Semiconductor physics: Resistivity,mobility and concentration.

Aerozeppelin
Messages
16
Reaction score
0

Homework Statement



If a sample of pure silicon at 300 Kelvin has a resistivity of 950Ωm, and if the electron-to-hole mobility ration is 3:1, with the electron mobility equal to 0.12m2V-1s-1, what are the intrinsic hole and electron concentrations?

Homework Equations



I know resistivity is equal to
\frac{ρL}{A}. But I don't think this is applicable. I cannot find any equations in books or on the internet connecting these properties (concentration, mobility and resistivity).

The Attempt at a Solution



Well considering the ratio of electron to hole mobility is 3:1, I presume that the hole mobility is 0.04m2V-1s-1 (ie. electron mobility divided by 3). After that, I'm completely lost!
Any help would be appriciated! :smile:
 
Physics news on Phys.org
You should be able to write the conductivity of the silicon (inverse of the resistivity) in terms of the concentration of carriers and their mobility. If you haven't learned this, Google it.
 
Can't believe I didn't think of that!
Thanks a million.

Here's my answer if anyone wants it.

σ=\frac{1}{ρ}

1.053x10-3 = q (nμn + pμp)
Where:

q is the charge of an electron.
μn is the mobility of electron. = 0.12
And μp is the mobility of holes. = 0.04

To solve for concentrations (n and p),

Rearranging,

0.12n + 0.04p = 6.579x1015

N = P in intrinsic.

∴ 0.16n = 6.579x1015

& p = n = 4.118x1016
 
Looks good, I think you did it all correctly. However, there are no units on the answer. I always take off points for a number with no units.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top