# Semiconductor doping

1. Oct 9, 2015

### orangeincup

1. The problem statement, all variables and given/known data
Si atoms get doped inside GaAs to a concentration of 1.5*10^8 . Assume that the silicon atoms are
fully ionized and that 35percent atoms replace gallium and that 65% of the added
atoms replace arsenic t=300k
Fnd the acceptor and donor concentrations

Calculate electron / hole concentrations and Ef-Efi
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
Nd=.35*(7*10^15) =2.4*10^15
Na=(.65)*(7*10^15)=4.5*10^15

For the next part(assuming the above is correct), should I use ni of the GaAs or Si for the next part? Or the given value of 1.5*10^8?

(2.4*10^15-4.5*10^15)/2 + sqrt((2.4*10^15-4.5*10^15/2)^2+1.5*10^8^2) =-8.99*10^14 = donor concentration

(-2.4*10^15+4.5*10^15)/2 + sqrt((-2.4*10^15+4.5*10^15/2)^2+1.5*10^8^2)=1.2*10^15 = acceptor concentration

2. Oct 10, 2015

### Staff: Mentor

You should explain what you are calculating. Where does the number 7*10^15 come from?

Negative concentrations don't make sense.
The material is still GaAs, so you needs its properties. Plus the new acceptors/donors.

3. Oct 10, 2015

### orangeincup

it's the effective density of states of GaAs

Should I be using the intrinsic carrier concentration instead is 1.8*10^8 which I have not used yet.

4. Oct 10, 2015

### Staff: Mentor

You do not replace .35 of your gallium atoms.
.35 of the silicon atoms replace a gallium atom, but you have a tiny amount of silicon atoms compared to the gallium atoms.

5. Oct 10, 2015

### orangeincup

Is the concentration of silicon 1.8*10^8? 1.I thought the question was saying GaAs was at a concentration of 1.5*10^8 after the Silicon was added, am I reading it wrong?

Si replacing GA > Nd=.35*1.5*10^8 = 5.25*10^7
Si replacing As= Na=.65*1.5*10^8 = 9.75*10^7

Or should I be using the intrinsic carrier concentration of GaAs, and calculating how much Silicon has changed it by, then getting a fraction of the result?

6. Oct 13, 2015

### orangeincup

Si replacing GA > Nd=.35*1.5*10^8 = 5.25*10^7
Si replacing As= Na=.65*1.5*10^8 = 9.75*10^7

Electron concentration
n0=(Nd-Na)/2 + sqrt((Nd-Na/2)^2+ni^2)
(5.25*10^7-9.75*10^7)/2 + sqrt((5.25*10^7-9.75*10^7/2)^2+1.8*10^10^2) =
1.29*10^8 electron concentration
Hole concentration
p0=(Na-Nd)/2 + sqrt((Na-Nd/2)^2+ni^2)
(-5.25*10^7+9.75*10^7)/2 + sqrt(((-5.25*10^7+9.75*10^7)/2)^2+1.5*10^8^2))=
1.74*10^8 hole concentration

Ef-Efi
n0=niexp[(Ef-Efi)/kT]

1.29*10^8=(1.5*10^8)exp([Ef-Efi]/(300*8.61*10^-6))

Ef-Efi=(-0.00389)

7. Oct 13, 2015

### Staff: Mentor

Those numbers look more reasonable.

8. Oct 13, 2015

### orangeincup

Okay thank you