Question 4 of 7: Current-Source-Biased Common-Emitter Amplifier
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Comp-A1, Electronics — National Exams, December 2016. Open-book, 3 hours; seven 20-mark questions, five to be marked (all seven answered below as a complete study resource, per the standing "answer all M" rule). Unless stated otherwise, diode drops $V_D=0.7\text{V}$.
Reference texts: Sedra & Smith, Microelectronic Circuits (diode limiters, MOSFET current-mirror biasing and CS amplifiers, active-RC matched-feedback filters, BJT current-source-biased CE amplifiers, RC-ladder (Wien-bridge type) sinusoidal oscillators, CMOS static logic synthesis and sizing, flash-ADC comparator/encoder networks) — the single reference text covering every question on this paper.
Given. $I_E=0.2\text{mA}$ (set by the ideal current source); $\beta=100$; $V_A=100\text{V}$; $R_C=5\text{k}\Omega$; $R_L=5\text{k}\Omega$; $R_S=100\,\Omega$; coupling caps $C_1,C_2$ ideal (short at signal frequency).
Given data
Quantity
Value
$I_E$
$0.2$ mA
$\beta$
$100$
$V_A$
$100$ V
$R_C$
$5$ k$\Omega$
$R_L$
$5$ k$\Omega$
Find. Small-signal model; $R_i$, $R_o$; open-circuit and loaded voltage gain.
Approach. Because $C_2$ AC-bypasses the emitter to ground, this is a plain common-emitter stage: $R_i$ is looked into the base ($r_\pi$ only, per the figure's own labelled arrow), $R_o$ is looked into the collector ($R_C\parallel r_o$), and $C_1$'s ideal short means $R_L$ simply appears in parallel with $R_o$ at the output.
Part (b) — small-signal parameters and $R_i$, $R_o$. With $I_E=0.2\text{mA}$,
$$I_C=I_E\cdot\frac{\beta}{\beta+1}=0.2\text{mA}\times\frac{100}{101}=0.198\text{ mA}$$
$$g_m=\frac{I_C}{V_T}=\frac{0.198\text{mA}}{25\text{mV}}=\boxed{7.92\text{ mA/V}},\qquad r_\pi=\frac{\beta}{g_m}=\frac{100}{7.92\text{mA/V}}=\boxed{12.6\text{ k}\Omega}$$
$$r_o=\frac{V_A}{I_C}=\frac{100}{0.198\text{mA}}=505\text{ k}\Omega$$
$$R_i=\boxed{r_\pi=12.6\text{ k}\Omega},\qquad R_o=R_C\parallel r_o=\frac{5\text{k}\times505\text{k}}{510\text{k}}=\boxed{4.95\text{ k}\Omega}$$
Part (c) — open-circuit and loaded gain. With $C_1$ an ideal short, the open-circuit gain (output unloaded, $R_L$ removed) is measured straight at the collector:
$$A_{v,\text{oc}}=-g_m R_o=-(7.92\text{mA/V})(4.95\text{k}\Omega)=\boxed{-39.2\text{ V/V}}$$
Loading with $R_L=5\text{k}\Omega$ in parallel with $R_o$:
$$R_o\parallel R_L=\frac{4.95\text{k}\times5\text{k}}{9.95\text{k}}=2.49\text{k}\Omega\;\Rightarrow\;A_{v,\text{ld}}=-g_m(R_o\parallel R_L)=\boxed{-19.7\text{ V/V}}$$
(equivalently $A_{v,\text{ld}}=A_{v,\text{oc}}\times\dfrac{R_L}{R_L+R_o}=-39.2\times0.503=-19.7$, confirming the voltage-divider view of loading.)
Fig. Q4(a) — small-signal model: source through $R_S$ into $r_\pi$ (input resistance $R_i=r_\pi$), dependent source $g_m v_\pi$ at the collector loaded by $r_o\parallel R_C$ ($C_1,C_2$ ideal shorts collapse the emitter to AC ground and pass the collector signal straight to $v_o$).
