A Microscopic View of Ohm's Law — Question 4

Question 4

Two mobile carrier species contribute conductivities with the same sign

A homogeneous medium has electrons of density $2n$, mass m and relaxation time tau, plus positive carriers of density n, mass $2m$ and relaxation time $3\tau$. Each has charge magnitude e.

1. Derive total conductivity by summing the drift currents.

2. Find the fraction of current carried by positive carriers and their speed relative to electron drift speed.

Question 4 - Solution

Two mobile carrier species contribute conductivities with the same sign

(a) Method and calculation

$$\sigma_e=\frac{2ne^2\tau}{m},\quad\sigma_+=\frac{ne^2(3\tau)}{2m}.$$$$\sigma_{\mathrm{total}}=\sigma_e+\sigma_+=\frac{7ne^2\tau}{2m}.$$

(b) Interpretation and further calculation

$$I_+/I=\sigma_+/\sigma_{\mathrm{total}}=3/7.$$$$|v_+|/|v_e|=\frac{e(3\tau)/(2m)}{e\tau/m}=3/2.$$

Physical check

Although the carriers drift in opposite directions, the charge sign makes their conventional currents add.