GATE 2025 EC Q.47: solution and answer

GATE 2025 · ECQ.47 · 2 marks · MCQ

In the circuit shown, the identical transistors Q1 and Q2 are biased in the active region with $β = 120$ . The Zener diode is in the breakdown region with $V_{Z} = 5 V$ and $I_{Z} = 25 mA$ .

If $I_{L} = 12 mA$ and $V_{E B 1} = V_{E B 2} = 0.7 V$ , then the values of $R_{1}$ and $R_{2}$ (in $k Ω$ , rounded off to one decimal place) are ______, respectively.

GATE 2025 EC question 47: A +20 V supply rail runs across the top. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
0.6 and 0.4
(B)
1.4 and 2.5
(C)
14.0 and 25.0
(D)
6.0 and 4.0

Answer and solution

ANSWER

$R_{1} \approx 0.6 k Ω$ and $R_{2} \approx 0.4 k Ω$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

Assumptions, as given: both PNP transistors are in the active region with $V_{E B 1} = V_{E B 2} = 0.7 V$ and $β = 120$ , and the Zener is in breakdown with $V_{Z} = 5 V$ . Q2 has its base tied to its collector (diode-connected), and its collector/base node connects only to the Zener cathode, so Q2's whole emitter current is the Zener current $I_{Z} = 25 mA$ . Q2 simply drops $V_{E B 2} = 0.7 V$ .
Node voltages from the 20 V rail downward (KVL through Q2 and then the Zener): $V_{C 2} = 20 - 0.7 = 19.3 V$ , so the Zener anode, which is also Q1's base, sits at $V_{B 1} = 19.3 - V_{Z} = 19.3 - 5 = 14.3 V$ .
Emitter of Q1 (KVL across its emitter-base junction): $V_{E 1} = V_{B 1} + V_{E B 1} = 14.3 + 0.7 = 15.0 V$ . The voltage across $R_{2}$ is therefore $20 - 15 = 5 V$ .
Q1's collector current flows through $R_{L}$ , so $I_{C} = I_{L} = 12 mA$ . Active-region relations: $I_{B} = \frac{I _{C}}{β} = \frac{12}{120} = 0.1 mA$ and $I_{E} = I_{C} + I_{B} = 12.1 mA$ .
Ohm's law for $R_{2}$ , which carries $I_{E}$ : $R_{2} = \frac{5 V}{12.1 mA} = 0.413 k Ω \approx 0.4 k Ω$
KCL at the node joining the Zener anode, Q1's base and the top of $R_{1}$ : the Zener current enters from above, and the PNP base current $I_{B}$ flows out of Q1's base into this node, as the figure's arrow shows. Both leave through $R_{1}$ , so $I_{R 1} = I_{Z} + I_{B} = 25 + 0.1 = 25.1 mA$ .
Ohm's law for $R_{1}$ (top node at $14.3 V$ , bottom at ground): $R_{1} = \frac{14.3 V}{25.1 mA} = 0.570 k Ω \approx 0.6 k Ω$
So $R_{1} \approx 0.6 k Ω$ and $R_{2} \approx 0.4 k Ω$ , which is option (A).

KEY IDEA

The diode-connected Q2 and the Zener form a fixed voltage stack below the 20 V rail. That stack pins Q1's base at $20 - 0.7 - 5 = 14.3 V$ , and so its emitter at $15 V$ . $R_{2}$ then sets the emitter current $I_{E} = (1 + β) I_{L} / β$ across a fixed 5 V drop. $R_{1}$ carries the Zener current plus the PNP base current that flows out of Q1's base.

WATCH OUT FOR

A common slip is measuring the Zener voltage from ground and putting Q1's base at 5 V. The Zener actually sits under Q2, so what $R_{1}$ sees is $20 - 0.7 - 5 = 14.3 V$ . The other trap is units: volts divided by milliamps gives kilo-ohms directly, and a slip of a factor of 10 lands on option (D). The base current (0.1 mA) barely changes either value at one decimal place, but its direction matters. For a PNP it flows out of the base, so it adds to the Zener current in $R_{1}$ .

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

GATE 2025 EC Q.47: solution and answer

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