Question 3 of 5: Common-Source Amplifier — Gain and High-Frequency Response

Question 3: Common-Source Amplifier — Gain and High-Frequency Response (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 mid-band gain with the source fully bypassed, the mid-band gain once the bypass capacitor is removed, and the upper 3 dB frequency of the degenerated stage.

Approach. Recognise that the ideal bias current source is an ac open, so the drain load is $R_L\,\|\,r_o$; get part (d) from the standard common-source result, part (e) from the degenerated gain expression that retains $r_o$ (essential here, since $r_o$ and $R_L$ are equal), and part (f) from the method of open-circuit time constants with driving-point resistances taken from the node equations rather than from an $r_o\to\infty$ formula.

1. Part (d) — identify what the drain actually drives. The bias current source feeding the drain is ideal, so it is an open circuit to signals and contributes nothing; the drain therefore sees $R_L$ in parallel with the transistor's own $r_o$: $$R_{load}=R_L\,\|\,r_o=\frac{(20)(20)}{20+20}=10\ \text{k}\Omega.$$ With $C_2=\infty$ the source terminal is at ac ground, and because the gate draws no current there is no drop across $R_i$ at mid-band, so $v_{gs}=v_i$ exactly. Hence $$\boxed{\frac{v_{OUT}}{v_i}=-g_m(R_L\,\|\,r_o)=-(2\ \text{mA/V})(10\ \text{k}\Omega)=-20.0}$$
2. Part (e) — put $R_S$ back into the signal path. Removing $C_2$ leaves 100 Ω of unbypassed degeneration. Because $r_o$ is the same size as $R_L$ here, the full expression must be used rather than any $r_o\to\infty$ shortcut: $$\frac{v_{OUT}}{v_i}=\frac{-g_m r_o R_L}{r_o+R_L+(1+g_m r_o)R_S}.$$ Substituting $g_m r_o = (2\ \text{mA/V})(20\ \text{k}\Omega)=40$, $$\frac{v_{OUT}}{v_i}=\frac{-(2\times10^{-3})(20\times10^{3})(20\times10^{3})}{20\times10^{3}+20\times10^{3}+(41)(100)} =\frac{-8.00\times10^{5}}{4.41\times10^{4}}$$ $$\boxed{\frac{v_{OUT}}{v_i}=-18.14}$$ so a mere 100 Ω costs about 9 percent of the gain.
3. Check that number against the shortcuts, and against a direct solve. The two familiar approximations both fail here: $-g_mR_L/(1+g_mR_S)=-33.3$ ignores $r_o$ altogether, and $-g_m(r_o\|R_L)/(1+g_mR_S)=-16.7$ degenerates the wrong quantity. A three-node solution of the small-signal network — gate, source and drain, with the controlled source $g_mv_{gs}$ and $r_o$ between drain and source — returns $-18.1406$, matching the boxed expression to every digit. The physical reason the shortcuts miss is that $r_o$ bridges the drain to the source node, so degeneration and output resistance interact and cannot be handled one at a time.
4. Part (f) — set up the open-circuit time-constant method. With three capacitors and no dominant pole guaranteed, the upper cut-off is estimated as $$f_H\approx\frac{1}{2\pi\,\tau_H},\qquad \tau_H=R_{gs}C_{gs}+R_{gd}C_{gd}+R_{C_L}C_L,$$ where each $R$ is the resistance seen by that capacitor with the signal source nulled and the other two capacitors open-circuited. Each is obtained by injecting a test current between the relevant node pair and solving the same three-node conductance matrix used in the previous step.
5. Evaluate the three driving-point resistances. The results are $$R_{gs}=18.28\ \text{k}\Omega,$$ $$R_{gd}=R_i\big(1+|A_v|\big)+R_d=20(1+18.14)+10.93=393.7\ \text{k}\Omega,$$ $$R_{C_L}=R_d=R_L\,\big\|\,\big[r_o+R_S(1+g_mr_o)\big] =20\ \|\ 24.1=10.93\ \text{k}\Omega.$$ The middle expression is the Miller statement of $R_{gd}$: the gate–drain capacitor sees the source resistance multiplied by one plus the stage gain, plus the resistance at the drain. Note that the textbook degenerated form $R_{gs}=(R_i+R_S)/(1+g_mR_S)=16.75$ kΩ is 8 percent low, again because $r_o$ is comparable with $R_L$; the value above comes from the node equations and is exact.
6. Sum the time constants and take the cut-off. $$\tau_{gs}=(18.28\ \text{k}\Omega)(20\ \text{fF})=365.5\ \text{ps},\quad \tau_{gd}=(393.7\ \text{k}\Omega)(5\ \text{fF})=1968.7\ \text{ps},$$ $$\tau_{C_L}=(10.93\ \text{k}\Omega)(5\ \text{fF})=54.6\ \text{ps} \;\Rightarrow\;\tau_H=2388.9\ \text{ps}=2.389\ \text{ns},$$ $$\boxed{f_H\approx\frac{1}{2\pi(2.389\ \text{ns})}=66.6\ \text{MHz}}$$
7. Read the design message and check the estimate. The smallest capacitor in the circuit, $C_{gd}$ at 5 fF, contributes 82 percent of the whole time constant, purely because Miller multiplication makes it look back through $R_i$ nearly twenty times over — halving $R_i$ would buy far more bandwidth than halving $C_L$. As a check on the approximation, sweeping the complete three-capacitor network and bisecting for the true $-3$ dB point gives 67.1 MHz, so the open-circuit estimate is 0.7 percent low, comfortably inside its usual conservative margin.

Final results.