GATE 2023 EE Q.59: solution and answer

GATE 2023 · EEQ.59 · 2 marks · NAT

The Zener diode in circuit has a breakdown voltage of 5 V. The current gain $β$ of the transistor in the active region in 99. Ignore base-emitter voltage drop $V_{B E}$ . The current through the 20 $Ω$ resistance in milliamperes is ________ (Round off to 2 decimal places).

GATE 2023 EE question 59: A 25 V battery (+ terminal at top) on the left. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

The current through the 20 Ω resistor is 250.00 mA. The Zener stays below breakdown (only 2.5 V across it), so the whole emitter current of 250 mA flows through the 20 Ω resistor.

Official answer: 245 to 255Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Assumptions: the NPN transistor is in the active region and $V_{B E} = 0$ , so $V_{E} = V_{B}$ and $I_{E} = (β + 1) I_{B} = 100 I_{B}$ . Call the right-hand end of the 10 $Ω$ resistor node L. L is also the top of the 20 $Ω$ resistor. The Zener has its cathode at the base B and its anode at L. Because $V_{B E} = 0$ , the Zener sits directly across the 10 $Ω$ emitter resistor: $V_{B} - V_{L} = 10 I_{E} \geq 0$ . So the Zener can never be forward biased. It is either in breakdown or off.
Try breakdown first. Then $V_{B} - V_{L} = 5 V$ , which gives $I_{E} = 5/10 = 0.5 A$ and $I_{B} = I_{E} /100 = 5 mA$ . Take the Zener current $I_{Z}$ as flowing from cathode to anode (B to L). KCL at B gives $I_{Z} = I_{7 k} - I_{B}$ . KCL at L gives $I_{20} = I_{E} + I_{Z} = 0.495 + I_{7 k}$ . With $V_{L} = 20 I_{20}$ and $V_{B} = V_{L} + 5$ , the 7 k $Ω$ current is $I_{7 k} = \frac{25 - V _{B}}{7000} = \frac{20 - 20 I _{20}}{7000}$ . Substituting for $I_{20}$ : $7000 I_{7 k} = 20 - 20 (0.495 + I_{7 k})$ , so $7020 I_{7 k} = 10.1$ and $I_{7 k} \approx 1.44 mA$ .
The Zener current would then be $I_{Z} = I_{7 k} - I_{B} \approx 1.44 - 5 = - 3.56 mA < 0$ . Breakdown current must flow from cathode to anode, so it cannot be negative. The breakdown assumption fails, and the Zener is OFF ( $I_{Z} = 0$ ).
With the Zener open, all of the 7 k $Ω$ current is base current. Apply KVL from the 25 V rail through the 7 k $Ω$ resistor, the base-emitter junction ( $0$ V), and the 10 $Ω$ and 20 $Ω$ resistors, which both carry $I_{E} = (β + 1) I_{B}$ : $25 = 7000 I_{B} + (10 + 20) (β + 1) I_{B} = (7000 + 3000) I_{B}$
$I_{B} = \frac{25}{10 000} A = 2.5 mA, I_{E} = 100 \times 2.5 mA = 250 mA$
Check the Zener: $V_{B} - V_{L} = 10 I_{E} = 10 \times 0.25 = 2.5 V < 5 V$ . It is below breakdown, so it is off, as assumed. Check the active region: $V_{E} = V_{B} = 25 - 7000 (0.0025) = 7.5 V$ and $V_{C} = 25 V$ , so $V_{C E} = 17.5 V > 0$ . The transistor is active.
With $I_{Z} = 0$ , the only current into node L is the emitter current. So $I_{20 Ω} = I_{E} = 250.00 mA$ . As a check, $V_{L} = 20 \times 0.25 = 5 V$ and $V_{E} - V_{L} = 7.5 - 5 = 2.5 V$ , which matches the drop across the 10 $Ω$ resistor.

KEY IDEA

With $V_{B E} = 0$ , the Zener is connected straight across the 10 $Ω$ emitter resistor and is reverse biased by $10 I_{E}$ . Assume one Zener state, solve the emitter-follower bias with the emitter resistance reflected into the base loop as $(β + 1) (10 + 20) Ω$ , then check the assumption. Here breakdown gives a negative Zener current, so the Zener is off. That leaves $V_{Z} = 2.5 V < 5 V$ and $I_{20} = I_{E} = 250 mA$ .

WATCH OUT FOR

Seeing a 5 V Zener and assuming it is in breakdown. That puts 5 V across the 10 Ω resistor, so $I_{E} = 0.5 A$ and $I_{20} \approx 496.44 mA$ (or 500 mA if the Zener current is ignored), and the Zener current comes out negative, which is impossible. Always check a Zener's state after solving. Other slips: leaving out the $(β + 1)$ factor on the 20 Ω resistor in the base loop, and reporting $I_{C}$ instead of $I_{E}$ as the current in the 20 Ω resistor.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 13 Biasing questions, with notes

GATE 2023 EE Q.59: solution and answer

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