Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Phys-A5-B — Analog and Digital Electronic Circuits — National Exams, December 2018
3 hours duration. Open book exam. Answer all SIX (6) questions.
Reference texts: A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. (BJT/MOSFET biasing and small-signal amplifiers Ch. 6–7, current-mirror loads and cascodes Ch. 7–8, feedback amplifiers Ch. 10, high-frequency response and open-circuit time constants Ch. 9, D/A converters Ch. 17); M. M. Mano & M. D. Ciletti, Digital Design, 6th ed. (Boolean algebra Ch. 2, static CMOS logic gates Ch. 10).
Given. BJT cascode: $Q_1$ common-emitter (base driven by $v_i$, emitter grounded), $Q_2$ common-base (its base is tied to the node between $R_L$ and a very large bypass capacitor, so the base is an AC ground); $Q_2$'s collector is the output node $v_O$, loaded by $R_L=10\,\text{k}\Omega$; $I_{bias}=0.5\,\text{mA}$ flows into $v_O$ from $V_{CC}$. $V_{BE}=0.7\,\text{V}$, $\beta=100$, $V_A=\infty$ (so $r_{o1}=r_{o2}=\infty$).
Find. The midband small-signal gain $A_M=v_o/v_i$.
[Figure not reproduced: Fig. 1 from the exam paper. See the official exam paper or the cited reference text.]
Fig. 1 (reproduced from the exam paper) — BJT cascode amplifier: $Q_1$ (common-emitter) drives $Q_2$ (common-base, base AC-grounded through the $\infty$ capacitor); output taken at $Q_2$'s collector through $R_L$.
Approach. Find the DC bias current $I_{C1}$ from KCL at the cascode node, then combine $g_{m1}$ with the current gain from $Q_1$'s collector to $Q_2$'s collector to get $A_M$.
Part (a) — DC bias, find $I_{C1}$. The bypass capacitor blocks DC, so in steady state $R_L$ carries only $Q_2$'s base current down into the $Q_2$-base node. KCL at $v_O$: $I_{bias}=I_{C2}+I_{B2}=I_{C2}\left(1+\tfrac{1}{\beta}\right)=I_{E2}$. The node joining $Q_1$'s collector and $Q_2$'s emitter has only these two branches, so $I_{C1}=I_{E2}=I_{bias}=0.5\,\text{mA}$.
Transconductance of $Q_1$. With $V_T=25\,\text{mV}$: $$g_{m1}=\frac{I_{C1}}{V_T}=\frac{0.5\,\text{mA}}{25\,\text{mV}}=20\,\text{mA/V}$$
Current transfer through the cascode. Because $r_{o1}=r_{o2}=\infty$ and $Q_2$'s base is AC-grounded, every small-signal current that $Q_1$ pulls from the cascode node is supplied by $Q_2$'s emitter, i.e. $i_{e2}=g_{m1}v_i$; only the fraction $\alpha_2=\beta/(\beta+1)$ of an emitter current reaches the collector, so $i_{c2}=\alpha_2\, g_{m1}v_i$.
Assemble the gain. $v_o=-i_{c2}R_L$, so $$A_M=\frac{v_o}{v_i}=-\alpha_2\, g_{m1}R_L=-\frac{100}{101}(20\,\text{mA/V})(10\,\text{k}\Omega)=\boxed{-198.0\ \text{V/V}}$$
Check
Thermal voltage $V_T=25\,\text{mV}$ assumed (not stated in the source), the usual room-temperature value.
