Current Density — Question 4

Question 4

A taper and a carrier-density gradient compete to change drift speed

In a slowly tapered steady wire, $A(x)=A_0(1+x/L)$ and $n(x)=n_0/(1+x/L)^2$, for $0\le x\le L$. Current I is positive and carriers are electrons. Use cross-section averages.

1. Find current density and electron drift velocity versus position.

2. Find their endpoint ratios and the drift transit time from L to 0.

Question 4 - Solution

A taper and a carrier-density gradient compete to change drift speed

(a) Method and calculation

$$J(x)=I/A(x)=\frac{I}{A_0(1+x/L)}.$$$$v_d(x)=-\frac{J}{en}=-v_0(1+x/L),\quad v_0=I/(en_0A_0).$$

(b) Interpretation and further calculation

$$J(L)/J(0)=1/2,\quad |v_d(L)|/|v_d(0)|=2.$$$$t_{L\to0}=\int_0^L\frac{dx}{v_0(1+x/L)}=L\ln2/v_0.$$

Physical check

A smaller current density does not guarantee slower drift if carrier density falls faster.