In a transistor biased at the midpoint of its DC load line, if the device’s current gain (beta) increases while base bias remains unchanged, how does the Q-point shift along the load line?

Difficulty: Medium

Correct Answer: up

Explanation:


Introduction / Context:
The operating point (Q-point) of a transistor is where the device resides on its load line under quiescent (no-signal) conditions. Understanding how changes in current gain affect this point is essential for bias stability and linear amplifier design.


Given Data / Assumptions:

  • Transistor initially biased near the midpoint of the load line.
  • Base bias is fixed (no change in bias network values).
  • Current gain beta increases due to device variation or temperature.


Concept / Approach:
Collector current Ic ≈ beta * Ib (for a given base current Ib). If beta rises while Ib is constant, Ic increases. On a load line (defined by Vcc and RC), higher Ic moves the Q-point toward higher current and lower collector-emitter voltage, i.e., up the current axis toward saturation.


Step-by-Step Solution:

1) Assume fixed Ib due to fixed bias network.2) Increase beta → Ic increases (Ic = beta * Ib).3) On the load line, larger Ic corresponds to a point higher on the vertical axis.4) Hence Q moves up (toward saturation), away from cutoff.


Verification / Alternative check:
Graphically, draw the same load line and intersect it with the new Ic for the unchanged base current; the intersection is above the previous Q-point.


Why Other Options Are Wrong:

Down: Would imply reduced Ic, opposite to an increase in beta.Nowhere: Ignores the dependence Ic = beta * Ib.Off the load line: DC operation must remain on the load line defined by Vcc and RC unless supply or RC changes.


Common Pitfalls:
Assuming feedback or emitter degeneration compensates automatically; without such stabilization, beta shifts do move the Q-point.


Final Answer:
up.

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