2021 A/L Physics Past Paper · Question 22 (English Medium)

Answer: (2) $n_E > n_C > n_B$

What the question asks

What we know:

A transistor has three parts: the emitter (E), the base (B) and the collector (C). Each is doped to a different level.

What each letter means:

$n_E$ ($\text{m}^{-3}$): doping concentration of the emitter, the number of impurity atoms added in each cubic metre.

$n_B$ ($\text{m}^{-3}$): doping concentration of the base.

$n_C$ ($\text{m}^{-3}$): doping concentration of the collector.

We want: the right order of the three doping concentrations.

The plan: look at the job of each part. Then see how much doping that job needs.

Relevant theory

the region which emits charge carriers is known as the emitter and the region where charge carriers are collected is known as the collector.

The base region in the middle is a very thin layer of semiconductor and it has been doped with a very low doping level.

The collector region is doped with a moderate level and its area has to be made large.

Reading the options

Think of a crowded school gate. Many children are waiting to come out. That gate is the emitter. It must send out a great many carriers, so it is doped the most.

The children cross a thin, quiet corridor with few people in it. That is the base. It is very thin and doped very low.

At last they reach a big hall. That is the collector. It is large, and it is doped to a moderate level: more than the base, less than the emitter.

Emitter (E)

$Most$

Sends out the carriers, so it needs the most of them. Doped the most.

Base (B)

$Least$

A very thin layer, doped very low.

Collector (C)

$In between$

Doped to a moderate level, between the emitter and the base.

Put the three in order: the emitter most, then the collector, then the base.

$$n_E > n_C > n_B$$

This says: the emitter has the biggest doping concentration and the base the smallest. It is option (2).

Why the other options fail

The emitter is doped the most. The base is doped the least.

The base is doped very low, not the most.

The collector is doped only to a moderate level, less than the emitter.

The emitter and the collector are not equal. The collector is only moderate.

Conclusion

Does it make sense?

The emitter must send many carriers, so it needs the most doping. The base is thin and lightly doped, so most carriers pass through it to the collector.

So $n_E > n_C > n_B$ fits the job of each part.

$n_E > n_C > n_B$

The emitter is doped the most, the base very low, and the collector to a moderate level in between.

So $n_E > n_C > n_B$. The answer is (2).

Remember the three parts by their jobs: the emitter emits (most doping), the base controls (least, and thin), the collector collects (in between, and large).

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