Longrightarrow\; \frac{R_2}{R_1} = 9$$and with $R_1 = 1\ \text{k}\Omega$ this gives $\boxed{R_2 = 9\ \text{k}\Omega}$, so $\beta = 0.1$.
4. Part (c) — build the $A$ circuit by loading with the feedback two-port. The divider must be accounted for twice. Looking into the divider from the mixing point with the output shorted gives a resistance $R_1\parallel R_2$, which appears in the source of $M_1$ and degenerates the first stage; looking in from the output with the input loop opened gives $R_1+R_2$, which appears in shunt with $R_{D2}$. The first stage is then a degenerated common-source stage and the second a plain common-source stage, so$$A = \frac{v_O}{v_i} = \left[\frac{g_{m1}R_{D1}}{1+g_{m1}(R_1\parallel R_2)}\right]\Big[g_{m2}\left(R_{D2}\parallel (R_1+R_2)\right)\Big]$$(the two inversions cancel, so $A$ is positive, as it must be for negative feedback).
5. Write the loop gain. Multiplying by the feedback factor,$$A\beta = \left[\frac{g_{m1}R_{D1}}{1+g_{m1}(R_1\parallel R_2)}\right]\Big[g_{m2}\left(R_{D2}\parallel (R_1+R_2)\right)\Big]\frac{R_1}{R_1+R_2}$$and the closed-loop gain follows as $A_f = A/(1+A\beta)$.
6. Part (d) — insert the numbers, starting with the loading resistances. With $R_1 = 1\ \text{k}\Omega$ and $R_2 = 9\ \text{k}\Omega$,$$R_1\parallel R_2 = 0.9\ \text{k}\Omega, \qquad R_1+R_2 = 10\ \text{k}\Omega$$so the second stage sees $R_{D2}\parallel(R_1+R_2) = 10\parallel 10 = 5\ \text{k}\Omega$.
7. Evaluate the two stage gains. The degeneration of the first stage is substantial, $g_{m1}(R_1\parallel R_2) = (4\ \text{mA/V})(0.9\ \text{k}\Omega) = 3.6$:$$|A_1| = \frac{(4)(10)}{1+3.6} = 8.70, \qquad |A_2| = (4\ \text{mA/V})(5\ \text{k}\Omega) = 20.0$$Multiplying, $A = 8.70 \times 20.0 = 173.9$ V/V.
8. Close the loop. With $\beta = 0.1$,$$A\beta = (173.9)(0.1) = 17.39, \qquad A_f = \frac{A}{1+A\beta} = \frac{173.9}{18.39} = \boxed{9.46\ \text{V/V}}$$The amount of feedback is $1+A\beta = 18.4$, so the realised gain falls 5.4 per cent short of the ideal 10 V/V — the unavoidable gain error of a loop whose loop gain is only about 17.
9. Cross-check against a direct nodal solution. Solving the three unknown node voltages of the actual circuit (source of $M_1$, drain of $M_1$, output) without invoking feedback theory at all returns $v_O/v_S = 9.458$ V/V, within 0.02 per cent of the value obtained above. That agreement is the evidence that the two-port loading has been applied correctly, since omitting either termination shifts $A$ by tens of per cent.