Question 5 of 5: BJT DC Bias — Solve for the Node Voltages

Question 5: BJT DC Bias — Solve for the Node Voltages (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. Five small bias cells (mixed NPN/PNP).

Find. the indicated node voltage in each of the five cells (a)–(e).

Approach. For each cell assume forward-active, solve, then check the region: an NPN needs $V_C\gt V_B$ and a PNP needs $V_C\lt V_B$ — if the active solution violates this, redo in saturation with $\lvert V_{CE}\rvert=0.3\text{ V}$.

1. (a) NPN, emitter grounded. $V_B=V_{BE}=0.6\text{ V}$, so $I_B=(2-0.6)/20\text{k}=0.07\text{ mA}$ and $I_C=\beta I_B=3.5\text{ mA}$. Then $$V_C=5-I_C(1\text{k})=5-3.5=\boxed{1.5\text{ V}}.$$ Active check: $V_C=1.5\gt V_B=0.6$. Valid.
2. (b) PNP, emitter to +5 V through 100 $\Omega$. An active solution demands $I_C\approx 7.9\text{ mA}$, i.e. $V_C\approx 7.9\text{ V}\gt V_E$ — impossible for a PNP, so the device saturates ($V_{EC}=0.3$, $V_C=V_E-0.3$). Writing KCL $I_E=I_B+I_C$ with the base fed through 10 kΩ to +2 V and the collector through 1 kΩ to ground gives $111\,V_E=505.6$, so $$V_E=\boxed{4.56\text{ V}}\quad(V_C=4.26\text{ V},\ V_B=3.96\text{ V}).$$ Saturation check: $I_C/I_B\approx 22\lt\beta$. Valid.
3. (c) PNP with base grounded. The base sits at 0 V, so a conducting emitter-base junction pins the emitter one $V_{EB}$ above it: $$V_E=0+V_{EB,on}=\boxed{+0.6\text{ V}},$$ independent of the 470 $\Omega$/1 kΩ loads. (The collector runs to $-5\text{ V}$, so the device is on; whether it is active or saturated does not change $V_E$.)
4. (d) NPN, base biased through 5 kΩ to ground. Assuming active gives $V_C\approx-2.86\text{ V}\lt V_B$, i.e. the collector-base junction is forward biased — the device saturates ($V_C=V_E+0.3=V_B-0.3$). KCL $I_E=I_B+I_C$ with $I_B=-V_B/5\text{k}$, $I_C=(5-V_C)/2\text{k}$, $I_E=(V_E+5)/1\text{k}$ gives $17\,V_B=-17.5$, hence $V_B=-1.03\text{ V}$ and $$V_C=V_B-0.3=\boxed{-1.33\text{ V}}.$$ Saturation check: $\beta I_B=10.3\text{ mA}\gt I_C=3.2\text{ mA}$. Valid.
5. (e) Two-NPN feedback pair. $Q_1$ has its emitter grounded, so its base sits at $V_{BE}=0.6\text{ V}$; $Q_2$’s base is the $V_{C1}$ node and its emitter drives $Q_1$’s base, so $V_{C1}=V_{BE,Q2}+V_{BE,Q1}$: $$V_{C1}=2\,V_{BE,on}=\boxed{1.2\text{ V}}.$$ The 2 kΩ then carries $(5-1.2)/2\text{k}=1.9\text{ mA}=I_{C1}$, consistent with $V_{C1}=5-1.9\text{ mA}\cdot2\text{k}=1.2\text{ V}$; both transistors are active.