Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
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9780078028182
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Solutions Manual to
Principles of Electronic Materials and Devices
Fourth Edition
© 2018 McGraw-Hill
CHAPTER 5
Safa Kasap
University of Saskatchewan
Canada
Check author's website for updates
http://electronicmaterials.usask.ca
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
A SiC based Schottky photodiode used for UV detection (Courtesy of sglux, Germany)
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
Fourth Edition (© 2018 McGraw-Hill)
Chapter 5
5.1 Bandgap and photodetection
a. Determine the maximum value of the energy gap that a semiconductor, used as a photoconductor,
can have if it is to be sensitive to yellow light (600 nm).
b. A photodetector whose area is 5  10−2 cm2 is irradiated with yellow light whose intensity is 2 mW
cm−2. Assuming that each photon generates one electron-hole pair, calculate the number of pairs
generated per second.
c. From the known energy gap of the semiconductor GaAs (Eg = 1.42 eV), calculate the primary
wavelength of photons emitted from this crystal as a result of electron-hole recombination.
d. Is the above wavelength visible?
e. Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why?
Solution
a. We are given the wavelength  = 600 nm, therefore we need Eph = h = Eg so that,
Eg = hc/ = (6.626  10−34 J s)(3.0  108 m s−1) / (600  10−9 m)

Eg = 3.31  10−19 J or 2.07 eV
b. Area A = 5  10−2 cm2 and light intensity Ilight = 2  10−3 W/cm2. The received power is:
P = AIlight = (5  10−2 cm2)(2  10−3 W/cm2) = 1.0  10−4 W
Nph = number of photons arriving per second = P/Eph

Nph = (1.0  10−4 W) / (3.31  10−19 J) = 3.02  1014 Photons s−1
Since each photon contributes one electron-hole pair (EHP), the number of EHPs is then:
NEHP = 3.02  1014 EHP s−1
c. For GaAs, Eg = 1.42 eV and the corresponding wavelength is
 = hc/Eg = (6.626  10−34 J s)(3.0  108 m s−1) / (1.42 eV  1.602  10−19 J/eV)

 = 8.74  10−7 m or 874 nm
The wavelength of emitted radiation due to electron-hole pair (EHP) recombination = 874 nm.
d. It is not in the visible region (it is in the infrared).
e. From Table 5.1, for Si, Eg = 1.10 eV and the corresponding cut-off wavelength is,
g = hc/Eg = (6.626  10−34 J s)(3.0  108 m s−1) / (1.1 eV  1.602  10−19 J/eV)

g = 1.13  10−6 m or 1130 nm
Since the 874 nm wavelength of the GaAs laser is shorter than the cut-off wavelength of 1130 nm, the
Si photodetector can detect the 874 nm radiation (Put differently, the photon energy corresponding to
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
874 nm, 1.42 eV, is larger than the Eg, 1.10 eV, of Si which means that the Si photodetector can indeed
detect the 874 nm radiation).
5.2 Intrinsic Ge Using the values of the density of states effective masses me* and mh* in Table 5.1,
calculate the intrinsic concentration in Ge. What is ni if you use Nc and Nv from Table 5.1? Calculate
the intrinsic resistivity of Ge at 300 K.
Table 5.1 Selected typical properties of Ge, Si, InP, and GaAs at 300 K
Solution
From Table 5.1, we get
me = 0.56me
Now,
and
 2mekT 

N c = 2
2
h


3/ 2
mh = 0.4me
 2 (0.56  9.110 −31 kg)(1.38 10 −23 J K −1 )(300 K ) 
= 2

(6.626 10 −34 J s) 2


3/ 2
= 1.05×1025 m−3 or 1.05 ×1019 cm−3
 2mvkT 

N v = 2
2
 h

3/ 2
 2 (0.4  9.110 −31 kg)(1.38 10 −23 J K −1 )(300 K) 
= 2

(6.626 10 −34 J s) 2


3/ 2
= 6.35×1024 m−3 or 6.35 ×1018 cm−3
The intrinsic concentration is
 Eg 

ni = ( N c N v )1/ 2 exp −
 2kT 


ni = (1.049 1019 cm −3 )(6.33 1018 cm −3 )
 exp − 2(8.62 100.66eVeVK )(300 K) 
1/ 2
−5

−1

= 2.33 ×1013 cm−3  2.3 ×1013 cm−3
From Table 5.1, Nc = 1.04×1019 cm−3 and Nv = 6.0×1018 cm−3


ni = (1.04 1019 cm −3 )(6.0 1018 cm −3 )
 exp − 2(8.62 100.66eVeVK )(300 K) 
1/ 2

−5
−1

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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
= 2.26 × 1013 cm−3  2.3 ×1013 cm−3
The intrinsic conductivity is  = en e + ep h = eni (  e +  h )

 = (1.6 10 −19 C)(2.34 1013 cm −3 )(3900 + 1900) cm 2 V −1s −1 = 0.022 Ω−1 cm−1
And the intrinsic resistivity is ρ = 1/σ = 45.45 Ω cm
5.3 Fermi level in intrinsic semiconductors Using the values of the density of states effective
masses me* and mh* in Table 5.1, find the position of the Fermi energy in intrinsic Si, Ge and GaAs
with respect to the middle of the bandgap (Eg/2).
Solution
EFi = Ev +
 m 
1
3
E g − kT ln  e 
2
4
 mh 
For Si, me = 1.08 me and mh = 0.6 me

EFi = Ev +
 1.08 me 
1
3

E g − (8.62 10 −5 eVK −1 )(300 K ) ln 
2
4
 0.6 me 

EFi = Ev +
1
E g − 0.011 eV
2
So, the intrinsic Fermi level of Si is 0.011 eV below the middle of the bandgap (Eg/2).
For Ge, me = 0.56 me and mh = 0.4 me

EFi = Ev +
 .56 me 
1
3

E g − (8.62 10 −5 eVK −1 )(300 K ) ln 
2
4
0
.
4
m
e 


EFi = Ev +
1
E g − 0.0065 eV
2
So, the intrinsic Fermi level of Ge is 0.0065 eV below the middle of the bandgap (Eg/2).
For GaAs, me = 0.067 me and mh = 0.5 me

 0.067 me 
1
3

EFi = Ev + E g − (8.62 10 −5 eV K −1 )(300 K ) ln 
2
4
 0.5 me 

EFi = Ev +
1
E g + 0.039 eV
2
So, the intrinsic Fermi level of GaAs is 0.039 eV above the middle of the bandgap (Eg/2).
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
5.4 Extrinsic Si A Si crystal has been doped with P. The donor concentration is 1015 cm−3. Find the
conductivity, and resistivity of the crystal.
Solution
Nd = 1015 cm−3
Given,
Therefore the conductivity is
 = eNd e = (1.6 10−19 C)(1015 cm−3 )(1350cm2 V −1s −1 ) = 0.216 Ω−1cm−1
The resistivity is
ρ = 1/σ = 4.63 Ω−1cm−1
5.5 Extrinsic Si Find the concentration of acceptors required for a p-type Si crystal to have a
resistivity of 1  cm.
Solution
The resistivity,  =

1
eN a  h
Na =
1
e h 
=
(1.6  10
−19
1
= 1.38×1016 cm−3
C)(450 cm 2 V −1s −1 )(1  cm)
It is assumed that the doping concentration does not affect the drifty mobility. At this concentration,
from Figure 5.19 (see Figure 5Q05-1), the hole drift mobility is roughly 400 cm2 V−1 s−1 rather than
450 cm2 V−1 s−1. Using 350 cm2 V−1 s−1 we find,
Na =
1
e h 
=
(1.6  10
−19
1
= 1.21×1016 cm−3
2 −1 −1
C)(400 cm V s )(1  cm)
Author's Note: Question 5.7 provides a better example for calculating the required dopant
concentration for a given resistivity.
Figure 5Q05-1 The hole drift mobility at a dopant concentration of ~1.4×1016 cm−3 for Si.
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
5.6 Minimum conductivity
a. Consider the conductivity of a semiconductor,  = ene + eph. Will doping always increase the
conductivity?
b. Show that the minimum conductivity for Si is obtained when it is p-type doped such that the hole
concentration is
pm = ni
e
h
and the corresponding minimum conductivity (maximum resistivity) is
 min = 2eni e  h
c.
Calculate pm and min for Si and compare with intrinsic values.
Solution
a. Doping does not always increase the conductivity. Suppose that we have an intrinsic sample with n
= p but the hole drift mobility is smaller. If we dope the material very slightly with p-type then p > n.
However, this would decrease the conductivity because it would create more holes with lower mobility
at the expense of electrons with higher mobility. Obviously with further doping p increases sufficiently
to result in the conductivity increasing with the extent of doping.
b. To find the minimum conductivity, first consider the mass action law:
np = ni2
isolate n:
n = ni2/p
Now substitute for n in the equation for conductivity:
 = ene + eph
en 
 = i e +  h ep
p
2

To find the value of p that gives minimum conductivity (pm), differentiate the above equation with
respect to p and set it equal to zero:
en 
d
= − i 2 e + he
dp
p
2
en 
− i 2 e + he = 0
pm
2

Isolate pm and simplify,
pm = ni
e
h
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
Substituting this expression back into the equation for conductivity will give the minimum
conductivity:
 min =
eni  e
eni  e
e
+  h epm =
+  h eni
pm
h
ni  e  h
2
2
h
+ eni  e  h = eni e  h + eni  e  h
e

 min = eni e

 min = 2eni e  h
c. From Table 5.1, for Si: e = 1350 cm2 V−1 s−1, h = 450 cm2 V−1 s−1 and ni = 1.0  1010 cm−3.
Substituting into the equations for pm and min:
pm = ni
e
1350 cm 2 V −1 s −1
= (1.0 1010 cm −3 )
= 1.73  1010 cm−3
h
450 cm 2 V −1 s −1
 min = 2eni e  h

 min = 2(1.602 10 −19 C)(1.0 1010 cm−3 ) (1350 cm2 V −1 s −1 )(450 cm2 V −1 s −1 )

min = 2.5  10−6 −1 cm−1
The corresponding maximum resistivity is:
max = 1 / min = 4  105  cm
The intrinsic value corresponding to pm is simply ni (= 1.0  1010 cm−3). Comparing it to pm:
pm 1.73 1010 cm−3
=
= 1.73
ni
1.0 1010 cm−3
The intrinsic conductivity is:
int = eni(e + h)

int = (1.602  10−19 C)(1.0  1010 cm−3)(1350 cm2 V−1 s−1 + 450 cm2 V−1 s−1)

int = 2.88  10−6 −1 cm−1
Comparing this value to the minimum conductivity:
 min 2.5 10 −6  −1 cm−1
=
= 0.87
 int 2.88 10 −6  -1 cm-1
Sufficient p-type doping that increases the hole concentration by 73% decreases the conductivity by
15% to its minimum value.
5.7 Ionized impurity scattering and extrinsic Si The drift mobility of electrons and holes due
scattering from ionized impurities such as donors or acceptors, at room temperature, can be
empirically represented by a simple equation of the form
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   min +
 max −  min
1 + ( N d / N ref )
Chapter 5
Ionized dopant scattering controlled mobility [5.95]
in which Nd is the total ionized dopant concentration (ionized donors and acceptors summed together),
and min, max, Nref and a set of parameters that depend on whether  is for electrons or holes, the
semiconductor material and the dopant type. Table 5.4 lists typical values. Equation 5.94 is usually
restricted to the range Nd < 1019 cm−3. (Note that the scattering by thermal vibrations is also included
in Equation 5.95 through max.)
a
Find the donor (P) concentration for an n-type Si crystal whose resistivity should be 0.1  cm
b
Find the acceptor (B) concentration for an p-type Si crystal whose resistivity should be 0.1 
cm
Table 5.4 Ionized dopant scattering controlled drift mobility parameters in
  min + (max − min ) /[1 + ( N d / N ref ) ]
Material
Si electrons
Si holes
GaAs electrons
GaAs holes
InP electrons
InP holes
min
max
cm2 V−1 s−1
cm2 V−1 s−1
68.5
44.9
500
20
0
10
1414
470.5
9400
491.5
5000
170
Nref
cm−3
9.2×1016
2.23×1017
6.0×1016
1.48×1017
4.0×1017
4.87×1018

0.711
0.719
0.394
0.38
0.45
0.62
Notes: Data selectively combined from various sources. Room temperature values
Solution
a. The conductivity of n-Si is given by

 max −  min 
 = eN d e  eN d   min +

1 + ( N d / N ref ) 

We can define x = Nd/Nref and substitute and rearrange this equation to find
( −  )
 = eN ref x max  min + eN ref  min x
1+ x
We need to solve the above equation for  = 1/ = 1 / (0.1 −1 cm−1). That is, we need to solve
( −  )
y =  − eN ref x max  min + eN ref  min x = 0
1+ x
We can substitute  = 1/ = 1/ (0.1 −1 cm−1), Nref = 9.2×1016 cm−3,  = 0.711, min = 68.5 cm2 V−1 s−1
and max = 1414 cm2 V−1 s−1, we can plot y vs. x as in Figure 5Q07-1. The solution is at x  0.88 so that
the donor concentration must be
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
Nd = xNref = (0.88)(9.2×1016 cm−3) = 8.1×1016 cm−3
1
0
0.7
x
0.8
0.9
1
1.1
-1
y
-2
Figure 5Q07-1 The plot of y vs. x, which shows that y crosses the x-axis at x  0.88. (Livemath, originally
Theorist, used.)
b. The conductivity of p-Si is given by

 max −  min 
 = eN a  h  eN d   min +

1 + ( N a / N ref ) 

We can define x = Na/Nref and substitute and rearrange this equation to find
 = eN ref x
(  max −  min )
+ eN ref  max x
1 + x
We need to solve the above equation for  = 1/ = 1/0.1 −1 cm−1. That is, we need to solve
( −  )
y =  − eN ref x max  min + eN ref  max x = 0
1+ x
0.5
0
1
x
1.1
1.2
-0.5
y
-1
Figure 5Q07-2 The plot of y vs. x, which shows that y crosses the x-axis at x  1.13.
We can substitute  = 1/ = 1/0.1 −1 cm−1, Nref = 2.23×1017 cm−3,  = 0.719, min = 44.9 cm2 V−1 s−1
and max = 470.5 cm2 V−1 s−1, we can plot y vs. x as in Figure 5Q07-2. The solution is at x  1.13 so
that the acceptor concentration must be
Na = xNref = (1.13)(2.23×1017 cm−3) = 2.52×1017 cm−3
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Chapter 5
Conclusion: It is clear that we need almost 3 times more p-type doping than n-type to attain the same
conductivity in Si. The reason is that the hole mobility is approximately 3 times smaller than the
electron mobility
5.8 Intrinsic and Extrinsic III-V semiconductors InP is a III-V semiconductor. Calculate the
intrinsic concentration ni from Nc, Nv and Eg in Table 5.1. What is the intrinsic conductivity? Consider
a p-type InP crystal has been doped Zn (acceptors) with concentration 2×1017 cm−3. Find the
conductivity of this p-InP. If instead of Zn we had used Te (donors) with the same concentration, what
would be the conductivity? Use Table 5.4 for the electron and hole drift mobilities in InP.
Solution
The intrinsic concentration is
1
ni = ( N c N v ) 2 exp(− E g / 2kT )
1


(1.34 eV)

= [(5.2  1017 cm −3 )(1.11019 cm −3 )] 2 exp −
−5
−1
 2(8.617 10 eV K )(300 K ) 
which gives
ni = 1.33×107 cm−3.
We can use the drift mobility of electrons and holes given in Table 5.1, that is e = 4600 and h = 190
cm2 V−1 s−1, to calculate the intrinsic conductivity, which is given by
 = eni (e + h)
= (1.602×10−19 C)(1.33×107 cm−3)(4600 + 190 cm2 V−1 s−1) = 1.010−8 −1 cm−1.
If we use the values in Table 5.4, then we need to use max values for electrons and holes, that is, e =
5000, h = 170 cm2 V−1 s−1 so that
 = eni (e + h)
= (1.602×10−19 C)(1.33×107 cm−3)(5000 + 170 cm2 V−1 s−1) = 1.110−8 −1 cm−1.
InP is doped with dopants (acceptors) to make it p-type, and Na = 2×1017 cm−3. Thus, the hole drift
mobility is
 max −  min
(170 − 10)
   min +
= 10 +
= 150.6 cm2 V−1 s−1

1 + ( N a / N ref )
1 + (2 1017 / 4.87 1018 ) 0.62
so that
 = eNa = (1.602×10−19 C)(2×1017 cm−3)(150.6 cm2 V−1 s−1) = 4.82 −1 cm−1.
Suppose we doped it n-type with Nd = 2×1017 cm−3 then the electron drift mobility is
   min +
 max −  min
(5000 − 0)
= 0+
= 2887 cm2 V−1 s−1

1 + ( N a / N ref )
1 + (2 1017 / 4  1017 ) 0.45
so that
 = eNd = (1.602×10−19 C)(2×1017 cm−3)(2887 cm2 V−1 s−1) = 92.5 −1 cm−1.
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Chapter 5
5.9 Extrinsic III-V semiconductors GaAs is a III-V semiconductor. Suppose an p-type GaAs crystal
has been doped with Zn acceptor atoms in the amount 1017 cm−3. Find the resistivity of this p-GaAs.
Consider now an n-type GaAs dope with Se donor atoms. What should be the Se concentration so that
n-GaAs and p-GaAs have the same resistivity. Use Table 5.4 for the electron drift mobility in GaAs.
Solution
GaAs is doped with dopants (acceptors) to make it p-type, and Na = 1×1017 cm−3. Thus, the hole drift
mobility is
 max −  min
(491.5 − 20)
   min +
= 20 +
= 273.3 cm2 V−1 s−1

1 + ( N a / N ref )
1 + (11017 / 1.48  1017 ) 0.38
so that the conductivity is
 = eNa = (1.602×10−19 C)(1×1017 cm−3)(273.3 cm2 V−1 s−1) = 4.38 −1 cm−1

 = 1/ = 0.22  cm
We need an n-GaAs doped with donors of concentration Nd (to be determined) that has the same
conductivity. Using prime for electrons we need


 −  min

 max
−1 cm−1
 +
 = eN d    eN d   min
  = 4.38 
 ) 
1 + ( N d / N ref

We can use x = Nd/ N′ref to find
 x
y =  − eN ref
 −  min
 )
(  max
  min
 x=0
+ eN ref

1+ x
and then we can substitute N′ref = 6×1016 cm−3, ′ = 0.394, ′min = 500 cm2 V−1 s−1 and ′max = 9400
cm2 V−1 s−1, and ploy y vs. x as in Figure 5Q09-1.
2
1
0
0.04
x
0.05
0.06
0.07
0.08
0.09
-1
y
-2
Figure 5Q09-1 The plot of y vs. x which crosses the x-axis at x = 0.064.
Thus, the solution is
Nd = xN′ref = (0.064)(6×1016 cm−3) = 3.8×1015 cm−3
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Chapter 5
Notice that the electron concentration is 26 times less, roughly the same as the room temperature drift
mobility ratio of electrons to holes, which is 29.
5.10 Thermal velocity and mean free path in GaAs Given that the electron effective mass me* for
the GaAs is 0.067me, calculate the thermal velocity of the conduction band (CB) electrons. The
electron drift mobility μe depends on the mean free time e between electron scattering events (between
electrons and lattice vibrations). Given e = ee/me* and μe= 8500 cm2 V−1 s−1 for GaAs, calculate e,
and hence the mean free path of CB electrons. How many unit cells is if the lattice constant a of
GaAs is 0.565 nm? Calculate the drift velocity v d =  eE of the CB electrons in an applied field E of
104 V m−1. What is your conclusion?
Solution
1  2 3
me v th = kT
2
2
1/ 2

 3kT 
v th =   
 me 
Consider
e =

e =
1/ 2
 3(1.38 10 −23 JK −1 )(300 K ) 
 = 4.5 ×105 m s−1
= 
-31
0.067  9.110 kg


e e
me
e me
e
=
(8500 10 −4 m 2 V −1s −1 )(0.067  9.110 −31 kg)
= 3.24 ×10−13 s
1.6 10-19 C
Mean free path, ℓ = τevth = (3.239 ×10−13 s) (4.5 ×105 m s−1) = 1.46 ×10−7 m = 146 nm
No of unit cell = 1.46 ×10−7 m/ 0.565 ×10−9 m = 258
The drift velocity is, vd = μeE = (8500 ×10−4 m2V−1s−1)(104 Vm−1) = 8500 m s−1
So the drift velocity is much less than the thermal velocity.
5.11 Compensation doping in Si
a. A Si wafer has been doped n-type with 1017 As atoms cm−3.
1. Calculate the conductivity of the sample at 27 C.
2. Where is the Fermi level in this sample at 27 C with respect to the Fermi level (EFi) in intrinsic
Si?
3. Calculate the conductivity of the sample at 127 C.
b. The above n-type Si sample is further doped with 9  1016 boron atoms (p-type dopant) per centimeter
cubed.
1. Calculate the conductivity of the sample at 27 C.
2. Where is the Fermi level in this sample with respect to the Fermi level in the sample in (a) at 27
C? Is this an n-type or p-type Si?
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Chapter 5
Solution
a. Given temperature T = 27 C = 300 K, concentration of donors Nd = 1017 cm−3, and drift mobility e
 725 cm2 V−1 s−1; from Figure 5.19 shown in Figure 5Q11-1. At room temperature the electron
concentration n = Nd >> p (hole concentration).
Figure 5Q11-1 The variation of the drift mobility with dopant concentration in Si
for electrons and holes at 300 K from Figure 5.19.
(1) The conductivity of the sample is:
 = eNde - (1.602  10−19 C)(1017 cm−3)(725 cm2 V−1 s−1) = 11.6 −1 cm−1
(2) In intrinsic Si, EF = EFi,
ni = Ncexp[−(Ec − EFi)/kT]
(1)
In doped Si, n = Nd, EF = EFn,
n = Nd = Ncexp[−(Ec − EFn)/kT]
(2)
Equation (2) divided by (1) gives,
Nd
 E − EFi 
= exp Fn

ni
 kT 

 N  E − EFi
ln  d  = Fn
kT
 ni 

EF = EFn − EFi = kT ln(Nd/ni)
(3)
(4)
Substituting (ni = 1.0  1010 cm−3 from Table 5.1) we find
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
EF = (8.617  10−5 eV/K)(300 K)ln[(1017 cm−3) / (1.0  1010 cm−3)]

EF = 0.416 eV above EFi
Figure 5Q11-2 Figure 5.18 in the textbook, log-log plot for drift mobility versus
temperature for n-type Ge and n-type Si samples. Various donor concentrations for Si
are shown, Nd are in cm−3. The upper right insert is the simple theory for lattice limited
mobility whereas the lower left inset is the simple theory for impurity scattering limited
mobility.
(3) At T = 127 C = 400 K, e  450 cm2 V−1 s−1 from Figure 5Q11-2. The semiconductor is still ntype, then
 = eNde  (1.602  10−19 C)(1017 cm−3)(450 cm2 V−1 s−1) = 7.2 −1 cm−1
We need to check that Nd >> ni at 400 K. From Figure 5.16 in the textbook, ni  51012 cm− << Nd.
b. The sample is further doped with Na = 9  1016 cm−3 = 0.9  1017 cm−3 acceptors. Due to
compensation, the net effect is still an n-type semiconductor but with an electron concentration given
by,
n = Nd − Na = 1017 cm−3 − 0.9  1017 cm−3 = 1  1016 cm−3 (>> ni)
We note that the electron scattering now occurs from Na + Nd (1.9  1017 cm−3) number of ionized
centers so that e  600 cm2 V−1 s−1 from Figure 5Q11-1.
(1)
 = e(Nd − Na)e  (1.602  10−19 C)(1016 cm−3)(600 cm2 V−1 s−1) = 0.96 −1 cm−1
(2) Using Equation (2) with n = Nd − Na we have,
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
n Nd − Na
 E  − EFi 
=
= exp Fn

ni
ni
 kT 
so that
EF = EFn − EFi = (0.02586 eV)ln[(1  1016 cm−3) / (1.0  1010 cm−3)]

EF = 0.357 eV above EFi
The Fermi level from (a) and (b) has shifted “down” by an amount 0.059 eV. Since the
energy is still above the Fermi level, this is an n-type Si.
5.12 Temperature dependence of conductivity An n-type Si sample has been doped with 1015
phosphorus atoms cm−3. The donor energy level for P in Si is 0.045 eV below the conduction band
edge energy.
a. Calculate the room temperature conductivity of the sample.
b. Estimate the temperature above which the sample behaves as if intrinsic.
c. Estimate to within 20 percent the lowest temperature above which all the donors are ionized.
d. Sketch schematically the dependence of the electron concentration in the conduction band on the
temperature as log(n) versus 1/T, and mark the various important regions and critical temperatures.
For each region draw an energy band diagram that clearly shows from where the electrons are excited
into the conduction band.
e. Sketch schematically the dependence of the conductivity on the temperature as log() versus 1/T and
mark the various critical temperatures and other relevant information.
Solution
a. At room temperature (T = 300 K), e = 1400  10−4 m2 V−1 s−1 from Table 5.1. The conductivity
temperature T = 300 K is
 = eNde

 = (1.602  10−19 C)(1  1021 m−3)(1400  10−4 m2 V−1 s−1) = 22.4 −1 m−1
b. At T = Ti, the intrinsic concentration ni = Nd = 1  1015 cm−3. From Figure 5.16, as shown in Figure
5Q12-1, the graph of ni(T) versus 1/T,
1000 / Ti = 1.85 K−1

Ti = 1000 / (1.85 K−1) = 540 K or 267 C
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
Figure 5Q12-1 Figure 5.16, the temperature dependence of the intrinsic concentration. At ni = 1015 cm−3,
1000/Ti = 1.85.
c. The ionization region ends at T = Ts when all donors have been ionized, i.e. when n = Nd. From
Example 5.8, at T = Ts:
1
 − E 
1
2

n = N d =  N c N d  exp
2

 2kTs 

Ts =

Ts =
− E

2k ln 



Nd

1

N
N
c d 
2
=
− E
 2Nd 

2k ln 

N
c


E
 N 
k ln  c 
 2Nd 
Take Nc = 2.8  1019 cm−3 at 300 K from Table 5.1, and the difference between the donor energy level
and the conduction band energy is E = 0.045 eV. Therefore our first approximation to Ts is:
Ts =
(
)
(0.045 eV ) 1.602 10 −19 J/eV
E
= 54.68 K
=
 Nc 
 2.8 1019 cm −3 
− 23
 1.381 10 J/K ln 

k ln 
15
−3
 2 10 cm

 2Nd 
(
) ( (
)
)
Find the new Nc at this temperature, Nc:
3
(
)
3
 T 2
 54.68 K  2
N c = N c  s  = 2.8 1019 cm −3 
 = 2.179  1018 cm−3
 300 K 
 300 
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
Find a better approximation for Ts by using this new Nc,
Ts =
(
) ( (
)
(0.045 eV ) 1.602 10 −19 J/eV
E
= 74.64 K
=
 N c 
 2.179 1018 cm −3 
− 23
 1.38110 J/K ln 

k ln 
15
−3
 2 10 cm

 2Nd 
(
3
)
)
3
 T 2
 74.64 K  2
N c = N c  s  = 2.8 1019 cm −3 
 = 3.475  1018 cm−3
 300 K 
 300 

(
)
A better approximation to Ts is:
Ts =
(
) ( (
)
(0.045 eV ) 1.602 10 −19 J/eV
E
= 69.97 K
=
 N c 
 3.475 1018 cm −3 
− 23
 1.38110 J/K ln 

k ln 
15
−3
 2 10 cm

 2Nd 
(
3
)
)
3

 T   2
 69.97 K  2
N c = N c  s  = 2.8 1019 cm −3 
 = 3.154  1018 cm−3
 300 K 
 300 

Ts=
(
)
(
) ( (
)
(0.045 eV ) 1.602 10 −19 J/eV
E
= 70.89 K
=
 N c 
 3.154 1018 cm −3 
− 23
 1.38110 J/K ln 

k ln 
15
−3
 2 10 cm

 2Nd 
(
)
)
We can see that the change in Ts is very small, and for all practical purposes we can consider the
calculation as converged. Therefore Ts = 70.9 K = −202.1 C.
d. and e.
See Figures 5.15 and 5.20
Figure 5.15: The temperature dependence of the electron concentration in an n-type semiconductor.
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
Figure 5.20: Schematic illustration of the temperature dependence of electrical
conductivity for a doped (n-type) semiconductor.
*5.13 Ionization at low temperatures in doped semiconductors
Consider an n-type semiconductor.
The probability that a donor level Ed is occupied by an electron is
fd =
1
1
 E − EF 
1 + exp d

g
 kT 
[5.96]
where k is the Boltzmann constant, T is the temperature, EF is the Fermi energy, and g is a constant
called the degeneracy factor; in Si, g = 2 for donors, and for the occupation statistics of acceptors
g = 4. Show that
n2 +
nN c
Nd Nc
−
=0
 E 
 E 
g exp
 g exp

 kT 
 kT 
[5.97]
where n is the electron concentration in the conduction band, Nc is the effective density of states at the
conduction band edge, Nd is the donor concentration, and ΔE = Ec − Ed is the ionization energy of the
donors. Show that Equation 5.97 at low temperatures is equivalent to Equation 5.19. Consider a p-type
Si sample that has been doped with 1015 gallium (Ga) atoms cm−3. The acceptor energy level for Ga in
Si is 0.065 eV above the valence band edge energy, Ev. Estimate the lowest temperature (C) above
which 90 percent of the acceptors are ionized by assuming that the acceptor degeneracy factor g = 4.
Solution
Assume the electrons are donated by the donors, that is ni < n < Nd, so that
n = Nd × (Probability of not finding an electron at Ed)
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
n = N d 1 − f d ( Ed ) = N d

1
 (E − Ed ) 
1 + g exp  F

 kT
1
= Nd
 (E − Ed ) 
 (E − E F ) 
1 + g exp  c
exp − c


kT
 kT 


Nd
−1
 ( Ec − E F ) 
n
exp −
=

kT

 g exp  (Ec − Ed ) 
 kT 
Now,
 (E − E F ) 
n = N c exp − c

kT



 (E − E F )  n
exp− c
=
kT  N c

Chapter 5
(1)
(2)
We can now use Equation (1) to substitute for exp− (Ec − EF ) / kT  in Equation (2)
Nd
−1
n
 ( Ec − E F ) 
n
= exp −
=
Nc
kT 
 (E − Ed ) 

g exp  c
 kT 

 Nd 


n
1
n 

+
−
=0
Nc
 E 
 E 
g exp   g exp  
 kT 
 kT 
Simplifying the above expression, we get
n2 +
nN c
Nd Nc
−
=0
 E 
 E 
g exp
g
exp



 kT 
 kT 
(3)
At low temperature most of the donors remain unionized, so n << Nd . Thus, neglecting the second
term we get (nNc << NcNd)
n2 −
Nd Nc
=0
 E 
g exp

 kT 
1/ 2

1

n =  N d N c 
g

 E 
exp

 2kT 
(4)
which, for Si and ionization of donors with g = 2, is equivalent to
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1/ 2
1

n =  Nd Nc 
2

 E 
exp

 2kT 
Chapter 5
[5.19]
Consider 90% ionization of acceptors.
fa(Ea) = Probability of not finding an hole at Ea (holes have been ionized)
OR Probability of finding an electron at Ea.

1
1
=
= 0.9
 (Ea − E F ) 
 ( E d − Ev )  p
1 + g exp 
 1 + g exp  kT  N
 kT
v

1
= 0.9
15
 0.065 eV  0.9 10
1 + 4 exp 
19
 kT  1.2  10
Solving this equation, we get T = 128 K.
Author's Note: If we use Equation (4), we would find 91.8 K. If we use Equation (3), we would find
127.5 K. The reason is that Equation (4) assumes p << Na.
5.14 Compensation doping in n-type Si An n-type Si sample has been doped with 1 × 1017
phosphorus (P) atoms cm−3. The drift mobilities of holes and electrons in Si at 300 K depend on the
total concentration of dopants Ndopant (cm−3) approximately as follows:
e  88 +
1252
cm2 V −1s −1
−18
1 + 6.984 10 N dopant
and
h  54.3 +
407
cm2 V −1s −1
−18
1 + 3.745 10 N dopant
a. Calculate the room temperature conductivity of the sample.
b. Calculate the necessary acceptor doping (i.e., Na) that is required to make this sample p-type with
approximately the same conductivity.
Note that the above empirical drift mobility expressions in which Nd in the denominator is linear (not
raised to any power) enables the calculation of the dopant concentration needed for a given
conductivity analytically straightforward.
Solution
a.
 e  88 +
1252
cm 2 V −1s −1 = 825 cm 2 V −1s −1
1 + 6.984  10 −18 1017
σ = enμe = (1.6×10−19 C)(1×1017 cm−3)(825 cm2 V−1 s−1) = 13.2 Ω−1 cm−1.
b. p- type doping with conductivity 13.2 Ω−1 cm−1
σ = epμh
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
13.2  −1cm −1 = (1.6 10 −19 C)( N a − 1017 ) cm −3 (54.3 +
Chapter 5
407
cm 2 V −1s −1
1 + 3.745 10 −18 ( N a + 1017 )
Solving this quadratic equation, we get
Na = 5.92 ×1017 cm−3
Author's Note: You can show that this is a solution by substituting Na = 5.92 ×1017 cm−3 into the above
equation, and demonstrate that the equation is indeed satisfied.
Important Note: Figure 5.56 is correct in the 3rd Edition (as below) but not in the 4th Edition. It was
redrafted and an error occurred during redrafting. This will be corrected in the second reprint. It is
correct in the e-version.
5.15 GaAs Ga has a valency of III and As has V. When Ga and As atoms are brought together to form
the GaAs crystal, as depicted in Figure 5.54, the three valence electrons in each Ga and the five
valence electrons in each As are all shared to form four covalent bonds per atom. In the GaAs crystal
with some 1023 or so equal numbers of Ga and As atoms, we have an average of four valence electrons
per atom, whether Ga or As, so we would expect the bonding to be similar to that in the Si crystal: four
bonds per atom. The crystal structure, however, is not that of diamond but rather that of zinc blende
(Chapter 1).
a. What is the average number of valence electrons per atom for a pair of Ga and As atoms and in the
GaAs crystal?
b. What will happen if Se or Te, from Group VI, are substituted for an As atom in the GaAs crystal?
c. What will happen if Zn or Cd, from Group II, are substituted for a Ga atom in the GaAs crystal?
d. What will happen if Si, from Group IV, is substituted for an As atom in the GaAs crystal?
e. What will happen if Si, from Group IV, is substituted for a Ga atom in the GaAs crystal? What do
you think amphoteric dopant means?
f.
Based on the above discussion, what do you think the crystal structures of the III-V compound
semiconductors AlAs, GaP, InAs, InP, and InSb will be?
Figure 5.56 The GaAs crystal structure in two dimensions. Average number of valence electrons per atom is
four. Each Ga atom covalently bonds with four neighboring As atoms and vice versa.
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
Solution
Explanation of bonding in GaAs: Figure 5Q15-1 illustrates how Ga and As are bonded in GaAs.
The one s and three p orbitals hybridize to form 4 hyb orbitals. In As there are 5 valence electrons. One
hyb has two paired electrons and 3 hyb have 1 electron each as shown. In Ga there are 3 electrons so
one hyb is empty. This empty hyb of Ga can overlap the full hyb of As. The overlapped orbital, the
bonding orbital, then has two paired electrons. This is a bond between Ga and As even though the
electrons come from As (this type of bonding is called dative bonding). It is a bond because the
electrons in the overlapped orbital are shared by both As and Ga. The other 3 hyb of As can overlap 3
hyb of neighboring Ga to form "normal bonds". Repeating this in three dimensions generates the GaAs
crystal where each atom bonds to four neighboring atoms as shown. Because all the bonding orbitals
are full, the valence band formed from these orbitals is also full. The crystal structure is reminiscent of
that of Si. GaAs is a semiconductor.
a. The average number of valence electrons is 4 electrons per atom.
b. Se or Te replacing As will have one additional electron that cannot be involved in any of the four
bonds. Hence Se and Te will act as donors.
c. Zn or Cd replacing Ga will have one less electron than the substituted Ga atom. This creates a hole
in a bond. Zn and Cd will act as acceptors.
d. The Si atom has 1 less electron than the As atom and when it substitutes for an As atom in GaAs
there is a "hole" in one of the four bonds. This creates a hole, or the Si atom acts as an acceptor.
e. The Si atom has 1 more electron than the Ga atom and when it substitutes for a Ga atom in GaAs
there is an additional electron that cannot enter any of the four bonds and is therefore donated into the
CB (given sufficiently large temperature). Si substituting for Ga therefore acts as a donor.
f. All these compounds (AlAs, GaP, InAs, InP, InSb) are compounds of III elements and V elements
so they will follow the example of GaAs.
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Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
Figure 5Q15-1 Bonding between Ga and As atoms and the formation of the GaAs crystal
5.16 Doped GaAs Consider the GaAs crystal at 300 K.
a. Calculate the intrinsic conductivity and resistivity.
b. In a sample containing only 1015 cm−3 ionized donors, where is the Fermi level? What is the
conductivity of the sample?
c. In a sample containing 1015 cm−3 ionized donors and 9  1014 cm−3 ionized acceptors, what is the
free hole concentration? Where is the Fermi level?
Solution
a. Given temperature, T = 300 K, and intrinsic GaAs, from Table 5.1, ni = 2.1  106 cm−3, e  8800
cm2 V−1 s−1 and h  400 cm2 V−1 s−1. Then
 = eni(e + h)

 = (1.602  10−19 C)(2.1  106 cm−3)(8800 cm2 V−1 s−1 + 400 cm2 V−1 s−1)

 = 3.10  10−9 −1 cm−1

 = 1/ = 3.23  108  cm
b. Donors are now introduced. At room temperature, n = Nd = 1015 cm−3 >> ni >> p.
n = eNde  (1.602  10−19 C)(1015 cm−3)(8800 cm2 V−1 s−1) = 1.41 −1 cm−1

n = 1/n = 0.71  cm
In the intrinsic sample, EF = EFi,
ni = Ncexp[−(Ec − EFi)/kT]
(1)
In the doped sample, n = Nd, EF = EFn,
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n = Nd = Ncexp[−(Ec − EFn)/kT]
Chapter 5
(2)
Equation (2) divided by (1) gives,

Nd
 E − EFi 
= exp Fn

ni
 kT 
(3)
EF = EFn − EFi = kT ln(Nd/ni)
(4)
Substituting we find,
EF = (8.61710−5 eV/K)(300 K)ln[(1015 cm−3)/(2.1106 cm−3)]

EF = 0.516 eV above EFi (intrinsic Fermi level)
c. The sample is further doped with Na = 9  1014 cm−3 = 0.9  1015 cm−3 acceptors. Due to
compensation, the net effect is still an n-type semiconductor but with an electron concentration given
by,
n = Nd − Na = 1015 cm−3 − 0.91015 cm−3 = 11014 cm−3 (>> ni)
The sample is still n-type although there are less electrons than before due to the compensation effect.
From the mass action law, the hole concentration is:
p = ni2 / n = (2.1106 cm−3)2 / (11014 cm−3) = 0.044 cm−3
On average there are virtually no holes in 1 cm3 of sample.
The Intrinsic Fermi level is determined by
ni = Ncexp[−(Ec − EFi)/kT]
(5)
In the compensation doped sample, n = Nd − Na,

n = Ncexp[−(Ec − EFn)/kT]
(6)
EF = EFn − EFi = kT ln(n/ni)
(7)
Substituting we find,
EF = (8.61710−5 eV/K)(300 K)ln[(1014 cm−3)/(2.1106 cm−3)]

EF = 0.457 eV above EFi (intrinsic Fermi level)
Conclusion: As expected, lower n results in EFn being further away from Ec.
ADDENDUM
We can also estimate the new conductivity. We note that electron scattering now occurs from Na + Nd
number of ionized centers though we will assume that e  8800 cm2 V−1 s−1.
 = ene  (1.60210−19 C)(1014 cm−3)(8800 cm2 V−1 s−1) = 0.14 −1 cm−1
NOTE: We can use Equation 5.95 and Table 5.4 to get a more accurate drift mobility and hence
conductivity.
   min +
 max −  min
(9400 − 500)
2 −1 −1
= 500 +
s

15
15
16 0.394 = 7582 cm V
1 + ( N d / N ref )
1 + [(1.0 10 + 0.9 10 ) / 6 10 ]
so that
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Chapter 5
 = ene  (1.60210−19 C)(1014 cm−3)(7582 cm2 V−1 s−1) = 0.12 −1 cm−1
Notice that the conductivity is an order of magnitude less that the case (a) above.
5.17 Extrinsic III-V semiconductor GaAs is a III-V semiconductor. Suppose a GaAs crystal has
been doped with Te atoms in the amount 1×1017 donors cm−3 and Zn atoms in the amount 7×1015 cm−3.
Is this an n or and p-type GaAs? The electron and hole drift mobilities in GaAs are given in Table 5.4.
Find its resistivity.
Solution
Te has a valency of 6, 1 more than As (5) so that when it substitutes for As, it donates an electron. Te
is a donor in GaAs. Zn has a valency of 2, 1 less than Ga (3) so it acts like an acceptor in the structure.
Since, Nd  Na and Nd − Na >> ni, due to compensation doping, it is n-type with
n = Nd − Na = 1×1017 − 0.07×1017 = 0.93×1017 cm−3
Both ionized donors and acceptors scatter the electrons so the total concentration of ionized scatters is
Nd + Na = 1.07×1017 cm−3. We can estimate the drift mobility by using Equation 5.95
 max −  min
(9400 − 500)
2 −1 −1
   min +
= 500 +
s

17
16 0.394 = 4445 cm V
1 + ( N total / N ref )
1 + (1.07 10 / 6 10 )
so that the conductivity is
 = en = (1.602×10−19 C)(0.93×1017 cm−3)(4445 cm2 V−1 s−1) = 66.2 −1 cm−1

 = 1/ = 0.015  cm
5.18 Compensation doping in GaAs Consider an n-type GaAs crystal that has been doped with
1×1016 donors cm−3. Find the acceptor concentration you need to turn this n-type GaAs to p-type with
the same resistivity as the n-GaAs. Use Table 5.4 for the electron and hole drift mobilities in GaAs.
Solution
First calculate the conductivity of n-GaAs. We can use Equation 5.95 and Table 5.4
 max −  min
(9400 − 500)
2 −1 −1
 e   min +
= 500 +
s

16
16 0.394 = 6459 cm V
1 + ( N d / N ref )
1 + (1 10 / 6 10 )
which gives
 = eNde = (1.602×10−19 C)(1×1016 cm−3)(6459 cm2 V−1 s−1) = 10.4 −1 cm−1
Next, we dope it p-type by compensation. We need, Na  Nd and Na − Nd >> ni, so that
p = Na − Nd
The variable (what we seek) here is Na. Both ionized donors and acceptors scatter the electrons so the
total concentration of ionized scatters, Ntotal = Nd + Na. Since the sample is p-type we can estimate the
drift mobility by
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 h   min +
Chapter 5
 max −  min
1 + ( N total / N ref )
using the GaAs drift mobility parameters for holes in Table 5.4. Thus,

 max −  min 
 = e(Na − Nd)h  e( N a − N d )   min +

1 + ( N total / N ref ) 




491.5 − 20
  e( N a − 11016 ) 20 +
= 10.4  −1 cm −1
16
17 0.38 
1
+
[(
N
+
1

10
)
/(
1
.
48

10
)]
a


which we can solve for Na. This is can be done by plotting (on Excel for example)


491.5 − 20
y = 10.4 − (1.602 10 −19 )( N a − 11016 ) 20 +

1 + [( N a + 11016 ) /(1.48 1017 )]0.38 

and finding the x-axis crossing. Let's substitute x = Na / Nd = Na / (1×1016 cm−3) and then plot y vs. x as
in Figure 5Q18-1. The x-axis crossing is at x  30 so that Na = 30Nd = 3×1017 cm−3.
2
0
20
x
30
40
-2
y
-4
Figure 5Q18-1 y vs. x. The zero crossing is at x = 30.
5.19 Varshni equation and the change in the bandgap with temperature The Varshni equation
describes the change in the energy bandgap Eg of a semiconductor with temperature T in terms of
AT 2
E g = E go −
B +T
where Ego is the bandgap at T = 0 K, and A and B are material-specific constants. For example, for
GaAs, Ego = 1.519 eV, A = 5.405 × 10−4 eV K−1, B = 204 K, so that at T = 300 K, Eg = 1.42 eV. Show
that
dE g
dT
=−
( E go − E g )  T + 2 B 
AT (T + 2 B)
=−


2
T
(B + T )
 T+B 
What is dEg/dT for GaAs? The Varshni equation can be used to calculate the shift in the peak emission
wavelength of a light emitting diode (LED) with temperature or the cutoff wavelength of a detector. If
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Chapter 5
the emitted photon energy from an electron and hole recombination is hf ≈ Eg + (1/2)kT, find the shift
in the emitted wavelength from 27°C down to −30°C from a GaAs LED.
Solution
E g = E go −
AT 2
B +T
( B + T )2 AT − AT 2
AT (T + 2 B)
AT 2  T + 2 B 
=−
=−
=−


dT
(B + T )2
(B + T )2
( B + T )T  T + B 

dE g

dE g
dT
=−
( Ego − Eg )  T + 2 B 


T
 T +B 
=−
(1.519 − 1.42)eV (300 + 408)K
= −4.64 ×10−4 eV K−1
300 K
(300 + 204)K
For GaAs
dE g
dT
GaAs at 27°C, Eg = 1.42 eV
hf =

=
hc

 E g + 12 kT
hc
(6.626 10 −34 J s)(2.998 108 ms −1 )
= 8.652×10−7 m or 865.2 nm
=
E g + 12 kT (1.42 + 12 0.02585 eV)(1.602 10 -19 J/eV)
GaAs at −30°C, kT = 0.0209 eV
Eg = 1.519 eV −

=
(5.405 10 −4 eVK −1 )(243 K) 2
= 1.447 eV
(204 + 243)K
hc
(6.626 10 −34 J s)(2.998 108 m s −1 )
= 8.507×10−7 m or 850.7 nm
=
E g + 12 kT (1.447 + 12 0.0209 eV)(1.602 10 -19 J/eV)
Shift in peak emission wavelength = 865.2 nm – 850.7 nm = 14.5 nm.
5.20 Varshni equation and the intrinsic concentration The intrinsic concentration ni as a function
of temperature can be calculated from Equation 5.11 but we have to remember that Nc, Nv and Eg
actually depend on the temperature. The Varshni equation in Question 5.19 with coefficient A and B
can be used to find the bandgap Eg at any temperature.
a
Given ni =1×1010 cm−3 for Si, calculate ni at 400 C by assuming first a constant bandgap of
1.11 eV. Then recalculate ni by using Eg at 400 C. For Si, the Varshi parameters are: Ego =
1.169 eV, A = 4.9×10−4 eV K−1, B = 655 K.
b
Given that electron and hole drift mobility follow e  T−2.4 and h  T−2.2 type of behavior,
what is the intrinsic resistivity of Si at 400 C?
Solution
a
The intrinsic concentration at T1 is given by
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Chapter 5
 E g1 

ni1 = ( N c N v )1/ 2 exp −
 2kT1 
in which Nc  T3/2 and Nv  T3/2 so that
ni1 = CT1
3/ 2
 E g1 

exp −
 2kT1 
where C is a constant. At T
ni = CT
3/ 2
 Eg 

exp −
 2kT 


The ratio ni to ni1 eliminates C so that
T 
ni = ni1  
 T1 
3/ 2
 1  E g E g1  
exp −
−
 
 2k  T
T
1 



First assume that Eg = Eg1 = 1.11 eV,
T 
ni = ni1  
 T1 
3/ 2
 E g1  1 1  

exp −
−
 2k  T T1  



Substituting the values, T1 = 300 K, T = 673 K and ni = 1×1010 cm− ,

 673 
ni = (1 10 

 300 
3/ 2
10 )
 (1.11)(1.602  10 −19 )  1
1 
 = 4.94×1015 cm−.
exp −
−
− 23


2
(
1
.
3807

10
)
673
300



Now, we can improve on this by using the Varshni equation
AT 2
(4.9 10 −4 eV K −1 )(673 K ) 2
E g = E go −
= (1.169 eV) −
= 1.002 eV
B +T
655 K + 673 K
so that
T 
ni = ni1  
 T1 

3/ 2
 1  E g E g1  
exp −
−
 
 2k  T
T

1 


 673 
ni = (1 1010 ) 

 300 
3/ 2

1.602 10 −19 1.002 1.11 
 = 1.25×1016 cm−,
exp −
−
− 23 

 2(1.3807  10 )  673 300  
about a factor of 2.5 higher.
Let's calculate the electron and hole drift mobilites at the temperature of interest (673 K)
Given e = C T−2.4 and e =1400 cm2 V− s− at room temperature (300 K) in Table 5.1, we
have
T 
e (T ) = e (T1 ) 
 T1 
−2.4
 673 
= (1400)

 300 
−2.4
= 201.4 cm2 V− s−
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Chapter 5
Similarly for the hole the drift mobility,
T 
 h (T ) = h (T1 ) 
 T1 
−2.2
 673 
= (450)

 300 
−2.2
= 76.1 cm2 V− s−
The conductivity at 400 C (673 K) is then,
 = eni(h + h) = (1.60210− C)( cm−)(201.4 + 76.1 cm V− s−)
= 0.56 − cm−
5.21 Degenerate semiconductor Consider the general exponential expression for the concentration
of electrons in the CB,
 ( E − EF ) 
n = N c exp − c
kT 

and the mass action law, np = ni2. What happens when the doping level is such that n approaches Nc and
exceeds it? Can you still use the above expressions for n and p?
Consider an n-type Si that has been heavily doped and the electron concentration in the CB is
10 cm−3. Where is the Fermi level? Can you use np = ni2 to find the hole concentration? What is its
resistivity? How does this compare with a typical metal? What use is such a semiconductor?
20
Solution
Consider
n = Ncexp[−(Ec− EF)/kT]
(1)
and
np = ni2
(2)
These expressions have been derived using the Boltzmann tail (E > EF + a few kT) to the
Fermi − Dirac (FD) function f(E) as in Section 5.1.4. Therefore the expressions are NOT valid when
the Fermi level is within a few kT of Ec. In these cases, we need to consider the behavior of the FD
function f(E) rather than its tail and the expressions for n and p are complicated.
It is helpful to put the 1020 cm−3 doping level into perspective by considering the number of atoms per
unit volume (atomic concentration, nat) in the Si crystal. From Appendix C, the density is 2.33 g cm−,
so that
nat =
i.e.
(Density ) N A (2.33 103 kg m −3 )(6.022 10 23 mol −1 )
=
M at
(28.09 10 −3 kg mol −1 )
nat = 4.995  1028 m−3 or 5.0  1022 cm−3
Given that the electron concentration n = 1020 cm−3, we see that
n/nat = (1020 cm−3) / (5.0  1022 cm−3) = 0.00200
which means that if all donors could be ionized we would need 1 in 500 doping or 0.2% donor doping
in the semiconductor. We cannot use Equation (1) to find the position of EF. The Fermi level will be in
the conduction band. The semiconductor is degenerate. The electrical properties are more like that of a
metal than that of a semiconductor. The conductivity would be very high compared to a nondegenerate
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Chapter 5
semiconductor. Highly doped (degenerate) semiconductor regions in devices can replace metals by
providing sufficiently high conductivity, and without needing to deposit actual metals. Degenerate
semiconductors are widely used in a variety of devices, as will be apparent in Chapter 6.
Figure 5.21 (a) Degenerate n-type semiconductor. Large number of donors form a band that overlaps the CB.
(b) Degenerate p-type semiconductor.
5.22 Degenerate semiconductors and the Fermi level Consider a degenerate n-Si doped with a
donor concentration Nd = 3×1020 cm−. Where is the Fermi level with respect to the bottom of the
conduction band at room temperature? Where is the Fermi level in a similar degenerate p-Si doped
with acceptors in the amount Na = 3×1020 cm−. What use are such semiconductors?
Solution
Following Example 5.12, taking n = Nd = 3×1026 m− and using the density of states related effective
mass for the electron (me*), then at 0 K,
2
h 2  3n  3
(6.626 10 −34 ) 2  3(3 10 26 ) 
EFn (0) = *   =


8me   
8(1.08  9.1110 −31 ) 


2/3
= 0.151 eV
Notice that we used me*. At T = 300 K, we have
2
2
  2  kT  2 
 = (0.151)1 −   0.02585   = 0.148 eV
EFn = EFn (0)1 −
 12  0.151  
 12  EFn (0)  




which is a small change. For a Si sample doped with acceptors, p = Na = 3×1026 cm− and we need to
use mh* = 0.60me,
2
h2  3 p  3
(6.626 10 −34 ) 2  3(3 10 26 ) 
EFp (0) = *   =


8me   
8(0.60  9.1110 −31 ) 


2/3
= 0.272 eV
and at 300 K
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Chapter 5
  2  kT  2 
  2  0.02585  2 



EFn = EFn (0) 1 −
= (0.272)1 −
 12  0.272   = 0.270 eV
 12  EFn (0)  




again only a small change with temperature.
Note: More accurate calculations using the Joyce-Dixon Equations give EFn = 0.159 eV and EFn =
0.311 eV.
5.23 Photoconductivity and speed Consider two p-type Si samples both doped with 1015 B atoms
cm−3. Both have identical dimensions of length L (1 mm), width W (1 mm), and depth (thickness) D
(0.1 mm). One sample, labeled A, has an electron lifetime of 1 s whereas the other, labeled B, has an
electron lifetime of 5 s.
a. At time t = 0, a laser light of wavelength 750 nm is switched on to illuminate the surface (L  W) of
both the samples. The incident laser light intensity on both samples is 10 mW cm−2. At time t = 50
s, the laser is switched off. Sketch the time evolution of the minority carrier concentration for both
samples on the same axes.
b. What is the photocurrent (current due to illumination alone) if each sample is connected to a 1 V
battery?
Solution
a. Consider the photoconductor shown in Figure 5.29
Figure 5.29 A semiconductor slab of length L, width W, and depth D is illuminated with light of wavelength λ.
Iph is the steady-state photocurrent.
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Chapter 5
Figure 5Q23-1 Schematic sketch (not to scale) of the excess carrier concentration in the samples A and B as a
function of time from the laser switch-on to beyond laser switch-off.
Figure 5Q23-2 Relative excess carrier concentration versus time from excitation onwards. Blue is A; red is B.
Gph is assumed to be 1 for simplicity and time is in microseconds.
Note: The rise in the excess concentration is given by
n(t) = Gph [1 − exp(−t/)]
which gives the impression that the initial slope is . The initial rate of rise turns out to be proportional
to the photogeneration rate Gph. Consider expanding the exponential for small t,
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n(t)  Gph [1 − 1 + (t/)] = Gpht
so that the initial rate of rise (the initial slope) is dn/dt = Gph.
The initial tangential line has the equation n(t) = Gpht, which cuts the n(t) = Gph line at t = .
The rise time to 63% of the final value is .
Define t = t – toff as time lapsed from the cessation of illumination and toff is the time at which
the illumination (photogeneration) is switched off (toff = 50 s). The excess carrier concentration is
given by
n(t) = Gphexp(−t′/)]
and for small t << 
n(t)  Gph(1 − t′/ + …) = Gph(1 − t′/)
Thus, the initial slope is
dn(t)/dt = −Gph
which is a straight line. The initial line, that is the tangent at t′ = 0, is n(t) = Gph(1 − t′/ ). This
tangent cuts the t-time axis at t = 
Consider the decay portion t > 0. After a time t =  , the excess concentration has decayed down to
exp(−) or to a value 0.36Gph.
Conclusion: Both the initial rise and fall rates are determined by the photogeneration rate Gph. The
tangent during the initial rise intersects the saturation line at t =  . The tangent to the initial fall
intersects the t-time axis at t = 
b. From the mobility versus dopant graph (Figure 5.19), shown in Figure 5Q23-3, h = 450  10−4 m2
V−1 s−1 and e = 1350  10−4 m2 V−1 s−1. Given are the wavelength of illumination  = 750  10−9 m,
light intensity I = 100 W/m2, length L = 1 mm, width W = 1 mm, and depth (thickness) D = 0.1 mm.
Figure 5Q23-3 Figure 5.19 shows the variation of the drift mobility with dopant concentration in Si for
electrons and holes at 300 K. The present case corresponds to the red lines.
The photoconductivity is given by (see Example 5.14),
 =
eI (e +  h )
hcD
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Chapter 5
where  = 1 is the quantum efficiency. Assume all incident light is transmitted in to the
semiconductor. Assume all light intensity is absorbed (correct assumption as the absorption coefficient
at this wavelength is large). If E is the applied electric field, then the photocurrent density and hence
the photocurrent is given by,
Jph = J = I/A = E

Iph = I = AE = (WD)E
Substitute for the photoconductivity
 V  eI ( e +  h )
I ph = I = (W  D ) 
hcD
L

I ph = I =
WVe I (e +  h )
Lhc
The photocurrent will travel perpendicular to the WD direction, while the electric field E will be
directed along L. Substituting the given values for sample A (electron lifetime A = 10−6 s, voltage V =
1 V):
I A =


2
2


Vs
Vs 
(0.001 m )(1 V )(1.602 10 −19 C)(1)(100 W m −2 )(750 10 −9 m )(10 −6 s ) 0.135 m + 0.042 m 
(0.001 m )(6.626 10
−34
)(
J s 3.0 10 m s
8
−1
)
IA = 1.07  10−5 A or 10.7 A
The photocurrent in sample B can be calculated with the same equation, using B = 5  10−6 s. The
result is
IB = 5.35  10−5 A or 53.5 A
Comment: Most important assumptions are (i) all incident radiation is transmitted into the
semiconductor (so that the semiconductor has an antireflection coating), (ii) all incident radiation is
then absorbed within the semiconductor and absorbed uniformly
5.24 Einstein relation The Fermi level EF for a semiconductor in equlibrium and in the dark is
uniform through the crystal, that is dEF/dx = 0. Consider a semiconductor in open circuit and the total
current due to electrons, which must be zero
J e = en( x) eE + eDe
dn( x)
=0
dx
[5.98]
where n = n(x) is the electron concentration at a point x. Given that, by definition, the field E = −dV/dx,
show that
De
d ln n
dV
= e
dx
dx
[5.99]
A small change V in voltage across x means a change Ec = −eV in Ec. For a nondegerate
semiconductor, we can write,
Ec ( x ) − EF = −kT ln( n / N c )
[5.100]
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Chapter 5
Differentiate Ec with respect to x, and substitute into Equation 5.99 to derive the Einstein relation.
(Remember that dEF/dx = 0 in equilibrium.) What is your conclusion?
Solution
Consider Equation 5.100. Differentiating this with respect to x
dEc ( x) dEF
d
−
= −kT ln( n / N c )
dx
dx
dx
Using Ec = −eV and dEF/dx = 0 in equilibrium

edV
d
= kT
ln n
dx
dx

dV  kT  d ln n
= 
dx  e  dx
and substituting this into the right hand side of Equation 5.99 we find that

De
 kT  d ln n 
d ln n
dV
= e
= e  

dx
dx
 e  dx 
De
=
e
kT
e
which is the Einstein relation.
5.25 Diffusion length and coefficient Let 1/L be the mean probability per unit distance that an
electron disappears by recombination in a semiconductor. Then the probability that an electron
recombines with a hole in a small distance x is x/L. The change n in the electron concentration is
−nx/L. Thus, n = −nx/L, or n/n = −x/L. We can integrate this from n = no at x = 0 to n = n(x) at x
to find,
n(x) = noexp(−x/L)
Exponential electron distribution
[5.100]
Suppose that the total number of electrons per unit area, N = noL. Show that

 xn( x)dx = L
x = x =
0
N

 x n( x)dx = 2 L
x = x =
2
and
2
2
2
0
N
Average and Mean square in exponential probability distribution
[5.101]
What is your conclusion? What is L? Usually, the diffusion coefficient D is written as D = L2/,
whereas the derivation in Section 5.6 has D = L2/2. Can you explain the difference?
Solution
Consider



 xn( x)dx =  xn exp(− x / L)dx = n  x exp(− x / L)dx
x = x =
0
0
N
o
o
N
0
N
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Chapter 5
From Integration Tables
1
1
x
exp(
ax
)
dx
=
x
−

 exp(ax )
0
a
a


x=
no
L( L − x) exp(− x / L)xx==0 = no L( L − x) exp(− x / L)xx==0 = L
N
no L
Now consider


 x n( x)dx = n  x exp(− x / L)dx
x = x =
2
2
2
o
0
2
0
N
N
From Integration Tables

1 2
2x
2
 x exp(ax)dx = a  x − a + a  exp(ax)
2
2
0



x =
no
L(2 L2 − 2 Lx + x 2 ) exp(− x / L) x =0 = 2 L2
N
x2 =
The problem with the simple derivation leading to Equation 5.34 is discussed just above Equation 5.40
on page 461 in the text. The derivation D = /2 in Equation 5.34 should really have x 2 instead of
, so that the diffusion coefficient is given as
D=
x 2 2( x) 2  2
=
=
2
2

5.26 Hall effect in semiconductors Consider a slab of length l, width w and thickness t as shown in
Figure 5.57. We pass a current Ix along the length of the slab, taken along x from 1 to 4. In Hall effect
experiments, we need to measure the voltage difference between two points on opposite faces (top and
bottom) of the slab under an applied magnetic field Bz along z. Before we apply the field, the Hall
voltage should be zero, which is achieved by using a potentiometer between 2 and 3 so that the voltage
different between 5 and 6 can be set to zero before the application of Bz. (The potentiometer places
point 6 electrically opposite point 5.) When Bz is applied, V65 gives the Hall voltage VH and is
measured using a voltmeter with a high input resistance. For an n-type semiconductor V65 is negative
(6 is negative with respect to 5). Show that the Hall coefficient is given by
RH =
V65 t
I x Bz
Hall coefficient
Consider an n-type Si doped with 1014 donor cm−3 (Nd). Calculate the Hall voltage if t = 500 m, the
magnetic field is 0.01 T and the current is 0.1 mA. What is the voltage drop between 1 and 4 if l = 5
mm and w = 2 mm and what is the power dissipated in the semiconductor? Is there any advantage in
increasing the dopant concentration to reduce the voltage drop and power dissipated in the sample?
Consider a Hall effect sensor. The sensitivity SH is the magnitude of the Hall voltage per unit magnetic
field, SH = VH/Bz. Power dissipated within the semiconductor is Ix2R, which we would like to keep as
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Chapter 5
low as possible. A figure of merit MH can be defined for a Hall effect sensor as the Hall sensitivity per
unit power dissipated,
MH =
VH / Bz
I x2 R
Hall effect figure of merit
MH =
w e
Ixl
Hall effect figure of merit
Show that
What is your conclusion? If the Hall sensor is integrated into an integrated circuit, there is a further
limitation. Can the voltage drop along l (between 1 and 2 in Figure 5.57) be of any magnitude?
Figure 5.57 Hall voltage measurement is done in two steps. A current Ix is passed along the
semiconductor slab. Without the magnetic field, the potentiometer is adjusted until the voltage
between 6 and 5 is zero. Then a magnetic field Bz is applied and the Hall voltage V65 is measured. V65
is negative for an n-type semiconductor.
Solution
Consider the definition of the Hall coefficient
RH =
Ey
J x Bz
=
V65 / w
V t
= 65
( I x / tw) Bz I x Bz
We have an n-type Si doped with 1014 donor cm−3 (Nd) which means that the Hall coefficient is
RH = −
1
1
=−
= 0.0624 C m3
−19
eN d
(1.602 10 )(110 20 )
so that
V65 t
V65 (500 10 −6 m)
RH = −0.0624 C m =
=
I x Bz (0.110 −3 A)(0.01T)
and solving
V65 = −0.125 mV, which is measurable.
3
Consider the figure of merit
MH =
w e
VH / Bz RH I x
R
1 / en
w / en
= 2 = 2H =
=
=
2
Ix R
tI x R tI x R tI x (l / tw) tI x (l / en e ) I x l
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Chapter 5
The parameters w, Ix and l are given. The electron drift mobility from Figure 5.19 at Nd = 1014 cm− is
roughly 1400 cm2 V− s−
MH =
we (2 10 −3 m)(1400 10 −4 m 2 V −1 s −1 )
= 560 m2 V− s− A−
=
Ixl
(0.110 −3 A)(5 10 −3 m)
MH = 560 m2 J−
or
Consider the merit of figure MH. We can increase this by using a semiconductor that has a higher drift
mobility. There are semiconductors that have much higher electron drift mobility than Si.
*5.27 Hall Effect in Semiconductors The Hall effect in a semiconductor sample involves not only
the electron and hole concentrations n and p, respectively, but also the electron and hole drift
mobilities, e and h. The hall coefficient of a semiconductor is (see Chapter 2),
RH =
where b =
p − nb 2
2
e( p + nb )
[5.103]
e
h
a. Given the mass action law, pn = ni2, find n for maximum RH (negative and positive RH). Assume
that the drift mobilities remain relatively unaffected as n changes (due to doping). Given the electron
and hole drift mobilities, e = 1400 cm2 V−1 s−1, h = 450 cm2 V−1 s−1 for silicon determine n for
maximum RH in terms of ni. Find the maximum magnitude of RH.
b. Taking b = 3.1, plot RH as a function of electron concentration n/ni from 0.1 to 10.
c. Show that when n ≫ ni, RH = −1∕en and when n ≪ ni, RH = +1∕ep.
Solution
a. Substituting the mass action law (p = ni2/n) into Equation 5.103,
ni2
− nb 2
p − nb 2
u
RH =
= n2
=
2
e( p + nb)
n
 v
e i + nb 2 
 n

(1)
where u and v represent the numerator and denominator in Equation (1) as a function of n.
Then
dRH uv − uv
=
=0
dn
v2
where primes are derivatives with respect to n. This means that uv − u v = 0, so that,
2
2

 ni2
   ni2
  n2
 n 2

2   ni
uv − uv = − 2 − b  e + nb   −  − nb 2  2e i + nb  − i2 + b  = 0
 n
   n
   n
  n
 n

We can multiply through by n3 and then combine terms and factor to obtain,
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
Chapter 5

b3n 4 − 3ni2b(1 + b) n 2 + ni4 = 0
or,
b3x2 + [−3b(1 + b)]x + 1 = 0
where x = (n/ni)2. This is a quadratic equation in x. Its solution is,
n 2 3b(1 + b)  9b 2 (1 + b) 2 − 4b 3
x= 2 =
ni
2b 3
For Si, b = 3.1, and we have two set of solutions corresponding to,
n/ni = 1.117
or
p/ni = 0.895
n/ni = 0.163
or
p/ni = 1/0.163 = 6.13
The maximum magnitude of RH occurs for n/ni = 1.117, so that
RH =

(0.895ni ) − (1.117 ni )(3.1) 2
p − nb 2
(0.895) − (1.117)(3.1) 2
=
=
e( p + nb) 2 e[0.895ni + (1.117 ni )(3.1)]2 eni [0.895 + (1.117)(3.1)]2
RH = −
0.518
3

10
−3 = − C cm .
(1.602 10 C)(110 cm )
−19
b. Substituting the mass action law p = ni2/n into Equation 103 and using a normalized electron
concentration y = n/ni, we get,
1
ni2
− yb 2
− nb 2
2
p − nb
y
RH =
= n
=
2
2
2
e( p + nb)
 ni2

1

eni  + yb 
e + nb 
y

 n

or
z=
1
− yb 2
y
RH
=
2
(1 / eni )  1

 + yb 
y

Normalized Hall coefficient (eni)RH versus normalized concentration n/ni as z versus y is shown in
Figure 5Q27-1 for b = 3.1.
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Chapter 5
Figure 5Q27-1: Normalized Hall coefficient (eniRH) versus normalized electron concentration (n/ni).
Values 0.16, 1.12 and 0.32 shown are n/ni values when the magnitude of RH reaches maxima and zero
respectively. b = 3.1
c. From part (b) above consider small y, that is y → 0 and 1/y being large
RH =

RH =
1
− yb 2
y
1

eni  + yb 
y

2

1
−0
y
1

eni  + 0 
y

2
=
1
y
1
eni  
 y
2
=
1
1
eni  
 y
1
1
1
=
=
(p-type semiconductor)
 ni 
 ni2  ep
eni   e 
 n
n
 
where we have used np = ni2.
Now consider y being large and 1/y → 0,
RH =

RH =
1
− yb 2
y
1

eni  + yb 
y


2
− yb 2
−1
=
2
eni y
eni ( yb)
−1
−1
=
(n-type semiconductor)
eni (n / ni ) en
Note that both cases above assume that b is neither zero or infinite.
Author's Note: While the plot of eniRH versus n/ni locales the optimal doping for the best RH, which is
n = 1.12ni in this case, this optimal doping depends on the mobility ratio b. Figure 5Q27-2 shows
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Chapter 5
eniRH versus n/ni for various b ratios from 1 to 100. Notice that the best RH corresponds to a large b. It
is therefore important to choose a material in which the mobility ratio is high. (See Question 5.28)
Figure 5Q27-2: Normalized Hall coefficient RH(eni) vs. n/ni for various b = e/h. The effect of the mobility
ratio b on the Hall effect coefficient RH. b = 3 corresponds to Si, whereas b = 100 is close to the mobility ratio
(92) in InSb.
5.28 Hall effect in semiconductors Most Hall-effect high-sensitivity sensors typically use III–V
semiconductors, such as GaAs, InAs, InSb. Hall-effect integrated circuits with integrated amplifiers,
on the other hand, use Si. Consider nearly intrinsic samples in which n ≈ p ≈ ni, and calculate RH for
each using the data in Table 5.5. What is your conclusion? Which sensor would exhibit the worst
temperature drift? (Consider the bandgap, and drift in ni).
Table 5.5 Hall effect in selected semiconductors
Solution
We can write the Hall coefficient under ambipolar conduction in an intrinsic semiconductor, p = n = ni
from Equation 5.103 as
RH =
p − nb 2
1 − b2
=
2
2
e( p + nb )
eni (1 + b )
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Chapter 5
1 − b2
2
eni (1 + b )

RH =
where
b =  e /  h and p = n = ni.
(1)
We can then fill in Table 5.5 as shown in Table 5Q28-1
Table 5Q28-1 Hall effect in selected semiconductors
Eg (eV)
ni (cm−3)
μe(cm2V−1s−1)
μh(cm2V−1s−1)
b
RH (m3A−1s−1)
Si
1.10
1×1010
1,400
450
3.1
−320
GaAs
1.42
2×106
8,500
400
21.25
−2.84×106
InAs
0.36
1×1015
33,000
460
71.71
−6.1×10−3
InSb
0.17
2×1016
78,000
850
91.76
−3.1×10−4
Both ni and b are important in Equation (1). The examination of Table 5Q28-1 shows that the highest
coefficient is for GaAs that has the lowest ni, which provides the most sensitive sensor. RH is sensitive
to intrinsic concentration. Lower ni means a wider bandgap material.
Consider an intrinsic semiconductor, n = p = ni, and the drift in RH with temperature. Let ni =
Cexp(−Eg/2kT), where C is a constant.
dRH
d
1 − b2
=
dT
dT e[C exp(− E g / 2kT )](1 + b )2
Neglecting the temperature dependence of b and differentiating and substituting we obtain
dRH  E g  dT

= −
RH  2kT  T
Clearly, the largest bandgap leads to the greatest temperature drift. To avoid this drift, the
semiconductor is doped by a small amount to make it extrinsic over the temperature range of interest.
We would like the mobility ratio to be as large as possible because a larger ratio b leads to a
higher RH. When the ratio b = 1, RH is zero. For a large b, RH is maximum given by
RH =
1 − b2
− b2
−1

=
2
2
eni b
eni
eni (1 + b )
GaAs has the maximum RH in Table 5Q28-1. Remember that we are comparing intrinsic samples.
GaAs is widely used in Hall Effect elements, though Si obviously has the advantage of an established
microelectronics technology that can integrate the Hall element, the associated amplifiers and circuits
into a single chip. InAs and InSb have larger drift mobility ratios and hence higher Hall effect figures
of merit. (See Question 5.26.)
Author's Note: Let x = n/ni be the relative electron concentration in a semiconductor. Then p = ni2/n =
ni/x. The Hall coefficient becomes
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Chapter 5
ni
− xni b 2
 1  1 − b2 x2
p − nb
x

RH =
=
= 
2
2
en
e( p + nb )
n


 i  1 + b 2 x 3 + 2bx
e i + xni b 
x
x

2

RH (eni ) =
1 − b2 x2
1
+ b 2 x 3 + 2bx
x
RH has been expressed in terms of the relative electron concentration x, and the mobility ratio b. Figure
5Q28-1 shows a plot of (eni)RH versus x for various choices of b. The importance of b is clear. See
Appendix A.
Figure 5Q28-1: Normalized Hall
coefficient (eni)RH vs. n/ni for various b =
e/h.
Black, b = 1
Blue, b = 3.1, Si
Green, b = 21, GaAs
Red, b = 72, InAs
*5.29 Compound semiconductor devices Silicon and germanium crystalline semiconductors are
what are called elemental group IV semiconductors. It is possible to have compound semiconductors
from atoms in groups III and V. For example, GaAs is a compound semiconductor that has Ga from
group III and As from group V, so that in the crystalline structure we have an "effective" or "mean"
valency of IV per atom and the solid behaves like a semiconductor. Similarly GaSb (gallium
antimonide) would be a III-V type semiconductor. Provided we have a stoichiometric compound, the
semiconductor will be ideally intrinsic. If, however, there is an excess of Sb atoms in the solid GaSb,
then we will have nonstoichiometry and the semiconductor will be extrinsic. In this case, excess Sb
atoms will act as donors in the GaSb structure. There are many useful compound semiconductors, the
most important of which is GaAs. Some can be doped both n- and p-type, but many are one type only.
For example, ZnO is a II-VI compound semiconductor with a direct bandgap of 3.2 eV, but
unfortunately, due to the presence of excess Zn, it is naturally n-type and cannot be doped to p-type.
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Chapter 5
a. GaSb (gallium antimonide) is an interesting direct bandgap semiconductor with an energy bandgap
Eg = 0.67 eV, almost equal to that of germanium. It can be used as an LED (light-emitting diode) or
laser diode material. What would be the wavelength of emission from a GaSb LED? Will this be
visible?
b. Calculate the intrinsic conductivity of GaSb at 300 K taking Nc = 2.3  1019 cm−3, N = 6.1  1019
cm−3, e = 5000 cm2 V−1 s−1, and h = 1000 cm2 V−1 s−1. Compare with the intrinsic conductivity of
Ge.
c. Excess Sb atoms will make gallium antimonide nonstoichiometric, that is, GaSb1+, which will result
in an extrinsic semiconductor. Given that the density of GaSb is 5.4 g cm−3, calculate  (excess Sb)
that will result in GaSb having a conductivity of 100 −1 cm−1. Will this be an n- or p-type
semiconductor? You may assume that the drift mobilities are relatively unaffected by the doping.
Solution
a. Given Eg = 0.67 eV, the corresponding wavelength is:
 = hc / Eg = (6.626  10−34 J s)(3.0  108 m s−1) / (0.67 eV  1.602  10−19 J/eV)

 = 1.85  10−6 m or 1850 nm, not in the visible.
b. Assume temperature, T = 300 K.
For GaSb, we are given Nc = 2.3  1025 m−3; N = 6.1  1025 m−3; e = 5000  10−4 m2 V−1 s−1; h =
1000  10−4 m2 V−1 s−1, therefore the intrinsic concentration is:
ni = (NcN)1/2exp[−Eg/2kT]

ni = [(2.3  1025 m−3)(6.1  1025 m−3)]1/2exp[−(0.67 eV)/(28.617  10−5 eV/K  300 K)]

ni = 8.82  1019 m−3
The intrinsic conductivity is therefore:
 = eni(e + h)

 = (1.602  10−19 C)(8.822  1019 m−3)(0.5 m2 V−1 s−1 + 0.1 m2 V−1 s−1) = 8.48 −1 m−1
For Ge, from from Table 5.1, ni = 2.3  1019 m−3, e = 0.39 m2 V−1 s−1 and h = 0.19 m2 V−1 s−1. The
intrinsic conductivity is then:
 = eni(e + h)

 = (1.602  10−19 C)(2.3  1019 m−3)(0.39 m2 V−1 s−1 + 0.19 m2 V−1 s−1) = 2.14 −1 m−1
Intrinsic Ge and GaSb have comparable conductivities (same order of magnitude).
(Typically, intrinsic conductivities of semiconductors are largely determined by the bandgap).
c. Excess antimony will make the sample n-type. We will assume that the electron mobility is roughly
the same. In the extrinsic sample, n = Nd >> p.
Given that n = 100 −1 m−1, and n  eNde:
Nd = n / ee = (100 −1 cm−1  100 cm/m)/[(1.602  10−19 C)(0.5 m2 V−1 s−1)]

Nd = 1.248  1023 m−3 (>> ni)
Now consider the atomic concentration.
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Chapter 5
GaSb has the GaAs crystal structure so that the unit cell (cubic) has 4 Ga atoms and 4 Sb atoms. The
cube side is a. The density  is therefore given by,
 1   (4 M Ga + 4M Sb ) 

3 
NA
 a 

 =
which means that the Sb atom concentration is
nSb =
4
N A
=
3
a
( M Ga + M Sb )
or
(5.4 103 kg m −3 )(6.022 10 23 mol −1 )
nSb =
(10 −3 )(69.72 g mol −1 + 121.75 g mol −1 )

nSb = 1.698  1028 m−3
Thus,  = Nd/nSb = (1.248  1023 m−3) / (1.698  1028 m−3) = 7.35  10−6
5.30 Excess minority carrier concentration Consider an n-type semiconductor and weak injection
conditions. Assume that the minority carrier recombination time, h, is constant (independent of
injection─ hence the weak injection assumption). The rate of change of the instantaneous hole
concentration,  pn/ t, due to recombination is given by
pn
p
=− n
t
h
[5.104]
The net rate of increase (change) in pn is the sum of the total generation rate G and the rate of
change due to recombination, that is,
dpn
p
=G− n
dt
h
[5.105]
By separating the generation term G into thermal generation Go and photogeneration Gph and
considering the dark condition as one possible solution, show that
dpn
p
= G ph − n
dt
h
[5.106]
How does your derivation compare with Equation 5.27? What are the assumptions inherent in
Equation 5.106?
Solution
We are given G as total generation rate, that is,
G = thermal generation (Go) + photogeneration (Gph),
i.e.
G = Go + Gph.
Also, we know that pn = pn − pno and therefore pn = pn + pno. From Equation 5.104
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Chapter 5
dpn
p
=G− n
dt

Substitute for G,
dpn
p
= Go + Gph − n
dt

(1)
In the dark and in equilibrium we have no photogeneration (Gph = 0), an equilibrium concentration of
holes (pn = pno) and no net increase in the hole concentration (dpn/dt = 0).

0 = Go −
Hence,
Go =
pno

pno

Substitute this expression for Go into Equation (1).
dpn pno
p
=
+ Gph − n
dt


Substitute for pn in terms of excess hole concentration pn.
d (pn + pno ) pno
p + pno
=
+ Gph − n
dt



dpn
p
= Gph − n
dt

This is Eqn. 5.106 which is the same as Equation 5.27. The derivation above is more rigorous
whereas that in Section 5.5 is quite intuitive.
*5.31 Direct recombination and GaAs
Consider recombination in a direct bandgap p-type
semiconductor, e.g., GaAs doped with an acceptor concentration Na. The recombination involves a
direct meeting of an electron–hole pair as depicted in Figure 5.22. Suppose that excess electrons and
holes have been injected (e.g., by photoexcitation), and that Δnp is the excess electron concentration
and Δpp is the excess hole concentration. Assume Δnp is controlled by recombination and thermal
generation only; that is, recombination is the equilibrium storing mechanism. The recombination rate
will be proportional to np pp, and the thermal generation rate will be proportional to npo ppo. In the dark,
in equilibrium, thermal generation rate is equal to the recombination rate. The latter is proportional to
nno ppo. The rate of change of Δnp is
n p
t

= − B n p p p − n po p po

[5.107]
where B is a proportionality constant, called the direct recombination capture coefficient. The
recombination lifetime τr is defined by
n p
t
=−
n p
r
[5.108]
a. Show that for low-level injection, npo  np  ppo, τr is constant and given by
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r =
Chapter 5
1
1
=
Bp po BN a
[5.109]
b. Show that under high-level injection, Δnp  ppo,
n p
t
 − Bp p n p = − B (n p )
2
[5.110]
so that the recombination lifetime τr is now given by
r =
1
1
=
Bp p Bn p
[5.111]
that is, the lifetime τr is inversely proportional to the injected carrier concentration.
c. Consider what happens in the presence of photogeneration at a rate Gph (electron–hole pairs per
unit volume per unit time). Steady state will be reached when the photogeneration rate and
recombination rate become equal. That is,
 n p 

Gph = 
= B n p p p − n po p po
 t  recombinat ion


[5.112]
A photoconductive film of n-type GaAs doped with 1013 cm−3 donors is 2 mm long (L), 1 mm wide
(W), and 5 µm thick (D). The sample has electrodes attached to its ends (electrode area is therefore 1
mm × 5 µm) which are connected to a 1 V supply through an ammeter. The GaAs photoconductor is
uniformly illuminated over the surface area 2 mm × 1 mm with a 1 mW laser radiation of wavelength λ
= 850 nm (infrared). The recombination coefficient B for GaAs is 7.21 × 10−16 m3 s−1. At λ = 840 nm,
the absorption coefficient is about 5 × 103 cm−1. Calculate the photocurrent Iphoto and the electrical
power dissipated as Joule heating in the sample. What will be the power dissipated as heat in the
sample in an open circuit, where I = 0?
Solution
Excess carrier concentration must be equal that means n p = p p
And for p-type material, p po  n po
a. Start with Equation 5.107,
n p
t

n p
t



= − B n p p p − n po p po = − B (n po + n p )( p po + n p ) − n po p po
= − Bn p ( p po + n po + n p )

(1)
For low level of injection, n po  n p  p po


n p
t
r =
=−
n p
r
= − Bn p p po
1
1
=
Bp po BN a
[5.109]
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Chapter 5
b. Consider Equation (1),
n p
t
= − Bn p ( p po + n po + n p )
Under high level of injection, n p  p po  n po


n p
t
r =
=−
n p
r
= − B(n p )
2
1
1
=
Bn p Bp p
[5.111]
c. If Po is the incident radiation, then the transmitted power is
Poexp(−D).
The absorbed power is
Po[1 − exp(−D)].
The number of absorbed photons per second is

N ph =
Po 1 − exp(−D) Po 1 − exp(−D)
=
hf
hc
N ph =
(10 −3 J s −1 ) 1 − exp(−5 105 m −1  5 10 −6 m) (800 10 −9 m)
(6.626 10 −34 J s)(3 108 m s −1 )


= 3.7×1015 photons per second
Assuming unity quantum efficiency then each of these absorbed photons will generate an electron hole
pair (EHP) per second. If Gph is the photogeneration rate of EHP per unit volume then,
3.7 1015 s −1
Gph =
=
= 3.7×1026 m−3 s−1
-3
-3
-6
Volume (2 10 m)(110 m)(5 10 m)
N ph
Under steady state conditions (over a long time), the excess hole concentration is pn(∞). We can find
this from
dp n
p
= Gph − n = 0 as t → 
dt
r

pn =  r Gph (steady state)
where τr is the recombination lifetime of the photogenerated electron hole pairs.
Electrons and holes are created in pairs by photogeneration, and when they recombine directly, one
electron and one hole is removed each time. Thus,
 p n =  nn
Suppose the injection level is high, then
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r =
Chapter 5
1
1
=
Bn p Bp p

pn =

pn =
G ph
Bp p
G ph
B
=
3.7  10 26 m −3s −1
= 7.16×1020 m−3 or 7.16×1014 cm−3
7.21 10 −16 m 3s −1
It appears that Δpn >> nno = Nd = 1013 cm−3
If we assume weak injection, i.e.  r =
1
1
1
−4 s
=
=
−16
−3 −1
19
−3 = 1.39×10
Bnno BN d (7.2110 m s )(10 m )
Then, pn = (1.39 10−4 s)(3.7 1026 m−3s −1 ) = 5.14×1022 m−3 or 5.14×1016 cm−3 which is greater than
Nd .
The photoconductivity Δσ of the sample is then
 =  light −  dark = enn e + eph = epn (e + h )

 = (1.6 10 −19 C)(8500 cm 2 V −1s −1 + 400 cm 2 V −1s −1 )(7.16 1014 cm −3 ) = 1.02 Ω−1cm−1
The conductance G is then
G =
A WD  (0.1cm)(5 10 −4 cm)(1.02  −1 cm−1 )
=
=
= 2.55×10−4 Ω−1
L
L
(0.2 cm)
The photocurrent Iphoto is then
Iphoto = ΔI = VΔG = (1V)(2.55×10−4 Ω−1) = 2.55×10−4 A or 0.255 mA
The electrical power dissipated as joule heating is
Pelectrical = VI = (1V)(2.55×10−4 A) = 2.55×10−4 W or 0.255 mW
The energy of the incident photon is
Eph =
hc

=
(6.626  10−34 J s)(3  108 ms−1 )
= 1.476 eV
(840  10-9 m)(1.6  10−19 J eV-1 )
The excess energy, Eph – Eg = 1.476 eV – 1.420 eV – (3/2)(0.0259) eV = 0.017 eV is very small,
nearly zero, and lost as lattice vibration to heat. Thus,
Pthemalization = Nph(Eph – Eg)
= (3.7×1015 s−1)(0.017 eV)(1.6×10−19 J eV−1)
= 1×10−7 J s−1 or 1×10−7 W or 0.1 W
5.32 Piezoresistive strain gauge factor Piezoresistive coefficients for an n-type Si along [110] are L
= −31.2 × 10− Pa−1, and T = −17.6 × 10− Pa−1 whereas for p-type Si along the same crystal
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Chapter 5
direction, L = 71.8 × 10− Pa−, and T = −66.3 × 10− Pa−. Given the elastic modulus Y  170 GPa,
calculate the gauge factors for these n-type and p-type Si piezoresistors. What is your conclusion?
Solution
The strain L and stress L are related by L = YL. Further the transverse and longitudinal strains are
related by T = −L, where  is the Poisson ratio. Consider the change in the resistance due to strain
L,
R / R =  /  =  L L +  T  T = Y L ( L −  T )
The gauge factor is defined by
GF =
Fractional change in gauge property R / R Y L ( L −  T )
=
=
= Y ( L −  T )
Input signal
L
L
For Si, typically  is small, ~0.1. Further T is smaller than L so that T is even smaller. Thus,
GF = Y L L
For n-type Si,
| GF | = | Y L | = | (170 109 Pa)(−31.20 10 −11 Pa -1 ) | = 53.0
For p-type Si
| GF | = | Y L| = | (170 109 Pa)(−71.8 10 −11 Pa -1 ) | = 122.1
Generally, the sign of GF is normally ignored and GF is quoted as a positive quantity as in above
calculations.
From Ch. 2, Example 2.13, for a metal resistor based strain gauge GF ~1.7.
Conclusions: First, clearly p-Si is a better choice than n-Si. In addition, obviously, a semiconductor
based strain gauge is a much better choice than a metal resistor based gauge in terms of their GFs.
5.33 Piezoresistivity application to deflection and force measurement Consider the cantilever in
Figure 5.39c. Suppose we apply a force F to the free end, which results in a deflection h of the tip of
the cantilever from its horizontal equilibrium position. The maximum stress σm is induced at the
support end of the cantilever, at its surface where the piezoresistor is embedded to measure the stress.
When the cantilever is bent, there is a tensile or longitudinal stress σL on the surface because the top
surface is extended and the bottom surface is contracted. If L, W, and D are, respectively, the length,
width, and thickness of the cantilever (see Figure 5.39c), then the relationships between the force F
and deflection h, and the maximum stress σL are
 L (max) =
3YDh
2 L2
and
F=
WD 3Y
h
4 L3
where Y is the elastic (Young’s) modulus. A particular Si cantilever has a length (L) of 500 µm, width
(W) of 100 µm, and thickness (D) of 10 µm. Given Y = 170 GPa, and that the piezoresistor embedded
in the cantilever is along the [110] direction with πL ≈ 72 × 10−11 Pa−1, find the percentage change in
the resistance, ΔR/R, of the piezoresistor when the deflection is 0.1 µm. What is the force that would
give this deflection? (Neglect the transverse stresses on the piezoresistor.) How does the design choice
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Chapter 5
for the length L of the cantilever depend on whether one is interested in measuring the deflection h or
the force F? (Note: σL depends on the distance x from the support end; it decreases with x. Assume that
the length of the piezoresistor is very short compared with L so that σL does not change significantly
along its length.)
Figure 5.39c A force applied to a cantilever bends it. A piezoresistor at the support end (where the stress is
large) measures the stress, which is proportional to the force.
Solution
The change in the resistance,
R / R =  /  =  L L =  L 
3YDh
2 L2

ΔR/R = (72×10−11 Pa−1)×3(170×109 Pa)(10×10−6 m)(0.1×10−6 m) / [2(500×10−6 m)2]

ΔR/R = 7.344×10−4 or 0.073 %
Apply
F=
WD 3Y
h
4 L3

F=
(100 10 −6 m)(10 10 −6 m) 3 (170 109 N m −2 )
(0.110 −6 m) = 3.4×10−6 N
−6
3
4(500 10 m)
Measurement of deflection h
R / R =  /  =  L L =  L 
3YD
h
2 L2
depends on Y (elastic modulus) of the cantilever, and needs a small L to get a large fractional
resistance change per unit deflection (h).
Measurement of force F
3YD 4 L3
6L
R / R =  /  =  L L =  L  2
F =L 
F
3
2 L WD Y
WD 2
does not depend on Y of the cantilever, and needs a long L to get a large fractional resistance change
per unit applied force (F).
Obviously the two designs have distinctly different requirements.
5.34 Schottky junction
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Chapter 5
a. Consider a Schottky junction diode between Au and n-Si, doped with 1016 donors cm−3. The crosssectional area is 1 mm2. Given the work function of Au as 5.1 eV, what is the theoretical barrier
height B from the metal to the semiconductor?
b. Given that the experimental barrier height B is about 0.8 eV, what is the reverse saturation current
and the current when there is a forward bias of 0.3 V across the diode? (Use Equation 4.39.)
Solution
a. At room temperature, all donors are ionized and the electron concentration n = Nd = 1022 m−3, the
donor concentration, which is related to the energy difference from the Fermi level to Ec, that is E =
Ec− EF, by
 E 
N d = N c exp −

 kT 

N 
E = −kT ln  d 
 Nc 
From Table 5.1, Nc = 2.8  1025 m−3, and we are given and Nd = 1022 m−3. Assuming temperature T =
300 K,
 10 22 m −3 
1


E = − 1.38110 − 23 J/K (300 K ) ln 
 = 0.2053 eV
25
−3 
−19
 2.8 10 m  1.602 10 J/eV 
(
)
The work function of n-Si (nSi) must be less than that of Au (Au = 5.1 eV) in order to have a
Schottky junction. nSi is given as nSi = E + , where  = 4.01 eV (see Table 5.1) is the electron
affinity of Si.

nSi = 0.2053 eV + 4.01 eV = 4.215 eV
This is indeed less than Au and therefore we have a Schottky junction. The effective barrier height B
is then,
B = Au −  = 5.1 eV − 4.01 eV = 1.09 eV
b. The experimental potential barrier is given as B = 0.8 eV. (Less than the theoretical value.)
Assume temperature T = 300 K. The reverse saturation current (Io) is given by (see Example 5.22)
  
I o = ABeT 2 exp − B 
 kT 
where Be = 1.2  106 A K−2 m−2 is the effective Richardson-Dushman constant (see also Equation 4.40
in Section 4.9),
(
)

 (0.8 eV ) 1.602 10 −19 J/eV 
2

I o = 10 −6 m 2 1.2 10 6 A K − 2 m − 2 (300 K ) exp −
− 23
 1.38110 J/K (300 K ) 

Io = 3.97  10−9 A
(
)(
)
(
)
The forward bias voltage Vf is given as 0.3 V. The forward current If is then:
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(
Chapter 5
)
)
  1.602 10 −19 C (0.3 V )  
  eV f  
 − 1
 − 1 = 3.97 10 −9 A exp
I f = I o exp
− 23
(
)
kT
1
.
381

10
J/K
300
K
 
 
 
 
(

)
(
If = 0.000433 A or 0.43 mA
5.35 Schottky junction Consider a Schottky junction diode between Al and n-Si, doped with 5 ×
1016 donors cm−3. The cross-sectional area is 1 mm2. Given that the electron affinity χ of Si is 4.01 eV
and the work function of Al is 4.28 eV, what is the theoretical barrier height B from the metal to the
semiconductor? What is the built-in voltage? If the experimental barrier height B is about 0.6 eV,
what is the reverse saturation current and the current when there is a forward bias of 0.2 V across the
diode? Take Be = 110 A cm−2 K−2.
Solution
The barrier height B is
B = m – χ = 4.28 eV – 4.01 eV = 0.27 eV
 ( E − EF ) 
n = N c exp − c
kT 


Ec − EF = kT ln(Nc / n) = (0.02585eV) ln(2.8 1019 cm−3 / 5 1016 cm−3 ) = 0.16 eV
Now,
B = eVo + (Ec – EF)

eVo = B − (Ec – EF) = 0.27 eV – 0.16 eV = 0.11 eV
Alternatively,
nSi = E +  = 0.16 eV + 4.01 eV = 4.17 eV

eVo = m – nSi = 4.28 eV − 4.17 eV = 0.11 eV
Reverse saturation current is
I o = ABeT 2 exp(−


B
0.6 eV 

) = (110 −6 m 2 )(110 10 4 A m − 2 K − 2 )(300 K) 2 exp −
kT
0
.
02585
eV


Io = 8.23×10−6 A or 8.23 μA
The forward current is
  eV f  
  0.2 eV  
 − 1 = (8.23 10 −6 A) exp
 − 1 = 0.0188 A or 18.8 mA
I f = I o exp
kT
0
.
02585
eV

 





5.36 Schottky and ohmic contacts Consider an n-type Si sample doped with 1016 donors cm−3. The
length L is 100 m; the cross-sectional area A is 10 m  10 m. The two ends of the sample are
labeled as B and C. The electron affinity () of Si is 4.01 eV and the work functions  of four
potential metals for contacts at B and C are listed in Table 5.6.
Table 5.6 Work functions in eV
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Cs
Mg
Al
Au
2.14
3.66
4.28
5.1
Chapter 5
a. Ideally, which metals will result in a Schottky contact?
b. Ideally, which metals will result in an Ohmic contact?
c. Sketch the I-V characteristics when both B and C are ohmic contacts. What is the relationship between
I and V?
d. Sketch the I-V characteristics when B is ohmic and C is a Schottky junction. What is the relationship between I and V?
e. Sketch the I-V characteristics when both B and C are Schottky contacts. What is the relationship
between I and V?
Solution
We are given the concentration of donors, Nd = 1022 m−3. From Table 5.1, the electron affinity of Si is
4.01 eV and the effective density of states at the conduction edges is 2.8  1025 m−3. Assuming
temperature T = 300 K,
 E 
N d = N c exp −

 kT 

N 
E = −kT ln  d 
 Nc 

 10 22 m −3 
1


E = − 1.38110 − 23 J/K (300 K ) ln 

25
−3 
−19
 2.8 10 m  1.602 10 J/eV 

E = 0.2053 eV
(
)
The energy required to remove an electron from the Si semiconductor is the work function of the n-Si,
that is,
nSi = E +  = 0.2053 eV + 4.01 eV = 4.215 eV
a. For a Schottky contact you need m > nSi so Au will result in a Schottky junction. Note however
that for Al, m − nSi is 0.065 eV, of the order of the thermal energy so Al / Si “should not” be a
Schottky junction. This is not, however, necessarily the case as the junction depends very much on the
surface conditions (surface states) as well as the crystal plane on to which the contact is made (
depends on the crystal surface).
b. For an Ohmic contact you need m < nSi so Cs and Mg will result in Ohmic contacts.
c. This is a straight line with slope equal to the conductance, or the inverse of the resistance (I /V =
1/R) as shown in Figure 5Q36-1. The resistance can be found using the drift mobility of the electrons
e = 1160  10−4 m2 V−1 s−1 , from Figure 5.19, and the equation for conductivity,  = eNde
R=
L
L
100 10 −6 m
=
=
A eN d e A 1.602 10 −19 C 10 22 m −3 0.116 m 2 V −1 s −1 10 −5 m 2
(
)(
)(
)(
)
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
Chapter 5
R = 5381 
Figure 5Q36-1: I-V characteristic with B and C as Ohmic contacts.
d. When C is reverse biased (B is positive), current is limited by thermionic emission over B between
the metal and semiconductor at C. When C is forward biased (B negative), current is primarily limited
by the resistance of the semiconductor, unless this is very small indeed in which case the I-V is the
forward biased Schottky junction. The I−V is shown in Figure 5.36-2
Figure 5Q36-2: I-V characteristic with B as Ohmic
and C as Schottky.
Figure 5Q36-3: I-V characteristic with B and C as
Schottky contacts.
Since C is a Schottky junction, C must be Au. We can find the maximum saturation current (Io) by first
finding the barrier height, B = Au −  = 5.1 eV − 4.01 eV = 1.09 eV, and then assuming the
maximum effective Richardson-Dushman constant, Be = 1.2  106 A m−2 K−2 in
  
I o = ABeT 2 exp − B 
 kT 
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Chapter 5
(
)

 (1.09 eV ) 1.602 10 −19 J/eV 
2
2

I o = 10 10 −6 m 1.2 10 6 A K − 2 m − 2 (300 K ) exp −
1.38110 − 23 J/K (300 K ) 


Io = 5.36  10−18 A
(
)(
)
(
)
e. There is reverse saturation current in both directions. The current is saturated for even small
voltages (above a few kT/e) and is the thermionic emission current over B. The calculation for reverse
saturated current can be found as in part d.
5.37 Depletion region width in a Schottky junction Consider a metal to n-type semiconductor
Schottky junction as shown in Figure 5.58. Suppose that the donor concentration in the n-side is
constant and Nd.. There is a net positive space charge density net in this region, as shown in Figure
5.58, which is eNd. The gradient of the field, dE/dx =  netr where r is the relative permittivity of
the medium (Si). Integrate  net and then use the condition that at x = W, the field should be zero, E = 0,
and show that
E =−
eN d (W − x)
Electric field in depletion region
 o r
which is negative because it is in the −x direction. Show that this has a maximum amplitude at the
interface (x = 0) and is given by
E max = −
eN dW
Highest electric field magnitude
 o r
If V′ is the potential at any point x in Figure 5.59, then E = −dV′/dx. Show that
V = −
eN d x 2 eN dWx
+
2 o r
 o r
At x = W, V′ = Vo – V. Show that
 2  (V − V ) 
W = o r o

eN d


1/ 2
Depletion layer width with bias V
Show further that the maximum filed can also be written as
E max = −
2(Vo − V )
W
Highest electric field magnitude
Consider the Schottky junction between tungsten and an n-type Si in which Nd = 1017 cm−. Find the
depletion layer width under no applied bias, a forward bias of 0.2 V and a reverse bias of − V.
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Chapter 5
Figure 5.58 A Schottky junction that has been forward
biased. The depletion region width is W. x is measured
from the metal into the semiconductor. The voltage
across the depletion layer is Vo – V. There is a constant
net space charge density (x) = eNd in the depletion layer.
The field at any point is E(x) and the voltage is V′(x).
Solution
The gradient of the field, dE/dx =  netr where r is the relative permittivity of the medium (Si). We
can integrate  net and then use the condition that at x = W, the field should be zero, E = 0, and show
that
E=
eN d
 o r
dx =
eN d x
 o r
+C
where C is the integration constant. At x = W, E = 0, thus
E =0=
so that
C=−
Thus,
E=
eN dW
 o r
+C
eN dW
 o r
eN d x
 o r
−
eN dW
 o r
=−
eN d (W − x)
 o r
The field is maximum right at the junction at x = 0 so that
E max = −
eN dW
 o r
The voltage V′ at x can be found by integrating E(x),
V  = −  Edx = 
eN d (W − x)
 o r
dx =
eN dWx
 o r
−
eN d x 2
+ C
2 o r
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Chapter 5
where C is the integration constant. At x = 0, V′ = 0
0=
eN d (0) 2
+ C
2 o r
eN dW (0)
−
eN dWx
eN d x 2
2 o r
 o r
so that C = 0, and
V =
 o r
−
At x = W, V = Vo − V,
Vo − V =
eN dW 2
 o r
−
eN dW 2 eN dW 2
=
2 o r
2 o r
We can reorganize this for W and obtain
 2  (V − V ) 
W = o r o

eN d


1/ 2
Consider the maximum field
E max = −
eN dW
 o r
eN dW 2
eN d  2 o r (Vo − V ) 
2(Vo − V )
=−
−

=−
 o rW W o r 
eN d
W

Clearly, the voltage across the depletion region (Vo − V) divided by the width W underestimates the
maximum field by a factor of 2. On the other hand (Vo − V)/W does, however, as expected, give the
average field.
Consider the Schottky junction between tungsten and an n-type Si in which Nd = 1017 cm−. Find the
depletion layer width under no applied bias, a forward bias of 0.2 V and a reverse bias of − V.
The barrier height B is
B = m – χ = 4.55 eV – 4.01 eV = 0.54 eV
 ( E − EF ) 
n = N c exp − c

kT



Ec − EF = kT ln(N c / n) = (0.02585eV) ln(2.8 1019 cm−3 / 11017 cm−3 ) = 0.146 eV
Now,
B = eVo + (Ec – EF)

eVo = B − (Ec – EF) = 0.54 eV – 0.146 eV = 0.394 eV.
Alternatively
eVo = B − m = = B − (Ec – EF) + ] = 4.55 eV – (0.146 + 4.01) eV = 0.394 eV.
The depletion layer width W with no applied bias is
 2  V 
Wo =  o r o 
 eN d 
1/ 2
2(8.854 10 −12 F m −1 )(11.9)(0.394 V)
=
(1.602 10 −19 C)(110 23 m −3 )
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Chapter 5
= 7.2×10− m or 0.072 m.
Under a forward bias of V = 0.2 V, the new depletion layer width W is
 2  (V − V ) 
W = o r o

eN d


1/ 2
2(8.854 10 −12 F m −1 )(11.9)(0.394 − 0.2 V)
=
(1.602 10 −19 C)(110 23 m −3 )
= 5.1×10− m or 0.051 m.
Under a reverse bias of V = −5, the new depletion layer width W is
 2  (V − V ) 
W = o r o

eN d


1/ 2
=
2(8.854 10 −12 F m −1 )(11.9)(0.394 + 5 V)
(1.602 10 −19 C)(110 23 m −3 )
= 2.7×10− m or 0.27 m.
The depletion layer width increases with reverse bias voltage.
5.38 A practical Schottky diode A general equation for describing the I−V characteristics of
semiconductor diodes is
  eV  
 − 1
I = I o exp
  kT  
General diode equation
in which  is called the ideality factor1, and is unity for an ideal Schottky junction. The derivation
leading to Equation 5.68 for an ideal Schottky junction under forward bias assumes that electrons
(majority carriers) diffuse over the reduced built-in voltage (Vo – V) and are replenished by the
external current bringing electrons. But, if electrons are lost by recombination during diffusion, then
the external current will also have to replenish those recombined electrons, not only those that diffuse
over to the metal. A Schottky contact to a semiconductor as shown in Figure 5.58 has a neutral
semiconductor region, which has a certain bulk resistance Rs. In modeling a practical Schottky diode
we need to include Rs in series with a Schottky junction. The actual voltage across the junction is then
the voltage across V across the whole diode minus the voltage drop across Rs so that the I-V behavior
under forward bias (typically V > 3kT/e) for the diode is
 e(V − IRs ) 
I = I o exp

 kT 
General diode equation with a series resistance
Table 5.7 gives the I−V data on a commercial Schottky diode (CDF7621) at room temperature. Find Io,  and
Rs. What is your conclusion?
Table 5.7 Characteristics of a commercial Schottky diode (CDF7621)
V (V)
0.079
0.135
0.189
0.244
0.297
0.342
0.386
0.515
0.699
I (mA)
0.00102
0.0065
0.036
0.25
1.74
4.65
9.06
29.6
67.8
Solution
Many books use n for , but n can easily be confused with the electron concertation.
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1
Solutions to Principles of Electronic Materials and Devices: 4th Edition (10 April 2017)
Chapter 5
We can rearrange the general diode equation as follows
V=
kT
e
ln( I / I o ) + IRs
(1)
which allows us to calculate the current I from any diode voltage V, as long as we know Io,  and Rs.
Figure 5Q38-1 shows the I−V data in Table 5.7 plotted. There are two distinct regions. The dashed line
in Figure 5Q38-1 is drawn as the best fit to the first 4 or 5 points, before Rs begins to have an effect.
This line represents
 eV 

I = I o exp
 kT 
which, on a semilogarithmic plot, as in Figure 5Q38-1, is a straight line, i.e.
 e 
V
log I = log I o + 
 kT 
From the slope and intercept we can determine Io and . In Excel, of course, one can simply fit an
exponential function to the first 4−5 data points. In the present case, Io  7×10− A and   1.15. Then,
Equation (1) can be used with different Rs values, as trial and error, to calculate I from V and plot the
theory alongside the experimental points until the theoretical curve (purple) is as close as possible to
the data. The closeness can be easily judged by the eye , and gives Rs  4.2. There is some play in the
best-fit parameters. For example Rs = 4.1 − 4.3 and Io  (6.5 – 7.5)×10− A also show good fits. One
can also use math software such as Matlab to do a more rigorous curve it but the final set of
parameters Io,  and Rs would be close.
Best fit parameters
Quantity
Io (A)

Rs ()
"Best fit"
7.0×10−
1.15
4.2
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Chapter 5
Figure 5Q38-1 The I-V characteristics of the Schottky diode in Table 5.7 plotted as log(I) vs V. The purple
curve is the best fit based on general diode equation with a series resistance. The dashed line is drawn as the
best fit to the first 4 or 5 points, before Rs has an effect. This line allows Io and  to be determined. In the
present case, Io  7×10− A and   1.15. Then the general diode equation is used with different Rs values, as
trial and error, to bring the theoretical curve (purple) as close as possible to the data.
5.39 Peltier effect and electrical contacts Consider the Schottky junction and the ohmic contact
shown in Figures 5.40 and 5.44 between a metal and n-type semiconductor.
a. Is the Peltier effect similar in both contacts?
b. Is the sign in Q = I the same for both contacts?
c. Which junction would you choose for a thermoelectric cooler? Give reasons.
Solution
a. and b.
Peltier effect is based on change in the energy of an electron in passing through a junction of two
dissimilar materials. It will appear whether the junction is ohmic or Schottky if there is a change in the
electron energy in going through the junction.
Consider the Schottky junction case shown in Figure 5Q39-1 and 2. Consider the current flow from the
metal into the semiconductor in Figure 5Q39-1. The electron in the semiconductor is at Ec or, strictly
at Ec + (3/2)kT, but we will ignore the small (3/2)kT term. When it diffuses over into the metal, it is at
EF, at a lower energy. The electron has to loose energy, which is given off as heat (lattice vibrations).
Thus, heat is released.
Consider the current flow from the semiconductor to the metal as shown in Figure 5Q39-2. The
electron in the metal is at EF. When it flows into the semiconductor, it is at Ec. It has to gain energy
from lattice vibrations, which results in heat absorption.
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Chapter 5
Figure 5Q39-1 The electron in the semiconductor is at Ec.
When it flows (diffuses) over into the metal, it is at EF. It
has to loose energy, which is given off as heat (lattice
vibrations). Metal to semiconductor current flow releases
heat
Figure 5Q39-2 The electron in the metal is at EF. When it
flows into the semiconductor, it is at Ec. It has to gain
energy from lattice vibrations, which results in heat
absorption. Semiconductor to metal current flow absorbs
heat
Figure 5Q39-3 Current from the metal to the n-type
semiconductor results in heat release at the junction. The
electron at Ec in the semiconductor has to loose energy
when it flows into the metal where it is at EF. Metal to
semiconductor current flow releases heat
Figure 5Q39-4 Current from the n-type semiconductor to
the metal results in heat absorption at the junction. The
electron at EF in the metal has to absorb energy (from
lattice vibrations) to reach Ec in the semiconductor.
Semiconductor to metal current flow absorbs heat
Figure 5Q39-3 and 4 shown an ohmic contact between a metal and an n-type semiconductor. In the
Schottky junction in Figure 5Q39-3, the current is from the metal to the semiconductor. The electron at
Ec in the semiconductor has to loose energy when it enters the metal because it is at an energy EF in
the metal, which is lower. Heat is therefore released, that is, metal to semiconductor current flow
releases heat.
Thus, the sign in QP =  I is therefore the same for Schottky and ohmic contacts
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Chapter 5
c.
One can only pass a small current though a semiconductor with Schottky junctions at both ends
(M/SC/M where M = metal, SC = semiconductor) which is the reverse saturation current. This is
because one Schottky junction is always reverse biased. If ohmic contacts are used then the contacts do
not limit the current (which is determined by the bulk semiconductor) and Q can be substantial and
can be used practically as in a thermoelectric cooler. Note that it is possible to have a device with one
junction ohmic and the other Schottky if the current is not limited by the Schottky contact but rather
limited by the bulk resistance. (See also Question 5.36.)
*5.40 Peltier coolers and figure of merit (FOM) Consider the thermoelectric effect shown in Figure
5.45 in which a semiconductor has two contacts at its ends and is conducting an electric current I. We
assume that the cold junction is at a temperature Tc and the hot junction is at Th and that there is a
temperature difference of T = Th − Tc between the two ends of the semiconductor. The current I
flowing through the cold junction absorbs Peltier heat at a rate Q'P, given by
QP= I
[5.113]
where  is the Peltier coefficient for the junction between the metal and semiconductor. The current I
flowing through the semiconductor generates heat due to the Joule heating of the semiconductor. The
rate of Joule heat generated through the bulk of the semiconductor is
 L 
QJ =   I 2
 A 
[5.114]
We assume that half of this heat flows to the cold junction.
In addition there is heat flow from the hot to the cold junction through the semiconductor, given
by the thermal conduction equation
 A 
 =
QTC
T
 L 
[5.115]
The net rate of heat absorption (cooling rate) at the cold junction is then
Qnet = QP − 1/2QJ − QTC
[5.116]
By substituting from Equations 5.113 to 5.115 into Equation 5.116, obtain the net cooling rate
in terms of the current I. Then by differentiating Qnet with respect to current, show that maximum
cooling is obtained when the current is
 A
I m =   
L
[5.117]
and the maximum cooling is
 =
Qmax
A 1 2

  − T 

L 2

[5.118]
Under steady state operating conditions, the temperature difference, T, reaches a steady-state
value and the net cooling rate at the junction is then zero (T is constant). From Equation 6 show that
the maximum temperature difference achievable is
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Tmax =
1  2
2 
Chapter 5
[5.119]
The quantity 2/k is defined as the figure of merit (FOM) for the semiconductor as it
determines the maximum T achievable. The same expression also applies to metals, though we will
not derive it here.
Use Table 5.8 to determine the FOM for various materials listed therein and discuss the
significance of your calculations. Would you recommend a thermoelectric cooler based on a metal-tometal junction?
Table 5.8
Solution
The net rate of heat absorption or cooling, Q, is given by Equation 5.116
Qnet = QP − 1/2QJ − QTC
Substituting for Qp, QJ, and QTC as given by Equations 5.113 to 5.115 we find
 = I −
Qnet
LI 2 AT
−
2A
L
Differentiating with respect to the current I,

dQnet
LI
=−
dI
A
This rate is maximum at the maximum current (Im), when dQnet /dI = 0.
0=−

LI m
A
 A
I m =   
L
giving Equation 5.117. This expression can be then substituted back into the equation for Qnet to find
the maximum cooling rate
LI
AT
 = I m − m −
Qmax
2A
L
2
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
 =
Qmax
A 2 A 2 AT A 2 AT
−
−
=
−
L
2L
L
2L
L

 =
Qmax

A   2

− T 
L 2

Chapter 5
which is the desired equation. In the steady state, the net cooling rate will be zero and the maximum
temperature difference will be Tmax

A   2

− Tmax  = 0
L 2


 2
Tmax =
2
The figure of merit (FOM) of the materials listed can now be found. From Table 5.8, for n-Bi2Te3,  =
0.06 V,  = 1/ = 1/(10−5  m) = 105 −1 m−1, and  = 1.7 W m−1 K−1. Substituting,
FOM =
 2 (105  −1 m −1 )(0.06 V )2
=
(1.7 W m−1 K −1 ) = 212 K

The values for all the materials are listed in Table 5Q40-1,
Table 5Q40-1: Summarized values for FOM.
Material
FOM
K
n-Bi2Te3
212
p-Bi2Te3
338
Cu
0.0456
W
0.0119
Obviously metals have the worst FOM and semiconductors have the best FOM. Metal−metal junctions
would obviously not make practical Peltier coolers!
5.41 Seebeck coefficient of n-Si Thermoelectric power (Seebeck) measurements on an n-type Si
crystal doped with donors generate the results shown in Table 5.9. What can you do with this data and
how would you interpret the experiment? Consider also whether Equation 5.78 can be used for
degenerately doped semiconductors.
Table 5.9 Experimental Seebeck coefficients for an n-type Si
Nd (cm−)
×
× × × × ×
|Sn |(mV K−)
1.60
1.55
1.31
1.20
0.724
0.280
Note: Data extracted from T.H. Geballe and G.W. Hull, Phys. Rev., 98, 940, 1955
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Chapter 5
Solution
Consider
| S n |=
k  Ec − E F 5  k  kT ln( N d / N c ) 5  k
k 5

+ + r = 
+ + r  = ln( N d / N c ) +  + r 

e  kT
2  e
kT
2  e
e2

(1)
Figure 5Q41-1 shows a plot of |Sn| vs ln(Nd/Nc), which is nearly a straight line with a best line (linear
regression) given by (using Excel)
|Sn(103)| = −0.1076x + 0.3616
where (103) takes into account that Sn is in mV/K. The experimental slope is 1.08×10− (V/K). The
theoretical slope is k/e = 0.862×10− (V/K). The difference is ~25%.
We can do slightly better by noting that Sn expression for semiconductors is not expected to
apply under very heavy doping (the semiconductor is degenerate and metal-like). We therefore ignore
the most heavily doped sample and replot the data, except the last point, as in Figure 5Q41-2. The
slope becomes 0.945×10− (V/K). The theoretical slope is k/e = 0.86×10− (V/K) and the difference
now is 10%.
Figure 5Q41-1 Plot of |Sn| vs. ln(Nd/Nc) should be a straight line.
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Chapter 5
Figure 5Q41-2 Plot of |Sn| vs. ln(Nd/Nc) ignoring the most heavily doped sample.
Note: As the sample is doped, there is a change in the scattering mechanism (r). These changes are
ignored and do not greatly affect the analysis. No attempt is made to extract r from the intercept. The
reason is that the actual measured Seebeck voltage is larger than the predicted by Equation (1) because
of phonon drag. The phonon drag adds an additional term to Equation (1), which affects the intercept;
the slope is not affected by too much.
5.42 Seebeck coefficient of silicon and phonon drag Seebeck experiments on a p-type Si crystal
doped with × cm−3 of B atoms indicate that Sp = +1.13 mV K− at room temperature (300 K) for
this sample. If the B doping is increased to × cm−3, Sp = +0.98 mV K−. Assume that r = 1 and
calculate the expected Sp for these two p-type samples. What r values would make the theoretical Sp
agree with experiments? Phonon drag increases the magnitude of the Seebeck coefficient expected
from diffusion of carriers alone in Equations 5.80 and 5.81. What is the contribution of phonon drag to
Sp?
Solution
A Livemath (Theorist) solution is given below. Although the default calculations are shown to five
significant figures, we do not actually need the calculations to be at level of 5-significant figures.
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Chapter 5
Conclusions: The phonon drag is quite important in this case. The reason is that the samples are
heavily doped so the Seebeck coefficient from Equation 5.81 is small. The phonon drag contribution
tends to be constant at a given temperature. For example, in the two different samples above, the
phonon drag contribution is 0.48 mV K− and 0.52 mV K−, roughly the same. Thus, we would expect
that in sample with less doping, the relative contribution of the phonon drag term would be smaller
(e.g. 25−30% but not negligible).
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Chapter 5
5.43 Seebeck coefficient and pn junction drift Consider a pn junction Si device (a diode) which has
the p-side doped with 2×1017 acceptors cm−3 and the n-side doped 1014 cm−3. What will be the voltage
developed across this device if a temperature fluctuation gives rise to a 0.1 C temperature difference
across the pn junction? Assume the p-side and the n-side have the same width. What would be the
voltage if the p-side was very thin compared with the n-side? What is your conclusion? Assume that r
= − for the n-side and r =  for the p-side. Neglect phonon drag effects.
Solution
Sn in Equation 5.80 depends on (Ec − EF) which depends on the doping concentration Nd. From n = Nd
= Ncexp[−(Ec − EF)/kT] we have
Ec − EF = kTln(Nd/Nc) = (0.0259 eV)ln(1×1014/2.8×1019) = 0.324 eV
Equation 5.80
Sn = –
k  Ec − E F 5 
+ + r
e  kT
2 
Sn = –
(1.38110 −23 )  (0.324)(1.602 10 −19 ) 5 
+ − 2 = −1.123 mV K−.

(1.602 10 −19 )  (1.38110 − 23 )(300) 2 
with r = −2, gives
For the p-type Si, from p = Na = Nvexp[−(EF − Ev)/kT] we have
EF − Ev = kTln(Na/Nv) = (0.0259 eV)ln(2×1017/1.2×1019) = 0.106 eV
Equation 5.81,
Sp = +
k  E F − Ev 5 
+ + r
e  kT
2 
Sp = +
(1.38110 −23 )  (0.106)(1.602 10 −19 ) 5 
+ + 1 = +0.655 mV K−.

(1.602 10 −19 )  (1.38110 − 23 )(300) 2 
with r = 1, gives
Figure 5Q43-1 shows the pn junction with 0.1 C across it with p and n regions each having a
temperature drop of T1 = T2 = 0.05 C because the widths of the p- and n-sides are the same. Notice
that the Seebeck voltages actually oppose each other. We assume that Sp and Sn do not change
significantly over this small temperature range (good assumption). The net Seebeck voltage of the cold
with respect to the hot end is
V = SpT1 + SnT2 = (+0.655 mV K−)(0.05 K) + (−1.123 mV K−)(0.05 K)
= 0.03275 mV − 0.05615 mV = −0.0234 mV or −23 V,
with the cold side being negative as shown in Figure 5Q43-1. If this is at the input of an op amp, this
could be amplified and yield a significant output voltage drift.
If the p-side is much narrower than the n-side, then the whole temperature change of 0.1 C
appears across the n-layer, that is T1 = 0 and T2 = 0.1 C so that
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Chapter 5
V = SpT1 + SnT2 = (+0.655 mV K−)(0 K) + (−1.123 mV K−)(0.1 K) = −0.112 mV
Figure 5Q43-1 A temperature difference between the two ends of the pn junction is 0.1 C. If the
semiconductor layers have the same width, each has a temperature drop of 0.05 C. If the p-side is much
narrower than the n-side, then the temperature drops essentially across the n-side. Voltage shown represent the
voltage of the cold side with respect to the hot side.
Author's Note: How can we estimate the phonon drag effect? From Question 5.42, the phonon drag
term is roughly 0.5 mV K−. Thus, for Case I,
V = SpT1 + SnT2 = (+0.655 + 0.5 mV K−)(0.05 K) +[−(1.123 + 0.5) mV K−)(0.05 K)
= 0.05775 −  = − mV or − V
This is the same as before because the phonon drag contributions cancel.
For Case II
V = SpT1 + SnT2 = (+0.655 + 0.5 mV K−)(0 K) +[−(1.123 + 0.5) mV K−)(0.1 K)
= 0 −  = − mV or − V
This is ~50% more than before, and the phonon drag cannot be totally ignored.
5.44 Photogeneration and carrier kinetic energies Figure 5.35 shows what happens when a photon
with energy hf > Eg is absorbed in GaAs to photogenerate an electron and a hole. The figure shows
that the electron has a higher kinetic energy (KE), which is the excess energy above Ec, than the hole,
since the hole is almost at Ev. The reason is that the electron effective mass in GaAs is almost 10 times
less than the hole effective mass, so the photogenerated electron has a much higher KE. When an
electron and hole are photogenerated in a direct bandgap semiconductor, they have the same k vector.
Energy conservation requires that the photon energy hf divides according to
(k )2 + (k )2
hf = E g +
2me
2mh
where k is the wavevector of the electron and hole and me and mh are the effective masses of the electron
and hole, respectively.
a. What is the ratio of the electron to hole KEs right after photogeneration?
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Chapter 5
b. If the incoming photon has an energy of 2.0 eV, and Eg = 1.42 eV for GaAs, calculate the KEs of
the electron and the hole in eV, and calculate to which energy levels they have been excited with
respect to their band edges.
c. Explain why the electron and hole wavevector k should be approximately the same right after
photogeneration. Consider kphoton for the photon, and the momentum conservation.
Solution
a. Consider the KE of the electron and hole, given by
2
2
(
(
k )
k )
and KE =
KE =
e
2me
h
2mh
From Table 5.1 we have

KEe mh
0.4me
= =
6
KE h me 0.067 me
b. The excess energy of the photon over 1.42 eV goes to its kinetic energy.

KEe = (2 eV – 1.42 eV)[6/(6+1)] = 0.5 eV

KEh = (2 eV – 1.42 eV)[1/(6+1)] = 0.08 eV
c. From momentum conservation
kCB − kVB = kphoton
Since the momentum of a photon is small

k CB − k VB = 0

kCB = k VB

ke = k h
*5.45 The Four Probe Resistivity Measurement The four probe resistivity measurement allows the
resistivity of a semiconductor crystal to be conveniently measured without complications arising from
contacts effects and without the need for samples of known geometry. It is widely used in the
semiconductor industry to measure the resistivity of Si wafers. The technique is illustrated in Figure
5.59a. Four collinear and equally separated sharp probes (needles) are placed on the surface of the
sample. The probes are spring pressured to make good contact. A current  is passed through the
sample via the outer probes A and D. The applied voltage to A and D is not relevant to the
measurement as long as a known current is passed through the sample. Indeed, the contacts at A and D
may be Schottky contacts and the current may be limited by the Schottky junctions. The voltage drop
between the two inner probes B and C are read with a digital voltmeter which takes a negligible input
current. Thus, the current paths in the semiconductor and also the voltage drop along BC are not upset
by the voltmeter connected between B and C. Within the semiconductor, the current  and voltage drop
along the current between B and C, that is VBC, are related by the resistivity of the semiconductor and
some geometric factor taking into account various possible current paths from A to D and the locations
of the points B and C. At any point in the sample where the current density is J and the electric field is
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Chapter 5
E, we must have J = E   Consider point A as an independent point current source and point D as an
independent point current sink. We can find the potential drop between BC for the two independent
currents and then add them up.
Suppose that the sample dimensions are much larger than the separation s of the needles.
Consider the currents emanating from point A as shown in Figure 5.59b. Current density at radius r is
I
J=
2r 2
The surface area through which current flows is half of 4πr2 because current flows only below
dV
A. Using E = −
and J = E   show that the potential drop between B and C (r = s and r = 2s) due
dr
to currents from A is
I
VBC =
4s
There will be a similar potential drop between B and C when we consider point D as an
independent point sink. Thus, by the principle of superposition, the total voltage drop between B and
C must be

Four-probe resistivity measurement
VBC =
I
2s
What are some of the important assumptions in the derivation?
A particular four-probe instrument has s = 1.5 mm. Measurements on an n-Si wafer give a
voltage (VBC) of 0.27 mV at a current of 0.1 mA. What are the wafer resistivity and donor
concentration?
Figure 5.59 (a) The four probe resistivity measurement. A current I is passed through the probes A
and D and the voltage drop V along BC is read on a high resistivity voltmeter. (b) We consider point A
as an independent current source and similarly point D as an independent current sink.
Solution
dV
dr
Consider
E=−
which means
I  1 
 I 
VBC = −  E dr = −  (J ) dr = −   
dr = −
  dr
2 
2r 
2 s  r 2 
s
s
s
2s
2s
2s
2s
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Chapter 5
I 2 s  1 
I  1 
I  1 1 
I
VCB = −
dr =
=
− =−




2



2 s  r 
2  r  s
2  2s s 
4s
I

VBC =
4s
Assumptions: (a) The wafer has uniform resistivity ( is constant), otherwise we cannot take it outside
2s
the integral. (b) The semiconductor wafer thickness t >> 3s (or AD) (c) No temperature gradient,
otherwise there will be thermoelectric voltages between the probes. (d) The probes do not strain (flex)
the wafer, otherwise the piezoelectric effect will change the resistivity.
Given the measured values we have
2.7 10 −4 V = VBC =
solving
I
(0.110 −3 A) 
=
4s 4 (1.5 10 −3 m)
 = 0.025  m
The resistivity is given by
1

so that
1

= eN d  e
=
1
= eN d  e  (1.602 10 −19 C) N d (1400 10 − 4 m 2 V −1 s −1 )
0.025  m
where we did not account for the change in the mobility with doping. Solving the above we find
Nd  1.751021 m−3 or 1.751015 cm−3.
The examination of Figure 5.19 shows that e at this donor concentration should be about 0.13 m2 V−
s−, which means that
1


=
1
= eN d  e  (1.602 10 −19 C) N d (1300 10 − 4 m 2 V −1 s −1 )
0.025  m
Nd  1.891021 m−3 or 1.891015 cm−.
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Chapter 5
APPENDIX
USEFUL INSTRUCTOR NOTES
A. Hall Effect
The general Hall Coefficiant under ambipolar conduction (both electrons and holes drifts are involved)
in normalized is given by Equation 5.103,
RH =
p − nb 2
2
e( p + nb )
where b = e/h.
Let x = n/ni be the relative electron concentration in a semiconductor. Then p = ni2/n = ni/x. The
Hall coefficient becomes
ni
− xni b 2
 1  1 − b2 x2
p − nb 2
x


RH =
=
=
2
2
eni  1 + b 2 x 3 + 2bx
e( p + nb )
 ni


e + xni b 
x
x


RH (eni ) =
1 − b2 x2
1
+ b 2 x 3 + 2bx
x
RH has been expressed in terms of the relative electron concentration x, and the mobility ratio b. Figure
A.1 shows a plot of RH(eni) versus x for various choices of b. The importance of b is clear. The
maximum magnitude for RH occurs when Si is slightly n-type (n = 1.12ni) but when GaAs and InAs
are p-type; but not by much. Further, if we were to use an intrinsic semiconductor, there would be a
also strong temperature dependence in RH (as shown in Question 5.28). Most importantly, notice how
having a larger mobility ratio improves the maximum magnitude of the Hall coefficient. The worst
case has b = 1.
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Chapter 5
Figure A.1 Normalized Hall coefficient (eni)RH vs. n/ni for various b = e/h. Black, b = 1 | Blue, b = 3.1, Si |
Green, b = 21, GaAs | Red, b = 72, InAs
The above discussion has not considered the actual magnitude of RH. In an actual sensor, the
Hall voltage VH is proportional to RH, given all other factors being the same for a given magnetic field
(e.g. geometry and current). Figure A.2 shows the magnitude of RH vs n/ni for Si, GaAs and InAs.
Clearly GaAs is superior to Si and is widely used as a Hall element.
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Chapter 5
Figure A.2 Magnitude of the Hall coefficient |RH| vs. n/ni for various b = e/h. Blue, b = 3.1, Si | Green, b =
21, GaAs | Red, b = 72, InAs
Hall effect devices are characterized by their sensitivity, that is the Hall voltage generated per
unit magnetic field. Table 5Q28-1 summarizes the sensitivity of various Hall elements based on GaAs,
InAs and InSb. Note that many Hall effect ICs commonly provide gain (amplification) after the Hall
element so the sensitivity can appear to be much larger than the values in this table. The values in
Table 5Q28-1 are for the element itself.
Table A.1 Selected semiconductor Hall effect devices. The sensitivity is the Hall voltage generated per unit
field under a constant current
Semiconductor
Device Part
Sensitivity
mV / mT
Drift
(dVH/dT)/VH
Manufacturer
%/C
Si
AD22151
4
Analog Devices
GaAs
KSY10
1.1
−0.05 %
Infineon (Siemens)
GaAs
CYSJ106C
1.3
−0.06 %
Chen Yang Tehnologies
GaAs
HG-705
1.9
−0.06 %
Asahi Kasei
InAs
HQ-0811
2.2
−0.2 %
Asahi Kasei
InSb
HW-322B
6.0
−1.8 %
Asahi Kasei
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Chapter 5
The temperature dependence of RH for various semiconductors as used in sensors is given Figure 5.4,
p218 in Semiconductors, 2nd Edition, Infineon Technologies, 2004 (Publicis Corporate Publishing)
Before I came here I was confused about this subject. Having listened to your lecture I am still confused. But on
a higher level.
Enrico Fermi (1901-1954; Nobel Laureate, 1938)
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