GATE 2020 EC Q.44: solution and answer

GATE 2020 · ECQ.44 · 2 marks · NAT

In the voltage regulator shown below, $V_{I}$ is the unregulated input at 15 V. Assume $V_{B E} = 0.7 V$ and the base current is negligible for both the BJTs. If the regulated output $V_{O}$ is 9 V, the value of $R_{2}$ is ________ $Ω$ .

GATE 2020 EC question 44: Series transistor voltage regulator. — Figure from the GATE 2020 EC paper. It is part of the question.

Answer and solution

ANSWER

$R_{2} = 800 Ω$

Official answer: 800Source: Official answer key, GATE 2020 EC (IIT Delhi)

Assumptions stated in the question: both BJTs are in the active region with $V_{B E} = 0.7 V$ , and both base currents are negligible. So the base of $Q_{2}$ draws no current from the $R_{1}$ – $R_{2}$ junction.
The emitter of $Q_{2}$ sits on the cathode of the reverse-biased Zener, whose anode is grounded. So $V_{E 2} = V_{Z} = 3.3 V$ .
KVL across the base-emitter junction of $Q_{2}$ : $V_{B 2} = V_{Z} + V_{B E} = 3.3 + 0.7 = 4.0 V$
The base current of $Q_{2}$ is zero, so $V_{B 2}$ comes from an unloaded voltage divider across the output: $V_{B 2} = V_{O} \frac{R _{2}}{R _{1} + R _{2}}$
Substitute $V_{O} = 9 V$ and $R_{1} = 1 k Ω$ : $4 = 9 \cdot \frac{R _{2}}{1000 + R _{2}} \Rightarrow 4000 + 4 R_{2} = 9 R_{2} \Rightarrow R_{2} = 800 Ω$
Check that the operating point is consistent. $V_{B 1} = V_{O} + V_{B E} = 9.7 V$ , so $R_{3}$ carries $I = \frac{15 - 9.7}{1 k Ω} = 5.3 mA$ . The base current of $Q_{1}$ is negligible, so this whole current flows into the collector of $Q_{2}$ and down through the Zener. The Zener is therefore reverse-biased and conducting. $Q_{2}$ is active because $V_{C E 2} = 9.7 - 3.3 = 6.4 V$ , and $Q_{1}$ is active because $V_{C E 1} = 15 - 9 = 6 V$ . The assumptions hold.

KEY IDEA

In a series regulator with an error-amplifier transistor, the feedback loop settles where the sampled output at the divider tap equals the reference plus one base-emitter drop: $V_{O} \frac{R _{2}}{R _{1} + R _{2}} = V_{Z} + V_{B E}$ . This is the same as $V_{O} = (V_{Z} + V_{B E}) (1 + \frac{R _{1}}{R _{2}})$ . $R_{3}$ and $V_{I}$ only set the bias current, not the output.

WATCH OUT FOR

Setting the divider tap equal to $V_{Z} = 3.3 V$ and forgetting the $V_{B E}$ of $Q_{2}$ . That gives $R_{2} = 3300/5.7 \approx 579 Ω$ . Other mistakes are putting the $V_{B E}$ of the pass transistor $Q_{1}$ into the divider equation, or trying to use $V_{I} = 15 V$ and $R_{3}$ to find $R_{2}$ . Neither of those affects the divider.

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2020 EC Q.44: solution and answer

Answer and solution

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