117
Metal-Oxide-Semiconductor System
Now we substitute for f s = 0 to get
C
K
L
v
v
s
si
d
s
kT
s
kT
(
)
flat band ≅
− ( )(
)− ( )(
) +
− (
ε
φ
φ
0
2
1 1 2
16
1 1 6
) )(
) +
=
φ
ε
s
kT
si
d
v
K
L
1 2
0
/
(3.92)
Then from Equation 3.86, the total capacitance of an MOS structure at flat
band condition is given by
C C
C
L
K
fb
ox
d
si
≡
=
+
(
)
−
1
0
1
ε
flat band
(3.93)
Since L K
d
s i ε 0
(
) is a finite number, Equation 3.93 shows that C fb is somewhat
less than C ox .
3.5.1.3 Depletion
In the depletion regime (0 < f s < 2f B ), the general expression for C d is given by
Equation 3.87. However, we can derive an approximate expression from depletion approximation discussed in Section 3.4.2.1. We know that at the depletion
condition Q s ≅ Q b , then substituting for Q s = Q b in Equation 3.23, we get
V V
Q
C
g
f b
s
b
ox
=
+ −
φ
(3.94)
Again, from Equation 3.64 we get:
Q
q K N
Q
qK N
b
s i
b s
s
b
si
b
= −
=
2
2
0
2
0
ε
φ
φ
ε
or
(3.95)
Then substituting for f s from Equation 3.95 to Equation 3.94, we can show
after simplification
V V
Q
qK N
Q
C
Q
qK N
C
Q
qK
g
f b
b
si
b
b
ox
b
si
b
ox
b
s i
−
=
−
−
−
2
0
2
0
2
2
2
ε
ε
ε
or
0 0
0
N V V
b
g
fb
−
(
) =
(3.96)
Equation 3.96 is a quadratic equation in Q b with solution given by
Metal-Oxide-Semiconductor System
Now we substitute for f s = 0 to get
C
K
L
v
v
s
si
d
s
kT
s
kT
(
)
flat band ≅
− ( )(
)− ( )(
) +
− (
ε
φ
φ
0
2
1 1 2
16
1 1 6
) )(
) +
=
φ
ε
s
kT
si
d
v
K
L
1 2
0
/
(3.92)
Then from Equation 3.86, the total capacitance of an MOS structure at flat
band condition is given by
C C
C
L
K
fb
ox
d
si
≡
=
+
(
)
−
1
0
1
ε
flat band
(3.93)
Since L K
d
s i ε 0
(
) is a finite number, Equation 3.93 shows that C fb is somewhat
less than C ox .
3.5.1.3 Depletion
In the depletion regime (0 < f s < 2f B ), the general expression for C d is given by
Equation 3.87. However, we can derive an approximate expression from depletion approximation discussed in Section 3.4.2.1. We know that at the depletion
condition Q s ≅ Q b , then substituting for Q s = Q b in Equation 3.23, we get
V V
Q
C
g
f b
s
b
ox
=
+ −
φ
(3.94)
Again, from Equation 3.64 we get:
Q
q K N
Q
qK N
b
s i
b s
s
b
si
b
= −
=
2
2
0
2
0
ε
φ
φ
ε
or
(3.95)
Then substituting for f s from Equation 3.95 to Equation 3.94, we can show
after simplification
V V
Q
qK N
Q
C
Q
qK N
C
Q
qK
g
f b
b
si
b
b
ox
b
si
b
ox
b
s i
−
=
−
−
−
2
0
2
0
2
2
2
ε
ε
ε
or
0 0
0
N V V
b
g
fb
−
(
) =
(3.96)
Equation 3.96 is a quadratic equation in Q b with solution given by
