Thévenin's theorem

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In electrical circuit theory, Thévenin's theorem for electrical networks states that any combination of voltage sources, current sources and resistors with two terminals is electrically equivalent to a single voltage source V and a single series resistor R. For single frequency AC systems the theorem can also be applied to general impedances, not just resistors. The theorem was first discovered by German scientist Hermann von Helmholtz in 1853, but was then rediscovered in 1883 by French telegraph engineer Léon Charles Thévenin (1857-1926).

This theorem states that a circuit of voltage sources and resistors can be converted into a Thévenin equivalent, which is a simplification technique used in circuit analysis. The Thévenin equivalent can be used as a good model for a power supply or battery (with the resistor representing the internal impedance and the source representing the electromotive force). The circuit consists of an ideal voltage source in series with an ideal resistor.

Any black box containing only voltage sources, current sources, and resistors can be converted to a Thévenin equivalent circuit.

1 Calculating the Thévenin equivalent
2 Limitations to the Thévenin equivalent circuit
3 Conversion to a Norton equivalent
4 Example of a Thévenin equivalent circuit
5 In popular culture
6 See also
7 External links

[edit] Calculating the Thévenin equivalent

To calculate the equivalent circuit, one needs a resistance and a voltage - two unknowns. And so, one needs two equations. These two equations are usually obtained by using the following steps, but any conditions one places on the terminals of the circuit should also work:

Calculate the output voltage, V_AB, when in open circuit condition (no load resistor - meaning infinite resistance). This is V_Th.
Calculate the output current, I_AB, when those leads are short circuited (load resistance is 0). R_Th equals V_Th divided by this I_AB.

The equivalent circuit is a voltage source with voltage V_Th in series with a resistance R_Th.

Case 2 could also be thought of like this:

2a. Now replace voltage sources with short circuits and current sources with open circuits.

2b. Replace the load circuit with an imaginary ohm meter and measure the total resistance, R, "looking back" into the circuit. This is R_Th.

The Thévenin-equivalent voltage is the voltage at the output terminals of the original circuit. When calculating a Thevenin-equivalent voltage, the voltage divider principle is often useful, by declaring one terminal to be V_out and the other terminal to be at the ground point.

The Thévenin-equivalent resistance is the resistance measured across points A and B "looking back" into the circuit. It is important to first replace all voltage- and current-sources with their internal resistances. For an ideal voltage source, this means replace the voltage source with a short circuit. For an ideal current source, this means replace the current source with an open circuit. Resistance can then be calculated across the terminals using the formulae for series and parallel circuits.

[edit] Limitations to the Thévenin equivalent circuit

Many, if not most circuits will be non-linear, especially when short-circuited. The Thévenin equivalent is therefore often applicable only over a limited load range.

The Thévenin equivalent is only equivalent from the point of view of the load. The power dissipation within the Thévenin equivalent does not represent the power dissipation within the real system.

[edit] Conversion to a Norton equivalent

A Norton equivalent circuit is related to the Thévenin equivalent by the following equations:

$R_{Th} = R_{No} \!$

$V_{Th} = I_{No} R_{No} \!$

[edit] Example of a Thévenin equivalent circuit

Step 0: The original circuit

Step 1: Calculating the equivalent output voltage

Step 2: Calculating the equivalent resistance

Step 3: The equivalent circuit

In the example, calculating equivalent voltage:

$V_\mathrm{AB} = {R_2 + R_3 \over (R_2 + R_3) + R_4} \cdot V_\mathrm{1}$

$= {1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \over (1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega) + 2\,\mathrm{k}\Omega} \cdot 15 \mathrm{V}$

$= {1 \over 2} \cdot 15 \mathrm{V} = 7.5 \mathrm{V}$

(notice that R₁ is not taken into consideration, as above calculations are done in an open circuit condition between A and B, therefore no current flows through this part which means there is no current flow through R₁ and therefore no voltage drop along this part)

Calculating equivalent resistance:

$R_\mathrm{AB} = R_1 + \left ( \left ( R_2 + R_3 \right ) \| R_4 \right )$

$= 1\,\mathrm{k}\Omega + \left ( \left ( 1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \right ) \| 2\,\mathrm{k}\Omega \right )$

$= 1\,\mathrm{k}\Omega + \left({1 \over ( 1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega )} + {1 \over (2\,\mathrm{k}\Omega ) }\right)^{-1} = 2\,\mathrm{k}\Omega$