4.44
Plot
_______________________________________
4.45
so cm
cm
_______________________________________
4.46
(a) p-type
Majority carriers are holes
cm
Minority carriers are electrons
(c)
so cm
_______________________________________
4.48
For Germanium
(K)
(eV)
(cm )
and
cm
(K)
(cm )
(eV)
_______________________________________
4.49
(a)
For cm , eV
4.29
So
We find
eV
_______________________________________
4.30
(a)
Then
cm
(b)
cm
_______________________________________
4.31
For the electron concentration
The Boltzmann approximation applies, so
4.22
(a)p-type
(b) eV
cm
eV
cm
_______________________________________
4.23
(a)
cm
cm
(b)
cm
cm
_______________________________________
4.24
(a)
eV
(b)
eV
(c)
cm
(d)Holes
(e)
Chapter 4
4.1
where and are the values at 300 K.
(a) Silicon
(K)
(eV)
(cm )
(b) Germanium (c) GaAs
(K)
(cm )
(cm )
_______________________________________
4.2
Plot
cm
and
cm
_______________________________________
4.39
(a) n-type
(b)
cm
cm
(c)
cm
cm
_______________________________________
4.40
cm
n-type
_______________________________________
meV
(b)
meV
_______________________________________
4.13
Let constant
Then
Let
so that
We can write
so that
The integral can then be written as
which becomes
_______________________________________
4.55
(a)Silicon
(i)
eV
(ii) eV
cm
cm Additional
donor atoms
(b)GaAs
(i)
eV
(ii) eV
cm
cm Additional
donor atoms
_______________________________________
4.56
(a)
eV
_______________________________________
4.20
(a) eV
cm
eV
cm
(b)
eV
eV
cm
_______________________________________
4.21
(a) eV
cm
eV
cm
(b)
eV
eV
cm
_______________________________________
4.7
where
and
Then
or
_______________________________________
4.8
Plot
_______________________________________
4.9
Plot
_______________________________________
4.10
4.37
(a) For the donor level
or
(b) We have
Now
or
Then
or
_______________________________________
4.38
(a) p-type
(b)Silicon:
or
cm
Then
cm
Germanium:
or
cm
Then
cm
Gallium Arsenide:
eV
eV
_______________________________________
4.16
We have
For gallium arsenide, ,
Then
The ionization energy is
or
eV
_______________________________________
4.17
4.41
cm
So cm ,
Then cm
so that cm
_______________________________________
4.42
Plot
_______________________________________
4.43
Plot
_______________________________________
we find the maximum at
_______________________________________
4.32
(a)Silicon: We have
We can write
For
eV and eV
we can write
or
cm
We also have
Again, we can write
4.36
(a)Ge: cm
(i)
or
cm
cm
(ii)
cm
cm
(b)GaAs: cm
(i) cm
cm
(ii) cm
cm
(c) The result implies that there is only one minority carrier in a volume of cm .
_______________________________________
Then
To find the maximum value:
which yields
The maximum value occurs at
nd the maximum value
Same as part (a). Maximum occurs at
or
_______________________________________
cm
(b)Boron atoms must be added
So cm
cm
_______________________________________
4.47
(a) n-type
(b)
cm
electrons are majority carriers
cm
holes are minority carriers
_______________________________________
4.3
(a)
By trial and error, K
(b)
By trial and error, K
_______________________________________
4.4
At K,
eV
At K,
eV
or
or eV
For
and eV
Then
or
cm
(b)GaAs: assume eV
Then
or
cm
Assume eV
Then
or
cm
_______________________________________
4.33
Plot
_______________________________________
4.34
_____________________________________
4.27
(a)
cm
eV
cm
(b) eV
cm
cm
eV
eV
cm
_______________________________________