Question 4 of 5: Shunt–shunt feedback amplifier — input and output resistance

Question 4: Shunt–shunt feedback amplifier — 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 into the amplifier from v_IN and R_OUT looking back into the drain node, first with the feedback path removed and then with R₁ = 100 kΩ closing the loop.

Approach. Bias M₁ from I_bias to get g_m; with R₁ open the gate is purely resistive and the drain purely R_D; with R₁ in place, apply Miller's theorem to the feedback resistor (equivalently, solve the two node equations) to obtain the reduced input resistance, and kill v_IN to obtain the output resistance.

1. Part (a) — Bias the transistor. The tail current source fixes I_D = I_bias = 1 mA, so with $I_D=\tfrac12KV_{OV}^2$, $$V_{OV}=\sqrt{\frac{2(1\ \text{mA})}{1\ \text{mA/V}^2}}=1.414\ \text{V},\qquad V_{GS}=V_{TH}+V_{OV}=2.41\ \text{V},\qquad g_m=\sqrt{2KI_D}=1.414\ \text{mA/V}$$ C₁ blocks DC, so no current flows in R₁ or R₂ and the gate sits at 0 V; the source therefore rests at −2.41 V and the drain at $V_{DD}-I_DR_D=10-3=7\ \text{V}$, giving V_DS = 9.41 V, comfortably saturated.
2. Read off the open-loop resistances. With R₁ = ∞ the only element between the input terminal and ground is R₂, because a MOSFET gate draws no current: $$\boxed{R_{IN}=R_2=20\ \text{k}\Omega}$$ Looking back into the drain with v_IN = 0 the gate is held at zero, so the controlled source is dormant and λ = 0 makes r_o infinite: $$\boxed{R_{OUT}=R_D=3\ \text{k}\Omega}$$
3. Part (b) — Find the gain that the feedback resistor sees. Writing KCL at the drain with the gate at v_G and the source at signal ground, $$\frac{v_O-v_G}{R_1}+\frac{v_O}{R_D}+g_mv_G=0 \quad\Rightarrow\quad A_v=\frac{v_O}{v_G}=\frac{R_D\left(1-g_mR_1\right)}{R_1+R_D}$$ $$A_v=\frac{3\ \text{k}\Omega\left(1-(1.414\ \text{mA/V})(100\ \text{k}\Omega)\right)}{103\ \text{k}\Omega}=-4.09\ \text{V/V}$$ Note this is well below the −g_mR_D = −4.24 of the bare stage, and lower still because R₁ now loads the drain.
4. Miller the feedback resistor into the input. A resistor bridging a node pair with gain A_v appears at the input as $R_1/(1-A_v)$; carrying the drain loading exactly gives the closed form $$R_f=\frac{R_1+R_D}{1+g_mR_D}=\frac{100+3}{1+(1.414)(3)}\ \text{k}\Omega =\frac{103}{5.243}\ \text{k}\Omega=19.65\ \text{k}\Omega$$ The feedback path has turned a 100 kΩ resistor into a 19.65 kΩ load on the input node — the signature of shunt feedback.
5. Combine with R₂ for the closed-loop input resistance. $$\boxed{R_{IN}=R_2\parallel R_f=\frac{(20)(19.65)}{20+19.65}\ \text{k}\Omega=9.91\ \text{k}\Omega}$$ a factor of 2.02 below the open-loop 20 kΩ, which is exactly the amount of shunt feedback applied at that node.
6. Find the closed-loop output resistance. To measure R_OUT the input source is set to zero; because v_IN is an ideal voltage source connected directly to the gate node, that grounds the gate and disables the controlled source. The drain then sees only R_D in parallel with the feedback network back to that grounded node: $$\boxed{R_{OUT}=R_D\parallel R_1=\frac{(3)(100)}{103}\ \text{k}\Omega=2.91\ \text{k}\Omega}$$ The 3 % reduction is simply R₁ loading the drain; with an ideal voltage source driving the gate there is no loop left to reduce R_OUT further, which is worth saying explicitly.
7. Cross-check the whole result numerically. Solving the two-node conductance system directly (inject 1 A at the gate node, λ = 0, source at signal ground) returns $R_{IN}=9\,910.9\ \Omega$ and $A_v=-4.0899$, matching the closed forms above to five figures — the check that the Miller reduction was applied to the right node pair.