# Homework Help: Carrier concentration in a semiconductor

1. Oct 9, 2015

### orangeincup

1. The problem statement, all variables and given/known data
The question is in the picture attached. When Na=0, that means the acceptor concentration(i.e. open holes?) is zero?

2. Relevant equations
(Nd-Na)/2 + sqrt((Nd-Na/2)^2+ni^2)
(Na-Nd)/2 + sqrt((Na-Nd/2)^2+ni^2)

3. 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*10-5

Do these answers look right for the first one? Is it the same the whole way down?

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2. Oct 12, 2015

### orangeincup

Ok I calculated the rest the same way, here's how I did the Nd=10^14

majority carrier =
(Nd-Na)/2 + sqrt((Nd-Na/2)^2+ni^2)
(10^14)/2 + sqrt((10^14/2)^2+1.5*10^10^2)= 1.00*10^14
minority carrier=
(-10^14)/2 + sqrt((-10^14/2)^2+1.5*10^10^2)
2.24*10^6

log(10^14) = 32.32 donor concentration
log(1.00*10^14)=32.23 majority carrier

So does this make sense? My donor conentration is now equation to my majority carrier concentration
Where as my minority carrier is a lot lower

log(2.24*10^6)=14.62