Physical Electronics
Homework #3
Due on: 25th October, 2013
1. An n-type Silicon sample contains a donor concentration of Nd = 1016 cm-3.
The minority carrier hole lifetime is found to be p0= 20 s.
a. What is the lifetime of the majority carrier electrons?
b. Determine the thermal-equilibrium generation rate for electrons and
holes in this material.
c. Determine the thermal-equilibrium recombination rate for electrons and
holes in this material.
Solution:
(a) Recombination rates are equal:
=
10
.
2.25 10 So:
10
2.25 10
20 10
s.
.
(b) Generation Rate = Recombination Rate
Therefore:
.
= 1.125x
(c) R = G = 1.125x
2. (a) A sample of semiconductor has a cross-sectional area of 1cm2 and a
thickness of 0.1cm. Determine the number of electron-hole pairs that are
generated per unit volume per unit time by the uniform absorption of 1 Watt
of light at a wavelength of 6300Å. Assume each photon creates one electronhole pair. (b) If the excess minority carrier lifetime is I0s, what is the steadystate excess carrier concentration?
Solution:
a) E = hv =
=
.
= 3.15x10
J.
This is the energy of 1 photon.
Now,
1W = 1J/s => 3.17x10
Volume = (1)(0.1) = 0.1
Then,
.
g=
= 3.17x
.
e-h pairs/
b) n = p = g = (3.17x10 )(10x10 )
n = p = 3.17x
3. A sample of Ge at T = 300K has a uniform donor concentration of 2 x 1013
cm-3. The excess carrier lifetime is found to be p0= 24 s. Determine the ambipolar diffusion coefficient and the ambi-polar mobility. What are the electron
and hole lifetimes?
Solution:
For Ge: T = 300K,
= 2.4x10
=10
10
2.4 10
=3.6 10
Also,
.
=
p=
= 1.6x10
.
We have:
3900,
1900
101,
49.2
Therefore:
=
.
=
.
.
.
.
= 58.4
.
Also,
 =
 


.
==
.
.
.
= -868
Now,
=
=
.
=
.

= 54s.
4. Consider a Silicon material doped with 3x1016cm-3 donor atoms. At t=0, a
light source is turned on, producing a uniform generation rate of g' =
1x1020cm-3s-1. At t = 2x10-6s, the light source is turned off. The excess carrier
lifetime is found to be p0= 1s. Determine the excess minority carrier
concentration as a function of time for 0≤t≤∞.
Solution:
n-type Silicon, For 0≤t≤2 10


1
1 10
exp
1 10
1
exp

10
1
exp
where
= 1 10 s
2 10 s,
 2
10
 2
0.865 10
1
exp
For t>2 10 s,


.
:
=
s