Question 4 of 7: BJT Emitter-Follower with Current-Source Bias
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Comp-A1, Electronics — National Exams, December 2018. Open-book, 3 hours; seven 20-mark questions, five to be marked (all seven answered below as a complete study resource). All diodes $V_D=0.7\text{V}$ unless stated otherwise.
Reference texts: Sedra & Smith, Microelectronic Circuits (diode limiters, MOSFET/BJT small-signal amplifiers, active loads and current mirrors, op-amp imperfections, CMOS logic, oscillators) — the single reference text covering every question on this paper.
Question 4: BJT Emitter-Follower with Current-Source Bias (20 marks)
Given. NPN $Q_1$: collector tied directly to $+10\text{V}$ (AC ground); base fed from $V_s$ through $R_s=100\Omega$ and $C_1\to\infty$, biased to ground through $R_B=100\text{k}\Omega$; emitter set by an ideal $1\text{mA}$ current sink to $-10\text{V}$; output taken from the emitter through $C_2\to\infty$ into $R_L=5\text{k}\Omega$ — a classic common-collector (emitter-follower) stage.
Given data
Quantity
Value
$I$ (emitter current sink)
$1\text{mA}$
$\beta$
100
$V_A$
$100\text{V}$
$V_T$
$25\text{mV}$
$R_s,R_B,R_L$
$100\Omega,\ 100\text{k}\Omega,\ 5\text{k}\Omega$
Find. $V_C,V_B,V_E$; small-signal model; $R_i$, $R_o$; open-circuit and $R_L$-loaded voltage gain.
Fig. Q4-b — Hybrid-$\pi$ small-signal equivalent. The collector is AC ground (fixed $+10\text{V}$ supply); $r_\pi$ carries $v_\pi$ from base to emitter, and $g_mv_\pi$ together with $r_o$ return current from ground to the emitter node, which also feeds $R_L$.
Approach. Use the ideal current sink to fix $I_E=1\text{mA}$ directly, back out $I_C,I_B$ via $\beta$, then work down from the base ($V_B=-I_BR_B$, since $C_1$ blocks DC current from $V_s$) to find $V_E=V_B-V_{BE}$. For the AC analysis, recognize the emitter-follower topology and apply its standard $R_i$/$R_o$/gain formulas with the transistor's own $r_o$ providing the only "open-circuit" load.
Part (a) — DC bias point. The current sink fixes $I_E=1\text{mA}$ directly (independent of $\beta$):
$$I_C=\frac{\beta}{\beta+1}I_E=\frac{100}{101}(1\text{mA})=0.9901\text{ mA}$$
$$I_B=\frac{I_E}{\beta+1}=9.901\mu A$$
$C_1$ blocks DC, so the only DC path to the base is through $R_B$ to ground; the base current must be supplied from ground through $R_B$, which requires $V_B<0$:
$$V_B=-I_BR_B=-(9.901\mu A)(100\text{k}\Omega)=\boxed{-0.990\text{ V}}$$
$$V_E=V_B-V_{BE}=-0.990-0.7=\boxed{-1.690\text{ V}}$$
$$V_C=\boxed{+10.0\text{ V}}\text{ (tied directly to the supply)}$$
Part (b) — small-signal model parameters.
$$g_m=\frac{I_C}{V_T}=\frac{0.9901\text{mA}}{25\text{mV}}=\boxed{39.6\text{ mA/V}}$$
$$r_\pi=\frac{\beta}{g_m}=\boxed{2.53\text{ k}\Omega}\qquad r_o=\frac{V_A}{I_C}=\boxed{101\text{ k}\Omega}$$
Part (c) — input and output resistance. Looking into the base, the emitter sees $r_o\|R_L$ (the current sink is ideal, infinite $r_o$, so it drops out) reflected up by $(\beta+1)$:
$$R_{ib}=r_\pi+(\beta+1)(r_o\|R_L)=2.53\text{k}+101(101\text{k}\|5\text{k})$$
$$=2.53\text{k}+101(4.764\text{k})=483.6\text{ k}\Omega$$
$$R_i=R_B\|R_{ib}=100\text{k}\|483.6\text{k}=\boxed{82.9\text{ k}\Omega}$$
Looking back into the emitter (excluding $R_L$), the base is driven through $R_s\|R_B=100\|100\text{k}=99.9\,\Omega$:
$$R_o=\frac{r_\pi+(R_s\|R_B)}{\beta+1}=\frac{2525+99.9}{101}=\boxed{26.0\ \Omega}$$
Part (d) — open-circuit and loaded voltage gain. Open-circuit, the only emitter load is the transistor's own $r_o$ (the current sink is ideal):
$$A_{v,oc}=\frac{(\beta+1)r_o}{r_\pi+(R_s\|R_B)+(\beta+1)r_o}=\frac{101(101\text{k})}{2625+101(101\text{k})}=\boxed{0.9997\text{ V/V}}$$
Loaded by $R_L=5\text{k}\Omega$ (in parallel with $r_o$):
$$r_o\|R_L=101\text{k}\|5\text{k}=4.764\text{ k}\Omega$$
$$A_{v,L}=\frac{(\beta+1)(r_o\|R_L)}{r_\pi+(R_s\|R_B)+(\beta+1)(r_o\|R_L)}=\frac{101(4.764\text{k})}{2625+101(4.764\text{k})}=\boxed{0.9946\text{ V/V}}$$
Both are close to unity, as expected of an emitter follower — it buffers rather than amplifies.
