Question 12.4: Design the ratio of emitter doping to base doping in order t...

Equation (12.35b) reduces to

\gamma \approx \frac{1}{1+\frac{N_{B}}{N_{E}} \cdot \frac{D_{E}}{D_{B}} \cdot \frac{x_{B}}{x_{E}}}     (12.35b)

\gamma=\frac{1}{1+\frac{p_{E 0}}{n_{B 0}}}=\frac{1}{1+\frac{n_{i}^{2} / N_{E}}{n_{i}^{2} / N_{B}}}

so

\gamma=\frac{1}{1+\frac{N_{B}}{N_{E}}}=0.9967

Then

\frac{N_{B}}{N_{E}}=0.00331 \quad \text { or } \quad \frac{N_{E}}{N_{B}}=302

Comment

The emitter doping concentration must be much larger than the base doping concentration to achieve a high emitter injection efficiency.