Solution-Manual-of-Modern-Semiconductor-Devices-for-Integrated-Circuits-Chenming-Calvin-Hu-Z-lib-org
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Chapter 1
Visualization of the Silicon Crystal
1.1 (a) Please refer to Figure 1-2. The 8 corner atoms are shared by 8 unit cells and
therefore contribute 1 atom. Similarly, the 6 face atoms are each shared by 2 unit
cells and contribute 3 atoms. And, 4 atoms are located inside the unit cell.
Hence, there are total 8 silicon atoms in each unit cell.
(b) The volume of the unit cell is
Vunit cell 5.43 A 5.43 10 8 cm 1.60 10 22 cm3 ,
3
3
and one unit cell contains 8 silicon atoms. The atomic density of silicon is
N Si
8 silicon atoms
5.00 10 22 (silicon atoms) cm 3 .
Vunit cell
Hence, there are 5.001022 silicon atoms in one cubic centimeter.
(c) In order to find the density of silicon, we need to calculate how heavy an
individual silicon atom is
Mass1 Si atom
28.1 g/mole
4.67 10 23 g/atom .
23
6.02 10 atoms/mole
Therefore, the density of silicon ( Si ) in g/cm3 is
ρ Si N Si Mass1 Si atom 2.33 g / cm 3 .
Fermi Function
1.2 (a) Assume E = E f in Equation (1.7.1), f(E) becomes ½. Hence, the probability is ½.
(b) Set E = E c + kT and E f = E c in Equation (1.7.1):
f(E)
1
1 e
Ec kT Ec /kT
1
0.27 .
1 e1
The probability of finding electrons in states at E c + kT is 0.27.
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* For Problem 1.2 Part (b), we cannot use approximations such as Equations (1.7.2)
or (1.7.3) since E-E f is neither much larger than kT nor much smaller than -kT.
(c) f(E) is the probability of a state being filled at E, and 1-f(E) is the probability of a
state being empty at E. Using Equation (1.7.1), we can rewrite the problem as
f(E E c kT) 1 f(E E c 3kT)
1
1
E c kT E f /kT 1
Ec 3kT E f /kT
1 e
1 e
where
1
1
1 e
E
Ec 3kT E f /kT
3 kT E
/kT
E
3 kT E
/kT
f
f
1 e c
1
e c
E 3 kT E f /kT
E 3kT E f /kT
1 e c
1 e c
1
.
E c 3 kT E f /kT
1 e
Now, the equation becomes
1
1 e
E c kT E f /kT
1 e
1
E c 3 kT E f /kT
.
This is true if and only if
Ec kT E f Ec 3kT E f .
Solving the equation above, we find
E f Ec 2kT .
1.3 (a) Assume E = E f and T > 0K in Equation (1.7.1). f(E) becomes ½. Hence, the
probability is ½.
(b) f(E) is the probability of a state being filled at E, and 1-f(E) is the probability of a
state being empty at E. Using Equation (1.7.1), we can rewrite the problem as
f(E Ec ) 1 f(E Ev )
1
1
Ec E f /kT 1
Ev E f /kT
1 e
1 e
where
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1
E
1
1 e
Ec E f /kT
E
/kT
E
E
/kT
1 e v f
1
e v f
1
.
Ev E f /kT
Ev E f /kT
E v E f /kT
1 e
1 e
1 e
Now, the equation becomes
1
1 e
Ec E f /kT
1 e
1
E v E f /kT
.
This is true if and only if
Ec E f Ev E f .
Solving the equation above, we find
Ef
Ec Ev
.
2
(c) The plot of the Fermi-Dirac distribution and the Maxwell-Boltzmann distribution
is shown below.
Probability
1
0.9
Fermi-Dirac
Distribution
0.8
0.7
0.6
Maxwell-Boltzmann
Distribution
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
(E-E f )/kT
The Boltzmann distribution considerably overestimates the Fermi distribution for
small (E-E f )/kT. If we set (E-E f )/kT = A in Equations (1.7.1) and (1.7.2), we
have
1
e A 1.10
.
A
1 e
Solving for A, we find
eA
1 eA
1.10
e A 10.11
A ln 10.11 2.31 .
Therefore, the Boltzmann approximation is accurate to within 10% for (E-E f )/kT
2.31.
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1.4 (a) Please refer to the example in Sec. 1.7.2. The ratio of the nitrogen concentration
at 10 km above sea level to the nitrogen concentration at sea level is given by
E
/kT
N ( N 2 )10 km
e 10 km
(E
E
)/kT
ESea Level /kT e 10 km Sea Level
N ( N 2 ) Sea Level e
where
E10 km E Sea Level altitude mass of N 2 molecule acceleration of gravity
10 6 cm 28 1.66 10 24 g 980 cm s 2 4.56 10 14 erg.
The ratio is
N ( N 2 )10 km
N ( N 2 ) Sea Level
e ( 4 .5610
14
erg)/( 1.3810 16 erg K 1 273 K)
e 1.21 0.30 .
Since nitrogen is lighter than oxygen, the potential energy difference for nitrogen
is smaller, and consequently the exponential term for nitrogen is larger than 0.25
for oxygen. Therefore, the nitrogen concentration at 10 km is more than 25% of
the sea level N 2 concentration.
(b) We know that
N (O2 )10 km
N (O2 ) Sea Level
0.25 ,
N ( N 2 )10 km
N ( N 2 ) Sea Level
0.30 , and
N ( N 2 ) Sea Level
N (O2 ) Sea Level
4.
Then,
N ( N 2 )10 km
N (O2 )10 km
N ( N 2 )10 km
N ( N 2 ) Sea Level
0.30 4
N ( N 2 ) Sea Level
N (O2 ) Sea Level
N (O2 ) Sea Level
N (O2 )10 km
1
4.8 .
0.25
It is more N 2 -rich than at sea level.
1.5
1 f E f E 1
1
E f E E f /kT
1 e
E E E f /kT
e f
E E E f /kT
1 e f
1
E f E E f /kT
1 e
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1 e
1
E f E E f /kT
1
E f E E f /kT
1 e
f E f E
1.6 (a)
150K
1.0
f(E)
0.5
300K
0.0
Ef
E
(b) At 0K, the probability of a state below the Fermi level being filled is 1 and a state
above the Fermi level being filled is 0. So a total of 7 states are filled which
means there are 14 electrons (since 2 electrons can occupy each state) in the
system.
Density of States
1.7 Since the semiconductor is assumed to be, We are asked to use Equations (1.7.2) and
(1.7.4) to approximate the Fermi distribution. (This means that the doping
concentration is low and E f is not within a few kTs from E c or E v . A lightly doped
semiconductor is known as a non-degenerate semiconductor.) The carrier distribution
as a function of energy in the conduction band is proportional to
Distribution (E) E Ec e
1 / 2 E E f /kT
,
where e-(E-Ef)/kT is from Equation (1.7.2). Taking the derivative with respect to E and
setting it to zero, we obtain
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d
E Ec 1/ 2 e E E f /kT 1 1 e E E f /kT E Ec 1 e E E f /kT 0
2 E Ec
dE
kT
The exponential terms cancel out. Solving the remaining equation yields
1
E Ec 1/ 2 1 E Ec 1/ 2
kT
2
E Ec kT
2
E Ec
kT
.
2
So, the number of carriers in the conduction band peaks at E c +kT/2.
Similarly, in the valence band, the carrier distribution as a function of energy is
proportional to
Distribution (E) Ev E e
1 / 2 E f E /kT
,
where e-(Ef-E)/kT is Equation (1.7.2). Taking the derivative and setting it to zero, we
obtain
d
Ev E 1/ 2 e E f E /kT 1 Ev E 1/ 2 e E f E /kT Ev E 1/ 2 1 e E f E /kT 0 .
dE
2
kT
Again, the exponential terms cancel out, and solving the remaining equation yields
1
Ev E 1/ 2 1 Ev E 1/ 2
kT
2
Ev E kT
2
E Ev
kT
.
2
Therefore, the number of carriers in the valence band peaks at E v -kT/2.
1.8 Since it is given that the semiconductor is non-degenerate (not heavily doped), E f is not
within a few kTs from E c or E v . We can use Equations (1.7.2) and (1.7.4) to approximate the
Fermi-Dirac distribution.
(a) The electron concentration in the conduction band is given by
n
C.B.
Dc(E) f(E) dE A E Ec e
Ec
E E f /kT
dE .
In order to simplify the integration, we make the following substitutions:
E Ec
x E kT x Ec ,
kT
1
dE dx dE kT dx, and x : from 0 to .
kT
Now the equation becomes
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n A kTx e
kTx E c E f / kT
0
kT dx A kT
3/ 2
e
E c E f / kT
x e x dx
0
where
0
3
x e x dx x 2 e x dx 3 / 2
1
2
0
. (Gamma function)
Hence, the electron concentration in the conduction band is
n
2
A kT
3/ 2
e
E c E f / kT
.
Similarly, the hole concentration is given by
p Dv(E)1-f(E)dE B Ev E e
Ev
E f E /kT
dE .
-
V.B.
Again, we make the following substitutions to simplify the integration:
Ev E
1
x E kT x Ev ,
dE dx dE kT dx, and x : from to 0 .
kT
kT
Now the equation becomes
0
p B kTx e
E f kTx E v / kT
kT dx B kT
3/ 2
e
E f E v / kT
0
x e x dx
where
0
3
x e x dx x 2 e x dx 3 / 2
1
0
2
. (Gamma function)
Therefore, the hole concentration in the conduction band is
p
2
B kT
3/ 2
e
E f E v / kT
.
(b) The word “Intrinsic” implies that the electron concentration and the hole concentration
are equal. Therefore,
n p
2
A kT
3/ 2
e Ec Ei / kT
2
B kT
3/ 2
e Ei Ev / kT .
This simplifies to
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A e E c Ei / kT B e Ei Ev / kT .
Solving for E i yields
Ec Ev kT 1 Ec Ev
ln
0.009 eV ; k 8.62 10 5 eV K 1 , T 300 K .
2
2 2
2
Ei
Hence, the intrinsic Fermi level (E i ) is located at 0.009 eV below the mid-bandgap of the
semiconductor.
1.9 The unit step functions set the integration limits. D c (E) is zero for E < E c , and D v (E) is zero
for E > E v . Since it is given that the semiconductor is non-degenerate (not heavily doped), E f
is not within a few kTs from E c or E v . We can use Equations (1.7.2) and (1.7.4) to
approximate the Fermi-Dirac distribution.
(a) The electron concentration in the conduction band is given by
n
C.B.
Dc(E) f(E) dE A E Ec e
E E f /kT
Ec
dE .
In order to simplify the integration, we make the following substitutions:
E Ec
x E kT x Ec ,
kT
1
dE dx dE kT dx, and x : from 0 to .
kT
Now the equation becomes
n A kTx e
kTx E c E f / kT
0
kT dx A kT e
2
E c E f / kT
0
x e x dx
where
0
x e x dx 1 .
Hence, the electron concentration in the conduction band is
n A kT e
2
E c E f / kT
.
Similarly, the hole concentration is given by
p Dv(E)1-f(E)dE B Ev E e
Ev
V.B.
-
E f E /kT
dE .
Again, we make the following substitutions to simplify the integration:
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Ev E
1
x E kT x Ev ,
dE dx dE kT dx, and x : from to 0 .
kT
kT
Now the equation becomes
0
p B kTx e
E f kTx E v / kT
kT dx B kT e
2
E f E v / kT
0
x e x dx
where
0
x e x dx 1 .
Therefore, the hole concentration in the conduction band is
p B kT e
2
E f E v / kT
.
(b) The word “Intrinsic” implies that the electron concentration and the hole concentration are
equal. Therefore,
n p A kT e E c Ei / kT B kT e Ei E v / kT .
2
2
This simplifies to
A e Ec Ei / kT B e Ei Ev / kT .
If we solve for E i , we obtain
Ei
Ec Ev kT 1 Ec Ev
ln
0.009 eV ; k 8.62 10 5 eV K 1 , T 300 K .
2
2 2
2
Hence, the intrinsic Fermi level (E i ) is located at 0.009 eV below the mid-bandgap of the
semiconductor.
1.10
(a) The carrier distribution as a function of energy in the conduction band is proportional
to
1 / 2 E E f /kT
Distributi on (E) E Ec e
,
where e-(E-Ef)/kT is from Equation (1.7.2). Taking the derivative and setting it to zero, we
obtain
d
E Ec 1/ 2 e E E f /kT 1 1 e E E f /kT E Ec 1 e E E f /kT 0 .
2 E Ec
dE
kT
The exponential terms cancel out. Solving the remaining equation yields
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1
E Ec 1/ 2 1 E Ec 1/ 2
kT
2
E Ec kT
2
E Ec
kT
.
2
Hence, the number of carriers in the conduction band peaks at E c +kT/2.
(b) The electron concentration in the conduction band is given by
n
C.B.
Dc(E) f(E) dE
Top of the Conduction Band
Ec
Dc(E) f(E) dE .
We assume that the function f(E) falls off rapidly such that
Top of the Conduction Band
Top of the Conduction Band
Dc(E) f(E) dE
0.
Dc(E) f(E) dE
Ec
Now we may change the upper limit of integration from the Top of the Conduction Band
to ∞:
n A E Ec e
E E f /kT
Ec
dE .
Also, in order to simplify the integration, we make the following substitutions:
E Ec
x E kT x Ec ,
kT
1
dE dx dE kT dx, and x : from 0 to .
kT
The equation becomes
n A kTx e
kTx E c E f / kT
0
kT dx A kT
3/ 2
e
E c E f / kT
0
x e x dx
where
0
3
x e x dx x 2 e x dx 3 / 2
1
0
2
. (Gamma function)
Therefore, the electron concentration in the conduction band is
n
2
A kT
3/ 2
e
E c E f / kT
.
(b) The ratio of the peak electron concentration at E = E c +(1/2)kT to the electron
concentration at E = E c +40kT is
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/ kT
f
n( Ec 40kT ) AEc 40kT EC e c
1 / 2 E c 0.5 kT E f / kT
1
A
E
0
.
5
kT
E
e
c
c
n( Ec kT )
2
1 / 2 E 40 kT E f E c 0.5 kT E f / kT
40kT / 0.5kT e c
(40 / 0.5)e 39.5 5.60 1016 .
1 / 2 E 40 kT E
The ratio is very small, and this result justifies our assumption in Part (b).
(c) The kinetic energy of an electron at E is equal to E-E C . The average kinetic energy of
electrons is
K .E .
sum of the kinetic energy of all electrons
total number of electrons
E E D (E) f(E) dE
D (E) f(E) dE
E E A E E e
A EE e
c
C.B.
C.B.
c
c
E E f /kT
Ec
c
c
E E f /kT
c
Ec
dE
.
dE
In order to simplify the integration, we make the following substitutions:
E Ec
x E kT x Ec ,
kT
1
dE dx dE kT dx, and x : from 0 to .
kT
Now the equation becomes
A kTx
A kTx
0
0
3/ 2
1/ 2
e
e
kTx E c E f / kT
kTx E c E f / kT
kT dx
kT dx
A kT
e
A kT
e
5/ 2
3/ 2
0
/ kT
E c E f / kT
Ec E f
x
x
0
3/ 2 x
1/ 2
e dx
e x dx
where
5
3
3/ 2 x
1 x
x
e
dx
x
0
0 2 e dx 5 / 2
4
(Gamma functions)
and
3
1 x
1/ 2 x
x
e
dx
x
0
0 2 e dx 3 / 2
2
. (Gamma functions)
Hence, the average kinetic energy is (3/2)kT.
Electron and Hole Concentrations
1.11 (a) We use Equation (1.8.11) to calculate the hole concentration:
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2
n p ni2 p ni2 / n 1010 / 10 5 cm 3 1015 cm 3 .
(b) Please refer to Equations (1.9.3a) and (1.9.3b). Since N d -N a >> n i and all the impurities
are ionized, n = N d -N a , and p = (n i )2/(N d -N a ).
(c) Since the Fermi level is located 0.26 eV above E i and closer to E c , the sample is n-type.
If we assume that E i is located at the mid-bandgap (~ 0.55 eV), then E c -E f = 0.29 eV.
1
Ec
Ef
Ei
2
3
1: E c -E i =0.55 eV
2: E c -E f =0.29 eV
3: E f -E i =0.26 eV
Ev
Using Equations (1.8.5) and (1.8.11), we find
n Nc e
E c E f / kT
4.01 1014 cm 3 and p ni2 / n 2.49 105 cm 3 .
Therefore, the electron concentration is 4.011014 cm-3, and the hole concentration is
2.49105 cm-3.
* There is another way to solve this problem:
n ni e
E f Ei / kT
2.20 1014 cm 3 and p ni2 / n 4.55 105 cm 3 .
(d) If T = 800 K, there is enough thermal energy to free more electrons from siliconsilicon bonds. Hence, using Equation (1.8.12), we first calculate the intrinsic carrier
density n i at 800 K:
ni N c 800 K N v 800 K e
E g / 2 kT
2.56 1016 cm 3 .
where
2 mdn kT
N c (T 800 K ) 2
2
h
3/ 2
T
2.8 10
300 K
3/ 2
19
cm 3 1.22 1020 cm 3
and
2 mdp kT
N v (T 800 K ) 2
2
h
3/ 2
T
1.04 10
300 K
19
3/ 2
cm 3 4.53 1019 cm 3 .
Clearly, n i at 800K is much larger than N d -N a (which is equal to n from the previous
part). Hence the electron concentration is nn i , and the hole concentration is
p=(n i )2/nn i . The semiconductor is intrinsic at 800K, and E f is located very close to the
mid-bandgap.
Nearly Intrinsic Semiconductor
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1.12 Applying Equation (1.8.11) to this problem yields
ni2 / p n 2 p 2 p 2 ni2 p
1
ni 7.07 1012 cm 3 and n 1.41 1013 cm 3 .
2
1.13 (a) B is a group III element. When added to Si (which belongs to Group IV), it acts as an
acceptor producing a large number of holes. Hence, this becomes a P-type Si film.
(b) At T = 300 K, since the intrinsic carrier density is negligible compared to the dopant
concentration, p=N a =41016 cm-3, and n = (n i =1010cm-3)2/p = 2500 cm-3.
At T = 600 K,
ni N c 600 K N v 600 K e
E g / 2 kT
1.16 1015 cm 3
where
2 mdn kT
N c (T 600 K ) 2
2
h
3/ 2
T
2.8 10
300 K
3/ 2
19
cm 3 7.92 1019 cm 3
and
2 mdp kT
N v (T 600 K ) 2
2
h
3/ 2
T
1.04 10
300 K
19
3/ 2
cm 3 2.94 1019 cm 3 .
The intrinsic carrier concentration is no more negligible compared to the dopant
concentration. Thus, we have
p N a ni 4 1016 1.16 1015 cm 3 4.12 1016 cm 3 , and
2
n ni2 / p 1.16 1015 cm 3 / 4.12 1016 cm 3 3.27 1013 cm3 .
The electron concentration has increased by many orders of magnitude.
(c) At high temperatures, there is enough thermal energy to free more electrons from
silicon-silicon bonds, and consequently, the number of intrinsic carriers increases.
(d) Using Equation 1.8.8, we calculate the position of the Fermi level with respect to E v .
E f Ev kT lnN v T / pT 0.34 eV , T 600 K .
At 600 K, the Fermi level is located 0.34 eV above the valence band.
Incomplete Ionization of Dopants and Freeze-out
1.14 From Equation (1.9.1), we know that n + N a - = p + N d +. Since N d + is much larger than N a -,
all the samples are n-type, and n N d + - N a - = 31015 /cm3. This value is assumed to be
constant. Using the Equations (1.8.10) and (1.9.3b),
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p ni2 / N d N a N c N v exp E g / kT CT 3 exp E g / kT ,
where C is a temperature independent constant. Using the sensitivity of p defined by
p/T,
p / T 3 E g / kT CT 2 exp E g / kT
Therefore, the larger the energy gap is the less sensitive to temperature the minority carrier
is. For the definition of the sensitivity of p,
p / T / p 3 Eg / kT / T
The temperature sensitivity of the minority carrier is greater for larger E g .
1.15 (a) Let us first consider the case of n-type doping. The dopant atoms are located at
energy E d inside the bandgap, near the conduction band edge. The problem states that
we are considering the situation in which half the impurity atoms are ionized, i.e.
n=N d /2. In other words, the probability of dopant atoms being ionized is ½, or
conversely, the probability that a state at the donor energy E D is filled is ½.
From Problem 1.2 part (a), we know that if f (E D )=1/2, then E D =E f . From Equation
1.8.5,
n Nce
Ec E f / kT
.
We also know that E f =E D and E c -E D =0.05eV.
2 mdn kT
N c (T ) 2
2
h
N c (T )e
E c E f / kT
3/ 2
3/ 2
T
3
2.8 10
cm .
K
300
N
2 N c (T )
N c (T )e Ec E D / kT D
e Ec E D / kT .
Nd
2
19
This equation can be solved iteratively. Starting with an arbitrary guess of 100K for T,
we find T converges to 84.4 K.
Similarly, for boron
2 mdp kT
N v (T ) 2
2
h
N v (T )e
E f E v / kT
3/ 2
3/ 2
T
3
1.04 10
cm .
300
K
N
2 N v (T )
N v (T )e E a Ev / kT a
e E a Ev / kT .
Na
2
19
Starting from T =100K, we find T converges to 67.7K.
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(b) We want to find T where n i is 10N d . This can be written as
ni N c T N v T e
E g / 2 kT
T
1.71 10
300 K
19
3/ 2
e
E g / 2 kT
10 N d
where
2 mdn kT
N c (T ) 2
2
h
3/ 2
2 mdp kT
N v (T ) 2
2
h
3/ 2
T
2.8 1019
300 K
3/ 2
T
1.04 10
300 K
cm 3 and
3/ 2
19
cm 3 .
We need to solve the equation iteratively, as in part (a) for n i =10N d =1017cm-3. Starting
from T=300K, we get T=777 K for n i =10N d .
For n i =10N a , we simply replace N d in the equation above with N a . Starting from T
=300K, we find T=635 K.
(c) If we assume full ionization of impurities at T = 300 K,
For arsenic: n N d 1016 cm 3 ni , p
2
ni
2.1 104 cm 3
Nd
2
n
For boron: p N a 10 cm ni , n i 2.1 105 cm 3
Na
15
3
(d) Please refer to the example in Section 2.8. For arsenic,
E f E v kT ln
N v 1.04 1019 cm 3
0.88 eV .
p
2.1 10 4 cm 3
For boron,
E f Ev kT ln
N v 1.04 1019 cm 3
0.24 eV .
p
1015 cm 3
(e) In case of arsenic + boron,
2
n2
1010 cm 3
n N d N a 9 10 cm , and p i
1.11 10 4 cm 3 , and
n 9 1015 cm 3
1.04 1019 cm 3
Nv
0.90 eV .
E f Ev kT ln 0.026 eV ln
4
3
1
.
11
10
cm
p
15
3
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1.16 (a) If we assume full ionization of impurities, the electron concentration is n N d =
1017cm-3. The hole concentration is p=(n i )2/n=(1010 cm-3)2/1017cm-3=103 cm-3.
The Fermi level position, with respect to E c , is
Ec E f kT lnN c / n 0.026 ln 2.8 1019 cm 3 / 1017 cm 3 0.15 eV .
E f is located 0.15 eV below E c .
(b) In order to check the full ionization assumption with the calculated Fermi level, we
need to find the percentage of donors occupied by electrons.
ED E f Ec E f Ec ED 0.1eV , and
nD N d
1
1 e
E D E f / kT
1017 cm 3
15
3
0.1 eV / 0.026 eV 2.09 10 cm 2% of N d .
1 e
Since only 2% of dopants are not ionized, it is fine to assume that the impurities are
fully ionized.
(c) We assume full ionization of impurities, the electron concentration is n N d =
1019cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1019 cm-3 = 10 cm-3.
The Fermi level position, with respect to E c , is
Ec E f kT lnN c / n 0.026 ln 2.8 1019 cm 3 / 1019 cm 3 0.027 eV .
It is located 0.027 eV below E c .
Again, we need to find the percentage of donors occupied by electrons in order to
check the full ionization assumption with the calculated Fermi level.
E D E f E c E f E c E D 0.023 eV , and
nD N d
1
1 e
E D E f / kT
1019 cm 3
7.08 1018 cm 3 71% of N d .
1 e 0.023 eV / 0.026 eV
Since 71% of dopants are not ionized, the full ionization assumption is not correct.
(d) For T=30 K, we need to use Equation (1.10.2) to find the electron concentration since
the temperature is extremely low. First, we calculate N c and N v at T=30K:
2 m dn kT
N c (T 30 K ) 2
h2
3/ 2
T
2.8 10
300 K
3/ 2
19
cm 3 8.85 1017 cm 3
and
2 mdp kT
N v (T 30 K ) 2
h2
3/ 2
T
1.04 10
300 K
19
3/ 2
cm 3 3.29 1017 cm 3 .
The electron concentration is
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n
N c 30 K N d Ec ED / 2 kT
e
8.43 10 8 cm 3 .
2
And, the hole concentration is
p ni2 / n 0
where
ni
N c 30 K N v 30 K E g / 2 kT
e
2.32 10 75 cm 3 .
2
Since n i is extremely small, we can assume that all the electrons are contributed by
ionized dopants. Hence,
1
8.43 108 cm 3
8
3
n N d 1
8
.
43
10
cm
8.43 10 9 .
E D E f / kT
17
3
10 cm
1 e
The full ionization assumption is not correct since only 8.4310-7% of N d is ionized.
To locate the Fermi level,
1
n
1 0.048 eV .
ED E f kT ln 1
Nd
E c -E f = 0.05-0.048 = 0.002 eV. Therefore, the Fermi level is positioned 0.002 eV
below E c , between E c and E D .
1.17 (a) We assume full ionization of impurities, the electron concentration is n N d =
1016cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1016cm-3 = 104 cm-3.
The Fermi level position, with respect to E c , is
Ec E f kT lnN c / n 0.026 ln 2.8 1019 cm 3 / 1016 cm 3 0.21eV .
It is located 0.21 eV below E c .
We need to find the percentage of donors occupied by electrons in order to check the
full ionization assumption with the calculated Fermi level.
ED E f Ec E f Ec ED 0.16 eV , and
nD N d
1
1 e
E D E f / kT
1016 cm 3
13
3
1
0.16 eV / 0.026 eV 2.12 10 cm 2.12 10 % of N d .
1 e
Since only 0.21% of dopants are not ionized, the full ionization assumption is correct.
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(b) We assume full ionization of impurities, the electron concentration is n N d =
1018cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1018cm-3 = 102 cm-3.
The Fermi level position with respect to E c is
Ec E f kT lnN c / n 0.026 ln 2.8 1019 cm 3 / 1018 cm 3 0.087 eV .
It is located 0.087 eV below E c .
We need to find the percentage of donors occupied by electrons in order to check the
full ionization assumption with the calculated Fermi level.
ED E f Ec E f Ec ED 0.037 eV , and
nD N d
1
1 e
E D E f / kT
1018 cm 3
17
3
0.037 eV / 0.026 eV 1.94 10 cm 19% of N d .
1 e
Since 19% of dopants are not ionized, the full ionization assumption is not accurate but
acceptable.
(c) We assume full ionization of impurities, the electron concentration is n N d =
1019cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1019cm-3 = 10 cm-3.
The Fermi level position, with respect to E c , is
Ec E f kT lnN c / n 0.026 ln 2.8 1019 cm 3 / 1019 cm 3 0.027 eV .
It is located 0.027 eV below E c .
Again, we need to find the percentage of donors occupied by electrons in order to
check the full ionization assumption with the calculated Fermi level.
ED E f Ec E f Ec ED 0.023 eV , and
nD N d
1
1 e
E D E f / kT
1019 cm3
7.08 1018 cm 3 71% of N d .
1 e 0.023 eV / 0.026 eV
Since 71% of dopants are not ionized, the full ionization assumption is not correct.
Since N d is not fully ionized and N d (ionized) << N d (not-ionized),
n N d 1 f ED N d e
E D E f / kT
N ce
E c E f / kT
.
Solving the equation above for E f yields
Ef
ED Ec kT ln N d .
2
2
N
c
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Chapter 2
Mobility
2.1 (a) The mean free time between collisions using Equation (2.2.4b) is
n
q mn
mn
mn
n mn
q
2.85 10 13 sec
where n is given to be 500 cm2/Vsec (= 0.05 m2/Vsec), and m n is assumed to be
m0.
(b) We need to find the drift velocity first:
v d n ε 50000 cm / sec .
The distance traveled by drift between collisions is
d v d mn 0.14 nm .
2.2 From the thermal velocity example, we know that the approximate thermal velocity
of an electron in silicon is
vth
3kT
2.29 107 cm / sec .
m
Consequently, the drift velocity (v d ) is (1/10)v th = 2.29106 cm/sec, and the time it
takes for an electron to traverse a region of 1 m in width is
t
10 4 cm
4.37 10 11 sec .
6
2.29 10 cm / sec
Next, we need to find the mean free time between collisions using Equation (2.2.4b):
n
q mn
mn
mn
n mn
q
2.10 1013 sec
where n is 1400 cm2/Vsec (=0.14 m2/Vsec, for lightly doped silicon, given in Table
2-1), and m n is 0.26m 0 (given in Table 1-3). So, the average number of collision is
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t
mn
207.7 collision 207 collisions .
In order to find the voltage applied across the region, we need to calculate the electric
field using Equation (2.2.3b):
vd n ε
ε vd
n
2.29 10 6 cm / sec
1635.71 Vcm 1 .
1400 cm 2 / V sec
Then, the voltage across the region is
V ε width 1635.71 Vcm 1 10 4 cm 0.16V .
2.3 (a)
Log[]
10
4
1
10
2
3
2
10
100
200
300
400
500
600
700
400
500
600
700
T (K)
(b) If we combine 1 and 2 ,
Log[]
10
3
2
10
100
200
300
T (K)
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The total mobility at 300 K is
1
1
TOTAL (300 K )
1 (300 K ) 2 (300 K )
1
502.55 cm 2 / V sec .
(c) The applied electric field is
εV
l
1V
10V / cm .
1 mm
The current density is
J ndrift q n nε q n N d ε 80.41A / cm2 .
Drift
2.4 (a) From Figure 2-8 on page 45, we find the resistivity of the N-type sample doped
with 11016cm-3 of phosphorous is 0.5 -cm.
(b) The acceptor density (boron) exceeds the donor density (P). Hence, the resulting
conductivity is P-type, and the net dopant concentration is N net = |N d -N a | = p =
91016cm-3 of holes. However, the mobilities of electrons and holes depend on
the total dopant concentration, N T =1.11017cm-3. So, we have to use Equation
(2.2.14) to calculate the resistivity. From Figure 2-5, p (N T =1.11017cm-3) is
250 cm2/Vsec. The resistivity is
1
1
qN net p
1
0.28 cm .
q 9 10 cm 250 cm 2 / V sec
16
3
(c) For the sample in part (a),
N
E c E f kT ln c
Nd
2.8 1019 cm 3
0.026V ln
0.21 eV .
16
3
10
cm
0.21 eV
Ec
Ef
Ei
Ev
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For the sample in part (b),
N
1.04 1019 cm 3
0.12 eV
E f Ev kT ln v 0.026V ln
16
3
9 10 cm
N net
Ec
Ei
0.12 eV
Ef
Ev
2.5 (a) Sample 1: N-type Holes are minority carriers.
p = n i 2/N d = (1010cm-3)2/1017cm-3 = 102 cm-3
Sample 2: P-type Electrons are minority carriers.
n = n i 2/N a = (1010cm-3)2/1015cm-3 = 105 cm-3
Sample 3: N-type Holes are minority carriers.
p = n i 2/N net = (1010cm-3)2/(9.91017cm-3) 102 cm-3
(b) Sample 1: N d = 1017cm-3
n (N d = 1017cm-3) = 750 cm2/Vsec (from Figure 2-4)
= qN d n = 12 -1cm-1
Sample 2: N a = 1015cm-3
p (N a = 1015cm-3) = 480 cm2/Vsec (from Figure 2-4)
= qN a p = 12 -1cm-1
Sample 3: N T = N d +N a = 1.011017cm-3
n (N T = 1.011017cm-3) = 750 cm2/Vsec (from Figure 2-4)
N net = N d -N a = 0.991017cm-3
= qN net n = 11.88 -1cm-1
(c) For Sample 1,
N
Ec E f kT ln c
Nd
2.8 1019 cm 3
0.026V ln
0.15 eV .
17
3
10
cm
0.15 eV
Ec
Ef
Ei
Ev
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For Sample 2,
N
1.04 1019 cm 3
0.24 eV .
E f Ev kT ln v 0.026V ln
15
3
10 cm
Na
Ec
Ei
0.24 eV
Ef
Ev
For Sample 3,
2.8 1019 cm 3
Nc
0.026V ln
0.15 eV .
Ec E f kT ln
16
3
9.9 10 cm
N net N d N a
0.15 eV
Ec
Ef
Ei
Ev
2.6 (a) From Figure 2-5, n (N d = 1016cm-3 of As) is 1250 cm2/Vs. Using Equation
(2.2.14), we find
1
1
0.5 cm .
qn n
(b) The mobility of electrons in the sample depends not on the net dopant
concentration but on the total dopant concentration N T :
NT N d N a 2 1016 cm 3 .
From Figure 2-5,
n N T 1140 cm 2 / Vs and
p N T 390 cm 2 / Vs .
N net = N d -N a = 0. Hence, we can assume that there are only intrinsic carriers in
the sample. Using Equation (2.2.14),
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1
1
1
qni n qpi p qni n p
1
.
q 1 10 cm 1140 390cm 2 / V sec
10
3
The resistivity is 4.08105 -cm.
(c) Now, the total dopant concentration (N T ) is 0. Using the electron and hole
mobilities for lightly doped semiconductors (from Table 2.1), we have
n 1400 cm 2 / V sec and
p 470 cm 2 / V sec .
Using Equation (2.2.14),
1
1
1
qni n qpi p qni n p
1
.
q 1 10 cm 1400 470cm 2 / V sec
10
3
The resistivity is 3.34105 -cm. The resistivity of the doped sample in part (b)
is higher due to ionized impurity scattering.
2.7 It is given that the sample is n-type, and the applied electric field is1000V/cm. The
hole velocity dp is 2105cm/s.
(a) From the velocity and the applied electric field, we can calculate the mobility of
holes:
dp = p , p = dp / = 2105/1000 = 200cm2/V·s.
From Figure 2-5, we find N d is equal to 4.51017/cm3. Hence,
n = N d = 4.51017/cm3, and p = n i 2/n = n i 2/ N d = 1020 / 4.51017 = 222/cm3.
Clearly, the minority carriers are the holes.
(b) The Fermi level with respect to E c is
E f = E c - kTln(N d /N c ) = E c - 0.107 eV.
(c) R = L/A. Using Equation (2.2.14), we first calculate the resistivity of the sample:
= q( n n + p p) q n n = 1.610-19 400 4.51017 = 28.8/-cm, and
= -1 = 0.035 -cm.
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Therefore, R = (0.035) 20m / (10m 1.5m) = 467 .
Diffusion
2.8 (a) Using Equation (2.3.2),
J = qn = qD(dn/dx).
Therefore,
= D(1/n)(dn/dx) = -D/. (constant)
(b) J = q n n = qn and = n .
Therefore, = -D/ n = -(kT/q)/.
(c)
2.9 (a)
= -1000V/cm = -0.026/.
ε dV
dx
Solving for yields 0.25m.
1 dE v 1
.
q dx
q L qL
(b) E c is parallel to E v . Hence, we can calculate the electron concentration in terms of
Ec.
n( x ) n 0 e Ec ( x ) Ec ( 0 ) / kT
where E c ( x ) E c (0) / L x.
Therefore, n( x ) n 0 e x / LkT .
(c) J n qn n ε qDn
qni e x / LkT n
dn
0
dx
qDn ni e x / LkT
0
qL
LkT
Therefore,
n
q
Dn
kT
Dn
n .
kT
q
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Chapter 3
Terminology and knowledge
3.1
Integrated semiconductor companies
Companies that design and fabricate integrated circuits
Fabless
With no fabrication/processing facility
Foundries
Companies that specialize in processing wafers to produce silicon devices
Wafer fab
Fabrication facility where wafers are processed to produce silicon devices
Integrated circuits
System of transistors manufactured on silicon wafers
CPU
Central processor unit that is an active part of computer containing the datapath and
control
DRAM
Dynamic random access memory
Flat-panel displays
Display panels with flat screens
MEMS
Micro-electro-mechanical-system
DNA chips
Silicon chips used for DNA screening
Dry oxidation
Growth of SiO 2 using oxygen gas
Wet oxidation
Growth of SiO 2 using water vapor
Horizontal furnace
Horizontally oriented oxidation furnace
Vertical furnace
Vertically oriented oxidation furnace
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Photolithography/Optical lithography
Process in which the resist is optically patterned and selectively removed from
designated areas on a wafer
Wafer stepper
Equipment used in lithography process
Photoresist
Ultraviolet-light sensitive material
Photomask
Quartz photo-plate containing a copy of pattern to be transferred to Si or SiO 2
surface
Negative resist
Photoresist that becomes polymerized and resistant to a developer when exposed to
an UV light
Positive resist
Photoresist whose stabilizer breaks down when exposed to an UV light, leading to
the preferential removal of exposed regions in a developer
Strip
Removal of photoresist
Asher
System that removes the resist on a wafer by oxidizing it in oxygen plasma or UV
ozone system
Lithography field
Small area exposed to an UV light during the exposure through a photomask and an
optical reduction system
Stepper
Another name for lithography equipment
Step-and-repeat action
The process of exposing different parts of a wafer until the whole wafer has been
exposed
Phase-shift photomask
Photomask that produces 180 degree phase difference in neighboring clear features
so that their diffraction fringes cancel each other
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Optical Proximity Correction (OPC)
Printing a slightly different shape on the photomask to correct distortions resulting
from an exposure process
Overlay
Alignment between 2 separate lithography steps
Extreme UV lithography (EUVL)
Lithography that uses 13nm wavelength and is expected to result in much higher
resolution
Soft-x-ray lithography
Old name of EUVL
Electron-beam lithography (EBL)
Lithography using a focused stream of electrons
Electron projection lithography (EPL)
EBL that exposes a complex pattern simultaneously using a mask and a reduction
electron lens system
Wet etching
Removal of SiO 2 using hydrofluoric acid
Isotropic
Without preference in direction
Dry etching (also known as plasma etching or reactive-ion etching (RIE))
Removal of SiO 2 using plasma and reactive ions
Anisotropic
With preference in direction
Selectivity
The extent in which an etching process distinguishes between different materials
End-point detector
Detector that monitors the emission of characteristic light from the etching products
so as to signal when etching should end
Plasma process induced damage / Wafer charging damage
Damage to devices on wafers due to the use of plasma
Antenna Effect
Sensitivity of the damage to the size of the conductor
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Ion implantation
Method of doping in which ions of impurity are accelerated and shot into the
semiconductor surface
Gas-source doping
Method of doping in which a gas reacts with silicon and liberates phosphorus so
that phosphorous diffuses into the silicon substrate
Solid-source diffusion
Method of doping in which the dopants from the thin film coated on the silicon
surface diffuse into silicon
Anneal
Heating of wafers for dopant activation and damage removal
Dopant activation
Making dopants behave as donors and acceptors by heating the wafers
Implantation dose
Total number of implanted ions/cm2
Depth/ Range
The location of peak concentration below the surface of silicon
Straddle
Spread of dopant concentration profile
3.2
Diffusion
The movement of molecules from an area of high concentration to an area of low
concentration
Junction depth
Thickness of diffusion layer
Diffusivity
Constant that describes how quickly a given impurity diffuses in silicon for a given
furnace temperature
Predeposition
Portion of diffusion process step with the source present
Drive-in
Portion of diffusion process step without the source
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Furnace annealing
Heating of wafers in a furnace for dopant activation and damage removal
Rapid thermal annealing (RTA)
Annealing process in which a wafer is rapidly heated to high temperature and
cooled quickly down to the room temperature
Rapid thermal oxidation
Oxidation process in which a wafer is heated to the designated temperature quickly,
oxidized, and then cooled rapidly down to the room temperature
Rapid thermal chemical vapor deposition (CVD)
Chemical vapor deposition process in which a wafer is heated to the designated
temperature quickly, the material is deposited on the wafer, and the wafer is cooled
rapidly down to the room temperature
Laser annealing
Annealing process in which a silicon wafer is heated with a pulsed laser
Transient enhanced diffusion (TED)
Diffusion phenomena in which diffusion rate is increased by crystal damage due to
ion implantation
Interconnect
Metal connection between devices in integrated circuits
Inter-metal dielectrics
Materials used for electrical isolation between metal layers
Crystalline
Molecular structure with nearly perfect periodicity
Polycrystalline
Molecular structure composed of densely packed crystallites or grains of singlecrystals
Amorphous
Structure with no atomic or molecular ordering
Grain boundary
Interface between crystal grains
Thin-film transistors (TFT)
Transistors made of amorphous or polycrystalline silicon, widely used in flat-panel
displays
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Sputtering target
The source material for sputtering
Reactive sputtering
Sputtering accompanied by chemical reaction of sputtered ions. For example, Ti
sputtered in nitrogen gas forms TiN film on the Si wafer
Physical vapor deposition (PVD)
Another name for sputtering
Step coverage problem
The inability of sputtering to deposit uniform films in small holes or vertical
features on wafer
High-temperature oxide (HTO)
Very conformal oxide formed by a CVD process at a high temperature
Low-pressure chemical vapor deposition (LPCVD)
CVD process at low pressure, which yields good thickness uniformity and low gas
consumption
Plasma-enhanced chemical vapor deposition (PECVD)
CVD using plasma; low deposition temperatures
In-situ doping
Doping process in which dopant species are introduced during the CVD deposition
of polycrystalline silicon
Spin-on
Process in which liquid materials are spun onto the wafer
Epitaxy
Special type of thin-film deposition technology that produces a crystalline layer on
silicon surfaces that is an extension of the underlying semiconductor crystal
arrangement
Metallization
Interconnection of individual devices by metal lines
Via/ Plug
Electrical connection between adjacent metal layers
Electromigration
Migration of metal along the crystal-grain boundaries in a quasi-random manner,
causing voids to occur in metal interconnects
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Damascene process
Process used to form copper interconnect lines
Chemical-mechanical polishing (CMP)
Process in which a polishing pad and slurry are used to polish away material and
leave a very flat surface
Back-end process
Metallization; last step of IC fabrication
Front-end processes
Oxidation, diffusion
Planarization
Process to obtain a flat surface to improve lithography and etching
Multi-chip modules
Multiple chips put into one package
Solder bumps
Electrical connection between chip and package
Flip-chip bonding
Melting pre-formed solder bumps on IC pads to make all connections
simultaneously
Burn-in
Subjecting IC packages to higher than normal voltage and temperatures; identify
potential failures
Qualification
Routine used to verify the quality of manufacturing and reliability
Operation life test
Part of qualification process; to find out if the devices last over one thousand
operating hours
3.3
(a) Lithography field
A small area having the best optical resolution (The beam intensity is uniform
within a lithography field.)
(b) Misalignment
Layer to layer mismatch (Each new mask level should be aligned to one of the
previous levels)
(c) Selectivity
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The ratio of etching rate of the film to be etched to that of the substrate film
below
(d) End-point detection
Detecting the exposed substrate film after the removal of the desired film.
(e) Step coverage
The ratio of film thickness deposited on the flat surface to that deposited on the
non-flat surface.
(f) Electromigration
The movement of atoms in a metal film due to momentum transfer from the
electrons carrying the current.
3.4
(a) Wet. Otherwise, it would take too long to grow the thick oxide.
(b) Dry because it provides better control of the gate (channel) length and vertical
wall profile.
(c) Arsenic because
- it is a donor ion (group V),
- it reduces R p and R p, and
- it reduces diffusivity.
(d) PECVD
(e) Sputtering, dry etching, plasmas containing chlorine.
(f)
Nd
Na
Metallurgical junction
Oxidation
3.5
Wet oxidation is faster than dry oxidation because the solid solubility of H 2 O steam
into SiO 2 is higher than that of O 2 gas into SiO 2 . This creates a very sharp
concentration profile that causes H 2 O to diffuse towards the SiO 2 -Si interface much
more effectively than O 2 under the same conditions.
3.6
(a)
0.2um * 0.2um 0.165um * 0.2um
6.239hr
0.0117um 2 / hr
Solving the given equation for t ox and using quadratic formula,
A A 2 4 B(t )
Tox
0.256um .
2
B
(b) Linear approximation: Tox (t ) => 0.655um (156% error).
A
2
Quadratic approximation: Tox B(t ) => 0.320um (25% error).
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In this case A Tox , neither linear nor quadratic approximations would be
valid.
Deposition
3.7
Poly Silicon:
SiH 4 Si(s) + 2H 2
Silicon Nitride:
3SiH 2 Cl 2 + 4NH 3 Si 3 N 4 + 6HCl + 6H 2
LTO:
SiH 4 + O 2 SiO 2 + 2HCl + 2H 2
HTO:
SiH 2 Cl 2 + 2H 2 O SiO 2 + 2HCl + 2H 2
Diffusion
3.8 (a) x j Dt x j x j Dt (Dt )
x 2j 2 x j x j x j Dt ( Dt ) x 2j ( Dt )
2 x j x j x j ( Dt )
Since x j x j ,
2 x j x j ( Dt ) x j
( Dt )
.
2x j
(b) For boron,
D(500K) = 10.5Exp[-3.69/(8.61410-5773)]cm2/sec = 9.010-24 cm2/sec, and
500 years = 1.581010 sec.
Hence,
x j 1.42 10 13 cm 1.42 10 9 m .
Our assumption ( x j x j ) in part (a) is correct.
3.9 (a) Junction depth is distance from surface where the dopant concentration equals
the substrate concentration.
N sub N d N junction 1015 cm 3 (Required junction dopant concentration)
From Eq. (3.6.1),
2
N 0 x j 4 Dt
N junction
e
1015 cm 3
Dt
Using
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D D0 e Ea / kT , D0 10.5 cm 2 / sec, N 0 1015 cm 2 , and E a 3.7eV ,
X j 1.97 10 4 cm .
(b) X j has the form
1/ 2
k
X j 2 Dt ln
.
Dt
Taking the derivative,
1/ 2
dX j
Xj
1 k
1
1
.
ln
1/ 2
dDt
Dt Dt
k 2 Dt X j
2ln
Dt
dX j 1.98 10 -4
1
9.39 10 4 cm 1
9
4
dDt
2 10
1.98 10
3.70
373
. 10 6 3600 3.996 10 40 cm 2
X j
.Dt 3.75 10 35 cm 0.
Hence, X j
Dt
Dt 10.5e
Visualization
3.10 (a)
(b)
1.0m resist
1.0m oxide
1.0m oxide
Silicon substrate
(c)
1.0m resist
Silicon substrate
(d)
1.0m oxide
1.0m oxide
Silicon substrate
Silicon substrate
(e)
(f)
1.0m poly
1.0m oxide
1.0m oxide
0.3
0.3
Silicon substrate
0.3
0.3
Silicon substrate
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(g)
(h)
1.0m resist
1.0m poly
1.0m resist
1.0m poly
1.0m oxide
1.0m oxide
0.3
0.3
0.3
Silicon substrate
(i)
0.3
(j)
Silicon substrate
1.0m oxide
1.0m poly
1.0m poly
1.0m oxide
0.3
0.3
(k)
1.0m oxide
0.3
Silicon substrate
0.3
(l)
1.0m resist
1.0m oxide
Silicon substrate
1.0m resist
1.0m oxide
1.0m poly
1.0m poly
1.0m oxide
0.3
0.3
1.0m oxide
0.3
Silicon substrate
0.3
Silicon substrate
(n)
(m) 1.0m resist
1.0m oxide
1.0m oxide
1.0m poly
1.0m poly
1.0m oxide
1.0m oxide
0.3
0.3
0.3
Silicon substrate
(o)
0.3
(p)
Silicon substrate
1.0m poly
1.0m oxide
1.0m oxide
1.0m poly
1.0m poly
1.0m oxide
1.0m oxide
0.3
0.3
Si substrate0.3
0.3
0.3
0.3
0.3
Si substrate0.3
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(q)
(r)
1.0m resist
1.0m resist
1.0m poly
1.0m poly
1.0m oxide
1.0m oxide
1.0m poly
1.0m poly
1.0m oxide
1.0m oxide
0.3
0.3
(s)
0.3
Si substrate0.3
0.3
0.3
0.3
Si substrate0.3
1.0m poly
1.0m oxide
1.0m poly
1.0m oxide
0.3
0.3
0.3
Si substrate0.3
3.11
um
Poly 1 Mask
Contact 1 Mask
(a)
Poly 2 Mask
Contact 2 Mask
(c)
1.0um oxide
Si-sub
Si-s
ub
1.0 m oxide
1.0 m resist
1.0 m oxide
(a)
Silicon substrate
(b)
1.0um resist
1.0 m resist
Silicon substrate
(d)
1.0um oxide
1.0 m oxideSi-sub
1.0 m oxide
Silicon(b)substrate
Silicon substrate
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Changing the polarity of the contact 1 mask results in the same cross section as
problem 3.10 (d). The same cross-section is obtained by using a negative resist and
the reversed mask of contact 1, which is opaque in the inside of contact 1 area.
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Chapter 4. Problems
Part I: P Junction
Electrostatics of P Junctions
4.1 The linearly graded junction is P-type for x<0 and n-type for x>0 with the junction at
x = 0.
Nd-Na
Slope = a
N-type
x
0
P-type
Since the doping levels are symmetric on either side of the junction, the depletion
region widths into both sides must be the same (xn=xp=Wdep/2).
(a) To find the electric field distribution, we utilize the following relation:
dε ρ q ( d − a ) q × a × x
.
=
=
=
dx ε S
εS
εS
Integrating the equation with the boundary conditions ε(x = xn= xp = Wdep/2) =0,
we find
2
q × a 2 Wdep
ε( x) =
x −
.
2ε S
4
(b) The potential distribution is
dV
qa
= −ε ⇒ V = −
dx
2ε S
2
x 3 Wdep
x
+C.
−
3
4
For convenience, if we define V(x=0) =0 as a reference, we find C = 0. Hence,
dV
qa
= −ε ⇒ V = −
2ε S
dx
2
x 3 Wdep
x
−
.
3
4
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(c) The built-in potential is the same as (Eip-Ein)/q at thermal equilibrium.
Energy-Band Diagram
Eip
Ec
Ef
Ein
Ev
x
Wdep
-xp
xn
The carrier concentrations at the edges of the depletion region are
(E f − Ein )
n ( x n ) = ni e
kT
aWdep
n( x )
⇒ Ein = Ei ( x = x n ) = E f − kT ln n = E f − kT ln
ni
2n i
and
(Eip − E f )
p ( x p ) = pi e
kT
p (x p )
aWdep
⇒ Eip = Ei ( x = x p ) = E f − kT ln
= E f + kT ln
.
ni
2ni
Therefore, φbi is
φ bi =
E ip − Ein
q
=2
kT aWdep
.
ln
q
2 ni
(d) The total voltage drop takes place in the depletion region. From part (b), we
know that |V(xn)|=|V(xp)|=qaWdep3/(24εS). Remember that xn=xp=Wdep/2. Hence,
the potential drop is φbi-Va = qaWdep3/(12εS). Solving for Wdep gives
Wdep
12ε S
(φ bi − Va )
=
qa
1/ 3
.
Note that if we substitute this in part (c), we can iteratively solve for φbi.
4.2 (a)
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kT
q
φbi =
a d
ln 2
ni
(
) (
1016 cm − 3 × 5 × 1015 cm − 3
= (0.026 V ) × ln
2
1010 cm − 3
(
)
) = 0.7V
(b)
Wdep =
=
2ε sφbi 1
1
+
q a d
(
)
2 × 12 × 8.85 × 10 −14 × 0.7 1
1
× 16 +
−19
1.6 × 10
5 × 1015
10
= 5.28 × 10 − 5 cm
= 0.528 µm
= 528nm
In order to calculate xn and xp we need to use Wdep = xn+xp and xn Nd=xp Na .
xn = Wdep − x p = Wdep −
a
∴ xn =
a + d
d
xn
a
5 × 1015 cm − 3
Wdep =
15
−3
16
−3
5 × 10 cm + 10 cm
(
) (
)
× (0.528 µm ) = 0.176 µm
Likewise,
d
x p =
a + d
(c)
ε
max
=
q d xn
εs
1016 cm − 3
Wdep =
15
−3
16
−3
5 × 10 cm + 10 cm
(
(1.6 × 10
=
−19
) (
) (
(
) (
)
× (0.528 µm ) = 0.352 µm
)
C × 1016 cm −3 × 1.76 × 10 −5 cm
= 2.652 × 104 V / cm
−14
12 × 8.85 × 10 F / cm
)
Since the built-in potential is the integration of the electric field, the maximum
electric field can also be calculated from the area of the triangle of the electric
field profile.
1
φbi = Wdep max
2
2φ
2 × (0.7V )
∴ max = bi =
= 2.652 × 10 4 V / cm
−5
Wdep 5.28 × 10 cm
ε
ε
(d)
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Фbi
Ec
Ef
Ev
x
Wdep
xn
-xp
Ф
Фbi
x
ε
x
xn
x
-εmax
ρ
qd
+
x
-
x
xn
-qa
(e)
kT
q
φbi =
a d
ln 2
ni
Wdep =
=
(
) (
1016 cm − 3 × 1018 cm − 3
= (0.026 V ) × ln
2
1010 cm − 3
2ε sφbi
q
(
)
) = 0.838V
1
1
+
a d
(
)
2 × 12 × 8.85 × 10 −14 × 0.838 1
1
× 16 + 18
−19
1.6 × 10
10
10
= 3.35 × 10 −5 cm
= 0.335 µm
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a
xn =
a + d
1018 cm −3
Wdep = 18 −3
× (0.335 µm ) = 0.332 µm
16
−3
(10 cm ) + (10 cm )
d
x p =
a + d
1016 cm −3
Wdep = 18 −3
× (0.366 µm ) = 0.003µm
16
−3
(10 cm ) + (10 cm )
ε
max
=
2φbi
2 × (0.838 V )
=
= 5.003 × 10 4 V / cm
−5
Wdep 3.35 × 10 cm
The depletion width has decreased due to a higher doping and since the acceptor
doping is now 100 times greater than the donor concentration, most of the
depletion region is on the n-side.
4.3 (a) If we apply reverse bias to the sample, the depletion width will increase. From
Equation (4.2.5), we know that Naxp = Ndxn. This means that the numbers of
ionized dopants on the N and the P sides are equal. The side that has its dopants
totally ionized is fully depleted. Hence, we need to find the total number of
dopants per unit area on each side. The side with the smaller value will become
depleted before the other.
a × W P = 5 × 1016 cm −3 × 1.2 × 10 −4 cm = 6 × 1012 cm −2 and
d × W = 1 × 1017 cm −3 × 0.4 × 10 −4 cm = 4 × 1012 cm −2 .
where WP and WN are the widths of the P-type region and N-type region,
respectively. Since the N-type region contains less dopants per unit area, the Ntype region will be fully depleted before the P-type region.
(b) Repeating the calculation for Nd = 1×1016cm-3,
a × W P = 1 × 1016 cm −3 × 1.2 × 10 −4 cm = 1.2 × 1012 cm −2 .
Hence, the P-type region will be fully depleted first.
(Note: You can also solve part (a) and (b) by finding the voltage at which each
side depletes. However, this will make the problem unnecessarily complicated.)
(c) From Equation (4.4.1), we know that
C dep / unit area =
εS
Wdep
.
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First, we need to find the total depletion region width.
For part (a), we know that the N-type region is fully depleted. And, from Equation
(4.2.5), Naxp=Ndxn. Hence,
d W = a x p ⇒ x p =
C dep / unit area =
d W
= 0.8 µm ⇒ Wdep = 0.4 µm + 0.8 µm = 1.2 µm .
a
εS
= 8.63 × 10 −9 F / cm 2 .
1.2 µm
For part (b), we know that the P-type region is fully depleted. And, from Equation
(4.2.5), Naxp=Ndxn. Hence,
d x n = aW P ⇒ x n =
C dep / unit area =
aW P
= 0.12 µm ⇒ Wdep = 0.12 µm + 1.2 µm = 1.32 µm .
d
εS
= 7.85 × 10 −9 F / cm 2 .
1.32 µm
4.4 (a) At thermal equilibrium, the Fermi level is constant throughout the system. Since
this is the case for the given figure, equilibrium conditions prevail.
(b) Carrier concentrations are given by:
n = ce
− ( Ec − E f ) / kT
and n = v e
− ( E f − EV ) / kT
.
Therefore,
Log n
Log p
pp1
nn1
pp2
pn1
np2
np1
x
0
L/4
L/2
3L/4
L
(c) The potential is the reverse of the band diagram. That is,
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1
q
φ = − ( Ei − E f ) .
For this problem, sketching the qualitative shape is sufficient.
φ ( x)
x
0
L/4
L/2
3L/4
L
(d) We assume that the total energy is constant: ETOTAL = EKE + EPE = Constant.
E
ETOTAL
KE
PE
x
0
L/4
L/2
3L/4
L
4.5 (a) The intrinsic region has no dopants. Consequently, there is no charge exposed in
the I-region. On the N-type side, the ionized donor atoms are exposed due to the
diffusion of the electrons. The same situation prevails on the P-type region due to
the diffusion of holes. The field lines that start from the ionized donor atoms on
the N-type side do not have any negative charge to terminate on until they reach
the P-type region where ionized acceptors are present. Thus, the field is constant
in the I-region. Alternatively, you may note that dε/dx=0 in the intrinsic region
since ρ = 0. The free-carriers that are present in the I-region will be swept away
by the built-in electric field. As a result, we may assume the I-region is fully
depleted. To find the depletion-region width, we first need to calculate the builtin potential φbi using Equation (4.1.2):
φ bi =
kT d a
ln
q ni2
= 0.78V .
The electric field distribution in the structure is
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ε
xp
L=0.5 µm
xn
x
0
Slope = -(qNa/εS),
from Eq (4.2.2)
Slope = (qNd/εS), from Eq (4.2.4)
|Emax| = (qNd/εS)(xn) = (qNa/εS)xp
P
I
N
φbi is the voltage drop across the P-I-N structure, and is the sum of the area under
the electric field in each region. That is,
1
ε max x p + ε max L + 1 ε max xn where
2
2
φ bi =
ε max = q d
εS
xn .
We also know from Equation (4.2.5) that
x p a = xn d ⇒ x p =
d
xn .
a
Hence, the equation becomes
q d
2ε S
d
q d L
+ 1 x n2 +
x n − φ bi = 0 .
εS
a
Solving for xn yields 0.032 µm. Consequently, xp is 0.019 µm. Then, the
depletion-region width Wdep is xn+xp+L = 0.551 µm.
(b) We found the maximum electric field in part (a):
ε max
=
q d
εS
xn =
q a
εS
x p = 148 V / cm .
(c) We also found the built-in potential in part (a). It was 0.78 V.
(d) The breakdown voltage is the sum of the areas under the electric field distribution
with εmax = εcrit = 2105 Vcm-1. The depletion-region width in each region can be
written in term of εcrit as
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ε crit
=
q d
εS
xn =
q a
εS
x p ⇒ xn =
ε crit ε S
q d
and x p =
ε crit ε S
q a
.
For P-N junction, the breakdown voltage is
VB =
1
1
2 1
= 6.91 V .
+
ε crit x p + 1 ε crit xn = ε S ε crit
2
2
2q
a
d
For P-I-N junction, the breakdown voltage is
1
ε crit x p + ε crit L + 1 ε crit xn =
2
2
ε 2 1
1
+ ε crit 0.5µm = 16.14 V .
= S ε crit
+
2q
d a
VB =
Therefore, the breakdown voltage of the P-I-N structure is ~ 10V higher than the
P-N structure with the same doping levels.
Diffusion Equation
4.6 (a) At x = -∞, all the holes generated by illumination in the region x > 0 are
recombined well before reaching x = -∞ due to the finite lifetime. Hence, there is
no excess holes exist at x = -∞, and the hole concentration is
p = p0 =
ni2
= 210.25 cm −3 .
a
(b) At x = +∞, excess hole concentration won’t have any position dependence.
According to the continuity equations, photo generation will be balanced by
thermal recombination (since the steady-state conditions prevail), therefore GL =
U = p’/τ.
p = p0 + p ' = 210.25 cm −3 + GLτ ≈ 109 cm −3 .
(c) “Low-level injection” implies that the minority carrier concentration is much
smaller than the majority carrier concentration. For this problem, since the
maximum minority carrier concentration (p = 109 cm-3) is much smaller than the
majority carrier concentration (n = Nd = 1018 cm-3), low-level injection conditions
do prevail.
(d) The continuity equation is
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∂ 2 p' ( x)
dp '
p'
= Dp
+ GL −
2
∂x
dt
τ
dp '
= 0 at steady state .
dt
and
For x ≥ 0,
∂ 2 p' ( x)
p'
dp '
= Dp
+ GL −
2
dt
∂x
τ
⇒ p ' ( x) = A1e
(x /
D pτ
)
+ A2 e
where GL = 1015 cm −3 sec −1
(− x /
D pτ
)
+ G Lτ .
Since p’≠∞ as x→∞, A2 = 0. Therefore,
p1' ( x) = A1e
(x /
D pτ
)
(x /
D pτ
)
+ G Lτ .
For x < 0,
p ' ( x) = C1e
+ C2e
(− x /
D pτ
)
(G L = 0) .
Similarly C1 = 0. Therefore,
p 2 ' ( x) = C 2 e
(x /
D pτ
)
.
Now, if we apply the boundary condition or continuity to the equations above, we
find
p1 ' ( x = 0 ) = p 2 ' ( x = 0 ) &
A1 =
G Lτ
= 5 × 10 8 cm −3
2
∂p1 ' (0) ∂p 2 ' (0)
=
⇒ A1 = C 2 + G Lτ & A1 = −C 2
∂x
∂x
Gτ
and C 2 = − L = −5 × 10 8 cm −3 .
2
Therefore,
p1 ' ( x) = 10 9 cm −3 − 5 × 10 8 e
p2 ' ( x ) = 5 × 108 e
(x /
D pτ
)
(− x /
for
D pτ
)
for x ≥ 0 and
x < 0 where D pτ = 4.07 × 10−6 cm 2 .
4.7 (a) Thermal equilibrium will be disturbed if any non-thermally generated excess
carriers are present. If L (the distance from the origin to the excess carrier
generation point) is much larger than the diffusion length of the holes (i.e., L
>> Dp τ p ), then almost all the excess carriers generated due to the light would
have recombined (i.e., pN’ ≈ 0).
In this case, we could assume thermal
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equilibrium. Otherwise, if L is not much larger than
Dp τ p , then significant
number of excess carriers will be present at x=0 and the origin will not be at
thermal equilibrium.
(b) pN’(x) = nN’(x) everywhere in the semiconductor because of charge neutrality
inside the semiconductor and all the excess carriers due to light are generated in
pairs of electrons and holes. Excess concentrations at x = -L is γΝd. Total
electron and hole concentrations at x=-L are n(x = -L) = d + γΝd and p(x = -L) =
ni2/d + γΝd.
(c) The excess carriers are generated at x = -L and x = L. In the rest of the
semiconductor the excess carriers decay due to recombination. That is,
pN’(-L) > pN’(x) for -L < x < L.
So the maximum excess carrier concentration is at x = -L and x = L. Now,
pN’(-L) = pN’(L) = γd. At γ = 10-3, pN’(-L) << d in which Nd is the majority
for
carrier concentration at thermal equilibrium. Therefore, pN’(x) << Nd
-L ≤ x ≤ L which means the excess carrier concentration is always much smaller
than the equilibrium majority carrier concentration. This is the condition for lowlevel injection.
(d) Inside the bar, carrier generation due to light is zero. In steady state, the
continuity equation simplifies to terms involving diffusion and recombination:
0 = Dp
d 2 p ' p '
−
.
τn
dx 2
Rearranging terms gives us
d 2 p '
p '
=
.
D pτ
dx 2
(e) The general solution of the differential equation in part (d) is
∆p ( x ) = A ⋅ e
where Lp =
−
x
Lp
+ B⋅e
x
Lp
Dp τ , and A and B are integration constants.
The boundary conditions for this problem are
pN’(-L) = pN’(L) = γd
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Although you do not have to solve the differential equation, the solution to this
particular problem is
x
p ' (x ) =
⋅ cosh
L
L
p
2 cosh
L
p
γ d
-W/2
(Please note that cosh( z) =
e z + e −z
.)
2
W/2
γ d
γNd
γNd
2 cosh (L / L p )
-L
L
This curve confirms what you would physically expect from carrier injection on both
ends of a silicon bar.
4.8 (a) From Figure 2-5, µp is approximately 450 cm2V-1sec-1. τp is given to be 1 µsec.
Using Equation (2.5.6b) and (4.7.6), we calculate Dp and Lp, respectively:
Dp =
kT
µ p = 11.7cm 2 / sec and
q
L p = D pτ p = 34.2 µm .
(b) Using Equation (4.6.3), we calculate the excess hole concentration:
[
]
[
]
ni2 qV / kT
e
−1.
d
Hence, pN’ = nN’ = 0 for (i), and pN’ = nN’ = 4.8×1010 cm-3 for (ii).
p ' (0) = p 0 e qV / kT − 1 =
4.9 (a) From Figure 2-8,
ρ = 0.2 Ωcm, n − type ⇒ d = 3 × 1016 cm −3 , and
ρ = 1 Ωcm, p − type ⇒ a = 1.5 × 1016 cm −3 .
Therefore, using Equation (4.1.2), we find that the built-in voltage is
φ bi = kT ln
d a
= 0.76 V .
ni2
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(b) Using Equation (4.8.2),
n P (0 ) =
ni2 (qV / kT )
e
− 1 = 0.65 × 1014 cm −3 , and
a
p (0 ) =
ni2 (qV / kT )
e
− 1 = 3.25 × 1013 cm −3 .
d
(
)
(
)
(c) P-type side:
Na = 1.5×1016cm-3 ⇒ µn = 1150 cm2V-1sec-1, τn=10-6sec, Dn=29.9 cm2/sec,
and Ln=5.47×10-3cm.
Using Equation (4.9.2), we need to derive an expression for Jn(x):
J n ( x) = −q
= −q
(
)
Dn
n P 0 e qV / kT − 1 e x / Ln =
Ln
Dn ni2 qV / kT
e
− 1 e x / Ln = 54e x / Ln mA / cm 2 .
Ln a
(
)
N-type side:
Nd = 3×1016cm-3 ⇒ µp = 400 cm2V-1sec-1, τp=10-8sec, Dp=10.4 cm2/sec,
and Lp=3.22×10-4cm.
Using Equation (4.9.1), we need to derive an expression for Jp(x):
J p ( x) = −q
Dp
Lp
(
)
p 0 e qV / kT − 1 e
− x / Lp
=
D p ni2 qV / kT
−x / L
−x / L
= −q
e
− 1 e p = 168e p mA / cm 2 .
Lp d
(
)
And the total current density J is
J = J n (0) + J p (0) = 0.222 A / cm 2 .
J
Jn
p
n
Jp
-xp
xn
x1
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(d) At the point (x1) where Jn and Jp cross each other, Jn = Jp = ½ J. Therefore,
Jp =
0.222
−( x ) / L
A / cm 2 = 0.111A / cm 2 = 0.168e 1 p ⇒ x1 = 1.38 µm .
2
4.10 For this problem, please note that:
(1) ∆pN = 0 for xb ≤ x ≤ xc because τp = 0. This yields the boundary condition ∆pN = 0
and x = xb.
(2) Because we have a P+N diode, we need only deal with the N-side of the junction
in establishing an expression for I. Moreover, the depletion width is all but totally
on the N-side of the junction given a P+N diode (i.e., xn ≅ Wdep).
(3) Because τp = ∞ for 0 ≤ x ≤ xb, there will be no IR-G current. (τp = ∞ implies that
there are no R-G centers.) Thus, we only need to develop an expression for the
diffusion current flowing in the diode.
Since we are interested in the static state, ∂∆pN/∂t =0. Also, GL = 0 (no light) and
∆pN/ τp 0 because τp = ∞. Thus the minority carrier diffusion equation reduces to
the form
∂2∆pN/∂x2 = 0 for W ≤ x ≤ xb
which is subject to the boundary conditions
2
∆pN(xb) = 0 and ∆pN(W) =
ni
d
qVkTA
e −1 .
The general (narrow-base type) solution is ∆pN(x) = A1 + A2 x.
Applying the boundary conditions
0 = A1 + A2 xb and ∆pN(W) = A1 + A2 W
giving
∆pN(W) = -A2(xb - W) or A2 = - ∆pN(W)/( xb - W) and A1 = -A2 xb = ∆pN(W)
xb
.
xb − W
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x − x ni 2 xb − x qVkTA
=
⋅ e − 1 for W ≤ x ≤ xb,
Thus ∆p ( x ) = ∆p (W ) b
xb − W d xb − W
d∆p n
n
J p ≅ − qDp
=q i
dx'
d
2
2
I ≅ AJ p = qA
ni
d
Dp
xb − W
Dp
xb − W
qVkTA
and
⋅ e
−
1
qVkTA
.
⋅ e
−
1
Please note that since ∆pN(x) is a linear function of x, Jp is constant throughout the
narrow-base (Wdep ≤ xb ≤ xb) region.
Proof of Minority Drift Current Being egligible
4.11 (a) 0.5 Ω-cm n-type Si => Nd = 1016cm-3, Dn = 30 cm2sec-1, Ln=5.5 µm, Dp = 10
cm2sec-1, Lp=3.2 µm
n i2 qVa / kT
− x / Lp
− x / Lp
e
−1 e
≈ 1014 e
cm − 3 .
d
(b) Minority current:
(
p( x ) = p 0 +
J p ( x) = −qD p
)
(
)
dp( x)
n2
−x / L
−x / L
= q i D p e qVa / kT − 1 e p = 0.5e p A / cm 2 .
dx
d Lp
Majority current:
(
J n ( x ) = J t ( x ) − J p ( x ) = J p ( 0) − J p ( x ) = 0.5 1 − e
J (Acm-2)
− x / Lp
) A / cm
2
0.6
Jt
0.5
0.4
Jn
0.3
0.2
Jp
0.1
0
0
2
4
6
8
10
12
14
16
18
20
x (µm)
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(c) J ndiff ( x) = qDn
d (n0 + n' ( x) )
dn' ( x)
D
−x / L
= qDn
= − J p ( x) n = 3J p ( x) = −1.5e p
dx
dx
Dp
(d) J ndrift ( x ) = J n ( x ) − J ndiff ( x ) = 0.5 + 1 × e
E ( x) =
− x / Lp
A / cm 2 .
J ndrift
−x / L
= 0.25 + 0.5 × e p Vcm −1 .
nqµ n
J pdrift ( x) = pqµ p E ( x) ≈ p' qµ p E ( x) = 1.6 × 10 −3 e
−x / Lp
(1 + 2e
−x / Lp
) A / cm 2 .
J pdrift << J pdiff ⇒ J p ≈ J pdiff .
J (Acm-2)
J (Acm-2)
0.5
0.03
0.45
0.025
0.4
0.35
Jpdiff
0.02
Jpdiff
0.3
0.015
0.25
0.01
Zoom In
0.2
0.005
Jpdrift
0.15
0
0.1
-0.005
0.05
Jpdrift
-0.01
0
0
2
4
6
8
10
12
14
16
18
20
0
2
4
6
8
10
x (µm)
12
x (µm)
(e) From E (x) in (d) and the Possion equation, Eq. (4.1.5), one the net
charge density, ρ to be
1.7 ×10-9 e-x/Lp C/ cm-3 .
This ρ is negligibly small compared with the excess hole charge density, qp' in (a), which
is
1.6 ×10-5 e-x/Lp C/ cm-3 .
This shows that p' = n' is a good assumption because ρ = qp' _ qn' .
Temperature Effect on IV
4.12 (a) From Equations (4.9.4) and (4.9.5),
Dp
Dn
≡ ηe − E g / kT .
+
L
p d Ln a
For simplicity, we will ignore the weak temperature dependence of η.
I ≈ I 0 (e qV / kT ) where
I 0 = Aq c v e
− E g / kT
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V≈
≈
k
q
=
V
T
kT I
ln ⇒
q I 0
qV kT d E g
+
kT
q dT kT
Eg
−
kT
dV k I
≈ ln
dT q I 0
E
kT d
ln I − ln η + g
+
kT
q dT
.
(b) Assuming V=0.5V and T=300K,
dV 0.5 − 1.1
=
= 0.002 V/K.
dT
300
(c) In Fig. 4-21, start at T=300K, V=0.5V, or I = 240 µA:
At that fixed I, ∆V = -0.2 V between -25C and 75C (∆T = 50 K). Hence, ∆V / ∆T
~ -0.2 V/ 100 ºK = -0.002 V/ºK, in excellent agreement with the value found in
part (b).
4.13
ni2 L p ni2 Ln
= Aq − p ' L p + − n' Ln .
+
I = Aq
τ
τ
τn
d p a τ n
p
4.14
(a) If we model the quasi-neutral regions as a resistor R, the voltage drop across
the neutral regions is Vneutral = IR and that across the junction is
V − Vneutral = V − IR .
Hence, Equation 4.9.4 becomes
I = I 0 (e q (V − IR ) / kT − 1) .
(b) Solving the above equation for V yields
kT I
V = IR +
ln + 1 .
q I0
(c) Let’s use I0 = 1.8×10-12A from Figure 4-22.
With R = 0,
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V (V)
R=0
I (A)
With R = 200Ω,
V (V)
R = 200Ω
R=0
I (A)
Clearly for the case R = 200Ω, the first term IR becomes dominant over
ln(I/I0+1), and the IV curve becomes very linear.
Charge Storage
4.15 Please refer to Section 4.9, 4.10, 8.2, 8.3, and 8.7.
(a) First, we need to find an expression for Qt:
WB
−xp
Qt = Aq ∫ p ' ( x)dx + ∫ nP ' ( x)dx =
x n
−W E
(
= Aq p 0 e qVa / kT
(
+ Aq nP 0 e
=
qVa / kT
W ' x − xn
− 1 − B 1 −
WB '
2
)
x + xp
W '
− 1 E 1 +
WE '
2
)
2 WB
xn
2 −xp
=
−W E
W ' W '
A ⋅ q qV A / kT
1
e
− 1 ( p 0WB '+ nP 0WE ') = Aqni2 e qVa / kT − 1 B + E .
2
2
d a
(
)
(
)
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WB’ and WE’ are the actual lengths of the base and emitter regions respectively
excluding the depletion regions. It for a short-base diode is given by
D 1
D 1
.
+ n
I t = Aqni2 e qVa / kT − 1 p
WB ' d WE ' a
(
)
Hence,
WB ' WE '
+
d a
Qt 1
(W 'W ')( aWB '+ dWE ') .
=
= B E
τS =
It 2 Dp 1
D 1 2(D pWE ' a + DnWB ' d )
+ n
WB ' d WE ' a
(b) s = 1.8 × 10 −10 sec (<< τS = τS = 1×10-6 sec)
(c) A short-based diode has a shorter storage time, thus it can operate at a higher
frequency.
Part II: Application to Optoelectronic Devices
IV of Photodiode/Solar Cell
4.16 (a) Let us examine the minority carrier diffusion equation for hole. In general,
∆∂p ∂ 2 ∆p ∆p
= Dp
−
+ GL .
∂t
∂x 2
τp
For the steady state problem at hand, ∂∆pN/∂t = 0. Also ∂2∆pN/∂x2 = 0 if one goes
far from the junction on the N-side where carrier perturbation introduced by the
junction has decayed to zero. Thus,
0=−
∆p ( x → ∞ )
τp
+ GL or ∆p ( x → ∞ ) = GLτ p . ⇐ boundary condition
(b) One simply parallels the ideal diode derivation to obtain the desired I-VA
relationship. Given a P+N junction, however, we need consider only the lightly
doped side of the junction. To simplify the mathematics, we set the origin of
coordinates to the n-edge of the depletion region. Under steady state conditions
and with x’ as defined as above, we must solve
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0 = Dp
∂ 2 ∆p ∆p −
+ GL
∂x 2
τp
subject to the boundary conditions
2
∆pN(x’=0) =
ni
d
qVkTA
e
and ∆p ( x → ∞ ) = GLτ p .
−1
The general solution is
∆p ( x' ) = GLτ p + A1e
− x'
Lp
+ A2e
x'
Lp
.
Because exp(x’/Lp) ∞ as x’ ∞, the only way the second boundary condition
can be satisfied is for A2 to be identically zero. With A2 = 0, the application of the
first boundary condition yields
2
∆p ( x' = 0) = GLτ p + A1 =
ni
d
qVkTA
e − 1 or
qVkTA
e
− GLτ p and
−
1
− x'
Lp
ni 2 qV A
kT
∆p ( x' ) = GLτ p + e − 1 − GLτ p ⋅ e .
d
2
A1 =
ni
d
The associated hole current density is then
D
d∆p J p ( x' ) = −qD p
=q p
dx'
Lp
ni 2
d
− x'
qVkTA
e − 1 − GLτ p ⋅ e L p
and for a P+N diode,
I = AJ = A[Jn(x = -xp) + Jp(x = xn)] ≅AJp(x’ = 0)
I = qA
D p ni 2
Lp d
or
qVkTA
Dτ
e
− qA p p GL .
−
1
Lp
Finally noting Dp τp = Lp2, we conclude
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qVA
I = I 0 e kT − 1 − I L
where
D p ni 2
and I L = qAL p G L .
I 0 = qA
Lp d
(c) Voltage that appears when the illuminated diode is open circuit is
V OC =
kT I L
ln 1 + .
q I0
Current that flows when the illuminated diode is short-circuited is I SC = − I L .
I
VOC
GL = 0
VA
ISC
GL = GL0
Part III: Metal-Semiconductor Junction
Ohmic Contacts and Schottky Diodes
4.17 (a) A very heavy P+ doping is important to forming an ohmic contact.
TiSi2
P+-Si
Ec
0.55 eV
Ef
Ev, Ef
0.55 eV
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(b) A very heavy N+ doping is important to forming an ohmic contact.
N+-Si
TiSi2
0.55 eV
Ef
Ec
0.55 eV
Ev, Ef
(c)
A very heavy doping is unacceptable to forming a rectifying contact.
P- Si
Ec
TiSi2
Ef
Ev
0.55 eV
2 eV
Ef
0.55 eV
4.18
(a)
χsi
Eo
q(φbi-|Va|) qφs
qϕM
qφBn
q|Va|
Ec
Ef
Ev
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ϕ M = 4.8V, χ Si = 4.05 eV
3.22 × 1019
= 4.26V
1016
= 4.8 − 4.05 = 0.75V
φ s = χ si + 0.026 ln
φ Bn
φ bi = φ M − φ s = 0.54V
φ bi − Va = 0.54 − 0.4 = 0.14V
(b)
ρ(x)
φ(x)
Va=0
Va=0,0.4
Va=0.4
ε(x)
Va=0.4
Va=0
4.19 (a)
1/C2
Voltage at which Wdep=1µm
(b)
ε
1µm
2µm
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2 µm
(c) V =
4
∫ εdx = 1. 5 ε
max
0
1 1
1 4
+ .1. ε max + .1. ε max = (1.3µm)ε max
2 5
2 5
To find εmax, we use the slope when xd<1um,
− q (−1.6 × 10 −19 )(1016 )
=
= 1.54 × 109 V/cm 2
εs
11.7ε o
Using this slope, we need xd=5um for ε=0,
ε max = (1.54 × 109 )(5 × 10−4 ) = 7.723 × 105V/cm
V = (1.3µm)(7.723 × 105V/cm) = 100.4V .
(d) Express V (areas under electric field profile) in terms of Wdep,
V = area = 30.9Wdep − 23.175
Wdep =
C=
εs
Wdep
V + 23.175
30.9
=
εs
V + 23.175
30.9
.
Depletion-Layer Analysis for Schottky Diodes
4.20 (a) ΦB(Platinum to Si) = 5.3 eV, d = 1016 cm-3 and A = 10-5 cm2
Now, Ec – Ef = kT ln(c/d) = 0.205 eV
∴ qΦs = 4.05 + 0.205 = 4.255 eV
∴ φbi = - (Φs - ΦM) = 1.045 V (Si to Pt)
The capacitance is give by
C = charge/voltage drop =
K
q d 2ε Si (φbi − VA )
=
(φbi − VA )
φ bi − VA
(
)
1/ 2
where K = 2.88×10-13 F V1/2
If VA = 0, C = 2.82×10-13 F = 0.282 pF
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φ bi
(b) For a 25% reduction C/C(0) = 0.75 ⇒
φ bi + VA
∴ VA = φbi(1- 0.75-2) = -0.813 V
1/ 2
= 0.75
4.21 The development of relationships for the electrostatic variables in a linearly graded
MS diode closely parallels the uniformly doped analysis.
(a) With d(x) = ax for x ≥ 0, invoking the depletion approximation yields
ρ(x) = qax
0 ≤ x ≤ Wdep.
Substituting into Poisson’s equation gives
dε
qa
ρ
=
=
x
0 ≤ x ≤ Wdep and
dx ε Si ε Si
0
qa
W
∫ d 'ε' = ε ∫ x' dx'
ε(x )
or
Si x
ε(x ) = −
(
qa
2
Wdep
− x2
2ε Si
)
0 ≤ x ≤ Wdep.
Turning to the electrostatic potential, we can write
dV
qa
2
= −ε( x ) =
Wdep
− x2
and
2ε Si
dx
(
0
qa
dV ' =
∫
2ε Si
V (x )
V (x ) = −
)
Wdep
∫ (W
2
dep
)
− x'2 dx'
or
x
(
qa
3
2
2Wdep
− 3Wdep
x + x3
6ε Si
)
0 ≤ x ≤ Wdep.
Finally, V = -(φbi - VA) at x = 0, and therefore
qa 3
− (φbi − VA ) = −
Wdep
3ε Si
Wdep
3ε
= Si (φbi − VA )
qa
1/ 3
.
(b) Paralleling the developments for the linearly graded P-N junction, the
expression for φbi will be
1
φbi = Φ B − (Ec − E f ) x =W
0
q
[
]
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where W0 is the depletion width when VA = 0. Since approximate charge
neutrality applies fro x > W0, it follows that
n x =W = ni e
[( E − E )
f
i
x =W 0
]/ kT
0
(E
c
− E f ) x =W ≅
0
≅ d ( x = W0 ) = aW0
or
Eg
aW
− kT ln 0
2
ni
Thus, to determine φbi, one must simultaneously solve the following two
equations employing numerical techniques.
1/ 3
3ε
W0 = Si φbi
qa
(c) C J =
ε Si A
Wdep
=
and φbi =
Eg
aW0
1
+ kTln
Φ B −
q
2
ni
ε Si A
3ε Si
qa (φbi − VA )
1/ 3
4.22 (a) From the graph, the built-in voltage is 0.8V.
(b) From Equation 4.16.7, we know that the slope of the line is inversely
proportional to the doping concentration. Hence, for Region I,
1
2
2
7 × 1014 cm 2 / F 2
=
+
⇒
≡
=
(
)
.
φ
V
Slope
bi
C 2 q d ε s
q d ε s
5V
Solving for Nd yields
d = 8.62 × 1016 cm −3 .
For Region II, the slope is 0. This implies that Region II is a very heavily doped
N++ region. Hence, Nd ≥1020cm-3.
In order to find the width of Region I, we calculate Wdep at Va = 5V since
Region I becomes fully depleted. Using Equation 4.3.1,
Wdep
≥ 1020cm-3
Nd
2ε s (φbi + 5V )
=
= 0.30 µm .
q d
8.62×1016cm-3
0
0.30µm
x
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Comparison between Schottky Diodes and P Junction Diodes
4.23 (a) Similarity:
Both have the same
slope of 1 decade / 60mV
at room temperature.
Log(I)
Schottky
Diode
Difference:
The schottky diode current
is several orders of magnitude
larger than the PN diode
current.
PN
Diode
V
(b) For Schottky diode, use Equations 4.18.2 and 4.18.6:
I 0 = AKT 2e − qφ Bn / kT = (0.01)(100 ) 300 2 e − q 0.55V / kT A = 58.5 µA
[
(
]
)
For PN diode, use Equation 4.9.5:
Dp
I 0 = Aqni2
L p d
= 0.01cm 2 q 1010 cm −3
(
)(
)
2
= 0.32 fA
4cm 2 / s (1µs )
(
4cm 2 / s
)
(c) For Schottky diode,
kT I
V=
ln + 1 = 0.36V .
q I0
For PN diode,
kT I
V=
ln + 1 = 1.03V .
q I0
(d) Use a metal or silicide that yields a smaller φBn.
(e) I0 may be so large, especially at elevated temperatures, that the reverse leakage
current
a. interferes with the rectification property of the diode, and
b. causes too large a power (heat) dissipation, I0×Vr, under reverse bias.
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Ohmic Contacts
4.24 (a) The semiconductor is uniformly doped with a doping concentration of d =
1017cm-3. Using depletion approximation, the energy band varies as
Ec (x) = ½ (qd/εSi) (x - x0)2
where x0 is a constant and x is positive in the semiconductor and zero at the MS
junction.
At x = 0, Ec(0) = 0.65 eV
(using the equilibrium Fermi-level as the reference)
and at x= 10 nm, Ec(0) ≈ 0 ⇒ x0 = 10 nm and d = 8.85×1018 cm-3
(b)
qφB
Ec
Ef
x
4.25 (a) Rc =
50mV
× 0.08µm 2 = 4 × 10 −8 Ω ⋅ cm 2
1mA
(b) Using Eq. 4.21.7,
Rc ≡
2
V
Hφ /
=
e Bn
J qvthx H d
d
= Be
HφBn / d
⇒ φ Bn =
(
) (m / m )(ε
= (5.4 × 10 )× 0.26 × 11.7
H = 5.4 × 1010 ×
n
0
s
d Rc
ln
H
B
/ ε0 )
10
= 9.418 × 1010 cm −3 / 2V −1
vthx = 2kT / πmn = 1.057 × 10 7 cm / s
B=
2
= 1.256 × 10 −9 Ωcm 2
qvthx H d
∴ φ Bn =
d Rc
ln = 0.3675 V
H
B
This is the maximum ϕBn allowed since a larger ϕBn will result in a voltage
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drop larger than 50 mV at the ohmic contact.
(c) According to the graph, φ Bn ≈ 0.36 V .
4.26
(a) φ Bn = 0.67V
∴ RC ≈ 6 × 10 −7 Ω ⋅ cm 2
φBp = 0.43V
∴ RC ≈ 5 × 10 −8 Ω ⋅ cm 2
(b) φBp = 0.23V
∴ RC ≈ 7 × 10 −9 Ω ⋅ cm 2
(c) φ Bn = 0.28V
∴ RC ≈ 1 × 10 −8 Ω ⋅ cm 2
(d) According to the graph, this contact resistance can be achieved with a barrier
height of 0.3V or lower and a doping concentration of 3 × 10 20 cm −3 or higher is
required. Looking at Table 4-2, we can see that this cannot be achieved using
only one type of metal silicide. So different silicides should be used for
contact on N+ and P+ silicon.
+ :
ErSi1.7
≥ 3.5 × 10 20 cm −3
P+ :
PtSi
≥ 2.5 × 10 20 cm −3
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Chapter 5
Energy Band Diagram
5.1
Ec
E c ,E f
Ef
Ev
Ev
E c ,E f
Ef
Ev
Accumulation
E c ,E f
Ev
Flat-band
Ec
Ef
Ev
Ec
Ev
Depletion
E c ,E f
Ec
Ev
Ef
Ev
Threshold
E c ,E f
Ev
Ec
Ef
Ev
Inversion
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5.2
(a) 1 -cm N-type silicon substrate : N d =51015 cm-3, E f = E c - 0.223eV
3.1eV
3.75eV
Ec
0.22eV
Ef=Ec
W
Ox
N-Si
n-Si
Ef
EE
Vv
3.1eV
3.75eV
Ec
0.22eV
Ef=Ec
W
Ox
Ef
EVv
E
n-Si
N-Si
VFB
V
fb = 3.75 - (3.1 + 0.22) = 0.43V
(b) 1 -cm P-type silicon substrate :N a =1.51016 cm-3, E f = E v + 0.168eV
Thermal Equil.
Flat Band
3.1eV
3.75eV
0.93eV
Ef = Ec
W
Ox
P-Si
p-Si
3.1eV
3.75eV
Ec
Ef
EV
E
Ec
Ef=Ec
v
0.93eV
Ox
W
P-Si
p-Si
Ef
EEV
v
VFB
fb = 3.75 - (3.1 + 0.93) = - 0.28V
(c) Heavily doped P+-polycrystalline silicon gate with 1 -cm N-type silicon
substrate.
Thermal Equil.
Flat Band
3.1eV
Ec
E
0.22eV c
Ef
P+ -poly
Ox
n-Si
N-Si
3.1eV
3.1eV
3.1eV
EV
v
0.22eV
EVv
Ef=E
P+ -poly
Ox
N-Si
n-Si
Ec
Ef
EEVv
VVFBfb = 3.1 + 1.12 - (3.1 + 0.22) = 0.88V
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(d) Heavily doped N+-polycrystalline silicon gate with 1 -cm P-type silicon
substrate.
Thermal Equil.
Flat Band
3.1eV
Ec
3.1eV
3.1eV
3.1eV
Ec
0.93eV
Ef
E
EV
EVv
E
E f= E c
v
N+ -poly Ox
E
EvV
P-Si
p-Si
0.93eV
N+-poly Ox
p-Si
P-Si
Ef
E
E vV
VVFBfb= 3.1 - (3.1 + 0.93) = -0.93V
MOS System: Inversion, threshold, depletion, and accumulation
5.3
(a) 3
(b) 2
(c) 1
(d) 4
(e) 5
5.4
(a)
1018 cm 3
Na
0.026 V ln 10 3 0.479 V
ln
ni
10 cm
1 E
1 E
V fb si si g B g B
q q 2 q
2 q
1.12V
0.479V 1.039V
2
kT
q
B
(b)
Wd max
2 si 2 B
qN a
2 11.7 8.84 10 14 2 0.479
3.521 10 6 cm
1.6 10 19 1018
(c)
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Vt V fb 2 B
2 si qN a 2 B
Cox
1.039V 2 0.479V
2 11.7 8.85 10 14 F / cm 1.6 10 19 C 1018 cm 3 2 0.479V
3.9 8.85 10 14 F / cm / 2 10 7 cm
0.2455 V
(d)
Only the flat-band voltage changes.
E 1 E
V fb si g si g
q q q 2 q
1.12V
0.479V 0.081V
2
Vt 0.081V 2 0.479V
1E
B g
2 q
B
2 11.7 8.85 10 14 F / cm 1.6 10 19 C 1018 cm 3 2 0.479V
3.9 8.85 10 14 F / cm / 2 10 7 cm
1.3655 V
5.5
(a) At V g -V fb = -1V, the MOS capacitor is in accumulation.
Cox
QS ox
F
4 10 7 2
Vox Tox
cm
Tox 8.62nm
(b) At threshold,
V g V fb 1V s Vox 2 B
B
1/ 2
2 s qN a 2 B
C ox
where
kT N a
.
ln
q ni
Solving iteratively, we get
N a 2.9 1016 cm 3 .
(c) Since s 0, Vox 1V .
(d) V g V fb 1V s Vox s
2 s qN a s 1 / 2
C ox
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s 0.245( s )1 / 2 0.5
Solving the equations above, we get
s 0.354V .
5.6
s
Vt
INV
DEP
V fb
W dep
ACC
Vg
W dmax
-2 b
INV
Vt
DEP
V fb
ACC
Vg
Q inv
Q acc
slope=C ox
V fb
Vt
INV
DEP
ACC
Vg
slope=C ox
Vt
Q dep
V fb
Vg
V fb
Vg
Qs
qN a W dmax
qN a W d
V fb
Vt
5.7
Vg
Vt
(a) From Equation 5.3.2,
V g V fb s
1
C ox
2qN a s s where s
.
2
s
Rearranging the terms, we obtain
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2
s
2qN a s
C ox
s V fb V g 0 .
Solving for s yields
2qN a s
C ox
s
2qN a s
4V fb V g
C ox2
2
.
Since s cannot be less than 0,
2qN a s
s
2C ox
2qN a s
2
V fb V g
V fb V g
2
4
4C ox2
where
2qN a s
C ox
.
Hence,
s
2
2
2
V fb V g
V fb V g
4 4
2
4
2
2
V fb V g
2
2
4
V fb V g .
Or,
2
2
V fb V g .
s
4
2
(b) From Equation 5.3.1,
Vox
2qN a s
2
s
V fb V g .
2
C ox
4
(c) We can qualitatively predict that s V g since | |<<1. Also, V ox V g .
s (V)
R
R
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V g (V)
V ox (V)
V g (V)
(d) From Equation 4.2.10,
2q s
Wdep
5.8 (a) Cox
C ox
ox
Tox
3.45 10 7
B kT ln
V fb
Eg
2
s
2q s
2
V fb V g .
C ox 2
4
F
cm 2
Na
0.4V
ni
B 0.96V
Vt V fb 2 B
1
2q s N a 2 B 0.96 0.8 0.34 V 0.18V
Cox
(b) Qacc C ox V g V fb 3.45 10 7
C
cm 2
(c) Qdep 2q s N a 2 B 1.1 10 7
C
cm 2
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Qinv C ox V g Vt 6.3 10 7
C
cm 2
(d)
Q S , 10-7 C/cm2
3.45
-1.96
-0.96
0.18
2
V g , volts
-1.1
-7.4
5.9
Parameters
a
b
c
d
e
Increase
Accumulation Region Capacitance
Flat-band Voltage, Vfb
Depletion Region Capacitance
Threshold Voltage, Vt
Inversion Region Capacitance
Decrease
Unchanged
X
X
X
X
X
(a) At accumulation: Accumulation capacitance
C Cox ox / Tox
(b) At flat-band: Flat-band Voltage
N
V fb g 4.05 0.56 0.026 ln a
ni
N a , | B |
(c) At depletion: Depletion Capacitance
1 / C 1 / Cox Wdep / s
N a ,Wdep , C
(d) At threshold: Threshold Voltage
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Vt V fb | 2 B | Qd / C ox
N a , | B |, Qd ,Vt
(e) At inversion: Inversion Capacitance
1 / C 1 / C ox Wd max / s
N a ,Wd max , C
5.10 To find the value of the oxide capacitance,
Cox
ox
Tox
1.15 10 7
F
.
cm 2
The capacitance at V g =V fb is given by
C fb
1
1
C ox
LD
7.9 10 8
s
F
cm 2
where
kT
LD 2s
q Na
1/ 2
4.09 10 6 cm .
To find the minimum value of the capacitance,
4 s B
kT N a
3 10 5 cm ; B
ln
0.348V
qN a
q
ni
Wd max
C min
1
C ox
1
Wd max
2.65 10 8
s
F
.
cm 2
V t is given by
Vt 2 B
1
C ox
2 s qN a 2 B V fb 1.11V
where V fb = g - s = -2.00 V.
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C ox
C fb
C min
V fb
Vt
To find the effective oxide charge, we need to find
Al 4.1V
s 4.05 0.562 0.3482 4.96V
Al s 0.86V
Vt V fb Al s 2 (0.86) 1.14V
Qox
C ox
Hence,
Qox 1.31 10 7
C
q
8.2 1011
.
2
cm
cm 2
Field Threshold Voltage
5.11 (a) V fb Al Si 4.1V ( Si E g / 2q
Ei E f
)
q
4.1V (4.05V 1.12 / 2V kT / q ln( N a / ni )) 0.80V
(b) B
Ei E f
q
Wd max
kT / q ln( N a / ni ) 0.290V
2 s 2 B
0.866um
qN a
Vt V fb 2 B qN aWd max / Cox 5V
C ox 2.65 10 9 F / cm 2
(c) C ox
Tox
0 / 1m 2.65 10 9 F / cm 2 where K 2.99
(d) In accumulation,
V g - V fb = -1V,
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C ox = Q s /V ox
Vt 5V
C ox 2.65 10 9 F / cm 2
K 2.99
It is the maximum allowable K.
(e) V g Vt
Qinv
C ox
Qinv C ox (V g Vt ) (2.65 10 9 F / cm 2 )(2V ) 5.3 10 9 C / cm 2
(f) 1 / C 1 / C ox 1 / C dep 1 / C ox Wd max / s 1 / 2.65 10 9 F / cm 2 0.866 m / s
C 2.17 10 9 F / cm 2
(g) V g V fb Vs Vox
(V g Vt ) V fb 2 B Vox
7V 0.8V 2(0.29V ) Vox
Vox 7.22V
Oxide Charge
5.12 V Q / C ox
C0
45 10 12
Q C ox V
V
0.05 3.52 10 8 C / cm 2
5
A
6.4 10
V fb g s Q f / C ox
It should be negative since negative charges increase V fb .
5.13 Oxide charge will change the device characteristics. Mobile charges are particularly
bad as they give the device instability. That is, as the charge moves from one side
of the oxide to the other, it will change the threshold voltage. This is undesirable.
Mobile charges are generally introduced into the oxide during wafer cleaning or
oxidation. Strict measures of cleanliness can be achieved by using ultra-clean
chemicals. And, the impurities such as sodium can be immobilized by introducing
a chlorine compound during oxidation.
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C-V Characteristics
5.14 V g V fb s Vox
2
qN aWd
W qN a
Using Vox
and s d
, we solve for Wd
C ox
2 s
W 2 W
V g V fb qN a d d
2 s C ox
V g V fb
2
2 s 4 s 4C ox 2 s C ox
qN a
Wd
2C ox
2C ox V g V fb
2
Wd
s
1 1
q s N a
Since
C ox
1
1
1
1 Wd
,
C C ox C dep C ox s
2V g V fb
1
1
.
2
C
q s N a
C ox
5.15 (a) The substrate doping is P-type since the threshold voltage is larger than V fb .
(b) Tox A
ox
C ox
10 4 0.345 1012
6.9 nm
50 10 12
(c) Vt V fb 2 B
2 s qN a 2 B
kT N
2 ln a
Cox
q ni
2 s qN a ln
Na
ni
Cox
Using V fb =-1V and V t = 0.5V and solving iteratively, we obtain
N a 3 1017 cm 2 .
(e) At position C, MOS has the minimum capacitance.
C min
1 Wd max
s
C ox
1
12.7 pF where Wd max
2 s 2kT N a
ln
qN a q
ni
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(e)
Ec
c) Vg =VT
a) Flat band
Ei
Ef
Ec
0.45eV
Ev
Substrate
Ei
Ef
Substrate Ev
Gate
Ec
b) Vg=0
Gate
Ec
d) & e) the same
Ei
Ef
Ef
Ev
Ev
Substrate
Substrate
Gate
Gate
(f) V g 0 V fb s
1
C ox
2 s qN a s
Substituting the numerical values and solving the quadratic equation, we obtain
s 0.644 s 1 0
s 0.53V .
5.16 (a) P-type, since V t > V fb .
A
(b) T ox :
V fb :
A
W dmax :
N sub : A
Vt : A
B
B
A
B
B
(Cmax= ox /T ox
(directly from the graph)
B
(1/C min = 1/C ox +W dmax / s and C ox A > C ox B)
(W dmax ~(N sub )-1/2)
(directly from the graph)
5.17 A = 100100 m2 = 10-4 cm-2
For the MOS capacitor, the field in the oxide max = 8106 =
10V
Tox
T ox = 1.2510-6 cm or 12.5 nm
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Cox
ox
A = 28.32 pF = C MOS
Tox
For P+N junction,
B
kT N d
ln
q n i
Wdep
C
s
Wdep
= 0.356 V for N d = 1016 cm-3
2 s B VB
= 8.3510-5 cm for V R = 5 V.
qN d
A = 1.25 pF = C P-N jn
C MOS / C P-N jn = 22.66
5.18 (a) The flatband voltage is 0V because the 2 silicon sides are equally doped
(i)Vg =0
(ii) Vg<0 and large
(iii) Vg>0 and large
(b)
As both sides are equally doped, the values of C 1 and C 2 will be equal.
1
C1 = C 2 C ox C dep
C dep
s
Wdep
1 1
where
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Wdep
2 s 2kT N d
ln
qN d q
ni
C dep = 34.5 nF cm-2 and C ox ox .
Tox
(c) When the left side is P-type, both the silicon layers go into accumulation and
depletion for the same type of bias.
The flat band voltage will be ( Si + E g /2 + BL ) - ( Si + E g /2 - BR ) = BL + BR
= 0.713 V
C
VGg
V
5.19
Bias
Condition
accumulation
flatband
threshold
inversion
Surface
Potential
~0
=0
= 2 B
MOS cap
(LF)
C ox
C ox
C ox
C ox
MOS Cap (HF)
MOSFET
C ox
C ox
(C ox -1 +W dep,max / s )-1
(C ox -1 +W dep,max / s )-1
C ox
C ox
C ox
C ox
5.20 (a) Accumulation -1 V < V g
Depletion -3 V < V g < -1 V
Inversion V g < -3 V
The substrate is N-type.
(b) C 0
ox
A
Tox
Thus, T ox = 205 nm.
1
(c) From the high-frequency C-V curve, C min C 0 C dep
1 1
Plugging in C min = 0.4C 0 C dep = 0.67C 0 = 54.9 pF
Now, Cdep
s
Wdep
A
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For uniform doping Wdep
kT N d
2 s 2 B
at threshold, where B
ln
qN d
q n i
4 s kT N d
ln
2
Wdep
q 2 ni
which can be solved by iteration and we get the value of N d ~ 1015 cm-3 .
Nd
(d) V fb ~ .55V + (kT/q)ln(1015/1010) = .85V
Since V fb from the plot is ~ -1V, then Q ox consists of positive charges. The
shift in V fb is 1.85V. Thus,
Q ox = 1.85V C 0 = 1.85V 82pF / 4.75x10-3 cm2 = 32 nC/cm2.
Poly-gate Depletion
5.21 (a) Using Gauss Law,
Q poly qN poly Wdpoly poly ε poly ox ε ox ,
where ox is the electric field inside the oxide and
Wdpoly
(b) poly
ox ε ox
qN poly
.
2
qN polyWdpoly
2 poly
2
2
qN poly ox2 ε ox
ox2 ε ox
.
2
2 poly q 2 N poly
2q poly N poly
(c) V g poly Vox s V fb .
Vox C ox Qs Q poly Vox
1
C ox
2q poly N poly poly
V g poly V fb s Vox poly V fb 2 B
poly
Tox
ox
poly
Tox
ox
2q poly N poly poly
2q poly N poly poly V g V fb 2 B 0
Tox
ox
2q poly N poly
Tox2
ox2
2q poly N poly 4V g V fb 2 B
2
Tox
2 ox
2q poly N poly
2
ox
2
ox
T
4
2q poly N poly V g V fb 2 B
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We know that poly >0. Also, if we set
2q poly N poly and V g V fb 2 B , then
Tox
ox
poly 2 and poly 2 2 1 1
Hence,
poly
V
(d) W dpoly
g
V fb 2 B
ox ε ox
qN poly
2
ox
2
ox
T
2
.
2 ox2 V g V fb 2 B
q poly N poly 1 1
Tox2 q poly N poly
2 poly poly
qN poly
2 poly
qN poly
2
.
2 poly Tox2
Tox
2
q
N
V
V
2
poly
poly
g
fb
B
2 ox
qN poly 4 ox2
2q poly N poly .
(e) Using the equation derived in part (c), we find poly 0.28 V .
And, using the equation derived in part (d), we find Wdpoly 2.46 nm .
N
B kT ln a 0.41 V , V fb 0.97 V , and poly s .
q ni
(f) Calculate V t using Equation 5.4.3:
Vt V fb 2 B
1
C ox
4q s N a B 0.97 0.82 0.095V 0.055 V
Using Equation 5.8.3 with poly 0.28 V ,
Qinv C ox V g Vt poly
ox
Tox
1.5 0.055 0.28V
2.20 10 6 coul / cm 2 .
(g) First, we calculate C oxe .
C oxe
ox
Tox Wdpoly / 3
1.22 10 6 F / cm 2 .
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Qinv C oxe V g Vt 1.90 10 6 coul / cm 2 .
This value is smaller than what we have found in part (f).
5.22 (a)
W dpoly
Ec
E v ,E f
poly
Ec
Ef
Ev
V g is negative, V fb is positive, st is negative, V ox is negative and poly is negative
5.23 (a) Q poly qN polyWdpoly qN poly
(b) C poly
s
Wdpoly
s qN poly
Q poly
1
1
(c) Ctotal (Cox C poly ) 1 (
2 sV poly
qN poly
2 s qN polyV poly
s qN poly
2V poly
2V poly 1
1
)
Cox
s qN poly
Threshold Voltage Expression
5.24 Vt V fb s Vox
V fb s
qN aWdep
V fb s
V fb 2B
Cox
qN a 2 ss
Cox
qN a 2 s 2B
Cox
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Chapter 6
MOSFET V t
6.1
15 -cm =1015 cm-3, E f = E v + 0.26eV oxide trap density = 8x1010 cm-2, Z = 50
m, L = 2 m, T ox = 5nm
(a) V’ fb = V fb + V. V fb = 3.1 - (3.1 + 0.86) = -0.86eV, V = Q f /C ox = 18.5 mV.
Therefore V’ fb V fb = -0.86 eV
When oxide thickness is thin, the trap charge effect can be ignored.
(b) V t = V’ fb + V ox + V s V fb + 2 B + (2 s qN a 2 B )1/2/C ox
= -0.86 +0.6 + 0.02V = -0.24V
(c) To make V t = 0.5V, one should implant boron into silicon substrate such that
V t = Q imp /C ox . Therefore ion implant dose should be
(0.5V+0.24V) C ox q = 3.21012 cm-2.
6.2
(a) Using Equation 4.16.4 and referring to Table 1-4, we find
N (GaAs)
4.7 1017
1V kT ln
0.96V .
bi Bn kT ln c
17
Nd
110
Then,
2 sbi 1
213 0 bi 1
Wdep
0.12 m .
q Nd
q
Nd
2
qN d Wdep
213 0 bi V 1
(b) Wdep 0.2 m
V
bi 1.82V .
q
Nd
213 0
A negative V g is need to increase W dep and turn-off the channel. (A metal/Ntype semiconductor Schottky diode exhibits the same forward/reverse bias
properties as an P+/N diode.)
(c) Yes. If the positive V g is kept small (say 0.5V), the forward current of the
Schottky gate maybe comparable to the subthreshold drain leakage current. A
positive V g would reduce W dep and therefore raise I ds .
(d) The channel thickness or doping concentration must be reduced so that
W dep channel thickness at V g = 0.
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6.3
kT N sub
F
, B
ln
0.47eV
2
q
ni
cm
1
V fb 2 B
2q s N sub 2 B
C ox
Eg
1
B 2 B
2q s N sub 2 B 0.09 0.61 0.52V
2
C ox
1
V fb 2 B
2q s N sub 2 B
C ox
Eg
1
B 2 B
2q s N sub 2 B 0.56 0.47 0.61 1.64V
2
C ox
1
V fb 2 B
2q s N sub 2 B
Cox
E
1
g B 2 B
2q s N sub 2 B 0.52V
2
Cox
Cox 6.9 10 7
(a)` Vt
Vt
(b) Vt
Vt
(c) Vt
Vt
(d) Vb
Vs
Vd
Vg
0V
0V
2.5V
2.5V
(e) Vb
Vs
Vd
Vg
2.5V
2.5V
0V
0V
(f) I dsat
I dsatc
I dsatb
nWC ox
(V gs Vt ) 2
2L
(2.5 (0.52)) 2
5.3
(2.5 (1.64)) 2
The transistor with the lower absolute value of threshold voltage has a higher
saturation current. That is why P+ poly-gate PMOSFETs are typically used in
IC.
(g) The ratio of the current is the ratio of the mobilities.
To find n ,
Vgs Vt 0.2/ 6Toxe 2.5 0.52 0.2V / 6 5 10 7 cm 1.07 MV / cm
and n = 250cm2V-1s-1.
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To find p ,
V gs 1.5Vt 0.25 / 6Toxe 2.5 1.5 (0.52) 0.25V / 6 5 10 7 cm 1.01MV / cm
and p = 63cm2V-1s-1.
I dsat ( c )
I dsat ( a )
p 1
n 4
Basic MOSFET IV Characteristics
6.4 (a) Due to the highly doped regions nearby, transistor C-V always approaches C ox
in inversion. Hence, it is impossible to determine the frequency. Either high or
low frequency could have been used.
(b) Since V t >V fb , this is a NMOS.
(c) From the I d -V g curve, V t is 0.55V. More precisely,
V
V
W
I d C ox Vds V g Vt ds 0.55 Vt ds 0
2
2
L
V t 0 .5 V
W
(d) Slope of curve I d - V g line C ox Vds 5 10 3 1 .
L
Vds 0.1V
From the CV curve,
W
F
C ox WL 1pF C ox
10 -4
L
cm 2
cm 2
5 10 3
Thus,
500
Vs
0.1 10 4
(e) Vdsat V g Vt
I dsat
Vg
V dsat
I dsat
C oxW
2L
(Vdsat ) 2 0.025
1V
0.5V
6.25mA
A
2
.Vdsat
2
V
2.5V
2V
100mA
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I d (mA)
V g =2.5V
100
V g =1V
6.25
0.5
6.5
2
V d (V)
(a) For V gs =4V, V dsat ~ 3V=(V gs -V t ) and V t =1V
(b) I dsat n C oxW / 2 L (V g Vt ) 2
C ox 3.45 10 7 F / cm 2
2 I dsat
n
361cm 2 / Vs
W
Cox (Vg Vt ) 2
L
(c) At V gs =3V, V dsat =(3-1)V=2V
I dsat 361 3.45 10 7 (10 / 2) (3 1) 2 / 2 1.25 10 3 A
6.6
(a) At saturation, V d = V dsat = V g -V t . V dd =2V. Therefore the transistor is in
saturation mode when V g <2.5V. I dsat = 125(V g -0.5)2A. When V g >2.5V, the
transistor is in linear region with I d = 500(V g -1.5) A.
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40
SQRT(Idsat, A)
35
30
25
20
15
10
5
0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Gate Voltage, Vi (V)
(b) & (c) Transconductance: solid line, Output Conductance: dotted line
gd , gm
gm
500 A/v
250 A/v
gd
0.5
6.7
1.5
2.5
3.0
Vi
(a) V gs Vt 2V
V gs 2.5V
(b) Qn C ox (V gs Vt Vc ) 0 (Pinch-Off)
(c) I ds @ Vds 4V
Vdsat V gs Vt 3V
(in saturation)
I ds (V gs Vt ) 2
I ds 10 3
32
2.25 10 3 A
2
2
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(d)
C
Cox
(Same for high and
low frequencies)
Vfb
6.8
Vg
Vt
kT N a
ln
0.297V
q
ni
(a) B
ox
C ox
7.08 10 8 F
t ox
Eg
V fb Si ( Si
cm 2
B ) 0.857 V
2
V g V fb Vs Vox Vt V fb 2 B
(b) I dsat
nCoxW
2L
V
2 s qN a 2 B
C ox
0.064V
Vt 1.21mA
2
g
(c) Since V d is less than (V g -V t ), it is in the linear region.
Id
gd
C oxW
L
V
g
V V
t
d
Vd2
2
C oxW
I d
Vg Vt Vd 1.17mS
V D
L
(d) Since V d is less than (V g -V t ), it is in the linear region.
2
C oxW
Vd
Id
V g Vt Vd
L
2
gd
C oxW
I d
Vd 1.13mS
V g
L
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Potential and Carrier Velocity in MOSFET Channel
6.9
I d Q n n
x
0
dVc
dV
W (V g Vt Vc )C ox n c W
dx
dx
Vc
Id
dx (V g Vt Vc )dVc
n C oxW
0
I d x /( n C oxW ) (V g Vt )Vc 1 / 2Vc2
Solving this quadratic equation of V c , we get
Vc ( x) (V g Vt ) (V g Vt ) 2
2I d x
n C oxW
Choosing “-” so that V c (0)=0,
2I d x
Vc ( x) (V g Vt ) 1 1
n C oxW (V g Vt ) 2
W
2 x n C ox
(V g Vt ) 2
2L
(V g Vt ) 1 1
n C oxW (V g Vt ) 2
(V g Vt ) 1 1 x / L .
6.10 (a) I ds WCoxe (Vgs mVcs Vt ) es d Vcs dx
x
I
0
Vcs
ds
dx WC oxe s (V gs mVcs Vt )dVcs
0
I ds x WC oxe s (V gs Vt
mVcs
)Vcs
2
Equating the expression above with
I ds
W
m
C oxe s (V gs Vt Vds )Vds ,
L
2
we get
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mVds
mVcs
x
)Vcs
V gs Vt
Vds (V gs Vt
L
2
2
x
2
mVcs 2(V g Vt )Vcs (2(V g Vt ) mVds )Vds 0
L
Solving the quadratic equation, we get
Vcs
Vcs
V gs Vt
m
V gs Vt
m
(V g Vt ) 2 m
(1 1
x
(2(V g Vt ) mVds )Vds
L
m
x
)
L
x
(b) Qinv ( x) C oxe (V gs mVcs Vt ) C oxe V gs Vt (V gs Vt )(1 1
L
x
x
C oxe V gs Vt 1 1 1 C oxe (V gs Vt ) 1
L
L
Vg Vt 1 1 1 Vg Vt 1
dVcs
ε( x)
(c)
dx
m 2 1 x L
2mL 1 x
L
L
n V g Vt 1
dVcs
n ε( x)
( x) n
dx
2mL
1 x
L
x n V g Vt
1
(d) WQinv n ε W .C oxe (V gs Vt ) 1 .
L
2mL
1 x
L
WC oxe n
Vgs Vt 2 I dsat
2mL
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(e)
V cs
x
IV Characteristics of Novel MOSFET
dVc
dV
W (V g Vt Vc )C ox ( x) n c W
dx
dx
dV
(V g Vt Vc ) 2ox n c W
dx
Ax B
6.11 (a) I d Qn n
L/2
I d ( Ax 2 B)dx
L / 2
Vds
(V g Vt Vc ) ox nWdVc
0
A
I d [ x 3 Bx] LL/ 2/ 2 ox nW [(V g Vt )Vc 1 / 2Vc2 ]V0ds
3
W ox n
Id
[(V g Vt )Vds 1 / 2Vds2 ]
2
L AL
B
12
I
(b) Vdsat Vds @ d |Vgs 0
Vdsat V g Vt
Vds
(c) It suggests a large W dmax . Vox Qn / C ox
6.12 (a) I d Qn n
dVc
dV
W ( x) (V g Vt Vc )C ox n c W ( x)
dx
dx
L
Vds
0
0
I d / W ( x)dx
(V g Vt Vc ) n C ox dVc
I d ln[(W0 L) / W0 ] n C ox [(V g Vt )Vds 1 / 2Vds2 ]
Id
n C ox
ln(1 L / W0 )
[(V g Vt )Vds 1 / 2Vds2 ]
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(b) Vdsat Vds @
I dast
I d
|V 0
Vds gs
n C ox
ln(1 L / W0 )
Vdsat V g Vt
(V g Vt ) 2
2
CMOS
6.13 (a) V fb,NMOS = -(E g /2) – (kT/q * ln(5e15/1e10)) = - 0.55 -0 .4 = -0.95V
V fb,PMOS = -0.55 + 0.4 = -0.15V
Not symmetrical
(b) V fb,NMOS = 0.55 - 0.4 = 0.15V
V fb,PMOS = 0.55 + 0.4 = 0.95V
Not symmetrical
(c) Since V ox and V s will be symmetrical, I would use a mid-gap gate material such
as tungsten.
So the workfuction will be 4.05 eV + E g,Si /2 = 4.6eV. However, processing
issues makes tungsten (or any metal gates for that matter) a challenge to
implement.
(d) In the same process, the NMOS and PMOS will have same oxide thickness. If
the substrate doping levels for n and p flavors are the same, then I would use P+
gates for PMOS devices and N+ gates for NMOS devices. In this way, the
flatband voltages will be symmetrical and the resulting |V t | small.
6.14 (a) PMOS, N-type substrate:
N
n kT ln d 0.38V
ni
Eg
Vfb m si
n 4.1 4.05 0.55 0.38 0.12V .
2
NMOS, N-type substrate:
N
p kT ln a 0.42V
ni
Eg
Vfb m si
p 4.1 4.05 0.55 0.42 0.92V
2
(b) C ox
ox
t ox
6.9 10 7
F
cm 2
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PMOS:
Vt V fb 2 n
1
C ox
2 s qN d 2 n 0.12 0.76 0.10 0.98V
NMOS:
V fb 2 p
1
C ox
2 s qN d 2 p 0.92 0.84 0.24 0.16V
(c) The threshold voltage must be changed by
Qimpl
Vt
0.82V .
C ox
Hence,
C
Qimpl 5.7 10 7
.
cm 2
6.15 I ds WC oxe (V gs mVcs Vt )
L
0
Vds
I ds dx
0
I ds L I ds
Vds
sat
WC
oxe
1
dVcs
dx
ε sat
s V gs mVcs Vt I ds
WC oxe s (V gs Vt
ε sat dVcs
m
Vds )Vds
2
V
L ds
ε sat
V
W
m
C oxe s (V gs Vt V ds )V ds 1 ds
L
2
Lε sat
I ds WC oxe s (V gs Vt
I ds
s dVcs dx
m
Vds )Vds
2
I ds (Long channel)
.
1 V ds ε sat L
6.16
A
B
C
D
NFET Operation Mode
Cut-off
Saturation
Linear
Linear
PFET Operation Mode
Linear
Linear
Saturation
Cut-off
A: Vgs<Vth for NFET, therefore it is cut off. For PFET |Vgs| > |Vth| and
|Vds|<|Vdsat| (|Vds|~0V, |Vdsat| ~ 1.05V), so it operates in linear mode.
B: For NFET Vgs > Vth and Vds>Vdsat (Vds~1.75V, Vdsat ~ 0.3V), so it operates
in saturation mode. For PFET |Vgs| > |Vth| and |Vds|<|Vdsat| (|Vds|~0.25V, |Vdsat|
~ 0.6V), so it operates in linear mode.
The answers to C and D can be worked out through the same procedure.
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6.17 (a)
cut-off
{ NMOS
PMOS linear
Vo
saturated
{ NMOS
PMOS linear
A
B
saturated
{ NMOS
PMOS saturated
C
linear
{ NMOS
PMOS saturated
linear
{ NMOS
PMOS cut-off
D
0
VTn
Vx
(VDD+VTp) V
DD
Vi
(b) At the point B where V i =V x , the NMOS is just becoming saturated from the
linear region. Since NMOS is in the linear region
V x Vtn 2
I dn K N V x Vtn V x Vtn
2
Since PMOS is saturated
K
2
I dp P Vdd V x Vtp
2
But I DN = I DP
2
Vx 1 35
2
2
5 Vx 1
40V x 1
2
2
40(V x - 1)2 = 40(4 - V x )2 V x = 2.45 V
Thus,
Point
A
B
C
D
V i (V)
1
2.45
2.45
4
V o (V)
5
3.45
1.45
0
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Body Effect
6.18 For a P-channel MOSFET, we have
2 s qN d (2 B Vbs )
Vt V fb 2 B
C ox
2 s qN d
( 2 B Vbs 2 B )
C ox
(a) For 100 nm oxide, C ox = 3.4510-8 F/cm2.
If V bs = 5V, V t = -0.8V.
By iteration, using initial guess of B = 0.3V, we obtain
N d = 8.91014 /cm3 and B = 0.284V.
V t
(b) If V sb is -2.5 V, V t = -0.497V.
V t = -1.5 - 0.497 = 2.0 V
Velocity-Saturation Effect
6.19 In all 3 cases, use the general equation I=WQ inv v drift .
Case A:
The NMOS is in the triode region.
On source side, Q inv =C ox (V g -V t ) = 138e-9(5-.7) = 593 nC/cm2.
So drift = I/(WQ inv ) = 1.5e-3/(15e-4 593e-9) = 1.7 x 106 cm/sec.
On drain side, Q inv = C ox (V g -V t -V d ) = 138e-9(5-.7-.5) = 524 nC/cm2.
Thus, dr = 1.5e-3/(15e-4 524e-9) = 1.9 x 106 cm/sec.
Case B:
The NMOS enters saturation region.
On source side, dr = 3.75e-3/(15e-4593e-9) = 4.2 x 106 cm/sec.
On drain side, the electron velocity is saturated.
Thus, dr = sat = 8 x 106 cm/sec.
Case C:
Similar to case B.
On source side, dr = 4e-3/(15e-4593e-9) = 4.5 x 106 cm/sec.
On drain side, dr = sat = 8 x 106 cm/sec.
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6.20
T ox
V dsat
I dsat
W
L
No change
Vt
Vg
Reducing T ox means smaller V t => larger V dsat (1/(V g -V t ) + 1/(E sat L))-1 & larger
I dsat (Q inv C ox ).
Reducing W has no effect on V dsat and decreases I dsat since I dsat = WQ inv v sat .
Reducing L reduces V dsat (as discussed in lecture) and increases I dsat . If you want to
consider very short-channel length devices (L => 0), then essentially I dsat is
independent of L.
Reducing V t => larger V dsat & larger I dsat .
Reducing V g => smaller V dsat & smaller I dsat .
V
6.21 I d s C oxW V gs Vt m ds Vds
2
WC ox V gs Vt mVds Vsat
WC ox V gs Vt mVds s
V
L ds
ε sat
ε sat
2
V
V
V gs Vt m ds Vds V gs Vt mVds ε sat / 2 L ds
2
ε sat
ε L
V gs Vt mVds sat V gs Vt
2
mVds
Vds
2
V gsVds VtVds V gs Vt mVds ε sat L
Vds V gs Vt mε sat L (V gs Vt )ε sat L
(V g Vt )ε sat L
1
m
Vds
Vgs Vt mε sat L (Vgs Vt ) ε sat L
6.22 I ds
1
1
W
W
m
C oxe ns V gs Vt
C oxe ns V gs Vt Vds Vds
L
2
L
Vds
V
1 1 1
1 ds
ε sat L
Vds2 ε sat L Vds
1
Vdsat
m
2
1
m
V gs Vt ε sat L
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6.23 (a) We know that
1
2
V
V
g
t
1
Vdsat ε c L 1 2
E c L
1
2
4
c L = 0.1 V Vdsat 0.1 1 2
1 = 0.54 V
0.1
(b) c L = 10 V Vdsat
1
2
4
10 1 2 1 = 1.83 V
10
(c) We know that
10fF
C Z
2
.
I dsat n0 ox Vdsat and Cox Z
L
2L
n0 10fF
7mA
0.54 2
2 L2
n0 = 480 cm2 V-1 s-1
6.24 V dsat = sat L(V g -V t ) / (V g - V t + sat L)
What is sat ? 2v sat / s = sat . s is given by the universal mobility curve.
At T ox =60A, (V g +V t +0.2)/6T ox = eff = .9MV/cm.
From the curve, s ~ 250 cm2V-1s-1.
This yields sat ~ 2(8x106)/250 V/cm = 6.4x104V/cm. Plug this back into the
expression for V dsat to get L ~ 0.19um.
I dsat /W = Q inv v sat = C ox (V g -V t -V dsat )v sat
= (3.9 o /60e-8) (2.5-.5-.75) 8106 = 575uA/um width.
Note: You will often find in literature that the saturation current is stated in units
of uA/um instead of amperes. Also, notice that the Q inv at V c =V dsat is not zero.
That is, I dsat is limited by velocity saturation instead of pinch-off.
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6.25 (a)
1
V gs Vt
mE sat L
2
V gs Vt mE sat L
2mv sat L
ns
2 1.2 8 10 6 cm / s 1 10 5 cm
0.64V
300cm 2 / V s
(b)
V gs Vt mE sat L
L
Vgs Vt
mE sat
ns Vgs Vt
2mv sat
300cm 2 / Vs 0.2V
3.13 10 6 cm 31.3nm
2 1.2 8 10 6 cm / s
Effective Channel Length
6.26 (a) For very small V ds ,
V
L
Rchannel ds
I ds s C oxW (V g Vt )
In a short-channel device, S/D resistance can seriously degrade saturation
current. Note that series resistance is worse for higher currents because R channel
is the lowest under these bias conditions.
(b)
Rtotal Rs Rd Rchannel Rsd Rchannel
Leff
Lgate L
)
Rsd
Rsd (
s C oxW (V g Vt )
s C oxW (V g Vt )
Think of R total as the y-value, L gate as the x-value, and ( s C ox W(V g -V t ))-1 as the
slope. This fits nicely into the standard equation of the line: y = mx + b. You
can choose devices with several gate lengths and measure the current from these
devices at discrete gate voltages. Remember, that you are assuming V ds is small
(<100mV) in these measurements.
From the current, you can plot R total versus L gate . One sample data line is taken
with the same V g at different gate lengths. For example, if you measure your
current at 5 different V g ’s, you will get 5 separate curves. Ideally, all the lines
will intersect at the same point on your plot. This intersection point occurs at
L gate = L and R total = R sd .
In practice, it is not always straightforward to make such a plot. For instance,
V t can be difficult to determine accurately. Also, there is a strong dependence
of mobility on gate voltage for thin-oxide MOSFETs. It’s a good idea to check
your data by taking measurements at several different V g instead of at 2 or 3
gate voltages.
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(c) I dsat k(Vgs -I dsat Rs -Vt ) , where k is a constant of proportionality
I dsat ( 1 kRs ) k(Vgs -Vt ) I dsat 0 , notice here that k I dsat0 /(V gs - Vt )
I dsat I dsat 0 / ( 1 kRs ) I dsat 0 / ( 1 I dsat 0 Rs /(V gs -Vt ))
(d)
sat
= (V gs +V t +0.2)/6T ox = 1.1 MV/cm.
s ~225cm2V-1s-1 is picked out from the universal mobility plot.
sat = 2v sat /( s ) = 7x104V/cm.
I dsat0 = (long channel I dsat ) / (1+ (V gs -V t )/( sat L))
= 1.6mA / (1 + (1.1/.7)) = 0.622mA.
Plug in 0.622mA into the expression derived in part c and get the following:
@ R s = 0ohms, I dsat = .62mA
@ R s = 100ohms, I dsat = .59mA
@ R s = 1000ohms, I dsat = .40mA
6.27 (a) Choose three transistors with same channel width, Z, and different channel
length, L 1 , L 2 ,and L 3 . Measure I dsat at saturation condition for the 3 transistors
to
get I d1 , I d2 , and I d3 . Solve the 3 equations to get C ox , and L eff .
(b)L = L- L eff when gate oxide thickness is 4.5nm. Z =10 m, = 300 cm2/Vs.
Using approach of (a), L 0.1 m.
3.0
-1
Idsat = const(Ldrawn - L)
Idsat
-1
2.5
2.0
1.5
1.0
0.1 m : L
0.5
0.0
0
1
2
3
4
5
Channel Length(m)
(c) 2.59mA(L 1 - L eff ) = ZC ox (V gs -V t ), V t = 0.5V.
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Chapter 7
Subthreshold Leakage Current
7.1 a) Cox=3.45E-13/11E-8 = 3.138uF/cm2
Vt = -1.04 + 0.95 + Qdep/Cox = -1.04 + 0.95 + 0.180 = 0.109V
b) S = 60(1+Cdep/Cox) = 60(1+2.953E-7/3.138E-6) = 65.64mV/dec
c) 1/S = [log(W/L*100nA) – log(I leakage )]/Vt
I leakage = 121nA
Field Oxide Leakage
7.2 a) Cox=3.45E-13/0.3E-4 = 1.151E-8F/cm2
Vt = -1.01 + 0.92 + Qdep/Cox = -1.01 + 0.92 + 33.96 = 33.87V
b) S = 60(1+Cdep/Cox) = 60(1+2.119E-7/1.151E-8) = 1164.6mV/dec
c) 1/S = [log(W/L*100nA) – log(I leakage )]/Vt
I leakage = 2.81E-26nA
Vt Roll-off
7.3
log(I ds )
L=0.2um, N a =1E17
L=0.2um, N a =1E15
L=1um, N a =1E15
L=1um, N a =1E17
Vg
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Trade-off between I off and I on .
7.4 i) Larger V t : Decrease I off and decrease I on
ii) Larger L : Decrease I off and decrease I on
iii) Shallower junction : Decrease I off and somewhat decrease I on
iv) Smaller V dd : Decrease I off and decrease I on
v) Smaller T ox : Decrease I off and increase I on
A smaller T ox contributes to leakage reduction and increases the precious I on .
7.5
There is a lot of concern that we will soon be unable to extend Moore’s Law. In your own words
explain this concern and the concerns for high I on and low I off .
(a) Answer this question using 1 paragraph of less then 50 words. : Methods of manufacturing
very small objects such as photolithography will run into limitations. Transistors may not be able
to provide circuit functions with high speed and low power. Specifically, means of increasing
speed (I on ) tend to worsen leakage (I off ). Other relevant comments include the following. V t must
not be too low because of subthreshold current but small L reduces V t . Since I on (speed) is
roughly proportional to V gs -V t (V dd -V t ), I on and speed can not be raised by lowering V t
especially because V dd needs to be reduced to reduce the power consumption.
(b) Support your description in (a) with 3 hand drawn sketches of your choice. : Possible choices
are Log(I ds ) vs. V gs , Vt vs. L and Vds, Ion vs. Ioff, gate leakage vs. Tox, etc.
(c) Why is it not possible to achieve high I on and small I off by picking optimal T ox , X j , W dep etc?
Please explain in your own words.: Decreasing T ox is good for Ion but bad for I off . So is
increasing X j . W dep is basically fixed by the choice of T ox and V t .
(d) Provide three equations that help to quantify the issues discussed in part (c). : Some relevant
equations include Eq. (6.9.14), Eq.(6..10.1), (7.2.8), Eq.(7.3.3), Eq.(7.5.2).
7.6 (a) Final equation:
ld 3
ox Vt V fb 2B
qN a
Xj 3
X j Tox Toxe 2 s 2B
ox Vt V fb 2B
(b) The minimum acceptable L is several times of ld. In order to support the reduction of L at
each new technology node, ld must be reduced in proportion to L. According to the
derived equation, in order to reduce the minimum acceptable channel length,
Vt V fb 2B should be increased by gate workfunction engineering, Xj can be
decreased, and the oxide permittivity can be increased .
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7.7 (a) A smaller W dep
Reduces DIBL (V t is not as low) decreases I off
(b) I dsat
W
2
Coxe ns Vgs Vt where m 1 1 3Toxe / Wdmax
2mL
Reducing W dep increases m which decreases I dsat .
MOSFET with Ideal Retrograde Doping Profile
7.8 (a)
3.1eV
V ox
P+
T rg
Ec
Eg
3.1eV
Vt
Ef, Ev
Ec, Ef
(b) The figure above assumes that the gate is N+ polysislicon and the substrate (below the
undoped layer) is P+ silison. In that case , it can be seen that V t = V ox . In general, V t is
determined by Eq. (5.2.2) and
V t = V fb + φ st + V ox
Note – You need to apply a positive gate bias to reach inversion. Thus the Fermi level of
the poly gate is below the Fermi level of the substrate and the difference is V t .
(c) ε ox
ε
ox
ox
= ε si
ε
si
Eg / q
Vox
si st si
Tox
Trg
Trg
The last equality is the result of a further approximation for st . Using the result of part (b),
Vt V fb st st
si Tox
ox Trg
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4
(d) First derive Eq. (7.5.2) using Eq. (7.5.1) and eq. (5.5.1). By comparing Eq. (7.5.2) and
Eq. (7.5.3) one concludes that, for the same Vt, the depletion region width of the “ideal
retrograde doping” MOSFET (Trg) is half of that of a MOSFET with uniformly doped
substrate.
(e) A small W dep reduces the V t roll-off caused by DIBL which in effect decreases I off .
(f) I dsat
W
2
Coxe ns Vgs Vt where m 1 1 3Toxe / Wdmax
2mL
Reducing W dep increases m which decreases I dsat .
A smaller I dsat causes a longer inverter delay. However, it is important to reduce W dmax
so that L can be reduced to increase the device integration density and to reduce chip cost.
I dsat and speed still get improved due to the reduction in L.
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Chapter 8
Energy Band Diagram of BJT
8.1
(a) & (b)
For the given doping concentrations, one computes E f -E i = -0.521 eV, 0.419 eV,
and –0.299 eV in the emitter, base and collector, respectively. Also with
N aE >>N dB , the E-B depletion width will lie almost exclusively in the base.
Likewise, the majority of the C-B depletion width will lie in the collector.
Energy Band Diagram
Ec
x
Ei
Ef
E
Charge Density
x
x
(c) The built-in potential between the collector and emitter is
VCE
kT N aE
ln
q N aC
5 1018
0.026 ln
0.221 V .
15
1 10
(d) W Wn xnEB xnCB
1
x nEB
2
Si VbiEB
qN dB
x nCB
2
N aC
Si
VbiCB
qN dB N aC N dB
2
0.112m
1
2
0.01m
Therefore,
W 2.878m .
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qN
( E B) dB x nEB 1.72 10 5 Vcm 1 .
Si
qN
Electric Field max (C B ) dB x nCB 1.54 10 4 Vcm 1 .
Si
(e) Electric Field
max
(f)
Ec
Energy Band Diagram
Ei
Ef
E
x
Charge Density
x
x
IV Characteristics and Current Gain
8.2
From Eq. 8.4.1 and Eq. 8.4.2,
I
I
F C IB C
F
IB
I C F I E F I C I B F I C F I B .
Substituting I C / F into I B , we obtain
I
IC F IC F C 1 F F .
F
F
Solving for F yields
IC
F
.
1F
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8.3
(a) N+: Emitter
E f,n+
E f,p
N: Collector
P: Base
V BE is forward biased and
V BC is reverse biased.
(b)
P
EMITTER
E f,n
N
P
BASE
COLLECTOR
The P,N,P on the diagram refer to the minority carrier type in each region. The
horizontal dotted lines refer to the equilibrium minority concentration (i.e. p N0 ,
n P0 ). The remaining dotted curves correspond to the excess minority carrier
concentrations.
Assumptions made here: W E & W B are shorter than the diffusion lengths of the
holes & electrons respectively, resulting in a linear decay of excess minority
carriers in the emitter and base. You should also notice that the scale for the yaxis differs for each region.
(c) Base current consists of injection of holes into the emitter and recombination
with a very small part of the collector current (remember that I E ~ I C ). The
collector current consists almost entirely of electrons emitted from the forwardbiased BE junction which travel across the CB junction. Incidentally, an easy
way to remember how a BJT works is to associate the names emitter and
collector with the physical emission and collection of the minority electrons in
the base.
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8.4
x
nB ( x) nB (0)1 and
xB
x'
pE ( x' ) pE (0' )1 .
xE
In the base region,
dn qDB nB (0)
.
J n ( x) qDB
xB
dx
In Emitter region,
dp qDE pE (0)
J p ( x) qDE
.
xE
dx
n B (0) = 1013 cm-3 , p E (0’) = 1011 cm-3, and p C (x B +) = p CO = 105 cm-3
D B = 20.8 cm2s-1, D E =1.8 cm2s-1, and D C =11.9 cm2s-1.
(a) J n , B ( x 0) qDB nB (0) / xB 0.666 A / cm 2 ,
J p , E ( x' 0' ) qDE pE (0) / xE 3.6 10 4 A / cm 2 ,
and
J p ,C ( x 0.5m ) qDC pCO / xC 8.7 1010 A / cm 2 .
Therefore,
J TOTAL , BE 0.666 A / cm 2 3.6 10 4 A / cm 2 0.666 A / cm 2
and
J TOTAL , BC J n ( xB ) J p ( xB ) .
For short diode approximation, we assume J n (x B -) = J n (x=0). More accurate
relationship is J n (X B -) = T J n (x=0). However, since T 1, we can still say
J n (X B -) = J n (x=0).
Hence,
J TOTAL , BC 0.666 A / cm 2
J n , E J TOTAL , BE J p (x' 0' ) 0.666 A / cm 2
J n , C J TOTAL , BC J p ( x' xB ) 0.666 A / cm 2
J p , E J p , B 3.6 104 A / cm 2
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Jn
1
1 0 -2
1 0 -4
1 0 -6
Jp
E
C
B
1 0 -8
m
1 0 -1 0
0 .8
1 .3
3 .5
1
0.99875 LB 10 m
1 WB2
1
2 L2B
0.666
J n ( 0)
E
0.999325
J n (0) J p (0) 0.666 3.6 10 4
(b) T
F
8.5
F
519
1F
T E
(a) To improve the emitter injection efficiency, and to reduce the back injected
carriers from B-E.
(b) “Small” means WB LB , typically 1m
(c) W B would become a larger percentage of W B . Therefore, it would increase the
slope of output characteristics.
(d) At very small values of I C , the recombination current (excessive base junction
current) in the E-B depletion region becomes a less significant part of I B as I C
increases.
(e) At larger values of I C , due to the high level injection in base, I C does not
increase exponentially as it does in the moderate level injection region.
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high level
injection in base
IC (A )
,IB(A)
10 -2
10
-4
10
-6
10
-8
10
-10
10
-12
IC
IB
F
Excessive base junction current
0.2
0.4
0.6
0.8
V
1.0
1.2
BE
(f)
Region
VEB
VCB
Active
Saturation
Cutoff
+
+
-
+
-
Schottky Emitter and Collector
8.6 (a) As you will find out in the latter part of this problem the injection of majority
carrier of the semiconductor into the metal is much higher than the injection of
minority carrier into the semiconductor region from the metal. This would make
the emitter efficiency in the BJT very small! Hence, it would not be desirable to
use a metal as an emitter in a BJT.
(b) We know that the hole diffusion current is given by
2
D n
I diff I diff 0 e qVA / kT 1 qA P i e qVA / kT 1 .
LP N d
From Section 4.17 and 4.18,
I te I te0 e qVA / kT 1 AKe q B / kT e qVA / kT 1 .
Noting that
DP
LP
DP
p
kT / q
p
p
,
we have
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I diff I diff 0
I te
I te0
q
kT / q p
p
2
ni
Nd
Ke B / kT
0.0259437
1.6 10
10 6
0.72 / 0.0259
140e
19
1/ 2
10 20
16
10
5.05 10 7 .
(c) A Schottky base-collector junction (the collector being the metal) would be
functional in a BJT. The energy diagram of the base-collector junction would be similar
to Fig. 4-34(b). It would be effective collecting the electrons arriving at this junction from
the P type base to the metal. The field at the Schottky junction sweeps the electrons into
the metal collector just as in the PN base-collector junction shown in Fig. 8-1(b).
Gummel Number and Gummel Plot
8.7 (a)
=J C /J B =100.
(b) Intercept of J C is 10-10 A/cm2= qni2 Dn / N BWB
qni2 Dn
N B 10
8 1016 cm 3 .
10 WB
(c) Peak concentration
qV
ni2 kTBE
e
1017
NB
qVBE
e kT 8 1013
V BE ln(8 1013 ) 26mV 832mV .
(d) n
WB2
(0.2 10 4 ) 2
2 10 11 sec.
2 Dn
2 10
Ebers-Moll Model
8.8 (a)
Consider that n’ = n P0 (eqVa/kT-1) and p’ = p N0 (eqVa/kT-1). Next, take a look at the
BC junction ,
n’/p’ = N B /N C
(1)
Similarly, at the BE junction,
p’/n’ = N E /N B
(2)
Multiply (1) and (2) to get N E /N C = (8/4)(10/2) = 10.
Thus, N C = 0.1N E = 1017cm-3.
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(b) The BJT is operating in saturation because both BE and BC junctions are
forward-biased and the resulting minority carrier concentrations are larger than
the equilibrium values.
(c) The stored minority charge is equal to the area under the curve in the base.
Aq
p' x 1m p' x 2m WB
2
10 5 1.6 10 19 0.5 14 1014 10 4 1.12 10 13 coul .
Q
(d) I E I nE I pE
Aq[ Dn
n' ( x 1m) n' ( x 0 m)
p' ( x 1m) p' ( x 2m)
Dp
]
WE
WB
2 1014
6 1014
mA 0.192 mA .
10 5 1.6 10 19 30
10
4
4
1 10
1 10
DW N
10 10 4 N E 1
5
(e) B E E
1 5 1.7 (Not much gain.)
4
DEWB N B 30 10 N B 3
3
8.9 If the NPN BJT is biased at the boundary between active mode and saturation mode,
then forward-biased emitter-base junction (V BE >0) and unbiased collector-base
junction (V BC =0). So I R =0.
(a) At the given operating point, we can simplify the Ebers-Moll model as follows:
IE
I F I ES ( e qVBE / kT 1)
FIF
IC
E
IB
B
(b) Since I C I B I F 0 I B I F (1 F )
I B I ES ( e qVBE 1)(1 F )
VBE
IB
kT
ln
q I ES (1 F )
V EC V BC V BE 0
kT I ES (1 F )
IB
kT
ln
ln
.
q I ES (1 F ) q
IB
Drift-Base Transistors
8.10 (a) I B is independent of changes in base parameters. I C is dependent on base
parameters. You should convince yourself that this is the case by referring to
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the current equations in the reader. If a SiGe base is used, I c increases as a
result of the base bandgap narrowing. Coupled with a graded base that shortens
base transport time, SiGe-base BJTs are a simple and attractive alternative to
conventional Si-base BJTs.
* This solution ignores the case of increased hole barrier between base and
emitter. If you wish to include this effect, then I B (SiGe) is increased by eEg/kT
over I B (Si)
(b) In this case, n iB (SiGe) varies along the base region. This requires that we do an
integration to find J c0 (SiGe).
qDB ni2
J c0 (SiGe) =
NB
=
1
WB
E g ,SiGe x
exp(
)dx
0
kT
WB
E g ,SiGe / kT
qDB ni2
N BWB 1 exp( E g ,SiGe / kT )
.
Divide J c0 (SiGe) by J c0 (Si) and you get the following:
(SiGe)/ (Si) =
8.11
(a)
E g ,SiGe / kT
1 exp( E g ,SiGe / kT )
= 4.
Find where N dE (x) = N aB (x) to obtain the first junction.
20
10 e
x
0.106
4 10 e
18
x
0.19
x1 0.77 m .
To obtain the second junction, equate N aB (x) to the background concentration.
4 1018 e
x
0.19
5 1019
x2 1.27 m .
Therefore, the base width is x 2 -x 1 = 0.5 m.
(Please note that the depletion widths have been ignored in this case. In
general, you must subtract the depletion region widths in the base in both the
junctions from the metallurgical base width.)
(b) Base Gummel Number:
x2
x1
N aB ( x)dx 4 1018 0.19 10 4 exp x
0.19
x2
x1
cm 2 1.23 1012 cm 2 .
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From Eq. 8.2.12, the base Gummel number is the number above divided by the
electron diffusivity in the base, which we shall assume to be around 30 cm2/s.
The result is 41010 s-cm-4
Emitter Gummel Number:
x1
0
N dE ( x)dx 10 20 0.106 10 4 exp x
cm
0.106
x1
2
0
1.06 1015 cm 2 .
From Eq. 8.3.2, the emitter Gummel number is the number above divided by the
hole diffusivity in the base, which we shall assume to be around 2 cm2/s. The
result is 21014 s-cm-4
(c) Since the doping level is not constant, we use the average doping densities to
estimate the diffusivities.
Average base doping density:
GN B 1.23 1012
cm 3 2.46 1016 cm 3 .
4
WB
0.5 10
Average emitter doping density:
GN E 1.06 1019
1.38 1019 cm 3 .
xE
0.77 104
Average electron diffusivity in the base:
D' nB n 2.46 1016 kT / q 1150 0.026 cm 2 s 1 29.9 cm 2 s 1 .
Average hole diffusivity in the emitter:
D' pE p 1.38 1019 kT / q 70 0.026 cm 2 s 1 1.82 cm 2 s 1 .
1
1
GN B D' pE
GN E D'nB
1
0.99993 .
1.23 1012 1.82
1
1.06 1015 29.9
(d) Diffusion current:
J diff q qD dp
dx
D dn
d ln n
D
.
n dx
dx
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p N B 4 1018 e x / , 0.19 m
D
( x direction) .
Drift current:
J drift q p pE qp J diff
q
q
D and D p E E D
kT
kT
p
Therefore,
Ebi
kt
0.06 V
3.16 105 V / m .
q 0.19 m
Note: E bi should be “-x” direction so that diffusion current and drift current will
balance.
Ec
Ef
Ev
Kirk Effect
-
8.12
Base
N
Collector
N+
Collector
Original
Base Witdth
x
Effective
Base Witdth
Depletion
Layer, W dep
Clearly, W B_Effective = W B_Original + (W C -W dep ). W B_Original and W C are assumed to
be known. So, in order to find W B_Effective , we need to calculate W dep .
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IC
where v sat is the saturation velocity. The length of the depletion
AE v sat
region becomes
2 s VBC i
Wdep
.
IC
qN C
AE v sat
qN C
Therefore,
WB _ Effective WB _ Original
WC
2 s V BC i
.
IC
qN C
AE v sat
Charge Control Model
8.13
The equation describing the system is
dQF (t )
Q (t )
I B (t ) F .
dt
F F
Since I B (t) = 0 for t<0 and I B (t) =I B0 for t≥0, the equation becomes
dQF (t )
1
Q F (t ) I B 0
dt
F F
with the initial condition Q F (0) = 0. Solving this equation yields
QF (t ) F F I B 0 1 e t / F F .
Hence,
I C (t ) QF (t ) / F F I B 0 1 e t / F F .
8.14
dQB
QB
t / F
iB
QB A Be
dt
F
Boundary Conditions: F
QB () A F iB 2 , QB (0 ) F iB1 F iB 2 B
Hence, B= F (i B1 -i B2 )
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Therefore,
QB F iB 2 F iB1 iB 2 e t / F
Q
iC B iB 2 iB1 iB 2 e t / F ,
F
F 100, F 1ns (Chracteristic time T )
1F
And,
iC 1 mA (9 mA)e t / 1ns for t 0, and
iC 10 mA
for t 0.
iC(t)
T=1s
T = 1ns
10mA
1mA
t
Cutoff Frequency
8.15
Consider the following figure:
ib
ic
+
Signal
Source
C
r
v be
g m v be
Load
-
i b and i c are given by
ib vbe / input impedance vbe input admittance vbe 1 / r jC
ic g m vbe .
The gain is
i
gm
1
c
ib
1 / r jC
1 / g m r j jC dBE / g m
1
.
1 / F j jC dBE kT / qI C
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If F >>1 so that 1/ F becomes negligible, the equation above shows that F ()
1/, and becomes 1 at
fT
1
2 F C dBE kT / qI C
.
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