In order to do this we need to introduce the concept of
bias, and large signal and small
signal device behavior. Consider the following circuit,
shown in Figure 1. We are applying
the sum of two voltages to the diode,
V
B
V
B
,
the bias voltage (which is assumed to be a DC
voltage) and vs the signal voltage (which is
assumed to be AC, or sinusoidal). By definition, we will assume
that
|
v
s
|
v
s
is much less than
|
V
B
|
V
B
As a result of these voltages, there will be a current
I
B
I
B
flowing through the diode which will consist of two currents,
I
B
I
B
the so-called bias current, and
i
s
i
s
,
which will be the signal current. Again, we
assume that
i
s
i
s
is much smaller than
I
B
I
B
.
What we would like to do is to see if we can find a linear
relationship between
v
s
v
s
and
i
s
i
s
which we could use in our signal analysis. There are two ways
we can attack the problem; a graphical approach, and a purely
mathematical approach. Lets try the graphical approach first,
as it is more intuitive, and then we will confirm what we find
out with a mathematical one.
Let's remind ourselves about the I-V characteristics of a diode.
In the present situation,
V
D
V
D
is the sum of two voltages, a DC bias voltage
V
B
V
B
and an AC signal,
v
s
v
s
Let's plot
V
D
t
V
D
t
on the
V
D
V
D
axis as shown in Figure 3. How are we going to
figure out what the current is? What we need to do is to
project the voltage up onto the characteristic I-V curve, and
then project over to the vertical current axis We do this in
Figure 4. Note that the output current signal is
somewhat distorted, which means we do not have linear behavior
yet. Let's reduce the amplitude of the signal voltage, as
shown in Figure 5. Now we see two things: a) the
output is much less distorted, so we must getting a more linear
behavior, and b) we could get the amplitude of the output signal
i
s
i
s
simply by multiplying the input signal
v
s
v
s
by the slope of the I-V curve at the point where the device is
biased. We have replaced the non-linear I-V curve of
the diode by a linear one, which is applicable over the range of
the signal voltage.
i
s
=dd
V
D
I
D
|
I
D
=
I
B
i
s
I
D
I
B
V
D
I
D
(1)
To get the slope, we need a few simple equations:
I
D
=
I
sat
ⅇq
V
D
kT-1≈
I
sat
ⅇq
V
D
kT
I
D
I
sat
q
V
D
k
T
1
I
sat
q
V
D
k
T
(2)
dd
V
D
I
D
=qkT
I
sat
ⅇq
V
D
kT
V
D
I
D
q
k
T
I
sat
q
V
D
k
T
(3)
When we evaluate the partial derivative at voltage
V
D
V
D
,
we note that
I
sat
ⅇq
V
D
kT=
I
B
I
sat
q
V
D
k
T
I
B
(4)
and hence, the slope of the curve is just
qkT
I
B
q
k
T
I
B
or
40
I
B
40
I
B
,
since
qkT
q
k
T
just has a value of
40V-1
40
V
at room temperatures. Note that current divided by voltage is
just conductance, (which is just the inverse of resistance) and
so we have found the
small signal linear conductance
for the diode.
As far as the AC signal generator is concerned, we could replace
the diode with a resistor whose value is the inverse of the
conductance, or
r=140
I
B
r
1
40
I
B
,
where
I
B
I
B
is the DC bias current through the diode.
Students are sometimes confused about how we can replace a
diode, which only conducts in one direction, with a resistor,
which conducts both ways. The answer is to look carefully at
Figure 5. As the AC signal voltage
rises and falls, the AC output current also increases and
decreases in the same manner. Over the limited range of the AC
signal parameters, the diode is indeed a linear signal element,
not a rectifying one, as it is for large signal applications.
Now let's get the same answer from a purely mathematical
approach.
I
D
=
I
B
+
i
s
=
I
sat
ⅇq
V
D
kT-1≈ⅇq
V
B
+
v
s
kT
I
D
I
B
i
s
I
sat
q
V
D
k
T
1
q
V
B
v
s
k
T
(5)
In the last expression, we dropped the
-1
-1 as it is very small compared to the exponential term
and can be neglected.
Now we note that:
ⅇq
V
B
+
v
s
kT=ⅇq
V
B
kTⅇq
v
s
kT
q
V
B
v
s
k
T
q
V
B
k
T
q
v
s
k
T
(6)
And, for the second exponential, if
q
V
B
q
V
B
is much less than
kT
k
T
,
ⅇq
v
s
kT≈1+q
v
s
kT+…
q
v
s
k
T
1
q
v
s
k
T
…
(7)
where we have used the power series expansion for the
exponential, but have only taken the first two terms. Thus
I
B
+
i
s
≈
I
sat
ⅇq
V
B
kT1+q
v
s
kT
I
B
i
s
I
sat
q
V
B
k
T
1
q
v
s
k
T
(8)
Obviously
I
B
=
I
sat
ⅇq
V
B
kT
I
B
I
sat
q
V
B
k
T
(9)
and
i
s
=
I
sat
ⅇq
V
B
kTqkT
v
s
=qkT
I
B
v
s
i
s
I
sat
q
V
B
k
T
q
k
T
v
s
q
k
T
I
B
v
s
(10)
which gives us the same result as before
i
s
v
s
=qkT
I
B
i
s
v
s
q
k
T
I
B
(11)
