GATE 2026 EE Q.33: solution and answer

GATE 2026 · EEQ.33 · 1 mark · NAT

In the linear regulator circuit shown, the base to emitter voltage $V_{B E}$ of the BJT is 0.6 V. The Zener diode clamps the base voltage to 5.4 V. Ignore the biasing current of the Zener diode and BJT.

The maximum possible efficiency of the regulator circuit is ________ %

(Round off to one decimal place)

(Figure data: $I_{L} = 100 mA$ , $V_{B E} = 0.6 V$ )

GATE 2026 EE question 33: Series-pass (emitter-follower) linear regulator. — Figure from the GATE 2026 EE paper. It is part of the question.

Answer and solution

ANSWER

Maximum efficiency = 48.0 %.

Official answer: 47.5 to 48.0Source: Official answer key, GATE 2026 EE (IIT Guwahati)

Output voltage (KVL around the base–emitter loop): the Zener holds the base at $V_{B} = 5.4 V$ , and the emitter sits one $V_{B E}$ below the base (emitter follower): $V_{o u t} = V_{B} - V_{B E} = 5.4 - 0.6 = 4.8 V$
Output power: the load is a constant current sink of $I_{L} = 100 mA$ , so $P_{o u t} = V_{o u t} I_{L} = 4.8 \times 0.1 = 0.48 W$
Input current: the question says to ignore the Zener bias current (through the $5 k Ω$ resistor) and the BJT base current. The only current left flowing out of the 10 V source is then the collector current, and with $I_{B} = 0$ we get $I_{C} = I_{E} = I_{L} = 100 mA$ .
Input power: $P_{in} = V_{in} I_{L} = 10 \times 0.1 = 1 W$ The difference, $(V_{in} - V_{o u t}) I_{L} = 5.2 \times 0.1 = 0.52 W$ , is dissipated in the series-pass transistor.
Efficiency: $η = \frac{P _{o u t}}{P _{in}} = \frac{V _{o u t} I _{L}}{V _{in} I _{L}} = \frac{V _{o u t}}{V _{in}} = \frac{4.8}{10} = 0.48 = 48.0 %$ This is the maximum possible value, because any bias current can only add to $P_{in}$ without adding to $P_{o u t}$ . (For comparison: if you include the resistor current $(10 - 5.4) /5 k Ω = 0.92 mA$ , then $P_{in} = 10 \times 100.92 mA \approx 1.009 W$ and $η \approx 47.6 %$ . That is why the official key accepts a range.)

KEY IDEA

In a series-pass (emitter-follower) linear regulator, the same load current flows from the input to the output through the pass transistor. With bias currents ignored, the efficiency is therefore just $V_{o u t} / V_{in}$ . Here $V_{o u t} = V_{Z} - V_{B E} = 4.8 V$ , which gives $4.8/10 = 48 %$ .

WATCH OUT FOR

One mistake is to take $V_{o u t} = 5.4 V$ , the Zener voltage, and forget the $V_{B E}$ drop, which gives 54 %. Another is to use $V_{Z} + V_{B E} = 6.0 V$ , which gives 60 %. A third is to include the Zener or resistor current even though the question says to ignore it: that gives about 47.6 %. That value happens to fall inside the key range, but it is not the "maximum possible" efficiency the question asks for.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2026 EE Q.33: solution and answer

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