GATE 2018 EE Q.41: solution and answer

GATE 2018 · EEQ.41 · 2 marks · NAT

In the circuit shown in the figure, the bipolar junction transistor (BJT) has a current gain $β = 100$ . The base-emitter voltage drop is a constant, $V_{B E} = 0.7 V$ . The value of the Thevenin equivalent resistance $R_{T h}$ (in $Ω$ ) as shown in the figure is ______ (up to 2 decimal places).

GATE 2018 EE question 41: An npn BJT is drawn on its side along the top wire. — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

ANSWER

$R_{T h} = \frac{10 k Ω}{101} ∥ 1 k Ω = 90.09 Ω$ .

Official answer: 89.0 to 91.5Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Identify the terminals. The emitter is node $a$ and ground is node $b$ . The base is fed from $10.7 V$ through $R_{B} = 10 k Ω$ . The collector goes to $15 V$ through $R_{C} = 10 Ω$ . The resistor across $a$ – $b$ is $R_{E} = 1 k Ω$ .
Check the active region (KVL around the base-emitter loop). $10.7 = I_{B} (10 k Ω) + 0.7 + (β + 1) I_{B} (1 k Ω) \Rightarrow I_{B} = \frac{10 V}{111 k Ω} = 90.09 μ A$ So $I_{C} = 9.009 mA$ , $I_{E} = 9.099 mA$ and $V_{E} = I_{E} R_{E} = 9.099 V$ . Then $V_{C} = 15 - 10 (9.009 mA) = 14.91 V$ , so $V_{C E} \approx 5.81 V > 0$ . The transistor is in the active region, where (with constant $V_{B E}$ and constant $β$ ) the circuit is linear, so a Thevenin equivalent exists.
Small-signal model with a constant $V_{B E}$ . A constant $V_{B E}$ means the base-emitter junction has no incremental resistance ( $r_{π} = 0$ ): the base and emitter move together. The collector acts as a controlled current source $β i_{b}$ ( $r_{o}$ neglected), so the $10 Ω$ collector resistor and the $15 V$ source have no effect on the emitter. Both DC sources are set to zero.
Resistance looking into the emitter (resistance reflection). Apply a test voltage $v_{x}$ at the emitter. Since $v_{b e} = 0$ , the base is also at $v_{x}$ , so a current $v_{x} / R_{B}$ flows out of the base into $R_{B}$ ; the base current into the transistor is $i_{b} = - v_{x} / R_{B}$ . The current flowing into the emitter from the test source is $- i_{e} = - (β + 1) i_{b} = (β + 1) v_{x} / R_{B}$ . So $R_{emitter} = \frac{R _{B} + r _{π}}{β + 1} = \frac{10 000}{101} = 99.01 Ω$
Combine with the 1 kΩ resistor across $a$ – $b$ . $R_{T h} = 99.01 Ω ∥ 1000 Ω = \frac{( 1 0 ^{4} /101 ) ( 1 0 ^{3} )}{1 0 ^{4} /101 + 1 0 ^{3}} = \frac{1 0 ^{7}}{111 000} = 90.09 Ω$
Cross-check with $V_{oc} / I_{sc}$ . Open circuit: $V_{T h} = V_{E} = 9.099 V$ (step 2). Short $a$ – $b$ : the emitter is grounded, so $I_{B} = (10.7 - 0.7) /10 k Ω = 1 mA$ , $I_{C} = 100 mA$ and $I_{sc} = I_{E} = 101 mA$ (the shorted $1 k Ω$ carries nothing). Then $V_{C} = 15 - 10 (0.1) = 14 V$ , so the transistor is still active. $R_{T h} = \frac{9.099 V}{0.101 A} = 90.09 Ω$

KEY IDEA

Because $V_{B E}$ is constant, the base-emitter junction has zero incremental resistance ( $r_{π} = 0$ ). Looking into the emitter you therefore see only the base resistance divided by $(β + 1)$ , which is $10 k Ω/101 = 99.01 Ω$ . The collector branch is a current source and drops out. $R_{T h}$ is this value in parallel with the $1 k Ω$ across $a$ – $b$ . Dividing the open-circuit voltage by the short-circuit current gives the same answer, because the circuit is linear while the BJT stays active.

WATCH OUT FOR

Common mistakes: (1) Taking $R_{T h} = 1 k Ω$ and ignoring the transistor. (2) Multiplying $R_{B}$ by $(β + 1)$ instead of dividing: that is the rule for looking into the base, not the emitter. (3) Including the $10 Ω$ collector resistor, which sits behind an ideal current source ( $r_{o}$ neglected). (4) Adding a nonzero $r_{π} = V_{T} / I_{B}$ , when the question says $V_{B E}$ is constant.

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 13 Biasing questions, with notes

GATE 2018 EE Q.41: solution and answer

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