Question 5 of 5: Common-source amplifier — bypassed and unbypassed source, and the new 3 dB frequency

Question 5: Common-source amplifier — bypassed and unbypassed source, and the new 3 dB frequency (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 R_S bypassed, the mid-band gain once C₂ is removed, and the upper 3 dB frequency of the un-bypassed circuit.

Approach. With C₂ in place the stage is a textbook common source loaded by r_o ∥ R_L. Removing C₂ introduces source degeneration, which must be handled with the exact expression because r_o is comparable with R_L. The bandwidth then follows from the open-circuit time constants of C_gs, C_gd and C_L.

1. Part (d) — Take the bypassed mid-band gain. C₁ = ∞ couples the gate and C₂ = ∞ grounds the source, so no signal current flows in R_i (a MOSFET gate is an open circuit at mid-band) and $v_{gs}=v_i$. The drain drives r_o in parallel with R_L: $$\boxed{\frac{v_{OUT}}{v_i}=-g_m\left(r_o\parallel R_L\right) =-(2\ \text{mA/V})\left(\frac{20\times20}{40}\ \text{k}\Omega\right)=-20\ \text{V/V}}$$
2. Part (e) — Degenerate the stage with R_S. Removing C₂ leaves R_S in the source lead. Because r_o = R_L here, the familiar shortcut $-g_mR_L/(1+g_mR_S)$ is not accurate enough; the exact result from a two-node solve is $$\frac{v_{OUT}}{v_i}=\frac{-g_mR_L}{1+g_mR_S+\dfrac{R_L+R_S}{r_o}} =\frac{-g_mR_Lr_o}{r_o+R_L+R_S\left(1+g_mr_o\right)}$$ $$=\frac{-(2\ \text{mA/V})(20\ \text{k}\Omega)}{1+2+\dfrac{21}{20}} =\frac{-40}{4.05}\quad\Rightarrow\quad\boxed{\frac{v_{OUT}}{v_i}=-9.88\ \text{V/V}}$$ The shortcut would have returned −13.3 V/V, a 35 % overestimate, because it silently assumes $r_o\to\infty$. Degeneration has cost slightly more than half the gain in exchange for a 4.05-fold reduction in sensitivity to g_m.
3. Part (f) — Set up the open-circuit time constants. With three capacitors and no dominant one obvious, the sum-of-open-circuit-time-constants estimate is the right tool: for each capacitor find the resistance seen across its terminals with all other capacitors open, then $f_H\approx 1/\left(2\pi\sum_k R_kC_k\right)$. Define the effective drain resistance, which is R_L in parallel with the degenerated drain resistance of the transistor: $$R_d=R_L\parallel\left[r_o+R_S\left(1+g_mr_o\right)\right] =20\ \text{k}\Omega\parallel61\ \text{k}\Omega=15.06\ \text{k}\Omega$$
4. Evaluate the three resistances. C_L sits from drain to ground and therefore sees R_d directly. C_gd bridges gate and drain and picks up the Miller multiplication of the stage gain found in part (e). C_gs requires the two remaining node equations to be solved, because the degenerated source node moves with the gate: $$R_{CL}=R_d=15.06\ \text{k}\Omega$$ $$R_{gd}=R_i\left(1+\left|A_v\right|\right)+R_d=20(10.88)+15.06=232.6\ \text{k}\Omega$$ $$R_{gs}=\frac{R_i\left(R_L+R_S+r_o\right)+R_S\left(R_L+r_o\right)}{R_L+R_S+r_o\left(1+g_mR_S\right)} =\frac{20(41)+1(40)}{20+1+20(3)}\ \text{k}\Omega=10.62\ \text{k}\Omega$$
5. Sum the time constants and invert. $$\sum R_kC_k=(10.62\ \text{k}\Omega)(20\ \text{fF})+(232.6\ \text{k}\Omega)(5\ \text{fF}) +(15.06\ \text{k}\Omega)(5\ \text{fF})$$ $$=0.212+1.163+0.075=1.451\ \text{ns}$$ $$\boxed{f_H\approx\frac{1}{2\pi\left(1.451\ \text{ns}\right)}=1.10\times10^{8}\ \text{Hz} =110\ \text{MHz}}$$ C_gd carries 80 % of the total even though it is the smallest capacitor present — the Miller multiplication by $1+|A_v|$ against a 20 kΩ source resistance is what dominates the bandwidth.
6. Check the estimate against an exact solve. Solving the full three-node admittance system and bisecting for the frequency at which the magnitude falls to $1/\sqrt{2}$ of its mid-band value gives 111.3 MHz, so the open-circuit time-constant estimate is 1.4 % low — the usual conservative bias of the method, and close enough that no further work is warranted. For reference, the same circuit with C₂ fitted has f_H = 61.6 MHz: removing the bypass capacitor halves the gain but nearly doubles the bandwidth, leaving the gain–bandwidth product almost unchanged.