Question 11 of 11: BJT bias-point analysis, five sub-circuits

Question 11: BJT bias-point analysis, five sub-circuits [4 × 5 = 20]

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 (all parts). $\beta=50$, $V_{BE,on}=V_{EB,on}=0.6\ \text{V}$, $V_{CE,sat}=V_{EC,sat}=0.3\ \text{V}$, $V_A=\infty$ (so $r_o=\infty$, irrelevant to these dc bias problems).

Find. The labelled node voltage in each of the five sub-circuits. Note: in parts (b) and (c) the source labels the probed node “$V_E$”; per the drawn schematics this node is physically the PNP’s emitter terminal in both cases (the 100 Ω/470 Ω resistor to +5V lands directly on the emitter, identified by the arrow pointing into the base) — so the printed label matches the physical terminal once the transistor polarity (PNP, not NPN) is read correctly from the arrow direction.

Approach. For each sub-circuit, first try active-region operation (fixed $V_{BE,on}$, $I_C=\beta I_B$); if that predicts $V_{CE}$ (or $V_{EC}$) below the saturation value, redo with both $V_{BE,on}$ and $V_{CE,sat}$ fixed and solve the resulting linear KCL for the remaining node voltage, then confirm $I_C<\beta I_B$ (a saturation check).

1. (a) NPN, emitter grounded. Try active: $$I_B=\frac{2-0.6}{20\,\text{k}}=70\ \mu\text{A},\quad I_C=\beta I_B=3.5\ \text{mA},\quad V_C=5-I_C(1\text{k})=1.5\ \text{V}.$$ Check: $V_{CE}=1.5-0=1.5\ \text{V}>V_{CE,sat}=0.3\ \text{V}$ — active is valid. $$\boxed{V_C=1.5\ \text{V}}.$$
2. (b) PNP, emitter via 100 Ω to +5V. Trying active gives an impossible negative $V_{EC}$, so the device is saturated: fix $V_{EB}=0.6$, $V_{EC,sat}=0.3$ and solve KCL at the base node $V_B$ (using $V_E=V_B+0.6$, $V_C=V_E-0.3$): $$\frac{4.4-V_B}{100}=\frac{V_B-2}{10\,\text{k}}+\frac{V_B+0.3}{1\,\text{k}}\ \Rightarrow\ V_B=3.955\ \text{V}.$$ Then $$V_E=V_B+0.6=\boxed{4.555\ \text{V}}.$$ Check: $I_E=4.45\ \text{mA}$, $I_B=0.1955\ \text{mA}$, $I_C=4.255\ \text{mA}<\beta I_B=9.77\ \text{mA}$ — saturation confirmed.
3. (c) PNP, base grounded. With $V_B=0$ fixed, $V_E=V_B+0.6=0.6\ \text{V}$ regardless of region (the constant-drop model fixes $V_{EB}=0.6$ whenever the junction conducts). Testing shows $V_C$ from active operation would exceed $V_E$ (impossible for a PNP), so it is saturated: $V_C=V_E-0.3=0.3\ \text{V}$. Check: $I_E=(5-0.6)/470=9.36\ \text{mA}$, $I_C=(0.3+5)/1\text{k}=5.30\ \text{mA}$, $I_B=I_E-I_C=4.06\ \text{mA}$; $\beta I_B=203\ \text{mA}\gg I_C$ — saturation confirmed. $$\boxed{V_E=0.6\ \text{V}}.$$
4. (d) NPN, base only via 5 kΩ to ground. With no other bias path, the base current must come entirely from ground through the 5 kΩ (so $V_B<0$). Fixing $V_{BE}=0.6$ (⇒ $V_E=V_B-0.6$) and testing active first also gives an impossible $V_{CE}$, so saturated: $V_C=V_E+0.3=V_B-0.3$. KCL ($I_E=I_B+I_C$): $$\frac{V_B+4.4}{1\,\text{k}}=\frac{-V_B}{5\,\text{k}}+\frac{5.3-V_B}{2\,\text{k}}\ \Rightarrow\ V_B=-1.029\ \text{V}.$$ $$V_C=V_B-0.3=\boxed{-1.329\ \text{V}}.$$ Check: $I_B=0.206\ \text{mA}$, $I_C=3.165\ \text{mA}<\beta I_B=10.29\ \text{mA}$ — saturation confirmed.
5. (e) Cross-coupled Q1–Q2 feedback pair. Q2’s base and Q1’s collector are the same node ($=V_{C1}$); Q2’s collector and Q1’s base are the same node ($=V_Y$). Guess Q1 (NPN, emitter grounded) saturated: $V_Y=V_{BE1}=0.6\ \text{V}$, $$V_{C1}=V_Y+V_{CE1,sat}=0.6+0.3=\boxed{0.9\ \text{V}}.$$ Then Q2’s emitter: $V_{E2}=V_{C1}+V_{EB2}=0.9+0.6=1.5\ \text{V}$, giving $I_{E2}=(5-1.5)/100=35\ \text{mA}$. Trying Q2 active: $I_{B2}=I_{E2}/(\beta+1)=0.686\ \text{mA}$, $I_{C2}=\beta I_{B2}=34.3\ \text{mA}=I_{B1}$. Check Q2: $V_{EC2}=V_{E2}-V_Y=1.5-0.6=0.9\ \text{V}>0.3\ \text{V}$ — active is valid for Q2. Check Q1: $I_{C1}=(5-0.9)/2\text{k}+I_{B2}=2.05+0.686=2.74\ \text{mA}\ll\beta I_{B1}=1715\ \text{mA}$ — deep saturation confirmed for Q1, self-consistent.