Supplementary Course Notes
EE201
Thevenin)“101”)
Summary:
The purpose of this note is to provide a brief guideline on when and how to use Thevenin and
Norton equivalents to help with circuit analysis. It provides some simple steps for determining the
equivalents as well as an example to demonstrate the process. Finally, it provides a couple of
cautions about the limitations of using equivalents.
Why use equivalents?
Simply stated, an equivalent circuit can be used to simplify determination of the voltage across, or
current through a specific element. In order to do this, we replace a rather complicated network of
components with a simple 2-element replacement that
acts exactly the same as far as the target component is
concerned. Using the equivalent network
to find the voltage or current through the
target component is usually not a
problem. The challenge is determining the equivalent itself, and sometimes when it
will be helpful or necessary to do so.
When to use equivalents:
There are many techniques that can be used for circuit analysis, and they will all give the same
result. While there is often some advantages in using one method versus another in a specific
circuit situation, which one you use will depend on the questions you are trying to answer and your
personal familiarity and expertise at applying them. This note will focus on the analysis you will
typically have to do for transient D.C. circuits. In these cases, a Thevenin or Norton equivalent is
often very useful. You will typically require an equivalent if the network between a source(s) and
the target component (i.e. the component of interest) has a combination of series and parallel
components. Another, but less obvious way to tell is if the circuit is such that varying the current
or voltage in the target component will affect the voltage or current at the component. For
example, if we’re interested in the current through the 470kΩ resistor in the circuit above, we can
see that changing the current in it will change the voltage across it and thus affect the contribution
for the various sources in the network. This is a good hint that an equivalent circuit will be
helpful! At worst, you can’t go wrong by using an equivalent circuit (it will always perform the
same as what it replaces), you may just do a bit more work than necessary. Of course this assumes
that you determine the equivalent correctly!
Definitions:
A Thévenin or Norton equivalent circuit consists of a source (voltage or current respectively) and
one resistor, forming a two-terminal network, that will behave exactly the same from the target
component’s perspective as the more complex network it replaces. It is very important to
understand that the source and the resistor in the equivalent network are generally not the same as
any original component, even though they may have the same values! This means, for example,
you cannot infer that the current through an original resistor is the same as the current going
through RTh, even though it may have the same value. In some cases some relevant information
can be deduced from the equivalent, but this should be done cautiously!
©Denard Lynch
Page 49 of 126
Sep , 2011
Supplementary Course Notes
EE201
How to Determine Equivalents:
A simple series of steps will help derive an equivalent circuit:
1. define two terminals isolating the component of interest (target component),
2. “remove” the target component, and determine the resistance “looking back” into the terminals
with the sources replaced by their internal resistance (a short for voltage sources and an open
for current sources),
3. determine the “open circuit” voltage (Thévenin) or short circuit current (Norton) at the
terminals,
4. use the source and resistance to build a series network (voltage source, resistor =, Thévenin) or
parallel network (current source, resistor = Norton) and attach it to the two terminals in place
of the original network.
Admittedly, steps two and three can cause some challenges with some circuits. In these cases,
some practice with circuit analysis will help determine the necessary component values.
Finding the equivalent resistance (2), RTh or RN, is usually just a matter of reducing resistor
networks. This topic is well covered elsewhere and not the prime target of this discussion.
Finding the open circuit voltage or short circuit current (3) can often be aided by using Thévenin<>Norton source conversion themselves. A brief example will illustrate how.
Consider the circuit example shown above. Let’s say we’re interested in the current through the
4.7kΩ resistor. We first define two terminals
1M !
200k !
3V
around the component. (Note: I’ve also “flipped”
the 3V source so that it is a +3V facing the other
9V
6µA
way.)
100k !
4.7k !
1M !
9V
200k !
RTh/RN
3V
100k !
6µA
Next we remove the component and replace the sources
by their “equivalent impedances”.
Looking back into the terminals, we “see”
(100k + 200k)//1MΩ, so RTh = 230.7kΩ
Next we need to determine the open circuit voltage at the terminals. Here we should recognize the
conversion relation between Thévenin and Norton equivalents:
RTh
RTh = RN
IN
ETh
+
ETh = I N R N
RN
RTh = RN
ETh
IN =
RTh
Note this is the same as between all equivalents: same VOC, same ISC!
First, notice a “Norton Equivalent” on the right that could be converted to a Thévenin Eq. and
combined in series with the other “Thévenin Eq. on the right side:
©Denard Lynch
Page 50 of 126
Sep , 2011
Supplementary Course Notes
EE201
200k !
100k !
3V
300k !
.6V
3.6V
100k !
6µA
Also note that the polarities between the equivalent’s terminals must also match!
At this point we have two choices: use the voltage diver rule to determine VOC; or convert both the
remaining Thévenin Eq. to Norton Eq., combine them into 1 Norton Eq. and convert back to 1
Thévenin Eq.
Voltage divider first:
Note the two sources are opposing in polarity which leaves
1M !
300k !
(9-3.6) 5.4V to drop across the two resistors. Using simple
3.6V
voltage divider rules, and observing appropriate polarities, we
+
+ 4.15V
1.25V
can determine that 4.15V drops across the 1MΩ resistor, and 9V
the rest across the 300kΩ resistor. This leaves us with a
voltage between our terminals of interest of -4.85V (assuming
the bottom terminal is the reference), and the following
equivalent circuit..
230.8k !
-4.85V+
4.7k !
Using Source Conversions:
9µA
230.8k !
combining:
230.8k !
-4.85V+
21 µ A
300k !
12 µ A
1M !
4.7k !
and converting back:
!
We can now simply calculate the required value: -20.6µA (assuming the positive direction is
defined as down through the resistor)
Final Summary and Cautions:
The steps suggested above will help you derive Thévenin and Norton equivalents for D.C. circuits
containing resistors and independent sources, and use them to simplify circuit analysis. The most
important word of caution, as stated above, is that once the equivalent substitution has been done,
the original components are not there. In general, you cannot use the equivalent components to
deduce any parameters (current through, voltage across) for the original components that were
replaced. This is also true of simple Thévenin <-> Norton source conversions; even though the
resistances are the same value, they are not the same resistors!!
©Denard Lynch
Page 51 of 126
Sep , 2011