A PNP transistor is operating in active region. If the emitter voltage decreases such that the emitter becomes less positive than the base while the biasing condition of the collector-base junction remains unchanged, the transistor will enter _______.
Cut-off region
To solve this question, we need to understand the operating regions of a PNP transistor and how changes in voltages affect these regions.
PNP Transistor Basic Operation
A PNP transistor in active mode typically has the emitter-base junction forward-biased and the collector-base junction reverse-biased. These conditions facilitate amplification.
- Active Region: Emitter-Base junction is forward-biased, Collector-Base junction is reverse-biased. The transistor amplifies the input signal.
- Cut-off Region: Both the Emitter-Base and Collector-Base junctions are reverse-biased. The transistor is effectively off, with no current flowing through.
- Saturation Region: Both junctions are forward-biased. The transistor conducts maximum current, but there's no amplification.
- Reverse Active Region: Emitter-Base junction is reverse-biased, Collector-Base junction is forward-biased. This is an unusual mode, with reverse direction current flow.
Analysis of Given Situation
The question states: "If the emitter voltage decreases such that the emitter becomes less positive than the base while the biasing condition of the collector-base junction remains unchanged."
This suggests that the forward-bias condition of the emitter-base junction is lost. With the emitter becoming less positive relative to the base, the emitter-base junction can become reverse-biased. Since the collector-base junction is mentioned to remain unchanged, it stays reverse-biased.
Conclusion
With both the emitter-base and collector-base junctions reverse-biased, the PNP transistor enters the Cut-off Region.
Therefore, the correct answer is: Cut-off region.
Similar Questions
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