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In a junction transistor, the collector cut-off current 'ICBO' reduces considerably by doping the
  • a)
    emitter with high level of impurity
  • b)
    emitter with low level of impurity
  • c)
    collector with high level of impurity
  • d)
    collector with low level of impurity
Correct answer is option 'A'. Can you explain this answer?
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In a junction transistor, the collector cut-off current 'ICBO'...
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In a junction transistor, the collector cut-off current 'ICBO'...
Explanation:
A junction transistor is a type of bipolar transistor that consists of two p-n junctions formed between three doped regions. These doped regions are the emitter, base, and collector. The emitter is heavily doped, while the base is lightly doped, and the collector is moderately doped.

When the transistor is biased in the forward direction, electrons flow from the emitter to the base and then to the collector. This causes a large current to flow from the emitter to the collector, which is controlled by the small current flowing from the base to the emitter.

However, when the transistor is biased in the reverse direction, a small current flows from the collector to the base, known as the collector cut-off current (ICBO). This current is caused by the minority carriers (holes) in the collector region that are injected into the base region and recombine with the majority carriers (electrons).

To reduce the collector cut-off current, the emitter is doped with a high level of impurity. This has the following effects:

1. Increase in emitter injection efficiency: The high level of impurity in the emitter increases the concentration of majority carriers (electrons), resulting in a higher emitter injection efficiency. This means that more electrons are injected from the emitter into the base, which reduces the number of holes in the collector region available for injection into the base.

2. Narrowing of the depletion region: The high level of impurity in the emitter also narrows the depletion region at the emitter-base junction. This reduces the width of the base region that minority carriers (holes) can diffuse into and recombine with, further reducing the collector cut-off current.

3. Increase in base current: The high level of impurity in the emitter also increases the amount of current flowing from the base to the emitter. This helps to reduce the collector cut-off current by diverting some of the minority carriers (holes) away from the collector and towards the emitter.

In summary, doping the emitter with a high level of impurity reduces the collector cut-off current by increasing the emitter injection efficiency, narrowing the depletion region, and increasing the base current.
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In a junction transistor, the collector cut-off current 'ICBO' reduces considerably by doping thea)emitter with high level of impurityb)emitter with low level of impurityc)collector with high level of impurityd)collector with low level of impurityCorrect answer is option 'A'. Can you explain this answer?
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