CE amplifier bias intuition: In a common-emitter BJT amplifier with a collector resistor, reducing base current (and thus collector current) will cause the measured voltage between emitter and collector (VCE) to __________.

Difficulty: Easy

Correct Answer: increase

Explanation:


Introduction / Context:
Biasing in a common-emitter (CE) amplifier sets the quiescent operating point (Q-point). Understanding how base current affects collector current and the resulting voltages across resistors is crucial for troubleshooting and designing linear amplifier stages.


Given Data / Assumptions:

  • Single-supply CE amplifier with a collector resistor RC and small emitter resistor (or referenced to ground).
  • Base current IB reduction reduces collector current IC (IC ≈ beta * IB in active region).
  • Supply voltage VCC is fixed.


Concept / Approach:
Collector voltage VC = VCC − IC * RC. If IC decreases due to reduced IB, the drop IC * RC decreases, so VC rises toward VCC. Since emitter is near ground (or a small bias), VCE = VC − VE increases. Thus, lowering base current moves the transistor toward cutoff and raises VCE.


Step-by-Step Solution:

Decrease IB ⇒ decreases IC in active region.Voltage drop across RC = IC * RC decreases.Collector node voltage VC rises closer to VCC.Therefore VCE = VC − VE increases.


Verification / Alternative check:
Load-line analysis on the IC–VCE plane shows movement along the line toward the cutoff point (IC ≈ 0, VCE ≈ VCC) when base drive is reduced.


Why Other Options Are Wrong:

  • Decrease / drop to near 0 V: Happens when IB increases (moving toward saturation), not when it decreases.
  • Remain unchanged / oscillate randomly: Bias responds deterministically to IB changes; no randomness assumed.


Common Pitfalls:
Confusing the behavior near saturation with behavior near cutoff; forgetting the sign of the voltage drop across RC.


Final Answer:
increase

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