Electronic Devices - Bipolar Junction Transistors - Discussion

According to me, C is the right option. Because base width is much lower than collector width, if base is less heavily doped than collector, then the depletion region width will be more towards base. Base being very thin, the depletion region may cover the entire base region. In that case, no carriers will be able to travel through base towards collector.

Hence, base is more heavily doped than collector in order to prevent is issue, also known as punch-through condition.

In most transistors, emitter is heavily doped. Its job is to emit or inject electrons into the base. These bases are lightly doped and very thin, it passes most of the emitter-injected electrons on to the collector. The doping level of collector is intermediate between the heavy doping of emitter and the light doping of the base. So, answer is e>c>b.

C is right option. Emitter is highly doped for high conductivity, then base is moderately doped because we need (Wb<Lb) and collector is lightly doped. Here Wb is width of depletion layer and Lb is diffusion length.

Doping level of emitter is highest, whereas base is lightly doped, and collector doping is intermediate between emitter and base doping level. So B is the right answer.

Base is lightly doped.
Collector is moderately doped.
Emitter is heavily doped.
So, the answer is B.

The answer is wrong. Doping level is emitter>base>collector.

Answer is wrong the doping level e>b>c.

E>B>C.

This is the correct answer.

The correct answer is e > b > c.

The right answer is option C.