Bias sensitivity — in a common-emitter amplifier, if the base current decreases (with bias still in the active region), what happens to the voltage measured between the emitter and the collector (V_EC or equivalently V_CE for an NPN referenced properly)?

Introduction / Context:
Understanding the bias relationships in a common-emitter (C-E) amplifier is crucial for predicting signal behavior and diagnosing faults. Base current (or, equivalently, base-emitter voltage) controls collector current. Changes in collector current alter the voltage drop across the collector load resistor, which then shifts the collector-to-emitter voltage seen at the transistor.

Given Data / Assumptions:

• C-E amplifier in active region (not saturated or cut off initially).
• Collector load resistor R_C connected to a positive supply (NPN case).
• Decrease in base current implies decrease in collector current (I_C ≈ β * I_B).

Concept / Approach:
Ohm’s law and transistor action set the relationship: V_C = V_CC − I_C * R_C. If I_C decreases, the drop across R_C shrinks, raising the collector voltage. Since the emitter is near a small bias above ground (or at a fixed reference), the collector-emitter voltage V_CE increases. Equivalently, measuring V_EC (emitter relative to collector) would show a decrease because V_EC = −V_CE; but in common usage, “voltage between emitter and collector increases” refers to V_CE for NPN, which indeed increases as base current drops.

Step-by-Step Solution:

Verification / Alternative check:
A load-line diagram on the I_C–V_CE plane shows that moving toward lower I_C shifts the operating point toward higher V_CE along the DC load line.

Why Other Options Are Wrong:

• “V_EC decreases” or “no change”: contradict the load-line relation.
• “Drops to nearly zero due to saturation”: saturation occurs with excessive base drive, not reduced base current.

Common Pitfalls:
Confusing the sign convention between V_CE and V_EC; always define polarity before interpreting statements.

Final Answer:
Correct — less base current reduces I_C and increases V_CE (thus the emitter–collector voltage, referenced as V_CE, increases).