Common-emitter amplifier — mid-band gain and the two 3 dB frequencies: 22-Elec-B5 Advanced Electronics, December 2015, Q2

Question 2: Common-emitter amplifier — mid-band gain and the two 3 dB frequencies (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.

Quantity	Symbol	Value
Current gain	β	100
Thermal voltage	V_T	25 mV
Tail bias current (emitter current)	I_bias	1 mA
Collector load	R_L	5 kΩ
Base bias resistor to ground	R₁	1 kΩ
Input coupling capacitor	C₁	5 μF
Emitter bypass capacitor	C₂	∞ (perfect AC short)
Collector-base capacitance	C_μ	2 pF
Neglected	r_x, r_o	—

Find. The mid-band voltage gain v_OUT/v_S, the lower 3 dB corner f_L set by C₁, the upper 3 dB corner f_H, and the second high-frequency pole.

Q2 — common-emitter stage. C₂ = ∞ places the emitter at signal ground, so I_bias sets the operating point only; C₁ couples the source to a base biased through R₁.

Approach. Find the operating point from I_bias and get g_m and r_π; take the mid-band gain of the bypassed common-emitter stage; put C₁ against the resistance seen at the base for f_L; and split C_μ by Miller's theorem to obtain the two high-frequency poles.

Part (a) — Establish the bias point and the hybrid-π parameters. The current source in the emitter fixes I_E = I_bias = 1 mA, so $$I_C=\alpha I_E=\frac{\beta}{\beta+1}I_E=\frac{100}{101}(1\ \text{mA})=0.990\ \text{mA}$$ $$g_m=\frac{I_C}{V_T}=\frac{0.990\ \text{mA}}{25\ \text{mV}}=39.6\ \text{mA/V}, \qquad r_\pi=\frac{\beta}{g_m}=\frac{100}{39.6\ \text{mA/V}}=2.53\ \text{k}\Omega$$ The DC base voltage is $-I_B R_1 \approx -10\ \text{mV}\approx 0$, so $V_E\approx-0.7\ \text{V}$ and $V_C=V_{CC}-I_C R_L=10-4.95=5.05\ \text{V}$: with $V_{CE}=5.75\ \text{V}\gg V_{CE(sat)}$ the device is comfortably active.
Take the mid-band gain. In the mid-band C₁ is a short and C₂ holds the emitter at signal ground, so the stage is a plain common emitter whose base is driven directly by v_S (the source is drawn ideal, and R₁ only shunts it). With r_o neglected the load is R_L alone: $$\boxed{\frac{v_{OUT}}{v_S}=-g_m R_L=-(39.6\ \text{mA/V})(5\ \text{k}\Omega)=-198\ \text{V/V}}$$ i.e. about $-200$ V/V, or 45.9 dB with a 180° phase inversion.
Part (b) — Find the resistance that C₁ works against. Looking out from the coupling capacitor, the left-hand side is the ideal source (zero ohms) and the right-hand side is the amplifier input: $$R_{C1}=R_{sig}+\left(R_1\parallel r_\pi\right)=0+\frac{(1)(2.525)}{1+2.525}\ \text{k}\Omega=716\ \Omega$$ C₂ is infinite and so contributes no low-frequency pole; C₁ alone sets the lower corner: $$\boxed{f_L=\frac{1}{2\pi C_1 R_{C1}}=\frac{1}{2\pi(5\ \mu\text{F})(716\ \Omega)}=44.4\ \text{Hz}}$$
Part (c) — Split C_μ with Miller's theorem. C_μ bridges base and collector across a gain of $-g_mR_L=-198$, so it appears as two grounded capacitors: $$C_{in}=C_\mu\left(1+g_mR_L\right)=2\ \text{pF}\,(199)=398\ \text{pF},\qquad C_{out}=C_\mu\left(1+\frac{1}{g_mR_L}\right)=2.01\ \text{pF}$$ The huge input capacitance is the classic penalty of the inverting stage: a 2 pF device capacitance behaves like 398 pF at the base.
Evaluate the dominant (input) pole. C_in sees the amplifier's own input resistance $R_1\parallel r_\pi=716\ \Omega$, so $$\boxed{f_H=\frac{1}{2\pi\left(R_1\parallel r_\pi\right)C_{in}} =\frac{1}{2\pi(716\ \Omega)(398\ \text{pF})}=5.6\times10^{5}\ \text{Hz}=558\ \text{kHz}}$$ As a check, the exact single-capacitor result for the same driving resistance uses the open-circuit time constant of C_μ, $R_{\mu}=R_{sig}(1+g_mR_L)+R_L=716(199)+5000=148\ \text{k}\Omega$, giving $1/(2\pi R_\mu C_\mu)=539\ \text{kHz}$ — within 3.5 % of the Miller estimate, which confirms the input node is genuinely dominant.
Part (d) — Evaluate the second pole at the collector. The output half of the Miller split works against R_L: $$\boxed{f_{p2}=\frac{1}{2\pi R_L C_{out}}=\frac{1}{2\pi(5\ \text{k}\Omega)(2.01\ \text{pF})} =1.58\times10^{7}\ \text{Hz}=15.8\ \text{MHz}}$$ That is 28 times above f_H, so the single-pole roll-off assumed in part (c) is justified; beyond f_p2 the response falls at 40 dB/decade. (The exact transfer function also carries a right-half-plane zero at $g_m/(2\pi C_\mu)=3.2\ \text{GHz}$, far outside the useful band.)

Q2 — straight-line magnitude response. The mid-band plateau of 45.9 dB (198 V/V) runs from f_L = 44 Hz to f_H = 558 kHz, with the second pole at 15.8 MHz steepening the roll-off to 40 dB/decade.

Part	Quantity	Result
—	Operating point	I_C = 0.990 mA, g_m = 39.6 mA/V, r_π = 2.53 kΩ
(a)	Mid-band gain v_OUT/v_S	−198 V/V (45.9 dB, inverting)
(b)	Lower 3 dB frequency f_L	44.4 Hz
(c)	Upper 3 dB frequency f_H	5.6 × 10⁵ Hz (558 kHz; 539 kHz by exact time constant)
(d)	Second high-frequency pole f_p2	1.58 × 10⁷ Hz (15.8 MHz)

Check — the source resistance used in parts (c) and (d). The schematic draws v_S as an ideal voltage source with no series resistance, and R₁ as a shunt from the base node to ground (which is what gives the base its DC path). Taken absolutely literally, the base node would then be driven from zero ohms, the Miller input pole would recede to infinity, and the circuit would have a single pole at $1/(2\pi R_L C_\mu)=15.9\ \text{MHz}$ — leaving part (d) with nothing to answer. The reading used here, and the only one that makes both (c) and (d) earnable while using the given R₁, is that the base node is driven through the amplifier's own input resistance $R_1\parallel r_\pi=716\ \Omega$. Part (a), which does not depend on that resistance, is unaffected either way.

Topic. Frequency response of a bypassed common-emitter stage: coupling-capacitor low corner, Miller-dominated high corner.

Key relations. $g_m=I_C/V_T$, $r_\pi=\beta/g_m$, $I_C=\alpha I_E$; mid-band $A_v=-g_mR_L$; low corner $f_L=1/(2\pi C_1 R_{C1})$ with R_C1 the resistance seen by the coupling capacitor; Miller's theorem $C_{in}=C_\mu(1+|A_v|)$ and $C_{out}=C_\mu(1+1/|A_v|)$; open-circuit time constant $R_\mu=R_{sig}(1+g_mR_L)+R_L$.

Why this works. A capacitor only matters where its impedance is comparable with the resistance it faces, so the whole of frequency-response analysis reduces to identifying one resistance per capacitor. The coupling capacitor faces the input resistance and therefore controls the bottom of the band. C_μ is different because it bridges input and output: as the collector swings 198 times harder than the base and in the opposite direction, the charge the source must supply to C_μ is 199 times what a grounded 2 pF would demand, which is exactly what Miller's theorem formalises. That is why a 2 pF parasitic, not the 5 μF coupling capacitor, is what limits this amplifier at the top of its band, and why the gain–bandwidth product of the stage is roughly constant.

Common pitfall. Using I_C = I_bias = 1 mA rather than $\alpha I_{bias}$ (a 1 % error here, but a real one when β is small), and forgetting that C₂ = ∞ removes the emitter degeneration entirely — students who leave an emitter resistor in the gain expression lose a factor of several. The other classic slip is applying the Miller multiplier to C_μ at the output node as well: the output side sees only $C_\mu(1+1/|A_v|)\approx C_\mu$, not $C_\mu(1+|A_v|)$.

Source. Sedra & Smith, Microelectronic Circuits, 8th ed., §10.1–10.5 (low- and high-frequency response, Miller effect, open-circuit time constants); Razavi, Fundamentals of Microelectronics, 2nd ed., §11.1–11.4.

Question 2 of 5: Common-emitter amplifier — mid-band gain and the two 3 dB frequencies

Question 2: Common-emitter amplifier — mid-band gain and the two 3 dB frequencies (20 marks)