Difference in Resitivity in P and N type semiconductors

[Puma24]

Homework Statement

I've been given some Bench Notes for a Physics experiment I am supposed to be writing a report up for. What the notes show is a graph of logp (log of resistivity) and logn (log of number density of charge carriers) for given N and P type semiconductors. What is clearly shown is that for two N and P type semiconductors with the same n, P types have a higher resistivity than for the N type. My question is, why?

Homework Equations

None. :)

The Attempt at a Solution

Well, I was thinking that since we are dealing with positive charge carriers in the P type, there is an inherent resistivity increase due to the fact that the charge carriers, rather than just move themselves, have to wait for electrons to fill and then leave their electron holes. Perhaps there is some energy loss when the electrons enter and then exit the holes, perhaps something to do with entering the conduction band, lowering to the valence band, then having to be excited back to the conduction band before moving on.

I've tried googling information pages on this phenomenon, and while a lot of them indicate there is this notiable resitivity difference, none attempt to explain it.

Any help is appreciated. Thanks!

 

[phyzguy]

Electrons and holes in semiconductors are quasi-particles with different properties than free electrons. The resistivity of the different types is different because the mobility of holes is lower than that of electrons, and this difference is due to the difference in effective mass between electrons and holes. This difference in effective mass is due to the detailed band structure, which is dependent on the crystal structure. The references below might help. Note that the effective mass of an electron in a crystal is different than the mass of a free electron.


http://en.wikipedia.org/wiki/Semiconductor_carrier_mobility

http://en.wikipedia.org/wiki/Effective_mass_(solid-state_physics)