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How to calculate the carrier concentration

by myousuf
Tags: carrier, concentration
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myousuf
#1
Sep26-07, 07:13 PM
P: 4
Can any please help me in solving the following two questions

Q1
A Si sample is doped with 10^16 per cm cube boron atoms and a certain
number of shallow donors. The fermi level (Ef) is 0.36 eV above Ei
(intrinsic energy level) at 300K. What is the donor concentration Nd?

For Si at 300K ni(intrinsic carrier concentration) = 1.5 x 10^10 per
cm cube




Q2
A Si sample contains 10^16 per cm cube In(indium) acceptor atoms and
a certain number of shallow donors. The In (indium) acceptor level is
0.16 eV above Ev(Valence band edge), and Ef is 0.26eV above Ev at
300K. How many in atoms in cm per cube are unionized (i.e. neutral)?

For Si at 300K ni(intrinsic carrier concentration) = 1.5 x 10^10 per
cm cube




There is no additional information available. Please state the
question number when answering and indicate any formulas used.

The following equations may prove useful

n(o) x p(o) = ni^2

n(o) = ni x e((Ef-Ei))/KT)

p(o) = ni x e((Ei-Ef)/KT)
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malawi_glenn
#2
Sep27-07, 01:01 AM
Sci Advisor
HW Helper
malawi_glenn's Avatar
P: 4,739
What have you done so far? If you want help, than you should say what you think and what you have tried.

Is there any more relations that you know of? How about "carge neutralisty condition of doped semi conductor" ? And Law of mass action?
myousuf
#3
Sep27-07, 06:25 PM
P: 4
This is what I have came up with so far for question number 1

using the relation n(o) = ni x e((Ef-Ei))/KT)

with Ef-Ei=0.36 x 1.6 x 10^-19 , ni=1.5 x 10^10, T=300k , K=1.38 x 10^-23

we get n(o) = 1.654 x 10^16 per cm cube

However there are 10^16 B atoms to neutralize these charges

Hence Nd = 1.654 x 10^16 - 10^16 = 6.5 x 10^15 per cm cube

myousuf
#4
Sep27-07, 07:27 PM
P: 4
How to calculate the carrier concentration

For question number 2

unionized atoms are left at the acceptor level

Ef-Ev=0.26eV

Ea-Ev=0.16eV

Ea-Ef=0.16-0.26= -0.10eV

Using fermi-driac statistics f(E)=1/(1+e((E-Ef)/KT)))

for E=Ea,T=300 and substituting all the constants

f(E)=1/(1+e((Ea-Ef)/KT))), gives = 0.9794

However fermi driac statistics give the probablity of occupance of an electron in an Energy state E. hence 0.9794 is the probablity of occupance of an electron.

hole probablity of occupance=1-electron probablity of occupancy

hole probablity of occupance of the energy state Ea = 1-0.9794 = 0.02053

unionized atoms are left at the acceptor level(Ea)

Hence, number of IN(indium) atoms left unionized = hole probabilty of occupancy of Energy state Ea x number of In acceptor atoms

=0.02053 x 10^16

=2.05 x 10^14
myousuf
#5
Sep27-07, 07:29 PM
P: 4
Anyway there are still questions that remain unanswered

For instance, what role does shallow donor impurities have to play in question number 2.

Any suggestions to the proposed solution above will be highly appreciated


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