OpenStax-CNX module: m11348
1
Small Signal Models
∗
Bill Wilson
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 1.0†
Abstract
Small Signal Models
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, VB , 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 denition, we will assume that |vs | is much less than |VB | As a result
of these voltages, there will be a current IB owing through the diode which will consist of two currents, IB
the so-called bias current, and is , which will be the signal current. Again, we assume that is is much
smaller than IB .
Figure 1: Putting together a large signal bias, and a small signal AC excitation
What we would like to do is to see if we can nd a linear relationship between vs and is 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 rst, as it is more intuitive, and then we
will conrm what we nd out with a mathematical one.
Let's remind ourselves about the I-V characteristics of a diode. In the present situation, VD is the sum
of two voltages, a DC bias voltage VB and an AC signal, vs Let's plot VD (t) on the VD axis as shown in
Figure 3. How are we going to gure 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.
∗ Version
1.1: Jun 20, 2003 12:00 am -0500
† http://creativecommons.org/licenses/by/1.0
http://cnx.org/content/m11348/1.1/
OpenStax-CNX module: m11348
2
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 is simply by multiplying the input signal vs 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.
is =
d
(ID ) |ID=IB
dVD
Figure 2: Diode I-V behavior
Figure 3: Bias and signal excitation of a diode I-V curve
http://cnx.org/content/m11348/1.1/
(1)
OpenStax-CNX module: m11348
3
Figure 4: Graphically nding the AC response
Figure 5: With a smaller signal, the response is more linear
To get the slope, we need a few simple equations:
qVD
qVD
ID = Isat e kT − 1 ' Isat e kT
(2)
qVD
q
d
(ID ) =
Isat e kT
dVD
kT
(3)
When we evaluate the partial derivative at voltage VD , we note that
Isat e
qVD
kT
= IB
(4)
q
q
and hence, the slope of the curve is just kT
IB or 40IB , since kT
just has a value of 40V −1 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.
http://cnx.org/content/m11348/1.1/
OpenStax-CNX module: m11348
4
As far as the AC signal generator is concerned, we could replace the diode with a resistor whose value is
1
IB , where IB is the DC bias current through the diode.
the inverse of the conductance, or r = 40
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.
qVD
q (VB +vs )
ID = IB + is = Isat e kT − 1 ' e kT
(5)
In the last expression, we dropped the −1 as it is very small compared to the exponential term and can be
neglected.
Now we note that:
q (VB +vs )
qVB
qvs
e kT
= e kT e kT
(6)
And, for the second exponential, if qVB is much less than kT ,
qvs
e kT ' 1 +
qvs
+ ...
kT
(7)
where we have used the power series expansion for the exponential, but have only taken the rst two terms.
Thus
qVB
qvs IB + is ' Isat e kT 1 +
(8)
kT
Obviously
IB = Isat e
and
is
= Isat e
=
which gives us the same result as before
http://cnx.org/content/m11348/1.1/
qVB
kT
qVB
kT
q
kT IB vs
is
q
IB
=
vs
kT
q
kT
(9)
vs
(10)
(11)