Electrical conductivity of a semiconductor

[Plasmosis1]

Homework Statement

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.

Homework Equations

σ=|e|(n_e*μ_e+n_h*μ_h)

σ=conductivity
e=1.6*10^-19
n_e=number of free electrons
μ_e=electron mobility
n_h=number of holes
μ_h=hole mobility

The Attempt at a Solution

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

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

 

[mfb]

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

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.

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

Don't forget the units.

 

[Plasmosis1]

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

 

[mfb]

n_e*n_h = ?

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

 

[Plasmosis1]

Is the equation
n_e*n_h=(1.5x10¹⁰cm^-3)²
where n_i=1.5x10¹⁰cm^-3

So n_e=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?

 

[mfb]

Where does the first equation come from?

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

 

[Plasmosis1]

No one actually said this but because n_e<<n_h you can just neglect n_e.

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