GATE 2025 IN Q.35: solution and answer

GATE 2025 · INQ.35 · 1 mark · NAT

In the circuit shown, assume that the BJT in the circuit has very high $β$ and $V_{B E} = 0.7 V$ , and the Zener diode has $V_{Z} = 4.7 V$ . The current $I$ through the LED is ________ mA (rounded off to two decimal places).

GATE 2025 IN question 35: A supply node V_CC = 12 V at the top feeds a horizontal rail. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

ANSWER

$I \approx 4.00 mA$ .

Official answer: 3.90 to 4.10Source: Official answer key, GATE 2025 IN (IIT Roorkee)

Assumptions. The BJT is NPN with $V_{B E} = 0.7 V$ and very high $β$ , so $I_{B} \to 0$ and $α = \frac{β}{β + 1} \to 1$ . The Zener diode is ideal in breakdown, with $V_{Z} = 4.7 V$ . The LED's forward drop is not given, and we will see that it is not needed.
Check the Zener. Its cathode is at the base node and its anode is at ground, so it is reverse-biased. Suppose it were off. Then $R_{1}$ would carry only the negligible base current, and the base node would rise far above $4.7 V$ toward $12 V$ . That contradicts 'off', so the Zener is in breakdown and holds $V_{B} = V_{Z} = 4.7 V$ . Its current is $I_{Z} = \frac{12 - 4.7}{2.2 k Ω} - I_{B} \approx 3.32 mA > 0$ , which confirms breakdown.
KVL across the base-emitter junction: $V_{E} = V_{B} - V_{B E} = 4.7 - 0.7 = 4.0 V$ .
Ohm's law on the emitter resistor: $I_{E} = \frac{V _{E}}{R _{E}} = \frac{4.0 V}{1 k Ω} = 4.0 mA$ .
Collector current. With very high $β$ , $I_{C} = α I_{E} \approx I_{E}$ . The LED is in series with the collector, so $I = I_{C} \approx I_{E} = 4.00 mA .$
Check the operating region. The collector voltage is $V_{C} = 12 - V_{L E D}$ . The BJT stays in the active region as long as $V_{C}$ stays above about $V_{E} + V_{C E (sat)} \approx 4.2 V$ , which means $V_{L E D} ≲ 7.8 V$ . Any single LED (forward drop roughly $1.8$ to $3.5 V$ ) meets this easily. So the assumption holds. The LED current does not depend on the LED's own drop, which is why this circuit works as a constant-current LED driver.

KEY IDEA

The Zener fixes the base at $4.7 V$ , so the emitter sits one $V_{B E}$ lower at $4.0 V$ . That fixes $I_{E} = V_{E} / R_{E}$ . With very high $β$ , $I_{C} = I_{E}$ , and the LED in the collector carries this current whatever its forward drop is. The circuit is a Zener-referenced constant-current sink.

WATCH OUT FOR

One mistake is to find the current through $R_{1}$ instead, $(12 - 4.7) /2.2 k Ω \approx 3.32 mA$ . That is the Zener current, not the LED current. Another is to forget to subtract $V_{B E}$ and get $4.7 V /1 k Ω = 4.70 mA$ . A third is to try to use an LED forward voltage. It is not given and does not affect the answer while the BJT stays active.

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 13 Biasing questions, with notes

GATE 2025 IN Q.35: solution and answer

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