# Thevenin analysis

I have a hard time understanding why Thevenin analysis works. I can see the math, and I can find the equivalent resistor, but I am not sure why it is a valid analysis from a practical point of view. We are computing the open-circuit voltage across the two points of interest, which basically we are calculating for the maximum voltage across the two points. And, when we short those two points to find the short-circuit current, we are finding the current for the minimum voltage drop across the two points.
So, technically, we are finding a maximum/minimum voltage range between the two points of interest. But why can this range can be locked down to one voltage source in series with a resistor to replace that part of the circuit?

An important point you are missing is that the short-circuit current is not measured at a minimum voltage but at exactly zero volts. It is an ideal short circuit.

These two measurements give you two points on a plot of current versus voltage. One is the voltage with zero current and the other is the current with zero voltage. If the circuit is linear then these two points can be connected with a straignt line to give the complete current/voltage output graph of the circuit. Therefore, the circuit is equivalent to an ideal voltage source (of the open circuit voltage) in series with a resistor (determined by the slope of V/I).

You can actually use any two measurements of current and voltage to find the slope of the line, the open circuit voltage and short circuit current are usually the easiest to determine by circuit analysis. If making actual measurements you just need to keep the points far enough apart that measurement error doesn't swamp the slope calculation.

• Thanks, now I can see why they picked the two points, so that they can ccalculate the slope. But now, by having the resistance, why the voltage source is the open circuit voltage? Why it can't be some random voltage on that straight line? Jan 3, 2014 at 20:52
• The open-circuit voltage is the highest voltage that the circuit can possibly provide. So, you need an ideal voltage source that provides that voltage and nothing less. Jan 3, 2014 at 20:54
• @Rudy01: The Thevenin voltage is the open-circuit (zero-current) voltage, because that's the only condition under which the voltage drop across the resistance is zero. Jan 3, 2014 at 20:55
• @JoeHass: It's misleading to call the Thevenin voltage the "highest" voltage. Thevenin analysis works even in circuits that contain other sources, which could potentially (pun intended) force current "backwards" through the Thevenin source. Jan 3, 2014 at 20:59
• @DaveTweed Thanks, I should have said the "largest voltage difference that can appear at the terminals of the Thevenin equivalent circuit". Internal voltages of the original, non-Thevenin equivalent, circuit can indeed be higher. Jan 3, 2014 at 21:03

but I am not sure why it is a valid analysis from a practical point of view

When you consider the Thevenin representation of a circuit, note that the output voltage, as a function of the output current is, by inspection:

$$V_O = V_{TH} - I_O \cdot R_{TH}$$

Thus, when the output current is zero, i.e., the load is an open-circuit, the corresponding open-circuit output voltage is

$$V_{OC} = V_{TH} - 0\cdot R_{TH} = V_{TH}$$

This establishes that the open-circuit output voltage of an arbitrary linear DC circuit must be the Thevenin equivalent voltage.

When the output voltage is zero, i.e., the load is a short-circuit, the corresponding short-circuit output current is:

$$V_O = 0 = V_{TH} - I_{SC} \cdot R_{TH} \Rightarrow I_{SC} = \dfrac{V_{TH}}{R_{TH}}$$

This establishes that the short-circuit output current of an arbitrary linear DC circuit must be the Thevenin equivalent voltage divided by the Thevenin equivalent resistance.

But why can this range can be locked down to one voltage source in series with a resistor to replace that part of the circuit?

Because we are dealing with a linear DC circuit (Thevenin's theorem holds only for linear circuits) and thus $\dfrac{\Delta V_O}{\Delta I_O}$ must be constant, i.e., the slope of the output IV curve must be constant.

Finally, recall that there is the Norton representation too which is dual to the Thevenin representation and consists of a current source in parallel with a resistor such that the output current is given by the equation:

$$I_O = I_N - \dfrac{V_O}{R_{TH}}$$

Clearly, we have

$$I_N = I_{SC}$$

Given a linear circuit with two terminals, the equation relating the voltage and current at its terminals has the form

v=v_1+R·i (1)

Clearly, there exists a circuit containing one voltage source v_1 and one resistance R in series that has the same i-v relationship. This is called the Thevenin equivalent. There is no simpler circuit with the same i-v characteristic (although there is another one, the Norton equivalent, which is as simple as this).

To find the Thevenin equivalent, you can match both i-v relations as you like. For the linear relationship in (1) you just need any two data points. For computational convenience, you can set i=0, because this often simplifies computing one pair of points: (i=0,v=v_oc). Setting v=0 also often simplifies the circuit because one node vanishes, easing the computation of the second point: (i=i_sc,v=0).

But you could also try to find points on the line making other experiments (for instance when the previous easy ones could destroy the circuit). In RF circuits you often put a 50 ohm resistance at the terminals of the circuit and try to match your circuit i-v line with that of the Thevenin equivalent. Or you try putting different resistance values at the circuit terminals and look when the voltage is v_oc/2: this is exactly the Thevenin resistance value...

So, we are not finding a maximum/minimum voltage range but matching two i-v curves or, better, straight lines.