Muller and Kamins Problem 1.13
The idea is this. Carriers move in random directions, and we take the mean velocity
to be the thermal velocity Vth (not to be confused with the threshold voltage). To solve
the problem, we envision a kind of “velocity space.” Which is just the velocity vectors of
all the particles pulled to the origin. Let us assign the mean velocity magnitude to ALL
the particles. In that case, the velocity vectors uniformly terminate on the surface of the
sphere, whose radius is Vth . This is shown in fig. 1 below. We wish to calculate the mean
velocity of carriers in the z direction. To do this, realize that in this “velocity space”
representation, the fraction of total carriers through a circular swath of angular width dθ
is:
dN =
2πrdθ
rdθ
=
4π
2
(1)
where r=Vth sin(θ).
The projection of the velocity in this swath along the z-axis is Vth cos(θ). Thus, we can
write the fractional contribution of this swath to the mean as:
sin(θ)dθ
(2)
2
And we get the total mean velocity by integrating over all of the contributions from each
swath which can contribute velocity to the positive z direction.
d < Vth >= Vth
1
< V >=
2
Z
0
2π
1
Vth sin(θ)cos(θ)dθ = Vth
2
2π
Z
0
1
cos(θ)dsin(θ) = Vth
2
Z
0
1
1
xdx = Vth x2 |10
4
(3)
and so:
1
< V >= Vth
(4)
4
And the current density for these carriers (if they’re electrons whose total volume
density is Nt ) is:
1
J = −eNt < V >= − eNt Vth
4
1
(5)
Figure 1: “Velocity space” representation of randomly moving charges.
2
The negative sign is due to the fact that electrons are moving in a positive direction
here.
3