Calc Electron/Hole Concentrations & Ef-Efi for Si Doping in GaAs

In summary, the silicon replaces a gallium atom in the GaAs, creating a concentration of 1.5*10^8 of acceptors and donors. The electron concentration is 1.29*10^8 and the hole concentration is 1.74*10^8.
  • #1
orangeincup
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Homework Statement


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

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


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
 
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  • #2
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
mfb said:
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.
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
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
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
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
Those numbers look more reasonable.
 
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  • #8
mfb said:
Those numbers look more reasonable.
Okay thank you
 

What is the concept of electron and hole concentrations in semiconductors?

The concept of electron and hole concentrations refers to the number of electrons or holes present in a semiconductor material. Electrons and holes are charge carriers that are responsible for the electrical properties of a semiconductor.

What is the difference between n-type and p-type doping in semiconductors?

N-type doping refers to the addition of donor atoms (such as phosphorus or arsenic) to a semiconductor material, which increases the number of free electrons and creates an excess of negative charge. P-type doping, on the other hand, involves adding acceptor atoms (such as boron or gallium) which creates an excess of positively charged holes.

How are electron and hole concentrations calculated in semiconductors?

Electron and hole concentrations can be calculated using the following equations:

n = ND * exp[(EF - Ei) / kT]

p = NA * exp[(Ei - EF) / kT]

where n and p are the electron and hole concentrations, ND and NA are the donor and acceptor concentrations, EF is the Fermi level, Ei is the intrinsic energy level, k is the Boltzmann constant, and T is the temperature.

What is the significance of the Fermi level in semiconductors?

The Fermi level is an important parameter in semiconductors as it represents the energy level at which there is a 50/50 probability of finding an electron. In other words, it is the energy level at which the electron and hole concentrations are equal, and the material is in equilibrium.

How does doping affect the Fermi level and the bandgap of a semiconductor material?

Doping a semiconductor material with donor or acceptor atoms affects the Fermi level by shifting it towards the energy level of the majority carriers (i.e. electrons for n-type doping and holes for p-type doping). This shift in the Fermi level also affects the bandgap of the material, causing it to decrease for n-type doping and increase for p-type doping.

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