Thevenin and Norton
Two powerful circuit
analysis techniques are Thevenin's theorem and Norton's theorem. Both convert
a
complex circuit
to a simpler series or parallel equivalent circuit for easier analysis.
Analysis involves
removing part of
the circuit across two terminals to aid calculation, later combining the
circuit with the
Thevenin or Norton
equivalent circuit. There are plenty of books on this subject, so
this is just a quick
look.
Thevenin
The top left diagram represents the circuit
for analysis at terminals A and B. The top right hand circuit is
the Thevenin equivalent circuit, a voltage
source Vth with a series resistance, Rth. The bottom left diagram
is the same circuit driving a load. The
load is NOT included in the thevenin equivalent circuit and must be
separated, this is why the terminals are
marked A and B.
Value of Vth and Rth
The value of Vth is found by either measuring
(if you don't know what's in the circuit) or be using circuit
analysis. To find Rth ( with load removed)
short circuit voltage supplies, open current sources and
calculate the equivalent resistance.
An Example
A demonstration of the thevenin technique
to find I1
in the diagram below :
The circuit to the right of points A and B
is converted to a Thevenin source and resistance. With the 30v
battery and left hand 10ohm resistor omitted,
the Thevenin voltage becomes:
Vth = 40 * 10/30 = 400/30
Rth = 10||20 = 200/30 (10 ohm parallel
with 20 ohm voltage source s/c )
I1 then becomes 30 - Vth / ( 10 +
Rth )
I1 = 30 - 400/30 / 10 + 200/30 = 900/30
- 400/30 / ( 300/30 + 200/30)
= 500/30 / 500/30 =
1amp
A messy numerical
example, but can also be solved using kirchoff or verified with a simulator
program.
The early results
are sometimes best kept as fractions to make the division easier.
Norton
The Norton theorem
converts an ordinary circuit to an equivalent parallel circuit which is
a current source
in parallel with
a resistor. The technique is similar to the thevenin theorem and two points
in a circuit must
be defined, this
is where the analysis will take place.
As with Thevenin, the equivaent circuit is
a current generator In and norton equivalent resistance, Rn.
These must be worked out to use the Norton
theorem. The analysis points using Norton are short circuited,
whereas using the Thevenin Method they
are open circuit.
Value of Vth and Rth
The value of Vth is found by either measuring
(if you don't know what's in the circuit) or be using circuit
analysis. To find Rth ( with load removed)
short circuit voltage supplies, open current sources and
calculate the equivalent resistance.
An Example
Norton's theorem is demonstrated to find
the current I1 in the diagram
below :
The points A and B is where the Norton conversion
takes place, the right 50 ohm resistor is removed,
A and B are short circuited, see below
:
In
First
the total current is calculated 100 / ( 50 + 100 || 50 ) = 1.2 amp.
Using the current division rule,
In becomes
1.2 * 100 / (50+100) = 0.8 amp
Rn is
50 + 100 || 50 = 83.3333333 ohm
The Norton equivalent
circuit can now be completed with the right hand 50 ohm resistor included
:
The current
I can now be found using the current
division rule :
I = 0.8 * (83.3333 / ( 50 + 83.3333)
= 0.5 amp
This can be verified using the Thevenin
method or a simulation program.
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