The Electrical Properties of Solids — Question 4

Question 4

Two carrier populations contribute currents in the same field direction

A solid has electrons $n=2.0\times10^{21}\,\mathrm{m^{-3}}$, holes $p=1.0\times10^{21}\,\mathrm{m^{-3}}$, and mobilities $\mu_n=.12$, $\mu_p=.040\,\mathrm{m^2/(V\,s)}$. Use $e=1.60\times10^{-19}\,\mathrm C$ and $E=50.0\,\mathrm{V/m}$.

1. Find each carrier drift velocity.

2. Calculate conductivity and each current-density contribution.

Question 4 - Solution

Two carrier populations contribute currents in the same field direction

(a) Method and calculation

$$\boxed{v_n=-\mu_nE=-6.00\,\mathrm{m/s}}.$$$$\boxed{v_p=+\mu_pE=+2.00\,\mathrm{m/s}}.$$

(b) Interpretation and further calculation

$$\sigma=e(n\mu_n+p\mu_p)=\boxed{44.8\,\mathrm{S/m}}.$$$$\boxed{J_n=en\mu_nE=1920,\quad J_p=ep\mu_pE=320\,\mathrm{A/m^2}}.$$

Physical check

Negative electrons move against the field but carry conventional current along it. Positive holes move and contribute current along the field.