207
Compact Models for Small Geometry MOSFETs
Again, from Gauss’s law we get K ox ε 0 E ox  = Q GATE ; therefore, from Equation 5.92
we get
E
K
qK N
ox
ox
si
GATE p
=
1
2
0
0
ε
ε
φ
(5.93)
Now, the applied gate voltage with additional voltage drop in the polydepletion region is given by
V
V
V
gs
fb
s
p
ox
=
+ + +
φ φ
(5.94)
Since V ox  = E ox T ox , we can simplify Equation 5.94 using Equation 5.93 as
V
V
T
K
qK N
gs
fb
s
p
ox
ox
si
GATE p
=
+ + +
φ φ
ε
ε
φ
0
2
0
(5.95)
After simplification we can show from Equation 5.95
a V V
gs
fb
s
p
p
−
− −
(
) − =
φ φ
φ
2
0
(5.96)
where we defined
a
qK N
T
ox
si
GATE ox
=
K
2 2
0
2
2
ε
ε
0
(5.97)
Now let us define that the effective gate voltage due to additional voltage
drop in the poly is given by V gseff  = (V gs  − f p ); then rearranging Equation 5.96
we get
a V
V
V
V
a V
V
gs
p
f b
s
gs
p
g s
gseff
f b
s
−
(
) − +
(
)



 +
−
(
) − =
−
+
(
)


φ
φ
φ
φ
2
0
or
 
 +
−
=
2
0
V
V
gseff
g s
(5.98)
After simplification of Equation 5.98, we can show
aV
a V
V
a V
V
gseff
f b
s
gseff
f b
s
gs
2
2
2
1
0
−
+
(
) −




+
+
(
) −

 

  =
φ
φ
(5.99)
Now, we solve the quadratic Equation 5.99 on V gseff due to poly gate depletion.
Solving V gseff we get
V
V
a
a
a V
a V
aV
gseff
f b
s
fb
s
f b
s
gs
=
+
(
) − ±
+
(
) −
(
) −
+
(
) +
φ
φ
φ
1
2
1
2
2
1
4
4
2
2
2
(5.100)
Since (V gs  − f p ) > 0, we consider the positive sign of Equation 5.100, to get