We have found that the way to think about circuits is to locate
and group parallel and series resistor combinations. Those
resistors not involved with variables of interest can be
collapsed into a single resistance. This result is known as an
equivalent circuit: from the viewpoint of a pair of
terminals, a group of resistors functions as a single resistor,
the resistance of which can usually be found by applying the
parallel and series rules.
This result generalizes to include sources in a very interesting
and useful way. Let's consider our simple attenuator circuit (shown in the figure)
from the viewpoint of the output terminals. We want to find the
v-i relation for the output terminal pair,
and then find the equivalent circuit for the boxed circuit. To
perform this calculation, use the circuit laws and element
relations, but do not attach anything to the output
terminals. We seek the relation between
vv and
ii that describes the kind of
element that lurks within the dashed box. The result is
v=(
R
1
∥
R
2
)i+
R
2
R
1
+
R
2
v
in
v
∥
R
1
R
2
i
R
2
R
1
R
2
v
in
(1)
If the source were zero, it could be replaced by a short
circuit, which would confirm that the circuit does indeed
function as a parallel combination of resistors. However, the
source's presence means that the circuit is
not well modeled as a resistor.
If we consider the simple circuit of Figure 2, we find it has the v-i
relation at its terminals of
v=
R
eq
i+
v
eq
v
R
eq
i
v
eq
(2)
Comparing the two
v-i relations, we find
that they have the same form. In this case the
Thévenin equivalent resistance is
R
eq
=
R
1
∥
R
2
R
eq
∥
R
1
R
2
and the
Thévenin equivalent source
has voltage
v
eq
=
R
2
R
1
+
R
2
v
in
v
eq
R
2
R
1
R
2
v
in
.
Thus, from viewpoint of the terminals, you cannot distinguish the
two circuits. Because the equivalent circuit has fewer elements,
it is easier to analyze and understand than any other
alternative.
For any circuit containing resistors and
sources, the v-i relation will be of the
form
v=
R
eq
i+
v
eq
v
R
eq
i
v
eq
(3)
and the
Thévenin equivalent circuit for any
such circuit is that of
Figure 2. This equivalence applies no
matter how many sources or resistors may be present in the
circuit. In the
example below,
we know the circuit's construction and element values, and
derive the equivalent source and resistance. Because
Thévenin's theorem applies in general, we should be able
to make measurements or calculations
only from the
terminals to determine the equivalent circuit.
To be more specific, consider the equivalent circuit of this figure. Let the
terminals be open-circuited, which has the effect of setting the
current ii to zero. Because no
current flows through the resistor, the voltage across it is
zero (remember, Ohm's Law says that
v=Ri
v
R
i
).
Consequently, by applying KVL we have that the so-called
open-circuit voltage
v
oc
v
oc
equals the Thévenin equivalent voltage. Now consider the
situation when we set the terminal voltage to zero
(short-circuit it) and measure the resulting current. Referring
to the equivalent circuit, the source voltage now appears
entirely across the resistor, leaving the short-circuit current
to be
i
sc
=−
v
eq
R
eq
i
sc
v
eq
R
eq
.
From this property, we can determine the equivalent resistance.
v
eq
=
v
oc
v
eq
v
oc
(4)
R
eq
=−
v
oc
i
sc
R
eq
v
oc
i
sc
(5)
Use the open/short-circuit approach to derive the
Thévenin equivalent of the circuit shown in Figure 3.
v
oc
=
R
2
R
1
+
R
2
v
in
v
oc
R
2
R
1
R
2
v
in
and
i
sc
=−
v
in
R
1
i
sc
v
in
R
1
(resistor
R
2
R
2
is shorted out in this case). Thus,
v
eq
=
R
2
R
1
+
R
2
v
in
v
eq
R
2
R
1
R
2
v
in
and
R
eq
=
R
1
R
2
R
1
+
R
2
R
eq
R
1
R
2
R
1
R
2
.
For the circuit depicted in Figure 4, let's derive its Thévenin
equivalent two different ways. Starting with the
open/short-circuit approach, let's first find the open-circuit
voltage
v
oc
v
oc
.
We have a current divider relationship as
R
1
R
1
is in parallel with the series combination of
R
2
R
2
and
R
3
R
3
.
Thus,
v
oc
=
i
in
R
3
R
1
R
1
+
R
2
+
R
3
v
oc
i
in
R
3
R
1
R
1
R
2
R
3
.
When we short-circuit the terminals, no voltage appears across
R
3
R
3
,
and thus no current flows through it. In short,
R
3
R
3
does not affect the short-circuit current, and can be eliminated. We
again have a current divider relationship:
i
sc
=−
i
in
R
1
R
1
+
R
2
i
sc
i
in
R
1
R
1
R
2
.
Thus, the Thévenin equivalent resistance is
R
3
(
R
1
+
R
2
)
R
1
+
R
2
+
R
3
R
3
R
1
R
2
R
1
R
2
R
3
.
To verify, let's find the equivalent resistance by reaching
inside the circuit and setting the current source to
zero. Because the current is now zero, we can replace the
current source by an open circuit. From the viewpoint of the
terminals, resistor
R
3
R
3
is now in parallel with the series combination of
R
1
R
1
and
R
2
R
2
.
Thus,
R
eq
=
R
3
∥
R
1
+
R
2
R
eq
∥
R
3
R
1
R
2
,
and we obtain the same result.
As you might expect, equivalent circuits come in two forms: the
voltage-source oriented Thévenin equivalent and the
current-source oriented Mayer-Norton equivalent
(Figure 5).
To derive the latter, the v-i relation for
the Thévenin equivalent can be written as
v=
R
eq
i+
v
eq
v
R
eq
i
v
eq
(6)
or
i=v
R
eq
−
i
eq
i
v
R
eq
i
eq
(7)
where
i
eq
=
v
eq
R
eq
i
eq
v
eq
R
eq
is the Mayer-Norton equivalent source. The Mayer-Norton
equivalent shown in
Figure 5 can be
easily shown to have this
v-i
relation. Note that both variations have the same equivalent
resistance. The short-circuit current equals the negative of the
Mayer-Norton equivalent source.
Find the Mayer-Norton equivalent circuit for the circuit below.
i
eq
=
R
1
R
1
+
R
2
i
in
i
eq
R
1
R
1
R
2
i
in
and
R
eq
=
R
3
∥
R
1
+
R
2
R
eq
∥
R
3
R
1
R
2
.
Equivalent circuits can be used in two basic ways. The first is
to simplify the analysis of a complicated circuit by realizing
the any portion of a circuit can be
described by either a Thévenin or Mayer-Norton
equivalent. Which one is used depends on whether what is
attached to the terminals is a series configuration (making the
Thévenin equivalent the best) or a parallel one (making
Mayer-Norton the best).
Another application is modeling. When we buy a flashlight
battery, either equivalent circuit can accurately describe
it. These models help us understand the limitations of a
battery. Since batteries are labeled with a voltage
specification, they should serve as voltage sources and the
Thévenin equivalent serves as the natural choice. If a
load resistance
R
L
R
L
is placed across its terminals, the voltage output can be found using
voltage divider:
v=
v
eq
R
L
R
L
+
R
eq
v
v
eq
R
L
R
L
R
eq
. If we have a load resistance much larger than the
battery's equivalent resistance, then, to a good approximation,
the battery does serve as a voltage source. If the load
resistance is much smaller, we certainly don't have a voltage
source (the output voltage depends directly on the load
resistance). Consider now the Mayer-Norton equivalent; the
current through the load resistance is given by current divider,
and equals
i=−
i
eq
R
eq
R
L
+
R
eq
i
i
eq
R
eq
R
L
R
eq
. For a current that does not vary with the load
resistance, this resistance should be much smaller than the
equivalent resistance. If the load resistance is comparable to
the equivalent resistance, the battery serves
neither as a voltage source or a current
course. Thus, when you buy a battery, you get a voltage source
if its equivalent resistance is much
smaller than the equivalent resistance of
the circuit to which you attach it. On the other hand, if you
attach it to a circuit having a small equivalent resistance, you
bought a current source.
He was an engineer with France's Postes,
Télégraphe et Téléphone. In 1883,
he published (twice!) a proof of what is now called the
Thévenin equivalent while developing ways of teaching
electrical engineering concepts at the École
Polytechnique. He did not realize that the same result had
been published by
Hermann
Helmholtz, the renowned nineteenth century physicist,
thiry years earlier.
After earning his doctorate in physics in 1920, he turned to
communications engineering when he joined Siemens & Halske
in 1922. In 1926, he published in a German technical journal
the Mayer-Norton equivalent. During his interesting career, he
rose to lead Siemen's Central Laboratory in 1936,
surruptiously leaked to the British all he knew of German
warfare capabilities a month after the Nazis invaded Poland,
was arrested by the Gestapo in 1943 for listening to BBC radio
broadcasts, spent two years in Nazi concentration camps, and
went to the United States for four years working for the Air
Force and Cornell University before returning to Siemens in
1950. He rose to a position on Siemen's Board of
Directors before retiring.
Edward
Norton was an electrical engineer who worked at Bell
Laboratory from its inception in 1922. In the
same month when Mayer's paper appeared,
Norton wrote in an internal technical memorandum a paragraph
describing the current-source equivalent. No evidence
suggests Norton knew of Mayer's publication.
"Electrical Engineering Digital Processing Systems in Braille."