GATE 2025 IN – Question 35
In the shown circuit, assume very high BJT current gain $\beta$, $V_{BE}=0.7$ V, and Zener voltage $V_Z=4.7$ V. The LED current $I$ is ______ mA (rounded to two decimal places).
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Correct answer: 3.90 to 4.10
Explanation
The Zener fixes the base voltage at 4.7 V.
The emitter is therefore at $V_E=4.7-0.7=4.0$ V.
With $R_E=1$ kΩ, $I_E=4.0/1000=4$ mA.
Very high $\beta$ makes $I_C\simeq I_E$, so the LED current is **4.00 mA**.