Question 4 of 5: DC Bias Points of Five BJT Circuits

Question 4: DC Bias Points of Five BJT Circuits (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. $\beta=50$, $V_{BE,on}=V_{EB,on}=0.6$ V, $V_{CE,sat}=V_{EC,sat}=0.3$ V, $V_A=\infty$ (so $r_o=\infty$). For each part: assume forward-active, solve, then verify the region (active needs $V_{CE}\gt V_{CE,sat}$ for NPN, $V_{EC}\gt V_{EC,sat}$ for PNP); if violated, re-solve in saturation.

Find. The labelled node voltage in each of the five sub-circuits.

Approach. Each part is a one- or two-transistor DC bias problem; write the base-loop and collector-loop equations, use $I_C=\beta I_B$ (active) or the saturation drops, and check the region before quoting the answer.

1. (a) Grounded-emitter NPN. Emitter at ground so $V_B=V_{BE}=0.6$ V. $I_B=\dfrac{2-0.6}{20\ \text{k}}=70\ \mu\text{A}$, $I_C=\beta I_B=3.5\ \text{mA}$. Then $V_C=5-I_C(1\ \text{k})=5-3.5=\boxed{1.5\ \text{V}}$. Active check: $V_C=1.5\gt V_B=0.6$ (and $I_C\lt I_{C,sat}=4.7$ mA), so the assumption holds.
2. (b) PNP with emitter to +5 V (100 Ω) and collector to ground (1 kΩ). Assuming active gives $V_C\approx 8$ V $\gt$ 5 V, which is impossible — the transistor is saturated. With $V_E=V_B+0.6$ and $V_C=V_E-0.3$, node KCL $\big(\tfrac{4.4-V_B}{100}=\tfrac{V_B-2}{10\text{k}}+\tfrac{V_B+0.3}{1\text{k}}\big)$ gives $V_B=3.96$ V, so $$V_E=V_B+0.6=\boxed{4.56\ \text{V}}\quad(V_C=4.26\ \text{V}).$$ Saturation confirmed: $I_C=4.25\ \text{mA}\lt\beta I_B=9.8\ \text{mA}$.
3. (c) PNP with grounded base. The base sits at 0 V, so the emitter is exactly one junction above it: $$V_E=V_{EB,on}=\boxed{0.6\ \text{V}},$$ independent of the surrounding resistors. (The device is on and in fact saturated, $V_C=0.3$ V, with $I_E=(5-0.6)/470=9.4$ mA, but $V_E$ is fixed at $0.6$ V regardless.)
4. (d) NPN on split ±5 V supplies, base to ground through 5 kΩ. Assuming active drives $V_C$ below $V_E$ (impossible), so the transistor saturates. With $V_E=V_B-0.6$, $V_C=V_E+0.3$ and base current $I_B=-V_B/5\text{k}$, KCL $\big(\tfrac{V_B+4.4}{1\text{k}}=-\tfrac{V_B}{5\text{k}}+\tfrac{5.3-V_B}{2\text{k}}\big)$ yields $V_B=-1.03$ V, hence $$V_C=V_E+0.3=\boxed{-1.33\ \text{V}}.$$ Saturation confirmed: $I_C=3.2\ \text{mA}\lt\beta I_B=10.3\ \text{mA}$.
5. (e) Complementary cross-coupled pair. $Q_2$ (PNP) has its emitter tied to +5 V through 100 Ω and its base tied to the same +5 V rail through $R_D=2\ \text{k}\Omega$ (the $V_{C1}$ node). With no current, both the emitter and the base of $Q_2$ float up to $+5$ V, so $V_{EB2}=0\lt0.6$ V and $Q_2$ cannot turn on. $Q_1$’s only base-drive path is $Q_2$’s collector, so $Q_1$ also stays off. With $Q_1$ off no current flows in $R_D$ and $$\boxed{V_{C1}=+5\ \text{V}}.$$