- #1

orangeincup

- 123

- 0

## Homework Statement

The question is in the picture attached. When Na=0, that means the acceptor concentration(i.e. open holes?) is zero?

## Homework Equations

(Nd-Na)/2 + sqrt((Nd-Na/2)^2+ni^2)

(Na-Nd)/2 + sqrt((Na-Nd/2)^2+ni^2)

## The Attempt at a Solution

Solving for Nd=10^8:

**majority carrier =**

(Nd-Na)/2 + sqrt((Nd-Na/2)^2+ni^2)

(10^8)/2 + sqrt((10^8/2)^2+1.5*10^10^2)= 1.505*10^10... almost same as ni?

**minority carrier=**

(-10^8)/2 + sqrt((-10^8/2)^2+1.5*10^10^2)=1.495*10^10... close to ni

**Log10(10^8)**=

8... is this all they want?

**log10(1.495*10^10)**=

10.17

**Fermi energy difference**

(Efi-Ev)

8.612*10^-5*300ln(1.505*10^10/1.5*10^10)=8.598*10-5Do these answers look right for the first one? Is it the same the whole way down?