Question 5 of 5: DC Bias of a Two-Stage NPN–PNP Amplifier
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2017 · 16-Elec-A5, Electronics. Closed-book, 3 hours. Answer all FIVE (5) questions; each worth 20 marks. Unless stated, op-amps are ideal and supply rails are ±15 V.
Reference texts: A. S. Sedra & K. C. Smith, Microelectronic Circuits, 7th ed. (Oxford) — Ch. 4 (diode circuits & rectifiers), Ch. 5–7 (MOSFET and BJT amplifiers, biasing and small-signal models), Ch. 2 & 9 (ideal op-amps, closed-loop bandwidth and slew rate). Cross-reference: R. C. Jaeger & T. N. Blalock, Microelectronic Circuit Design.
Question 5: DC Bias of a Two-Stage NPN–PNP Amplifier (20 marks)
Given. Stage 1 is an NPN ($Q_1$) with divider bias ($R_{B1}$ from $V_{CC}$, $R_{B2}$ to ground), collector load $R_{C1}$, emitter resistor $R_{E1}$. Stage 2 is a PNP ($Q_2$) whose base ties directly to the $Q_1$ collector node, emitter to $V_{CC}$ through $R_{E2}$, collector to ground through $R_{C2}$.
Given data — Question 5
$R_{B1}$
$R_{B2}$
$R_{C1}$
$R_{E1}$
$R_{C2}$
$R_{E2}$
$V_{CC}$
$\beta$
100 kΩ
50 kΩ
5 kΩ
3 kΩ
2.7 kΩ
2 kΩ
+15 V
100
Find. All node voltages and terminal currents for both transistors ($V_{BE}=0.7\text{ V}$ assumed).
[Figure not reproduced: Q5: two-stage DC-coupled amplifier - NPN Q1 (stage 1) driving PNP Q2 (stage 2); Q2 base ties directly to Q1 collector. See the official exam paper or the cited reference text.]
Q5: two-stage DC-coupled amplifier - NPN Q1 (stage 1) driving PNP Q2 (stage 2); Q2 base ties directly to Q1 collector.
Approach. Solve stage 1 with a Thévenin base equivalent (it is independent of stage 2). The $Q_1$ collector node also feeds $Q_2$’s base, so write one KCL at that node to find $V_{C1}=V_{B2}$, then propagate through the PNP stage; finish with active-region checks.
Stage-1 base equivalent. Thévenin of the $R_{B1}/R_{B2}$ divider:
$$V_{BB}=V_{CC}\frac{R_{B2}}{R_{B1}+R_{B2}}=15\cdot\frac{50}{150}=5\text{ V},\qquad R_{BB}=R_{B1}\parallel R_{B2}=33.3\text{ k}\Omega.$$
Couple to $Q_2$ — KCL at the $V_{C1}=V_{B2}$ node. The $Q_2$ base current (a PNP, so its conventional base current flows out of the base into this node) shares the node with $R_{C1}$ and the $Q_1$ collector:
$$\frac{V_{CC}-V_{C1}}{R_{C1}}+I_{B2}=I_{C1},\qquad I_{B2}=\frac{I_{E2}}{\beta+1},\quad I_{E2}=\frac{V_{CC}-(V_{C1}+0.7)}{R_{E2}}.$$
Substituting $I_{E2}$ and solving the single linear equation for $V_{C1}$ gives
$$\boxed{V_{C1}=V_{B2}=8.75\text{ V}.}$$
Stage-2 currents and voltages. For the PNP, $V_{E2}=V_{B2}+0.7=9.45\text{ V}$, hence
$$I_{E2}=\frac{V_{CC}-V_{E2}}{R_{E2}}=\frac{15-9.45}{2\text{ k}\Omega}=2.78\text{ mA},\quad I_{B2}=\frac{I_{E2}}{101}=27.5\ \mu\text{A},\quad I_{C2}=\beta I_{B2}=2.75\text{ mA}.$$
The collector sits at
$$V_{C2}=I_{C2}R_{C2}=(2.75\text{ mA})(2.7\text{ k}\Omega)=7.43\text{ V}.$$
Region checks. $Q_1$: $V_{C1}=8.75\text{ V}$ is well above $V_{B1}=4.57\text{ V}$, so its collector junction is reverse-biased — active. $Q_2$ (PNP): the terminal ordering is $V_{E2}=9.45\text{ V}$ above $V_{B2}=8.75\text{ V}$ above $V_{C2}=7.43\text{ V}$, i.e. emitter–base forward and base–collector reverse — active. Both assumptions hold, and the node KCL closes to zero.
