# Electrical conductivity of a semiconductor

1. Apr 28, 2014

### Plasmosis1

1. The problem statement, all variables and given/known data

Estimate the electrical conductivity, at 135°C, of silicon that has been doped with 3 x 1024 per meter cubed of aluminum atoms. Assume values for electron and hole mobilities of 0.03 and 0.007 m2/V-s, respectively.

2. Relevant equations

σ=|e|(nee+nhh)

σ=conductivity
e=1.6*10-19
ne=number of free electrons
μe=electron mobility
nh=number of holes
μh=hole mobility

3. The attempt at a solution

So far I have this:
σ=1.6*10-19*(ne*0.03+3*1024*0.007)

I don't know how to find ne. Is it the same value of nh?

2. Apr 28, 2014

### Staff: Mentor

Then doping would be pointless - no.
You should have some formula relating those two numbers (it also includes the temperature). Then the density of aluminium atoms (which type of doping is that?) allows to determine both.

Don't forget the units.

3. Apr 28, 2014

### Plasmosis1

The only other formula that I can think of is n=n0exp(-Eg/2kT)
But I can't use that because I don't know n0 or Eg.

Last edited: Apr 28, 2014
4. Apr 28, 2014

### Staff: Mentor

ne*nh = ?

You will have to look up at least one material constant of silicon.

5. Apr 28, 2014

### Plasmosis1

Is the equation
ne*nh=(1.5x1010cm-3)2
where ni=1.5x1010cm-3

So ne=1.5*1010*1003m-3/(3*1024)=1.5*1016m-3
∴ σ=1.6*10-19*(1.5*1016*0.03+3*1024*0.007)=3360 (Ωm)-1

Is that right?

6. Apr 29, 2014

### Staff: Mentor

Where does the first equation come from?

The other equations look good (apart from formatting issues).

7. Apr 29, 2014

### Plasmosis1

No one actually said this but because ne<<nh you can just neglect ne.

The original equation becomes:
σ=|e|*nhh
=1.6*10-19*3*1024*.007
=3360 (Ωm)-1