Common‑Source Amplifier — Effect of Removing the Source Bypass Capacitor: 22-Elec-B5 Advanced Electronics, December 2013, Q5

Question 5: Common‑Source Amplifier — Effect of Removing the Source Bypass Capacitor (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. A single common‑source stage driven from $v_i$ through $R_i$ and the infinite coupling capacitor $C_1$, degenerated by $R_S$ with the bypass $C_2$ across it, biased by an ideal current source $I_{bias}$ and loaded at the drain by $R_L$ in parallel with $C_L$.

Given data
Transconductance $g_m$	$2\ \text{mA/V}$	Output resistance $r_o$	$20\ \text{k}\Omega$
Source resistance of the driver $R_i$	$20\ \text{k}\Omega$	Drain load $R_L$	$20\ \text{k}\Omega$
Source degeneration $R_S$	$2\ \text{k}\Omega$	Gate‑source capacitance $C_{gs}$	$20\ \text{fF}$
Gate‑drain capacitance $C_{gd}$	$5\ \text{fF}$	Load capacitance $C_L$	$5\ \text{fF}$
Coupling and bypass $C_1$, $C_2$	$\infty$

Find. The mid‑band gain with the source bypassed, the mid‑band gain once $C_2$ is removed, and the upper 3 dB frequency in that unbypassed condition.

Question 5 — common‑source stage. With C₂ present the source sits at signal ground; removing it leaves R_S as series‑series feedback, which is what parts (e) and (f) explore.

Approach. Take the bypassed gain from the plain common‑source result, re‑derive the gain with $R_S$ in place from a single node equation that keeps $r_o$, then apply the open‑circuit time‑constant method to the three capacitors.

Part (d) — simplify the mid‑band circuit. $C_1$ is infinite, so it is a short; the gate draws no current, so nothing flows in $R_i$ and the whole of $v_i$ arrives at the gate. $C_2$ is infinite too, so the source is at signal ground. $I_{bias}$ is an ideal source, hence an open circuit to signals, and the drain sees only $R_L$ in parallel with $r_o$.
Write the bypassed gain. $$\frac{v_{OUT}}{v_i}=-g_m\left(r_o\parallel R_L\right)=-(2\times10^{-3})\left(\frac{(20)(20)}{20+20}\times10^{3}\right)=-(2\times10^{-3})(10\times10^{3})=\boxed{-20.0\ \text{V/V}}$$ i.e. 26.0 dB with an inversion. Note that $R_i$ does not appear — it costs nothing at mid‑band and everything at high frequency, as part (f) shows.
Part (e) — put $R_S$ back in the signal path. Removing $C_2$ leaves $R_S$ carrying the drain current, so the source rises with the signal and reduces $v_{gs}$. Keeping $r_o$ (it is only equal to $R_L$ here, so it cannot be dropped), one node equation gives $i_d\left[1+g_mR_S+(R_L+R_S)/r_o\right]=g_mv_i$, and with $v_{OUT}=-i_dR_L$, $$\frac{v_{OUT}}{v_i}=\frac{-g_mR_L}{1+g_mR_S+\dfrac{R_L+R_S}{r_o}}$$
Substitute. $g_mR_S=(2\times10^{-3})(2\times10^{3})=4$ and $(R_L+R_S)/r_o=22/20=1.1$, so the denominator is $1+4+1.1=6.1$ while $g_mR_L=40$: $$\frac{v_{OUT}}{v_i}=\frac{-40}{6.1}=\boxed{-6.56\ \text{V/V}}$$ a factor of 3.05 down on part (d), or 9.7 dB. Setting $R_S=0$ returns the denominator to $1+0+1$ and the gain to $-20$, which checks the algebra.
Part (f) — find the resistance at the drain. Degeneration raises the output resistance of the transistor to $R_{out}=r_o+R_S\left(1+g_mr_o\right)=20+2(1+40)=102\ \text{k}\Omega$, so the drain node resistance is $R_d=R_L\parallel R_{out}=20\parallel102=16.7\ \text{k}\Omega$. This is what $C_L$ sees: $\tau_{L}=R_dC_L=(16.7\times10^{3})(5\times10^{-15})=83.6\ \text{ps}$.
Resistance seen by $C_{gd}$. The bridging capacitor spans the gate and the drain, so its time constant carries the Miller multiplication: $$R_{gd}=R_i\left(1+\left|A_v\right|\right)+R_d=20\left(1+6.56\right)+16.7=167.9\ \text{k}\Omega$$ with $\left|A_v\right|=6.56$ from part (e). Hence $\tau_{gd}=(167.9\times10^{3})(5\times10^{-15})=839\ \text{ps}$ — the dominant contribution despite $C_{gd}$ being the smallest capacitor.
Resistance seen by $C_{gs}$. Holding the gate at $v_g=IR_i$ and drawing a test current $I$ out of the source node, the drain and source node equations give $v_s=12.46\ \text{k}\Omega\times I$, so $R_{gs}=(v_g-v_s)/I=20-12.46=7.54\ \text{k}\Omega$ and $\tau_{gs}=(7.54\times10^{3})(20\times10^{-15})=151\ \text{ps}$. The familiar $r_o\to\infty$ shortcut $(R_i+R_S)/(1+g_mR_S)=4.40\ \text{k}\Omega$ is a factor of 1.7 too small here, because $r_o$ is comparable with $R_L$.
Sum the time constants. $$\sum\tau=\tau_{gs}+\tau_{gd}+\tau_{L}=151+839+84=1074\ \text{ps}$$ $$f_H\simeq\frac{1}{2\pi\sum\tau}=\frac{1}{2\pi(1.074\times10^{-9})}=\boxed{148\ \text{MHz}}$$ A full nodal solution of the same three‑capacitor network puts the true $-3$ dB point at 151 MHz, so the open‑circuit estimate is 2 % conservative — its usual behaviour when no single pole dominates.
Read the trade. With $C_2$ fitted the same method gives $f_H=61.2\ \text{MHz}$ at a gain of 20. Removing it therefore buys a factor of 2.4 in bandwidth for a factor of 3.05 in gain: the gain–bandwidth product falls from about 1.22 GHz to 0.97 GHz, so series‑series feedback here is not free, but it also linearises the stage and makes the gain depend on the resistor ratio rather than on $g_m$.

Final results
Quantity	Symbol	Value
(d) Mid‑band gain, source bypassed	$v_{OUT}/v_i$	$-20.0\ \text{V/V}$ (26.0 dB)
(e) Mid‑band gain, $C_2$ removed	$v_{OUT}/v_i$	$-6.56\ \text{V/V}$ (16.3 dB)
Degenerated output resistance	$R_{out}$	$102\ \text{k}\Omega$
Drain node resistance	$R_d$	$16.7\ \text{k}\Omega$
Time constants $C_{gs}$ / $C_{gd}$ / $C_L$	$\tau$	$151$ / $839$ / $84\ \text{ps}$
Sum of open‑circuit time constants	$\sum\tau$	$1.074\ \text{ns}$
(f) Upper 3 dB frequency, $C_2$ removed	$f_H$	$148\ \text{MHz}$
Exact nodal result (cross‑check)	$f_{-3\text{dB}}$	$151\ \text{MHz}$

Check — the printed sub‑part letters run d), e), f) although Question 5 begins on its own page and has no parts a) to c); the marks quoted (6 + 6 + 8) total the 20 the paper allots each question, so the lettering is a typographical carry‑over and all three parts of Question 5 are answered here. $I_{bias}$ is treated as an ideal current source, hence an open circuit at signal frequencies, so it adds neither resistance nor capacitance at the drain.

Concept & Theory

Topic. Source degeneration in a common‑source stage: what it costs in gain, what it buys in bandwidth, and how to price both with open‑circuit time constants.

Key relations. $A_v=-g_m(r_o\parallel R_L)$ bypassed; $A_v=\dfrac{-g_mR_L}{1+g_mR_S+(R_L+R_S)/r_o}$ degenerated; $R_{out}=r_o+R_S(1+g_mr_o)$; $R_d=R_L\parallel R_{out}$; $R_{gd}=R_i(1+|A_v|)+R_d$; $f_H\simeq1/(2\pi\sum R_iC_i)$.

Why this works. Open‑circuit time constants work because the sum of the individual $R_iC_i$ products equals the coefficient of $s$ in the denominator of the exact transfer function, which in turn equals the sum of the reciprocal pole frequencies. That makes the estimate exact when one pole dominates and slightly pessimistic otherwise — a safe direction for design. The dominant term here belongs to $C_{gd}$ not because it is large but because it sees the Miller‑multiplied resistance $R_i(1+|A_v|)$: reducing gain reduces that multiplier, which is the mechanism by which removing $C_2$ widens the band at all.

Common pitfall. Using the $r_o\to\infty$ formulas for $R_{gs}$ and for the degenerated gain when $r_o$ is comparable to $R_L$, which here would report a 1.7× wrong $\tau_{gs}$ and a gain of $-40/5=-8$ instead of $-6.56$. And forgetting that degeneration raises the drain node resistance through $R_{out}$, which partly offsets the bandwidth gained at the input.

Source. Sedra & Smith, Microelectronic Circuits, 8th ed., §7.3 (the CS amplifier with a source resistance), §10.4 (open‑circuit time constants) and §10.5 (high‑frequency response of the CS stage); Razavi, Design of Analog CMOS Integrated Circuits, 2nd ed., §3.3 and §6.3.

Question 5 of 5: Common‑Source Amplifier — Effect of Removing the Source Bypass Capacitor

Question 5: Common‑Source Amplifier — Effect of Removing the Source Bypass Capacitor (20 marks)