Question 1 of 5: DC Bias, Gain and Terminal Resistances of a PNP–NPN Two-Stage Amplifier

Question 1: DC Bias, Gain and Terminal Resistances of a PNP–NPN Two-Stage Amplifier (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.

Given. A direct-coupled pair: a PNP common-emitter stage feeding an NPN emitter follower, both running from split ±5 V rails. All coupling and bypass capacitors are marked \(\infty\), so they are dc opens and ac shorts.

Find. The six dc bias quantities, then the mid-band voltage gain \(v_{out}/v_{in}\), the input resistance seen by the source and the output resistance seen by \(R_L\).

Approach. Solve the dc bias by writing one node equation at each base — keeping the finite base currents, because \(\beta = 100\) is not large enough to ignore them against a 30 kΩ divider — then build the hybrid-\(\pi\) model with \(r_o = \infty\) and cascade the common-emitter gain with the follower gain.

1. Part (a) — find the base voltage of \(Q_1\) from the divider, loaded by the base current. For a PNP the base current leaves the base and flows into the divider node, so KCL there reads \[\frac{V_{CC}-V_{B1}}{R_A}+\frac{I_{E1}}{\beta+1}=\frac{V_{B1}+V_{CC}}{R_B},\qquad I_{E1}=\frac{V_{CC}-(V_{B1}+0.7)}{R_{E1}}.\] Substituting \(V_{CC}=5\ \text{V}\), \(R_A=10\ \text{k}\Omega\), \(R_B=30\ \text{k}\Omega\) and \(R_{E1}=2\ \text{k}\Omega\) and solving the single linear equation gives \[\boxed{V_{B1}=2.564\ \text{V}}\] An unloaded divider would have put the base at exactly 2.50 V; the 8.6 µA of base current pulls it 64 mV higher.
2. Emitter voltage and emitter current of \(Q_1\). The PNP emitter sits one junction drop above its base, and the 2 kΩ resistor carries whatever the rail can push through it: \[V_{E1}=V_{B1}+0.7=3.264\ \text{V},\qquad I_{E1}=\frac{5-3.264}{2\ \text{k}\Omega}=0.868\ \text{mA}.\] With \(\alpha=\beta/(\beta+1)=0.9901\) the collector current follows as \[\boxed{I_{C1}=\alpha I_{E1}=0.859\ \text{mA}},\qquad I_{B1}=\frac{I_{E1}}{101}=8.59\ \mu\text{A}.\]
3. Collector node of \(Q_1\), which is also the base of \(Q_2\). The 0.859 mA delivered by the PNP collector splits between the 6 kΩ resistor and the follower's base current: \[I_{C1}=\frac{V_{C1}+V_{CC}}{R_{C1}}+\frac{1}{\beta+1}\cdot\frac{(V_{C1}-0.7)+V_{CC}}{R_{E2}}.\] Solving for the one unknown gives \[\boxed{V_{C1}=0.129\ \text{V}}\] The follower steals only 4.4 µA, so the 6 kΩ resistor carries almost all of the collector current — but the 0.03 V that the base current costs is still worth carrying.
4. Follower emitter and its collector current. \(Q_2\) is NPN, so its emitter sits one drop below the base: \[V_{E2}=V_{C1}-0.7=-0.571\ \text{V},\qquad I_{E2}=\frac{-0.571+5}{10\ \text{k}\Omega}=0.443\ \text{mA},\] \[\boxed{I_{C2}=\alpha I_{E2}=0.439\ \text{mA}}\] Both devices are comfortably active: \(V_{EC1}=3.264-0.129=3.14\ \text{V}\) and \(V_{CE2}=5-(-0.571)=5.57\ \text{V}\), each far above the 0.3 V saturation limit, so the small-signal model of part (b) is legitimate.
5. Part (b) — build the hybrid-\(\pi\) parameters. \[g_{m1}=\frac{I_{C1}}{V_T}=\frac{0.859\ \text{mA}}{25\ \text{mV}}=34.4\ \text{mA/V},\qquad r_{\pi1}=\frac{\beta}{g_{m1}}=2.91\ \text{k}\Omega,\] \[g_{m2}=\frac{I_{C2}}{V_T}=17.54\ \text{mA/V},\qquad r_{\pi2}=\frac{\beta}{g_{m2}}=5.70\ \text{k}\Omega.\] Because \(|V_A|=\infty\) every \(r_o\) is an open circuit, which is what keeps the algebra to one line per node.
6. Load the first stage with the resistance looking into the follower's base. The follower's emitter carries \(R_{E2}\parallel R_L = 10\parallel 1 = 0.909\ \text{k}\Omega\) at signal frequencies, and that resistance is reflected to the base multiplied by \(\beta+1\): \[R_{in2}=r_{\pi2}+(\beta+1)\left(R_{E2}\parallel R_L\right)=5.70+101(0.909)=97.5\ \text{k}\Omega.\] The ac load on \(Q_1\)'s collector is therefore \(R_{C1}\parallel R_{in2}=6\parallel 97.5=5.65\ \text{k}\Omega\) — the follower barely loads it, which is precisely why the follower is there.
7. Cascade the two stage gains. The 2 kΩ emitter resistor is bypassed by the infinite capacitor, so \(Q_1\)'s emitter is an ac ground and the stage is an undegenerated common emitter: \[A_{v1}=\frac{v_{b2}}{v_{in}}=-g_{m1}\left(R_{C1}\parallel R_{in2}\right)=-34.4\times 5.65=-194.3,\] \[A_{v2}=\frac{v_{out}}{v_{b2}}=\frac{(\beta+1)(R_{E2}\parallel R_L)}{r_{\pi2}+(\beta+1)(R_{E2}\parallel R_L)}=\frac{91.8}{97.5}=0.9415.\] Multiplying the two, \[\boxed{A_v=\frac{v_{out}}{v_{in}}=A_{v1}A_{v2}=-182.9\ \text{V/V}}\] which is 45.2 dB with a 180° inversion. A three-unknown nodal solve of the same network returns −182.90, confirming the cascade.
8. Part (c) — input resistance. Both divider resistors terminate on supply rails, which are ac grounds, so they appear in parallel with the base: \[R_{in}=R_A\parallel R_B\parallel r_{\pi1}=10\parallel 30\parallel 2.91=\boxed{2.10\ \text{k}\Omega}\] The 7.5 kΩ divider is what limits this figure; \(r_{\pi1}\) alone would give 2.91 kΩ.
9. Part (d) — output resistance looking back into the follower. Looking left from the load, the base circuit of \(Q_2\) is \(R_{C1}=6\ \text{k}\Omega\) (the PNP collector is an ideal current source with \(r_o=\infty\)), and the follower divides that plus \(r_{\pi2}\) by \(\beta+1\): \[R_{out}=R_{E2}\parallel\frac{r_{\pi2}+R_{C1}}{\beta+1}=10\ \text{k}\Omega\parallel\frac{5.70+6.00}{101}\ \text{k}\Omega=10\ \text{k}\Omega\parallel 115.9\ \Omega,\] \[\boxed{R_{out}=114.5\ \Omega}\] This is the whole point of the topology: a 5.65 kΩ high-impedance gain node is delivered to a 1 kΩ load through 115 Ω, so the load costs only 6 % of the available signal.