Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam — 07-Elec-A5 Electronics — December 2016.
Closed-book; any non-communicating calculator permitted; three hours. Five questions,
20 marks each — all five constitute a complete paper, so every question is solved
below. Per the paper’s notes, op-amps are ideal with supply rails of
±15 V unless a question states otherwise, and ground/chassis are common.
Reference texts (22-Elec-A5 Electronics): A. S. Sedra
& K. C. Smith, Microelectronic Circuits, 8th ed. (Oxford)
— diode circuits and rectifiers (Ch. 4), MOSFET biasing and amplifiers
(Ch. 5–7), op-amp circuits and offsets (Ch. 2/9), BJT amplifiers
(Ch. 6–7); R. C. Jaeger & T. N. Blalock,
Microelectronic Circuit Design (McGraw-Hill).
Given. A pnp transistor with its emitter (top) fed by a
1.5 mA current source from $+10$ V and bypassed to ground, base biased through
$R_B=50$ kΩ and driven by $v_{in}$, and collector (bottom) loaded by
$L=0.2$ mH to a $-10$ V rail (via $R_E=2$ kΩ) with the ac output taken
across a $40$ kΩ resistor.
Given data (Question 3)
Parameter
Value
Parameter
Value
$\beta$
100
$r_o$
40 kΩ
Emitter source $I$
1.5 mA
$R_B$
50 kΩ
$R_E$ (to $-10$ V)
2 kΩ
$L$
0.2 mH
Output $R$
40 kΩ
$v_{in}$
10 mV, 500 kHz
Find. (a) $V_B,V_E,V_C$; (b) a small-signal model and the approximate
amplitude of $v_o$.
Figure 3. pnp common-emitter stage: the
1.5 mA source sets the emitter current, the emitter is bypassed to ac ground, the base is
the input, and the collector is the output with an inductive load $L$ (dc short to the
$-10$ V rail through $R_E$).
Approach (a). The current source fixes $I_E=1.5$ mA; get $I_B$ and
$I_C$ from $\beta$, then walk the dc loops: the base current in $R_B$ sets $V_B$, the
emitter-base drop sets $V_E$, and $I_C$ through $R_E$ (the inductor is a dc short) sets $V_C$.
Terminal currents. The source sets $I_E=1.5$ mA, so
$$I_B=\frac{I_E}{\beta+1}=\frac{1.5}{101}=14.9\ \mu\text{A},\qquad
I_C=\beta I_B=1.485\ \text{mA}.$$
Base voltage. For the pnp, base current flows out of the base
through $R_B$ to ground, so $V_B$ sits above ground:
$$V_B=I_B R_B=(0.0149)(50)=\boxed{+0.74\ \text{V}}.$$
Emitter voltage. The emitter is one $V_{EB}$ above the base:
$$V_E=V_B+V_{EB}=0.74+0.7=\boxed{+1.44\ \text{V}}.$$
(The 1.5 mA source supplies whatever voltage this requires; the emitter node is at ac
ground through its bypass capacitor.)
Collector voltage. $I_C$ flows out of the collector down through the
inductor (a dc short) and $R_E$ to $-10$ V:
$$V_C=-10+I_C R_E=-10+(1.485)(2)=\boxed{-7.03\ \text{V}}.$$
Check: $V_{EC}=V_E-V_C=8.5$ V$\,\gt0.2$ V and $V_C\lt V_B$, so the pnp is
in the active region as assumed.
(b) Small-signal parameters.
$$g_m=\frac{I_C}{V_T}=\frac{1.485\ \text{mA}}{25\ \text{mV}}=59.4\ \text{mA/V},\qquad
r_\pi=\frac{\beta}{g_m}=1.68\ \text{k}\Omega.$$
The emitter is bypassed (ac ground) and $v_{in}$ drives the base directly, so
$v_\pi=v_{in}$ and this is a common-emitter stage. The model is drawn below.
Collector (output) load. The bottom of $L$ is bypassed to ac ground, so
the collector sees $j\omega L$ in parallel with $r_o$ and the $40$ kΩ output
resistor. At 500 kHz, $\omega L=2\pi(5\times10^5)(0.2\times10^{-3})=628\ \Omega$, which is
$\ll r_o\|40\text{ k}=20$ kΩ, so the load is essentially the inductor:
$$Z_C=j\omega L\,\|\,r_o\,\|\,40\text{ k}\approx j\,628\ \Omega,\qquad |Z_C|\approx628\ \Omega.$$
Output amplitude. For the common-emitter stage
$v_o=-g_m v_\pi Z_C$, so
$$|v_o|=g_m|Z_C|\,|v_{in}|=(0.0594)(628)(10\ \text{mV})=\boxed{\approx 0.37\ \text{V}}.$$
The gain magnitude is $g_m|Z_C|\approx37$; the low inductive impedance keeps it modest, and
the output leads because the load is essentially reactive.
Figure 3b. Small-signal (hybrid-π) model:
$v_\pi$ across $r_\pi$ (emitter at ac ground), the dependent source $g_m v_\pi$ into the
collector node, and the collector load $Z_C=j\omega L\|r_o\|40\text{ k}\Omega$.
