GATE 2026 EC Q.29: solution and answer

GATE 2026 · ECQ.29 · 1 mark · MSQ

Figure shows the output characteristics of two different Bipolar Junction Transistors (BJT), BJT 1 with magnitude of Early voltage $∣ V_{A 1} ∣$ , and BJT 2 with magnitude of Early voltage $∣ V_{A 2} ∣$ .

Which of the following options is/are correct regarding the Early voltages?

GATE 2026 EC question 29: A plot of collector current I_C (vertical axis) versus collector-emitter voltage V_CE (horizontal axis), both axes… — Figure from the GATE 2026 EC paper. It is part of the question.

(A)
$∣ V_{A 1} ∣ > ∣ V_{A 2} ∣$
(B)
$∣ V_{A 1} ∣$ is infinitely large
(C)
$∣ V_{A 1} ∣ < ∣ V_{A 2} ∣$
(D)
$∣ V_{A 2} ∣$ is finite

Answer and solution

ANSWER

Options (C) and (D) are correct: $∣ V_{A 1} ∣ < ∣ V_{A 2} ∣$ , and $∣ V_{A 2} ∣$ is finite.

Official answer: C, DSource: Official answer key, GATE 2026 EC (IIT Guwahati)

The Early effect gives $I_{C} = I_{C 0} (1 + \frac{V _{C E}}{V _{A}})$ at fixed $I_{B}$ , so each output curve's active-region segment is a straight line of slope $\frac{I _{C 0}}{V _{A}}$ .
Extrapolated backward, this line intersects the $V_{C E}$ -axis at $V_{C E} = - V_{A}$ . Since both curves branch from the same knee point (same $I_{C 0}$ ) in the figure, a steeper active-region slope corresponds to a smaller $∣ V_{A} ∣$ , and a flatter slope corresponds to a larger $∣ V_{A} ∣$ .
From the figure, BJT 1's curve (solid) rises noticeably faster with $V_{C E}$ than BJT 2's curve (dashed), which is nearly flat.
Steeper slope for BJT 1 $\Rightarrow$ smaller $∣ V_{A 1} ∣$ ; flatter slope for BJT 2 $\Rightarrow$ larger $∣ V_{A 2} ∣$ , so $∣ V_{A 1} ∣ < ∣ V_{A 2} ∣$ : option (C) is correct and (A) is wrong.
An ideal, perfectly horizontal output curve would correspond to $V_{A} \to \infty$ ; but BJT 1's curve visibly has a finite positive slope, so $∣ V_{A 1} ∣$ cannot be infinite: option (B) is wrong.
BJT 2's curve, though much flatter than BJT 1's, still has a small nonzero positive slope (it is not perfectly horizontal), so $∣ V_{A 2} ∣$ is large but finite: option (D) is correct.
Hence the correct options are (C) and (D).

KEY IDEA

In the Early-effect model, $I_{C} = I_{C 0} (1 + \frac{V _{C E}}{V _{A}})$ at fixed $I_{B}$ , so each output curve's active-region straight-line segment has slope $\frac{I _{C 0}}{V _{A}}$ ; extrapolated backward, that line meets the $V_{C E}$ -axis at $V_{C E} = - V_{A}$ . Since both curves in the figure branch from the same knee point (same $I_{C 0}$ ), a flatter active-region slope means a larger $∣ V_{A} ∣$ , and a steeper slope means a smaller $∣ V_{A} ∣$ .

WATCH OUT FOR

Assuming a nearly-flat curve on a printed figure must mean $V_{A} \to \infty$ , and so marking (B) or dismissing (D). The figure clearly shows BJT 2's line still has a small positive slope (it is not perfectly horizontal), so $∣ V_{A 2} ∣$ is large but finite, not infinite; only a truly horizontal line would signal $V_{A} \to \infty$ .

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 3 BJT questions, with notes

GATE 2026 EC Q.29: solution and answer

Answer and solution

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