Now we can go back now to our initial structure, shown in the
introduction to
MOSFETs, only this time we will add an oxide layer, a
gate structure, and another battery so that we can invert the
region under the gate and connect the two n-regions
together. Well also identify some names for parts of the
structure, so we will know what we are talking about. For
reasons which will be clear later, we call the n-region
connected to the negative side of the battery the
source, and the other one the
drain. We will ground the source, and also the
p-type substrate. We add two batteries,
Vgs
Vgs
between the gate and the source, and
Vds
Vds
between the drain and the source.
It will be helpful if we also make another sketch, which gives
us a perspective view of the device. For this we strip off the
gate and oxide, but we will imagine that we have applied a
voltage greater than
VT
VT
to the gate, so there is a n-type
region, called the channel which connects the
two. We will assume that the channel region is
LL long and
WW wide, as shown in Figure 2.
Next we want to take a look at a little section of channel, and
find its resistance
ⅆR
ⅆ
R
, when the
little section is
ⅆx
ⅆ
x
long.
ⅆR=dx
σ
s
W
ⅆ
R
dx
σ
s
W
(1)
We have introduced a slightly different
form for our resistance formula here. Normally, we would have a
simple σσ in the denominator,
and an area AA, for the
cross-sectional area of the channel. It turns out to be very
hard to figure out what that cross sectional area of the channel
is however. The electrons which form the inversion layer crowd
into a very thin sheet of surface charge which
really has little or no thickness, or penetration into the
substrate.
If, on the other hand we consider a surface conductivity (units:
simply mhos),
σs
σs
, where
σ
s
=
μ
s
Q
chan
σ
s
μ
s
Q
chan
(2)
then we will have an expression which we can evaluate. Here,
μs
μs
is a surface mobility, with units of
cm2Vsec
cm
2
V
sec
. We ran into
μμ in
earlier chapters, when we were
building our simple conduction model. It was the quantity which
represented the proportionality between the average carrier
velocity and the electric field.
v¯=μE
v
μ
E
(3)
μ=qτm
μ
q
τ
m
(4)
The surface mobility is a quantity which has to be measured for a
given system, and is usually just a number which is given to
you. Something around 300
cm2Vsec
cm
2
V
sec
is about right for silicon.
Qchan
Qchan
is called the surface charge density
or channel charge density and it has units of
Coulombscm2
Coulombs
cm
2
. This
is like a sheet of charge, which is different from the bulk
charge density, which has units of
Coulombscm2
Coulombs
cm
2
. Note that:
cm2VoltsecCoulombscm2=CoulsecVolt=IV=mhos
cm
2
Volt
sec
Coulombs
cm
2
Coul
sec
Volt
I
V
mhos
(5)
It turns out that it is pretty simple to get an expression for
Qchan
Qchan
, the surface charge density in the
channel. For any given gate voltage
Vgs
Vgs
, we know that the charge density on the
gate is given simply as:
Q
g
=
c
ox
V
gs
Q
g
c
ox
V
gs
(6)
However, until the gate voltage
Vgs
Vgs
gets larger than
VT
VT
we are not creating any mobile
electrons under the gate, we are just building up a depletion
region. We'll define
QT
QT
as the charge on the gate necessary to
get to threshold.
Q
T
=
c
ox
V
T
Q
T
c
ox
V
T
. Any charge added to the gate above
QT
QT
is matched by charge
Qchan
Qchan
in the channel. Thus, it is easy to
say:
Q
channel
=
Q
g
−
Q
T
Q
channel
Q
g
Q
T
(7)
or
Q
chan
=
c
ox
V
g
−
V
T
Q
chan
c
ox
V
g
V
T
(8)
Thus, putting Equation 7 and Equation 2
into Equation 1, we get:
ⅆR=ⅆx
μ
s
c
ox
V
gs
−
V
T
W
ⅆ
R
ⅆ
x
μ
s
c
ox
V
gs
V
T
W
(9)
If you look back at Figure 1, you will see that we
have defined a current
Id
Id
flowing into the drain. That current
flows through the channel, and hence through our little
incremental resistance
ⅆR
ⅆ
R
,
creating a voltage drop
ⅆ
V
c
ⅆ
V
c
across it, where
Vc
Vc
is the channel voltage.
ⅆ
V
c
x=
I
d
ⅆR=
I
d
ⅆx
μ
s
c
ox
V
gs
−
V
T
W
ⅆ
V
c
x
I
d
ⅆ
R
I
d
ⅆ
x
μ
s
c
ox
V
gs
V
T
W
(10)
Let's move the denominator to the left, and integrate. We
want to do our integral completely along the channel. The voltage on
the channel
V
c
x
V
c
x
goes from 0 on the left to
Vds
Vds
on the right. At the same time,
xx is going from 0 to
LL. Thus our limits of integration
will be 0 and
Vds
Vds
for the voltage integral
ⅆ
V
c
x
ⅆ
V
c
x
and from 0 to LL for the
xx integral
ⅆx
ⅆ
x
.
∫0VdsμscoxVgs−VTWdVc=∫0LIddx
Vc
0
Vds
μs
cox
Vgs
VT
W
x
0
L
Id
(11)
Both integrals are pretty trivial. Let's swap the equation
order, since we usually want
Id
Id
as a function of applied voltages.
I
d
L=
μ
s
c
ox
W
V
gs
−
V
T
V
ds
I
d
L
μ
s
c
ox
W
V
gs
V
T
V
ds
(12)
We now simply divide both sides by
LL, and we end up with an
expression for the drain current
Id
Id
, in terms of the drain-source voltage,
Vds
Vds
, the gate voltage
Vgs
Vgs
and some physical attributes of the
MOS transistor.
I
d
=
μ
s
c
ox
WL
V
gs
−
V
T
V
ds
I
d
μ
s
c
ox
W
L
V
gs
V
T
V
ds
(13)
"This course offers an introduction to solid state device including field effect and bipolar transistors. Properties of transmission lines and propagating E&M waves are also presented. It is […]"