17-Phys-A5, Undated paper, Q2

Question 2 (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.

Symbol	Value
$\beta$	50
$V_{EB}$	$0.7\,\text{V}$
$V_{CC}$	$+3\,\text{V}$ (both rails)
$R_E$	$1\,\text{k}\Omega$
$R_1,\,R_2$ (base divider)	$102\,\text{k}\Omega,\ 54\,\text{k}\Omega$
$R_B$	$200\,\text{k}\Omega$
$R_C$	$560\,\Omega$

$Q_1$ is a PNP in a common-base configuration: emitter at the top node (biased through $R_E$ from $+3\text{V}$, driven by $v_i$ through a large coupling capacitor, with $R_i$ the resistance looking into the emitter node); base bypassed to true AC ground by a large capacitor sitting directly at the base terminal, DC-biased through $R_B$ from the $R_1/R_2$ divider off $+3\text{V}$; collector loaded by $R_C$ to ground, with $v_o$ taken there.

Find. (a) $I_C$. (b) $R_i$ (input resistance looking into the emitter, i.e. $R_E\parallel r_e$).

Fig. 2 — PNP common-base stage: base bypassed to AC ground directly at its own terminal, $R_B$ isolating the DC bias divider from that AC ground.

Approach. Thevenize the base bias network (including $R_B$ in series with $R_1\parallel R_2$), write the emitter–base KVL loop, solve for $I_B$, then get $I_C=\beta I_B$; for (b), find the small-signal $r_e$ from $I_E$ and combine with $R_E$.

Part (a) — Thevenin equivalent at the base. $$\begin{aligned} V_{th}&=V_{CC}\frac{R_2}{R_1+R_2}=3\times\frac{54}{156}=1.038\,\text{V} \\ R_{th}&=R_B+(R_1\parallel R_2)=200+35.31=235.3\,\text{k}\Omega \end{aligned}$$
Emitter–base KVL. $V_E=V_{CC}-I_E R_E$, $V_B=V_E-V_{EB}$, and from the Thevenin side $V_B=V_{th}+I_B R_{th}$ (base current flows out of a PNP's base, through $R_{th}$, back toward the supply). With $I_E=(\beta+1)I_B$: $$V_{CC}-V_{EB}-(\beta+1)I_B R_E = V_{th}+I_B R_{th}$$ $$I_B=\frac{V_{CC}-V_{EB}-V_{th}}{R_{th}+(\beta+1)R_E}=\frac{3-0.7-1.038}{235.3\text{k}+51\times1\text{k}}=\frac{1.262\,\text{V}}{286.3\,\text{k}\Omega}=4.406\,\mu\text{A}$$
Collector current. $$\boxed{I_C=\beta I_B = 50\times 4.406\,\mu\text{A} = 0.2203\,\text{mA}}$$ and $I_E=(\beta+1)I_B=0.2247\,\text{mA}$, giving $V_E=3-0.2247=2.775\,\text{V}$, $V_B=2.075\,\text{V}$ — matching the Thevenin-side value $V_{th}+I_BR_{th}=1.038+4.406\mu\text{A}\times235.3\text{k}=2.075\,\text{V}$, which confirms the loop solution.
Part (b) — small-signal input resistance. With $V_T=25\,\text{mV}$: $$r_e=\frac{V_T}{I_E}=\frac{25\,\text{mV}}{0.2247\,\text{mA}}=111.3\,\Omega$$ The base is AC-grounded right at $Q_1$'s base terminal (the bypass cap sits before $R_B$), so this is a true common-base stage and the resistance looking into the emitter node is $R_E$ in parallel with $r_e$: $$\boxed{R_i = R_E\parallel r_e = \frac{(1000)(111.3)}{1000+111.3} = 100.1\,\Omega}$$

Quantity	Value
$I_B$	4.406 μA
$I_C$	0.2203 mA
$I_E$	0.2247 mA
$V_E,\ V_B$	2.775 V, 2.075 V
$r_e$	111.3 Ω
$R_i$	100.1 Ω

Concept & Theory

Topic: PNP common-base bias and small-signal input resistance.

Key relations: Thevenin base network ($V_{th}$, $R_{th}$); $I_E=(\beta+1)I_B$; $r_e=V_T/I_E$; common-base $R_{in}=R_E\parallel r_e$.

Why this works: putting the bypass capacitor directly at the base terminal (rather than at the far side of $R_B$) makes the base a true AC ground independent of $R_B$'s value; $R_B$ only ever appears in the DC bias loop, isolating the divider from loading that AC ground. The emitter then sees only $R_E$ in parallel with the transistor's own $r_e$.

Common pitfall: using $V_{BE}$ sign conventions from an NPN or forgetting the $V_1$-style offset (here $V_{EB}$, not $V_{BE}$) flips the KVL sign; also reporting $R_i=r_e$ alone and forgetting the parallel $R_E$.

Source: Sedra & Smith, Microelectronic Circuits, Ch. 6 (BJT biasing; common-base amplifier input resistance).

Question 2 of 6

Question 2 (20 marks)