GATE 2017 EC (EC2) – Question 24
Consider the circuit shown. Assume base-to-emitter voltage $V_{BE}=0.8$ V and common-base current gain $\alpha=1$. The collector-to-emitter voltage $V_{CE}$ (in volts) is ________.
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Correct answer: 5.5 to 6.5
Explanation
With $\alpha=1$, base current is zero, so the 44 kΩ/16 kΩ divider is unloaded. Its base voltage is $V_B=18(16)/(44+16)=4.8$ V.
The emitter is at $V_E=4.8-0.8=4.0$ V, giving $I_E=4/2=2$ mA. Since $I_C=I_E$, the 4 kΩ collector resistor drops 8 V and $V_C=18-8=10$ V.
Therefore $V_{CE}=V_C-V_E=10-4=$ **6.0 V**.