Question 5 of 5: Shunt feedback around a common-source stage — input and output resistance

Question 5: Shunt feedback around a common-source stage — input and output resistance (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.

Find. R_IN looking right from the v_IN terminal and R_OUT looking left into the v_OUT terminal, first with the feedback branch removed and then with R₁ = 100 kΩ closing the loop from drain to gate.

Approach. Recover g_m from the bias, evaluate the two open-loop resistances by inspection, then close the loop: because C₁ is a short, R₁ bridges the drain and the gate, so R_IN follows from Miller's theorem at the input and R_OUT from a test source at the output.

1. Recover the operating point. The tail source sets $I_D=I_{bias}=1$ mA, so from the square law $$V_{OV}=\sqrt{\frac{2I_{bias}}{K}}=\sqrt{\frac{2(1)}{1}}=1.414\ \text{V}, \qquad g_m=KV_{OV}=1.414\ \text{mA/V},$$ and $\lambda = 0$ makes $r_o$ infinite, so the drain sees only what is connected to it externally. C_S = ∞ places the source terminal at signal ground, so the forward path is a plain common-source stage with $v_{gs}=v_{gate}$.
2. Part (a) — input resistance with the loop open. With R₁ = ∞ the only element between the v_IN node and ground is R₂, since the gate itself draws no current: $$R_{IN}\big|_{R_1=\infty}=R_2=\boxed{20\ \text{k}\Omega}.$$
3. Part (a) — output resistance with the loop open. Looking back into the drain, the transistor is a current source of infinite resistance ($\lambda = 0$) and nothing else reaches the node, so $$R_{OUT}\big|_{R_1=\infty}=R_D=\boxed{3\ \text{k}\Omega}.$$ For reference the open-loop gain is $-g_mR_D=-4.24$ V/V, which is the number the feedback will act on. This completes part (a).
4. Part (b) — find the gain across the bridged element. With C₁ shorting, R₁ runs directly from the drain to the gate node. Applying KCL at the drain, where the current arriving through R₁ must leave through R_D and the transistor, $$A=\frac{v_{OUT}}{v_{gate}} =\frac{1/R_1-g_m}{1/R_D+1/R_1} =\frac{10^{-5}-1.414\times 10^{-3}}{3.333\times10^{-4}+10^{-5}}=-4.09\ \text{V/V}.$$ The feedback resistor has pulled the stage gain down slightly from the −4.24 of step 3, both by loading the drain and by feeding a little signal forward.
5. Part (b) — apply Miller's theorem at the input. A resistor R₁ bridging a node pair of gain A appears at the input as $R_1/(1-A)$ to ground: $$\frac{R_1}{1-A}=\frac{100\ \text{k}\Omega}{1+4.090}=19.65\ \text{k}\Omega.$$ That image sits in parallel with R₂, so $$R_{IN}=R_2\parallel\frac{R_1}{1+|A|} =20\parallel 19.65=\boxed{9.91\ \text{k}\Omega},$$ roughly half the open-loop value — the signature of shunt sampling at the input, which always lowers R_IN.
6. Part (b) — output resistance. To measure R_OUT the independent input source is set to zero. v_IN is drawn as an ideal voltage source connected directly to the gate node, so killing it holds the gate at signal ground; then $v_{gs}=0$, the controlled source is dead, and the test source at the drain sees only two resistors to ground: $$R_{OUT}=R_D\parallel R_1=\frac{(3)(100)}{103}=\boxed{2.91\ \text{k}\Omega}.$$
7. Say what that result does and does not mean. The 3 % reduction is pure resistive loading, not loop action: an ideal voltage source sitting on the summing node holds it rigid and so opens the shunt loop for this measurement. If instead the stage is driven from a current (Norton) source — the drive a shunt–shunt topology is designed for — the gate is free to move, the loop stays active, and a test source at the drain sees $$R_{OUT}'=\left[\frac{1}{R_D}+\frac{1}{R_1+R_2}+\frac{g_mR_2}{R_1+R_2}\right]^{-1}=1.73\ \text{k}\Omega,$$ a genuine 42 % reduction produced by the feedback. Both figures are reported below.
8. Cross-check the limits. Letting $R_1\rightarrow\infty$ in the part (b) expressions returns $A\rightarrow -g_mR_D=-4.24$, $R_{IN}\rightarrow R_2=20$ kΩ and $R_{OUT}\rightarrow R_D=3$ kΩ — exactly the part (a) answers, so the two parts are consistent. A full three-node solve of the closed-loop circuit reproduces 9.911 kΩ and 2.913 kΩ to four figures.

Final Results