GATE 2018 EC Q.26: solution and answer

GATE 2018 · ECQ.26 · 2 marks · MCQ

The circuit shown in the figure is used to provide regulated voltage (5 V) across the $1 k Ω$ resistor. Assume that the Zener diode has a constant reverse breakdown voltage for a current range, starting from a minimum required Zener current, $I_{Z m i n} = 2 mA$ to its maximum allowable current. The input voltage $V_{I}$ may vary by 5% from its nominal value of 6 V. The resistance of the diode in the breakdown region is negligible.

The value of $R$ and the minimum required power dissipation rating of the diode, respectively, are

GATE 2018 EC question 26: Zener shunt regulator. — Figure from the GATE 2018 EC paper. It is part of the question.

(A)
186 Ω and 10 mW
(B)
100 Ω and 40 mW
(C)
100 Ω and 10 mW
(D)
186 Ω and 40 mW

Answer and solution

ANSWER

$R = 100 Ω$ and a minimum Zener power rating of $40 mW$ , option (B).

Official answer: BSource: Official answer key, GATE 2018 EC (IIT Guwahati)

Assumptions: the Zener is ideal in breakdown ( $r_{z} = 0$ ), so the output stays at $V_{Z} = 5 V$ for as long as $I_{Z} \geq I_{Z m i n} = 2 mA$ . The load is fixed, so its current is fixed too: $I_{L} = \frac{5 V}{1 k Ω} = 5 mA$ .
Input range for a $\pm 5%$ swing about $6 V$ : $V_{I m i n} = 5.7 V$ and $V_{I m a x} = 6.3 V$ .
Choosing $R$ : by KCL at the output node, $I_{R} = I_{Z} + I_{L}$ . With $I_{L}$ fixed, $I_{Z}$ is smallest when $I_{R}$ is smallest, which happens at $V_{I m i n}$ . Regulation therefore needs $I_{R} \geq I_{Z m i n} + I_{L} = 2 + 5 = 7 mA$ at $V_{I} = 5.7 V$ .
Ohm's law across $R$ : $R \leq \frac{V _{I m i n} - V _{Z}}{I _{Z m i n} + I _{L}} = \frac{5.7 - 5}{7 mA} = 100 Ω$ A smaller $R$ passes more current and forces a larger diode rating. To keep the rating to a minimum, take the largest $R$ that still regulates: $R = 100 Ω$ .
Maximum Zener current, at $V_{I m a x}$ with the load unchanged: $I_{R m a x} = \frac{6.3 - 5}{100 Ω} = 13 mA$ , so $I_{Z m a x} = 13 - 5 = 8 mA$ .
Minimum power rating of the diode: $P_{Z} = V_{Z} I_{Z m a x} = 5 V \times 8 mA = 40 mW$ .
Checking the other options: with $R = 186 Ω$ , the current at $5.7 V$ would be only $I_{R} = 0.7/186 \approx 3.8 mA$ . That is less than the $5 mA$ the load draws on its own, so the Zener would leave breakdown and regulation would fail. This rules out (A) and (D). With $R = 100 Ω$ , a $10 mW$ rating is exceeded because the diode must dissipate $40 mW$ , which rules out (C). The answer is $100 Ω$ and $40 mW$ , option (B).

KEY IDEA

In a Zener shunt regulator, $R$ is set by the worst case for staying in breakdown: lowest input with maximum load current, where $R_{m a x} = \frac{V _{I m i n} - V _{Z}}{I _{Z m i n} + I _{L}}$ . The diode's power rating is set by the opposite worst case: highest input with minimum load current, where $P_{Z} = V_{Z} (I_{R m a x} - I_{L m i n})$ . Choosing the largest workable $R$ keeps that rating as small as possible.

WATCH OUT FOR

Sizing $R$ from the nominal $6 V$ or from $V_{I m a x}$ instead of $V_{I m i n}$ . Another mistake is using $I_{Z m i n}$ alone and forgetting the $5 mA$ load current, for example $0.7/2 mA$ or similar, which leads towards $186 Ω$ . A third is computing the power at $I_{Z m i n}$ ( $5 \times 2 = 10 mW$ ) rather than at the maximum Zener current that flows when $V_{I} = 6.3 V$ .

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2018 EC Q.26: solution and answer

Answer and solution

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