Question 4 of 5: Common-Source Amplifier — Mid-Band Gain With and Without the Source Bypass, and the New $f_H$

Question 4: Common-Source Amplifier — Mid-Band Gain With and Without the Source Bypass, and the New $f_H$ (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. (a) the mid-band gain with $C_2$ in place, (b) the mid-band gain once $C_2$ is removed, and (c) the resulting upper −3 dB frequency.

Approach. With $C_2$ present the source is an AC ground, so part (a) is the textbook common-source result. Removing $C_2$ degenerates the stage, and because $r_o$ is comparable with $R_L$ the exact expression — not the $r_o \to \infty$ shortcut — is required. Part (c) then uses open-circuit time constants, evaluating each driving-point resistance on the degenerated network.

1. Part (a): with $C_2$ fitted, the source is an AC ground. $C_2$ is specified as infinite, so it shorts $R_S$ at signal frequencies and the transistor sees its full transconductance into the parallel load. The gate node carries no current at mid-band (all capacitances are open), so no signal is lost across $R_i$ and $v_1 = v_i$: $$r_o \| R_L = \frac{20 \times 20}{40} = 10\ \text{k}\Omega, \qquad \frac{v_{OUT}}{v_i} = -g_m\,(r_o \| R_L) = -(2\ \text{mA/V})(10\ \text{k}\Omega) = \boxed{\,-20.0\ \text{V/V}\,}$$
2. Part (b): removing $C_2$ puts $R_S$ into the signal path. The source node is now free to move, so a series–series feedback loop closes around the device. Writing KCL at the drain and source nodes of the small-signal model and eliminating the source voltage gives the exact degenerated result, in which the denominator $D$ collects every path from drain to ground: $$\frac{v_{OUT}}{v_i} = -\,\frac{g_m r_o R_L}{\,r_o + R_L + R_S + g_m r_o R_S\,} \equiv -\,\frac{g_m r_o R_L}{D}.$$
3. Evaluate the degenerated gain. $$D = 20{,}000 + 20{,}000 + 3000 + (2\times10^{-3})(20{,}000)(3000) = 163\ \text{k}\Omega,$$ $$\frac{v_{OUT}}{v_i} = -\frac{(2\times10^{-3})(20{,}000)(20{,}000)}{163{,}000} = \boxed{\,-4.91\ \text{V/V}\,}$$ The 3 kΩ source resistor has cost a factor of four in gain. Note that the familiar shortcut $-g_m R_L/(1+g_m R_S) = -5.71$ V/V is 16 % high here, because it assumes $r_o \to \infty$ while the given $r_o$ is the same size as $R_L$; the exact form keeps the $(R_L + R_S)/r_o$ contribution and must be used.
4. Part (c): choose the method and list the three capacitors. The stage has three capacitances and no dominant one is obvious, so use the method of open-circuit time constants: $$f_H \;\approx\; \frac{1}{2\pi\left(R_{C_{gs}}C_{gs} + R_{C_{gd}}C_{gd} + R_{C_L}C_L\right)},$$ where each $R$ is the resistance seen by that capacitor with the other two open-circuited and the independent source killed (so the gate returns to ground through $R_i$).
5. Find the resistance seen by $C_{gd}$, which the Miller effect makes the largest. $C_{gd}$ bridges gate and drain, so it sees the gate-side resistance multiplied by one plus the stage gain, plus the drain-side resistance. Looking back into the drain of a degenerated device, $R_S$ is boosted by the intrinsic gain: $$R_d = R_L \,\Big\|\, \bigl[r_o + R_S(1 + g_m r_o)\bigr] = 20\text{k} \,\|\, \bigl[20\text{k} + 3\text{k}(41)\bigr] = 17.55\ \text{k}\Omega,$$ $$R_{C_{gd}} = R_i\bigl(1 + |A_v|\bigr) + R_d = 20\text{k}(1 + 4.908) + 17.55\text{k} = 135.7\ \text{k}\Omega .$$
6. Find the resistance seen by $C_{gs}$ from the exact nodal expression. With the source degenerated the textbook $(R_i + R_S)/(1 + g_m R_S)$ is again an $r_o \to \infty$ form and is wrong by a large factor. The exact driving-point resistance between gate and source reuses the same denominator $D$ as the gain: $$R_{C_{gs}} = R_i - \frac{R_S\,(g_m r_o R_i - R_L - r_o)}{D} = 20\text{k} - \frac{3\text{k}\,(800\text{k} - 40\text{k})}{163\text{k}} = 6.01\ \text{k}\Omega,$$ against 5.38 kΩ from the shortcut. A three-by-three nodal driving-point solve of the same network returns 6.0123 kΩ, confirming the closed form.
7. Find the resistance seen by $C_L$. $C_L$ sits from the drain to ground, and with $C_{gs}$ and $C_{gd}$ open the gate carries no current and rests at ground potential through $R_i$. The resistance is therefore the same drain-side quantity computed in Step 5: $$R_{C_L} = R_L \,\Big\|\, \bigl[r_o + R_S(1 + g_m r_o)\bigr] = 17.55\ \text{k}\Omega .$$
8. Sum the time constants and report $f_H$. $$\sum R_i C_i = (6012)(20\ \text{fF}) + (135{,}706)(5\ \text{fF}) + (17{,}546)(5\ \text{fF})$$ $$= 0.1202 + 0.6785 + 0.0877 = 0.8865\ \text{ns},$$ $$f_H \approx \frac{1}{2\pi(0.8865\ \text{ns})} = \boxed{\,180\ \text{MHz}\,}$$ $C_{gd}$ carries 76.5 % of the total even though it is the smallest capacitor, purely because the Miller multiplication puts 136 kΩ behind it — the practical lesson of the question.
9. Cross-check the estimate against an exact frequency sweep. Solving the full three-node network with all three capacitors and bisecting on $|H(f)|$ places the true −3 dB point at 184.8 MHz, so the OCTC estimate is 2.9 % low. That sign is expected and welcome: summing open-circuit time constants always over-estimates the total delay and therefore under-estimates $f_H$, making the design conservative. The mid-band value returned by the same network is 4.908, matching part (b) exactly.