Question 5 of 5: Common-Gate Amplifier inside a Voltage-Series Feedback Loop — Input and Output Resistance

Question 5: Common-Gate Amplifier inside a Voltage-Series Feedback Loop — 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. The resistance looking into the source terminal ($R_{IN}$) and the resistance looking into the drain terminal ($R_{OUT}$), first with the feedback network removed and then with it in place.

Approach. Get $g_m$ from the bias current, write the two resistances for the bare common-gate stage, then identify the feedback topology and solve the three-node small-signal circuit exactly — the closed forms that come out are then cross-checked against the feedback factor $1+A\beta$ and against the limit that must reproduce part (a).

1. Fix the operating point and the transconductance. The tail source forces $I_D = I_{bias} = 2$ mA through $M_1$. Using the paper's own saturation law $I_D = \tfrac12 K(V_{GS}-V_{TH})^2$, $$\begin{aligned} V_{ov} &= \sqrt{\frac{2I_D}{K}} = \sqrt{\frac{2\times 2\ \text{mA}}{1\ \text{mA/V}^2}} = 2\ \text{V},\\ g_m &= K V_{ov} = \sqrt{2KI_D} = 2\ \text{mA/V}. \end{aligned}$$ $$\boxed{g_m = 2\ \text{mA/V},\qquad \frac{1}{g_m} = 500\ \Omega}$$ Because $\lambda = 0$, the output resistance of the device itself is infinite, $r_o=\infty$, which simplifies every expression below.
2. Part (a) — the bare common-gate stage. With $R_1=\infty$ and $R_2=0$ the gate is grounded and there is no feedback. Driving the source with a test voltage $v_s$ makes $v_{gs}=-v_s$, so the drain current is $i_d = g_m v_{gs} = -g_m v_s$ and the current drawn from the test source is $i_x = g_m v_s$. Hence $$R_{IN} = \frac{v_s}{i_x} = \frac{1}{g_m} = 500\ \Omega .$$ Looking into the drain with the source grounded, the device presents $r_o=\infty$ in parallel with the drain resistor, so $$R_{OUT} = R_D\parallel r_o = R_D = 2\ \text{k}\Omega .$$ $$\boxed{R_{IN} = 500\ \Omega,\qquad R_{OUT} = 2\ \text{k}\Omega}$$ The low input resistance is the defining feature of the common-gate connection — it is a current buffer, well matched to a source that wants to see a low impedance.
3. Part (b) — identify the topology before computing anything. The divider $R_1$–$R_2$ is connected across the output node, so it samples the output voltage; it returns its tap to the gate, which is in series with the gate–source control loop rather than in parallel with the input node. That is voltage-series (series–shunt) feedback, and its signature is unambiguous: it should raise the input resistance and lower the output resistance, both by the same factor $1+A\beta$. The feedback factor is the divider ratio, $$\beta_f = \frac{R_2}{R_1+R_2} = \frac{100}{300} = \frac{1}{3},$$ and the open-loop gain of the common-gate stage is $A = g_m R_D = 2\times 2 = 4$, so the loop gain is $A\beta_f = 4/3$ and $1+A\beta_f = 7/3 = 2.333$.
4. Write the gate node, which is the only new equation. The gate draws no current, so $R_1$ and $R_2$ form an unloaded divider between the drain and ground: $$\begin{aligned} \frac{v_g-v_d}{R_1}+\frac{v_g}{R_2} &= 0,\\ v_g &= \beta_f\,v_d,\quad\text{where}\ \ \beta_f = \frac{R_2}{R_1+R_2}. \end{aligned}$$ Everything else follows from KCL at the drain and the source.
5. Solve for the output resistance exactly. For $R_{OUT}$ the input source is set to zero, which grounds the source terminal through $C_{IN}$, so $v_s=0$ and $v_{gs}=v_g$. Injecting a test current $i_t$ into the drain, the three paths that carry it away are the drain resistor, the feedback divider, and the channel: $$i_t = \frac{v_d}{R_D}+\frac{v_d-v_g}{R_1}+g_m v_g = v_d\left[\frac{1}{R_D}+\frac{1}{R_1+R_2}+\frac{g_m R_2}{R_1+R_2}\right],$$ where the middle term used $v_d-v_g = v_d R_1/(R_1+R_2)$. Therefore $$R_{OUT}=\left[\frac{1}{R_D}+\frac{1}{R_1+R_2}+\frac{g_mR_2}{R_1+R_2}\right]^{-1} = R_D\parallel(R_1+R_2)\parallel\frac{R_1+R_2}{g_mR_2}.$$ Substituting the numbers, the three conductances are $0.500\ \text{mS}$, $0.00333\ \text{mS}$ and $0.667\ \text{mS}$, summing to $1.170\ \text{mS}$, so $$\boxed{R_{OUT} = 855\ \Omega}$$ Compare $R_D/(1+A\beta_f) = 2000/2.333 = 857\ \Omega$: the two agree to within 0.3 per cent, the small difference being the direct shunting of the drain by the 300 kΩ divider, which the ideal feedback formula omits.
6. Solve for the input resistance exactly. Driving the source terminal with $v_s$, KCL at the drain gives $$\frac{v_d}{R_D}+\frac{v_d-v_g}{R_1}+g_m(v_g-v_s)=0 \quad\Longrightarrow\quad v_d\left[\frac{1}{R_D}+\frac{1}{R_1+R_2}+g_m\beta_f\right]=g_mv_s,$$ whose bracket is precisely $1/R_{OUT}$ from Step 5, so $v_d = g_m R_{OUT}\,v_s$. The current taken from the test source is the channel current, $i_x = g_m(v_s-v_g) = g_m(v_s-\beta_f v_d)$, and therefore $$R_{IN} = \frac{v_s}{i_x} = \frac{1}{g_m\left(1-\beta_f g_m R_{OUT}\right)} = \frac{1}{g_m}+\frac{R_DR_2}{R_D+R_1+R_2}.$$ Numerically $\beta_f g_m R_{OUT} = (1/3)(2\ \text{mA/V})(855\ \Omega) = 0.570$, so $$R_{IN}=\frac{1}{(2\ \text{mA/V})(0.430)}=1.162\ \text{k}\Omega,$$ and the equivalent additive form gives the same value, $500 + (2\times 100)/(302) \times 10^3\ \Omega = 500+662 = 1162\ \Omega$: $$\boxed{R_{IN} = 1.16\ \text{k}\Omega}$$ The ideal-feedback estimate $(1/g_m)(1+A\beta_f)=500\times 2.333=1167\ \Omega$ again agrees to better than half a per cent.
7. Check the result against the limit that must reproduce part (a). Letting $R_1\to\infty$ and $R_2\to 0$ sends $\beta_f\to 0$, so the closed forms collapse to $R_{OUT}\to R_D = 2\ \text{k}\Omega$ and $R_{IN}\to 1/g_m = 500\ \Omega$ — exactly the part (a) answers, which is the cheapest available check on the algebra of Steps 5 and 6. Both directions of change are also the ones the topology predicts: the input resistance rose by a factor of 2.32 and the output resistance fell by 2.34, both within one per cent of the predicted $1+A\beta_f = 2.333$, and the closed-loop voltage gain $v_{OUT}/v_{IN}=g_mR_{OUT}=1.71$ V/V has fallen from the open-loop 4 V/V by the same $1+A\beta_f$ — a textbook demonstration that feedback trades gain for terminal impedances.