Question

4.17 In an n- type semiconductor bar, there is an increase in electron concentration from left to right and an electric field pointing to the left. With a suitable sketch, indicate the directions of the electron drift and diffusion current flow and explain why. If we double the electron concentration everywhere, what happens to the diffusion current and the drift current

Answer 1

i) Jn = q Dn
`(dn)/(dx)`

ii) Diffusion current and drift current will be doubled

Step-by-step explanation:

i) Determine the directions of the electron drift and diffusion current flow

the electron concentration is higher on the right side of the N-type semiconductor hence any change in the concentration of electrons will cause diffusion of electrons from right to left and electron drift is from left to right and this is because the electrons in the semiconductor bar are attracted to the electric field.

Jn = q Dn
`(dn)/(dx)` ( electron diffusion current )

ii) Determine what happens to the diffusion current and drift current If we double the electron concentration everywhere

The electron diffusion current ( J'n ) = 2q Dn
`(dn)/(dx)` = 2Jn ( i.e. diffusion current doubles )

Drift current = ( Jdrift = nq*μn*ε)

when the electron concentration is double

Drift current = 2 ( nq*μn*ε )

hence the drift current will double