Chapter 14
14.1
m
(a) Si: m
(b) Ge: m
(c) GaAs: m
(d) InP: m
_______________________________________
14.2
(a) For nm,
eV
For nm,
eV
(b) For eV,
m
For eV,
m
For eV,
m
_______________________________________
14.3
(a) m
(i) From Figure 14.4, cm
(ii)
Fraction absorbed
(b) m
(i) From Figure 14.4, cm
(ii)
Fraction absorbed
_______________________________________
14.4
For eV, m
For silicon: cm
Then for W/cm, we obtain
or
cms
The excess concentration is
or
cm
_______________________________________
14.5
(a)
cms
For eV,
m
From Figure 14.4, cm
W/cm
(b)
cmm
_______________________________________
14.6
m
From Figure 14.4, cm
(a)
cmm
(b)
cmm
_______________________________________
14.7
GaAs:
For mcm, we have 50%
absorbed or 50% transmitted, then
We can write
or
cm
This value corresponds to
m , eV
_______________________________________
14.8
The ambipolar transport equation for minority
carrier holes in steady state is
or
where
The photon flux in the semiconductor is
and the generation rate is
so the differential equation becomes
The general solution is of the form
As , so that . Then
At , we have
so we can write
and
Then we have
Solving for A, we find
The solution can now be written as
_______________________________________
14.9
We have
or
where
The general solution can be written in the
form
For at means . Then
EMBED Equation.3 At ,
Now
and
so we can write
Solving for B, we obtain
The solution is then
where B was just given.
_______________________________________
14.10
cm
cm
Now
A/cm
A
(a)
We find
V
cm
Then
AmA
(b)
V
(c)
_______________________________________
14.11
From Problem 14.10, A
(a)
V
(b)
V
(c)
By trial and error, V
Now
AmA
mW
(d)
_______________________________________
14.12
From Problem 14.10, A
(a)
(i)
V
(ii)
By trial and error, V
Now
AmA
Then
mW
(b)
(i)
V
(ii)
By trial and error, V
Now
AmA
Then
mW
(c)
_______________________________________
14.13
where
which becomes
or
Then
(cm)(A/cm)(V)
_______________________________________
14.14
(a)
A
We have
or
which becomes
A/cm
or
A
We have
or
We see that when , V.
We find
(V)(mA)
0
0.1
0.2
0.3
0.4
0.45
0.50
0.55
0.57
0.5950
50
50
50
49.98
49.84
48.89
42.36
33.46
14.19
(b) The voltage at the maximum power point
is found from
By trial and error,
V
At this point, we find
mA
so the maximum power is
or
mW
(c) We have
or
_______________________________________
14.15
(a)
V
(b)
EMBED Equation.3 By trial and error, V
AmA
mW
(c)
(d)
Now
By trial and error, V
Then A
mW
_______________________________________
14.16
(a)
V
(b)
By trial and error, V
Then
AmA
mW
(c) cells
(d) Now V
A
Then
(e) Then A
So
_______________________________________
14.17
Let correspond to the edge of the space
charge region in the p-type material. Then in
the p-region
or
where
Then we have
The general solution is of the form
As , so that . Then
We also have ,
which yields
We then obtain
where is the incident flux at .
_______________________________________
14.18
For 90% absorption, we have
Then
or
For eV , cm
Then
cm
or
m
and for eV, cm.
Then
cm
or
m
_______________________________________
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