Question

Figure shows the output characteristics of two different Bipolar Junction Transistors (BJT), BJT 1 with magnitude of Early voltage \(|V_{A1}|\), and BJT 2 with magnitude of Early voltage \(|V_{A2}|\).

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

• A. \(|V_{A1}| > |V_{A2}|\)
• B. \(|V_{A1}|\) is infinitely large
• C. \(|V_{A1}| < |V_{A2}|\)
• D. \(|V_{A2}|\) is finite

Hint:

Extend each curve's straight part back to the \(V_{CE}\) axis; a flatter curve corresponds to a larger, but still finite, Early voltage.

Answer 1

The Correct Option is C, D

Step 1: Recall what the Early voltage means on this graph.
For a BJT biased at a fixed base current, \(I_C\) rises a little with \(V_{CE}\) once the transistor is in the active region (the Early effect). Extending the sloped part of the \(I_C\) versus \(V_{CE}\) line backward, it crosses the \(V_{CE}\) axis at \(V_{CE}=-V_A\), where \(V_A\) is that transistor's Early voltage.

Step 2: Connect the slope of the curve to the Early voltage.
Near a bias point, \(I_C\) can be written as
\[ I_C\approx I_{C0}\left(1+\frac{V_{CE}}{V_A}\right) \]
so the slope of the curve is
\[ \frac{dI_C}{dV_{CE}}\approx\frac{I_{C0}}{V_A} \]
A larger \(V_A\) gives a smaller slope (flatter curve), and a smaller \(V_A\) gives a larger slope (steeper curve).

Step 3: Read the two curves in the figure.
BJT 1's curve rises noticeably as \(V_{CE}\) increases (steeper slope). BJT 2's curve, drawn dashed, is much closer to flat over the same range of \(V_{CE}\) (smaller slope).

Step 4: Compare the Early voltages using the slopes.
Since BJT 1 is steeper, its slope is larger, so from Step 2 its Early voltage \(|V_{A1}|\) must be smaller. Since BJT 2 is flatter, its slope is smaller, so its Early voltage \(|V_{A2}|\) must be larger. Hence
\[ |V_{A1}|<|V_{A2}| \]
This makes option (C) correct and rules out option (A), which claims the opposite.

Step 5: Check whether either Early voltage could be infinite.
An infinite Early voltage would mean a slope of exactly zero, a perfectly horizontal line with \(I_C\) completely independent of \(V_{CE}\). BJT 1's line clearly slants upward, so its slope is not zero, ruling out option (B).

Step 6: Check option (D).
Even though BJT 2's line looks almost flat on this scale, no real transistor has a perfectly flat output characteristic; the Early effect always gives it some finite slope, however small. A truly flat line would need \(|V_{A2}|\to\infty\), which does not happen in an actual device. So \(|V_{A2}|\) is large but still finite, making option (D) correct.

Final Answer:
\[ \boxed{|V_{A1}|<|V_{A2}|\ \text{and}\ |V_{A2}|\ \text{is finite}} \]