In a Bipolar Junction Transistor (BJT), emitter injection efficiency is a measure of how effectively the emitter injects carriers (electrons in an NPN transistor or holes in a PNP transistor) compared to the base. The emitter injection efficiency ($\gamma$) is defined as:

$\gamma = \frac{\text{current injected by the emitter}}{\text{total current entering the base-emitter junction}} = \frac{I_E}{I_E + I_B}$

For the emitter injection efficiency to be close to unity ($\gamma \approx 1$), most of the carriers must be injected by the emitter, and only a small portion of carriers should come from the base. This implies the following:

• The emitter should be much more heavily doped than the base, i.e., $N_E >> N_B$, to ensure that a higher number of carriers are injected from the emitter into the base.
• The base should be lightly doped compared to the emitter to avoid recombination in the base.
• The collector's doping level is usually less than that of the base to facilitate easy collection of carriers, i.e., $N_C < N_B$.

Given this information, the correct answer is:

Ne >> Nb and Nb > Nc

This corresponds to the third option:
Ne >> Nb and Nb > Nc

Do you want details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How does base width affect the base transport factor in a BJT?
2. What is the relationship between doping concentration and breakdown voltage in a BJT?
3. How does emitter doping affect the forward current gain in a BJT?
4. Why is the base typically lightly doped in a BJT?
5. What factors affect the common-emitter current gain ($\beta$) of a BJT?

Tip: A higher emitter doping concentration reduces recombination in the base, improving efficiency and current gain in a BJT.