Our task now is to figure out how much voltage we need to get
V
g
V
g
up to
V
T
V
T
and then to figure out how much negative charge there is under
the gate, once
V
T
V
T
has been exceeded. The first part is actually pretty easy. It
is a lot like the problem we looked at, with the one-sided
diode, but with just a little added complication. To start out,
lets make a sketch of the charge density distribution under the
conditions of this image, just when we get to threshold. Well include the
sketch of the structure too, so it will be clear what charge we
are talking about. This is shown in Figure 1.
Now, we just use the equation we developed before for the electric field, which
came from integrating the differential form of Gauss' Law.
Ex=∫ρxεdx
E
x
x
ρ
x
ε
(1)
As before, we will do the integral graphically, starting at the
LHS of the picture. The field outside the structure must be
zero, so we have no electric field until we get to the delta
function of charge on the gate, at which time it jumps up to
some value we will call
E
ox
E
ox
.
There is no charge inside the oxide, so
ddxE
x
E
is zero, and thus
Ex
E
x
must remain constant at
E
ox
E
ox
until we reach the oxide/silicon interface.
If we were to put our little "pill box" on the oxide-silicon
interface, the integral of DD over
the face in the silicon would be
ε
Si
E
Si
ΔS
ε
Si
E
Si
Δ
S
, where
E
Si
E
Si
is the strength of the electric field inside the silicon. On
the face inside the oxide it would be
-
ε
ox
E
ox
ΔS
ε
ox
E
ox
Δ
S
, where
E
ox
E
ox
is the strength of the electric field in the oxide. The minus
sign comes from the fact that the field on the oxide side is
going into the pill box instead of out of it. There is no net
charge contained within the pill box, so the sum of these two
integrals must be zero. (The integral over the
entire surface equals the enclosed charge,
which is zero.)
ε
Si
E
max
ΔS-
ε
ox
E
ox
ΔS=0
ε
Si
E
max
Δ
S
ε
ox
E
ox
Δ
S
0
(2)
or
ε
Si
E
max
=
ε
ox
E
ox
ε
Si
E
max
ε
ox
E
ox
(3)
This is just a statement that it is the normal component of
displacement vector, DD, which must
be continuous across a dielectric interface, not the electric
field, EE. Solving Equation 2 for the electric field in the silicon:
E
Si
=
ε
ox
ε
Si
E
ox
E
Si
ε
ox
ε
Si
E
ox
(4)
The dielectric constant of oxides about one third that of the
dielectric constant of silicon dioxide, so we see a "jump" down
in the magnitude of the electric field as we go from oxide to
silicon. The charge density in the depletion region of the
silicon is just
-q
N
a
q
N
a
and so the electric field now starts decreasing at a rate
-q
N
a
ε
Si
q
N
a
ε
Si
and reaches zero at the end of the depletion region,
x
p
x
p
.
Clearly, we have two different regions, each with their own
voltage drop. (Remember the integral of electric field is
voltage, so the area under each region of
Ex
E
x
represents a voltage drop.) The drop in the little
triangular region we will call
Δ
V
Si
Δ
V
Si
and it represents the potential drop in going from the
bulk, down to the bottom of the drooping conduction band at the
silicon-oxide interface. Looking back at the earlier figure on threshold, you
should be able to see that this is nearly one whole band-gaps
worth of potential, and so we can safely say that
Δ
V
Si
≈0.8→1.0V
Δ
V
Si
0.8
1.0
V
.
Just as with the single-sided diode, the width of the depletion
region
x p
x p , is (which we saw in a previous equation):
x
p
=2
ε
Si
Δ
V
Si
q
N
a
x
p
2
ε
Si
Δ
V
Si
q
N
a
(5)
from which we can get an expression for
E
Si
E
Si
E
Si
=q
N
a
ε
Si
x
p
=2q
N
a
Δ
V
Si
ε
Si
E
Si
q
N
a
ε
Si
x
p
2
q
N
a
Δ
V
Si
ε
Si
(6)
by multiplying the slope of the
Ex
E
x
line by the width of the depletion region,
x
p
x
p
.
We can now use Equation 4 to find the
electric field in the oxide
E
ox
=
ε
Si
ε
ox
E
Si
=1
ε
ox
2q
ε
Si
N
a
Δ
V
Si
E
ox
ε
Si
ε
ox
E
Si
1
ε
ox
2
q
ε
Si
N
a
Δ
V
Si
(7)
Finally,
Δ
V
ox
Δ
V
ox
is simply the product of
E
ox
E
ox
and the oxide thickness,
x
ox
x
ox
Δ
V
ox
=
x
ox
E
ox
=
x
ox
ε
ox
2q
ε
Si
N
a
Δ
V
Si
Δ
V
ox
x
ox
E
ox
x
ox
ε
ox
2
q
ε
Si
N
a
Δ
V
Si
(8)
Note that
x
ox
x
ox
divided by
ε
ox
ε
ox
is simply one over
c
ox
c
ox
, the oxide capacitance, which we described earlier. Thus
Δ
V
ox
=1
c
ox
2q
ε
Si
N
a
Δ
V
Si
Δ
V
ox
1
c
ox
2
q
ε
Si
N
a
Δ
V
Si
(9)
And the threshold voltage
V
T
V
T
is then given as
V
T
=Δ
V
Si
+Δ
V
ox
=Δ
V
Si
+1
c
ox
2q
ε
Si
N
a
Δ
V
Si
V
T
Δ
V
Si
Δ
V
ox
Δ
V
Si
1
c
ox
2
q
ε
Si
N
a
Δ
V
Si
(10)
which is not that hard to calculate!
Equation 10 is
one of the most important equations in this discussion of field
effect transistors, as it tells us when the MOS device is turned
on.
Equation 10 has several "handles" available to the
device engineer to build a device with a given threshold
voltage. We know that as we increase
N
a
N
a
, the acceptor density, that the Fermi level gets
closer to the valance band, and hence
Δ
V
si
Δ
V
si
will change some. But as we said, it will always be
around 0.8 to 1 Volt, so it will not be the driving term which
dominates
V
T
V
T
. Let's see what we get with an acceptor concentration of
1017
10
17
. Just for completeness, let's calculate
E
f
-
E
v
E
f
E
v
.
p=
N
a
=
N
v
ⅇ
E
f
-
E
v
kT
p
N
a
N
v
E
f
E
v
k
T
(11)
thus
E
f
-
E
v
=kTln
N
v
N
a
E
f
E
v
k
T
N
v
N
a
In silicon,
N
v
N
v
is
1.08×1019
1.0819 and this makes
E
f
-
E
v
=0.117eV
E
f
E
v
0.117
eV
which we will call
ΔE
Δ
E
. It is conventional to say that a surface is inverted
if, at the silicon surface,
E
c
-
E
f
E
c
E
f
, the distance between the conduction band and the
Fermi level is the same as the distance between the Fermi level
and the valance band in the bulk. With a little time spent
looking at Equation 4, you should be
able to convince yourself that the total energy change in going
from the bulk to the surface in this case would be
qΔ
V
Si
=
E
g
-2ΔE=1.1eV-20.117eV=0.866eV
q
Δ
V
Si
E
g
2
Δ
E
1.1
eV
2
0.117
eV
0.866
eV
(12)
Using
N
A
=1017
N
A
10
17
,
ε
Si
=1.1×10-12Fcm
ε
Si
1.1-12
F
cm
and
q=1.6×10-19C
q
1.6-19
C
, we find that
2q
ε
si
N
a
Δ
V
Si
=1.74×10-7
2
q
ε
si
N
a
Δ
V
Si
1.74-7
(13)
We saw earlier that if we have an oxide thickness of
250Å, we get a value for
c
ox
c
ox
of
1.3×10-7Fcm2
1.3-7
F
cm
2
(
CoulombsVcm2
Coulombs
V
cm
2
), and so
Δ
V
ox
=1
c
ox
2q
ε
Si
N
a
Δ
V
Si
=11.3×10-71.74×10-7=1.32V
Δ
V
ox
1
c
ox
2
q
ε
Si
N
a
Δ
V
Si
1
1.3-7
1.74-7
1.32
V
(14)
and
V
T
=Δ
V
Si
+Δ
V
ox
=0.866+1.32=2.18V
V
T
Δ
V
Si
Δ
V
ox
0.866
1.32
2.18
V
(15)
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