Difficulty: Easy
Correct Answer: Forward bias the base/emitter junction and reverse bias the base/collector junction.
Explanation:
Introduction / Context:
For linear amplification, a bipolar transistor must be biased in the active region where collector current is controlled by base–emitter junction voltage while the collector–base junction remains reverse biased. Getting the junction polarities right is fundamental for predictable gain and low distortion.
Given Data / Assumptions:
Concept / Approach:
The base–emitter junction behaves like a diode that must be forward biased (~0.6–0.7 V for silicon) to inject carriers into the base. The collector–base junction must be reverse biased to sweep carriers into the collector and prevent saturation. This establishes the transistor in the forward-active region where IC ≈ beta * IB and small-signal models apply.
Step-by-Step Solution:
Ensure VBE is forward (~0.6–0.7 V), establishing emitter and collector currents.Maintain VCB > 0 (collector more positive than base for NPN) to keep the collector–base junction reverse biased.Verify that VCE remains well above saturation to allow signal swings.Operate around a stable Q-point for linear amplification.
Verification / Alternative check:
Measure DC voltages: for NPN in active region, expect VCE typically near mid-supply and VBE around 0.7 V, with collector voltage higher than base voltage.
Why Other Options Are Wrong:
Reversing junction biases forces cutoff or saturation, not linear region.Applying “positive to n and negative to p” is an oversimplified and often incorrect rule for three-terminal devices.Large base voltage alone may overdrive or damage the device; proper polarities and currents matter.
Common Pitfalls:
Allowing the collector–base junction to forward bias under large signals, which pushes the device toward saturation and distortion.
Final Answer:
Forward bias the base/emitter junction and reverse bias the base/collector junction.
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