5

I apologize if this is a very basic question. But I have always known it to be true that in Silicon, electrons have higher mobility than holes. From my semiconductor physics classes in first year, the thing I recalled was that mobility is inversely related to effective mass for the basic scattering model. So I assumed that electrons being more mobile must have lower effective mass. This is wrong as I learned going through the material again: electrons have HIGHER EFFECTIVE MASS than holes. So is there any explanation for this? Thanks!

First User
  • 289
  • 1
  • 5
  • I’m voting to close this question because it belongs on the Physics stack – Neil_UK Sep 08 '22 at 11:13
  • 1
    Here's the [wiki](https://en.wikipedia.org/wiki/Effective_mass_(solid-state_physics)) on your topic. There, you can find that the electron can have 'negative mass'. Are you referring to this description? – jonk Sep 08 '22 at 11:14
  • @jonk no the electron cannot have negative mass – Miss Mulan Sep 08 '22 at 14:02
  • 2
    @Neil_UK Sorry, but I don't agree. In the help center it's stated that *the theory and simulation of electromagnetic forces* are on topic, and arguably solid state physics as employed in electronic devices falls under that category. Moreover, semiconductor behavior is a topic commonly taught at university in Electronic Engineering courses (at least here in Italy). So closing is not the right action here. OTOH, I may agree that the OP could get better answers at PHY.SE just because more experts of the field may well be dwell there. – LorenzoDonati4Ukraine-OnStrike Sep 08 '22 at 20:13

1 Answers1

6

The mobility is proportional to a scattering time and inversely proportional to a quasiparticle mass. The explanation you are looking for is simple: in silicon, a lower hole mass is outweighed by a higher scattering time of electron quasiparticles in lattice.

V.V.T
  • 3,521
  • 7
  • 10
  • 1
    thanks. I was assuming a parameter to be constant when it is not. However, Im confused. You mean the scattering time of holes is LOWER right? because mobility is proportional to the scattering time? – First User Sep 08 '22 at 19:40
  • 1
    Oh, sorry, corrected. Thank you. – V.V.T Sep 08 '22 at 20:55