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It has been said that a picture is worth a thousand words, or equivalently, a
picture can replace a thousand words. Thevenin's theorem is a way to visualize a
complex network into a simpler "black box" and load to greatly simplify analysis.
This is especially helpful when doing analysis of a complex driving circuit under
different load conditions.
Thevenin's (pronounced THEV-eh-nin) Theorem states:
Any linear circuit an be arranged in the form of two networks, A and B connected by
two resistance-less conductors. The entire network A can be replaced by a single
generator of:
- voltage Vg equal to the voltage which appears across the terminals of A if B
were disconnected so that no current is drawn from A,
-
and internal resistance of Rg, equal to the resistance seen looking back into
the terminals of A with all voltage and current sources in A replaced by short
circuits and open circuits respectively.
As a practical example, looking at the circuit of figure 1, suppose we wanted to
determine the output voltage Vo under different RL load values.
We could write a series of mesh equations for the circuit, substitute different values
of RL into the circuit and use mesh analysis to solve for each variation.
Using Kirchoff's Voltage Law, let's write the five mesh equations for
figure 1:
[1] 10(i1) + 12(i1) + 8(i1 - i2) - 15 = 0 (mesh1)
[2] 8(i2 - i1) + 15(i2 - i3) + 24 = 0 (mesh2)
[3] 15(i3 - i2) + 8 (i3 - i4) - 24 = 0 (mesh3)
[4] 8(i4 - i3) + 9(i4 - i5) + 12 = 0 (mesh4)
[5] 9(i5 - i4) + 5(i5) + 10(i5) - 12 = 0 (mesh5)
There are five equations and five unknowns, so we can solve for i5. Working the
problem through and solving for i5, after rounding errors we get approximately
i5 = 0.45A.
V0 is thus:
10(i5) = 10(0.45) = 4.5V
Depending upon the complexity of the circuit and how many values we wanted to test,
this could become very tedious fast. A better solution would be to apply Thevenin's
Theorem to simplify the circuit to make the analysis easier.
First step is to break the circuit into parts A and B. The A and B division point is
completely arbitrary. Since we are interested of the effect of varying only the
load resistance RL, we will only place RL into the B network, and everything else
is placed into the A network as shown in figure 2.
The next step is to calculate the Thevenin equivalent for circuit A as shown in
figure 3.
In figure 4, The Thevenin Voltage, Vg, is calculated by looking into the circuit to
the left of RL. Since by definition, the circuit is open looking in and no current is flowing, R7 can be ignored for now.
Using Kirchoff's Voltage Law, let's write the four mesh equations for figure 4:
[1] 10(i1) + 12(i1) + 8(i1 - i2) - 15 = 0 (mesh 1)
[2] 8(i2 - i1) + 15(i2 - i3) + 24 = 0 (mesh 2)
[3] 15(i3 - i2) + 8 (i3 - i4) - 24 = 0 (mesh 3)
[4] 8(i4 - i3) + 9(i4) + 12 = 0 (mesh 4)
There are four equations, and four unknowns. Working the problem through and
solving for i4, after rounding errors we get approximately
i4 = -0.482A
Vg is thus:
9(i4) + 12 = 7.66V
Luckily you only have to do this once!
To calculate the Thevenin Resistance, Rg, short circuit all voltage sources, and
open circuit all current sources, then calculate the resultant resistance of the
circuit. Our circuit only has voltage sources, so these replaced by short circuits
in figure 5,. the reduced network. In this case R7 does matter! Rg is easily
calculated to be 7.1-Ohms
Going back to figure 1.
Now replace the circuit of figure.1 with the Thevenin Equivalent circuit of figure 6.
Using our much simpler voltage divider formula, we can now easily find Vo for any
RL value without resorting to complex analysis.
For our example, if RL = 10-Ohms
V0 = Vg(10/(10+Rg) = 7.66(10/17.1) = 4.5V
This is a much simpler method of solving the problem if you have anumber of different
values for RL that you want to investigate.
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