GATE 2024 EE Q.42: solution and answer

Official answer: ASource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Model: DC analysis with the NPN transistor in the active region, $V_{BE}=0.7\text{V}$, $I_C=\beta I_B$, $I_E=(\beta +1)I_B$, $r_o$ neglected.

2. Thevenin equivalent of the divider as seen from the base: $V_{TH}=V_{CC}\frac{R_2}{R_1+R_2}=12\times \frac{50}{150}=4\text{V}$ and $R_{TH}=R_1\parallel R_2=\frac{100\times 50}{150}=33.33\text{k}\Omega$.

3. Is the divider stiff? $\beta R_E=100\text{k}\Omega$ is only about $3R_{TH}$, not $\gg R_{TH}$, so the base current cannot be neglected. Use the exact loop equation.

4. KVL around the base-emitter loop: $V_{TH}=I_B{R}_{TH}+V_{BE}+(\beta +1)I_B{R}_E$, so $$I_B=\frac{V_{TH}-V_{BE}}{R_{TH}+(\beta +1)R_E}=\frac{4-0.7}{33.33+101\times 1}\text{mA}=\frac{3.3}{134.33}\text{mA}=24.57\mu \text{A}$$ (volts divided by kilohms gives milliamps).

5. Collector and emitter currents: $I_C=\beta I_B=100\times 24.57\mu \text{A}=2.457\text{mA}\approx 2.46\text{mA}$, and $I_E=(\beta +1)I_B=2.481\text{mA}$.

6. KVL around the collector-emitter loop: $$V_{CE}=V_{CC}-I_C{R}_C-I_E{R}_E=12-(2.457)(2)-(2.481)(1)=12-4.913-2.481=4.61\text{V}\approx 4.6\text{V}$$

7. Active-region check: $V_{CE}=4.6\text{V}$ is well above $V_{CE,sat}$ (about $0.2\text{V}$), so the active-region assumption holds. Q-point: $(V_{CE},I_C)=(4.6\text{V},2.46\text{mA})$, which is option (A).