PNP Common‑Emitter Stage — Mid‑band Gain and Both 3 dB Corners: 22-Elec-B5 Advanced Electronics, December 2013, Q2

Question 2: PNP Common‑Emitter Stage — Mid‑band Gain and Both 3 dB Corners (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 PNP common‑emitter stage: the source $v_S$ reaches the base through $R_S$ and the coupling capacitor $C_1$, $R_1$ returns the base to ground, the emitter is fed by the tail source $I_{bias}$ and bypassed to ground by $C_2=\infty$, and the collector drives $R_L$ down to $-V_{EE}$ with the output taken at the collector.

Given data
Current gain $\beta$	$100$	Emitter‑base drop $V_{EB}$	$0.7\ \text{V}$
Early voltage $V_A$ (with $r_o$ neglected)	$100\ \text{V}$	Collector‑base capacitance $C_\mu$	$2\ \text{pF}$
Source resistance $R_S$	$600\ \Omega$	Collector load $R_L$	$3\ \text{k}\Omega$
Base return $R_1$	$1\ \text{k}\Omega$	Coupling capacitor $C_1$	$10\ \mu\text{F}$
Emitter bypass $C_2$	$\infty$	Tail current $I_{bias}$	$2\ \text{mA}$
Thermal voltage $V_T$	$25\ \text{mV}$	Supplies $\|V_{CC}\|=\|V_{EE}\|$	$10\ \text{V}$

Find. The mid‑band voltage gain referred to the source, the lower 3 dB corner set by $C_1$, and the two high‑frequency poles produced by $C_\mu$.

Question 2 — PNP common‑emitter stage. The emitter is held at signal ground by C₂ = ∞, so the only finite capacitors in the problem are the coupling capacitor C₁ (low end) and C_μ (high end).

Approach. Bias first, then the hybrid‑π parameters; the mid‑band gain is an input divider times $-g_mR_L$; $f_L$ comes from the single time constant of $C_1$; the two high‑frequency poles follow from splitting $C_\mu$ by Miller’s theorem.

Part (a) — find the quiescent collector current. The tail source sets the emitter current directly, $I_E=I_{bias}=2\ \text{mA}$, so $$I_C=\alpha I_E=\frac{\beta}{\beta+1}I_E=\frac{100}{101}(2)=1.980\ \text{mA}$$
Build the hybrid‑π model. $g_m=I_C/V_T=1.980/25\times10^{-3}=79.2\ \text{mA/V}$ and $r_\pi=\beta/g_m=100/(79.2\times10^{-3})=1.26\ \text{k}\Omega$. The question instructs that $r_x$ and $r_o$ be neglected, so the collector sees only $R_L$. The PNP small‑signal model is identical in form to the NPN one; only the d.c. polarities differ.
Reduce the input network at mid‑band. $C_1$ is a short and $C_2=\infty$ grounds the emitter, so the base sees $R_1$ in parallel with $r_\pi$: $R_1\parallel r_\pi=(1000)(1262.5)/2262.5=558\ \Omega$. The source divider is therefore $$\frac{v_b}{v_S}=\frac{R_1\parallel r_\pi}{R_S+R_1\parallel r_\pi}=\frac{558}{600+558}=0.482$$
Combine with the stage gain. With the emitter at signal ground the collector delivers $-g_mv_\pi$ into $R_L$, i.e. $v_{OUT}/v_b=-g_mR_L=-(79.2\times10^{-3})(3000)=-238$, so $$\frac{v_{OUT}}{v_S}=0.482\times(-238)=\boxed{-114.5\ \text{V/V}}$$ which is 41.2 dB of magnitude with a 180° inversion. Almost half the available gain is lost in the source divider because $r_\pi$ is only twice $R_S$.
Part (b) — identify what sets the low end. $C_2$ is infinite and the tail source is ideal, so $C_1$ is the only capacitor that can produce a low‑frequency pole. Opening the loop at $C_1$, it looks into $R_S$ on one side and $R_1\parallel r_\pi$ on the other: $R_{C1}=R_S+R_1\parallel r_\pi=600+558=1.158\ \text{k}\Omega$.
Convert the time constant to a corner. $$f_L=\frac{1}{2\pi R_{C1}C_1}=\frac{1}{2\pi(1158)(10\times10^{-6})}=\boxed{13.7\ \text{Hz}}$$ Comfortably below the 20 Hz audio floor, which is what a 10 μF coupling capacitor into a kilohm‑class node is chosen for.
Part (c) — drive resistance seen by the input capacitance. At high frequency $C_1$ is a short, so the base node is driven from $R_{sig}=R_S\parallel R_1\parallel r_\pi=600\parallel1000\parallel1262.5=289\ \Omega$. Because $C_\pi$ is not specified it is taken as negligible and $C_\mu$ carries the whole high‑frequency behaviour.
Split $C_\mu$ by Miller’s theorem. The bridging capacitor sees the full stage gain from base to collector, so its input‑side image is $$C_{in}=C_\mu\left(1+g_mR_L\right)=2\left(1+238\right)=477\ \text{pF}$$ and the input pole becomes $$f_H=\frac{1}{2\pi R_{sig}C_{in}}=\frac{1}{2\pi(289)(477\times10^{-12})}=\boxed{1.15\ \text{MHz}}$$ A 2 pF capacitor has been multiplied into 477 pF — the Miller effect is the entire story of the upper corner in this stage.
Part (d) — the output‑side Miller image. On the collector side the same capacitor appears as $C_{out}=C_\mu\left(1+1/(g_mR_L)\right)=2(1+1/238)=2.01\ \text{pF}$, working against $R_L$ alone: $$f_{p2}=\frac{1}{2\pi R_LC_{out}}=\frac{1}{2\pi(3000)(2.01\times10^{-12})}=\boxed{26.4\ \text{MHz}}$$ It sits a factor of 23 above $f_H$, which retrospectively justifies calling the input pole dominant and reading $f_H$ straight off it.
Cross‑check the dominant pole exactly. With $C_\mu$ as the only capacitor the network is strictly first order, so its pole follows from the resistance the capacitor actually sees, $R_{eq}=R_{sig}+R_L+g_mR_{sig}R_L=289+3000+68\,700=72.0\ \text{k}\Omega$, giving $f_p=1/(2\pi R_{eq}C_\mu)=1.11\ \text{MHz}$ — 4 % below the Miller estimate, which is the usual accuracy of the approximation. The exact response also carries a right‑half‑plane zero at $f_z=g_m/(2\pi C_\mu)=6.3\ \text{GHz}$, far beyond any frequency of interest.

Question 2 — asymptotic magnitude response built from the four answers: 41.2 dB mid‑band between f_L = 13.7 Hz and f_H = 1.15 MHz, rolling off at 20 dB/decade until the second pole at 26.4 MHz doubles the slope.

Final results
Quantity	Symbol	Value
Quiescent collector current	$I_C$	$1.980\ \text{mA}$
Transconductance / input resistance	$g_m$, $r_\pi$	$79.2\ \text{mA/V}$, $1.26\ \text{k}\Omega$
(a) Mid‑band gain	$v_{OUT}/v_S$	$-114.5\ \text{V/V}$ (41.2 dB)
(b) Lower 3 dB frequency	$f_L$	$13.7\ \text{Hz}$
(c) Upper 3 dB frequency	$f_H$	$1.15\ \text{MHz}$
(d) Second high‑frequency pole	$f_{p2}$	$26.4\ \text{MHz}$
Exact dominant pole (cross‑check)	$f_p$	$1.11\ \text{MHz}$

Concept & Theory

Topic. Mid‑band gain and the two ends of the bandwidth of a bypassed common‑emitter stage, with the Miller split of $C_\mu$.

Key relations. $I_C=\alpha I_E$; $g_m=I_C/V_T$; $r_\pi=\beta/g_m$; $A_{mid}=\dfrac{R_1\parallel r_\pi}{R_S+R_1\parallel r_\pi}\left(-g_mR_L\right)$; $f_L=1/(2\pi R_{C1}C_1)$ with $R_{C1}=R_S+R_1\parallel r_\pi$; $C_{in}=C_\mu(1+g_mR_L)$, $C_{out}=C_\mu(1+1/g_mR_L)$; open‑circuit time constant $R_{eq}=R_{sig}+R_L+g_mR_{sig}R_L$.

Why this works. Coupling and bypass capacitors are open at d.c. and short at signal frequencies, so each one contributes a low‑frequency zero–pole pair whose corner is the reciprocal of the resistance it faces with every other capacitor at its mid‑band value. Device capacitances do the mirror‑image job at the top of the band. $C_\mu$ is awkward because it bridges input and output, and Miller’s theorem resolves that by replacing it with two grounded capacitors scaled by the voltage gain across it — large at the input, barely more than $C_\mu$ at the output. The approximation is good precisely when the resulting poles are well separated, which is what the 23:1 ratio here confirms.

Common pitfall. Forgetting that the emitter is at signal ground because $C_2=\infty$ and then dividing the gain by $(1+g_mR_E)$, which throws away a factor of a hundred; and applying the Miller multiplier to the source resistance $R_S$ alone instead of to $R_S\parallel R_1\parallel r_\pi$, which overstates the input time constant by more than a factor of two.

Source. Sedra & Smith, Microelectronic Circuits, 8th ed., §10.1–10.3 (low‑frequency response, Miller’s theorem) and §10.5 (high‑frequency response of the CE amplifier).

Question 2 of 5: PNP Common‑Emitter Stage — Mid‑band Gain and Both 3 dB Corners

Question 2: PNP Common‑Emitter Stage — Mid‑band Gain and Both 3 dB Corners (20 marks)