Thevenin Theorem - Finding a Thevenin Equivalent Circuit | CircuitBread Circuits 1
Thevenin’s Theorem (or Thévenin’s Theorem, if you know the right shortcuts on your keyboard) is a fan favorite of circuits professors around the world. Used in conjunction with Kirchhoff’s Current Law and Kirchhoff's Voltage Law, Thevenin’s Theorem can simplify the analysis of any circuit. In this tutorial, we’ll focus on DC circuit analysis but just as KCL and KVL apply to both DC and AC, Thevenin’s does as well.
What Thevenin’s does is identify the load that you care about, and simplify or reduce everything else. Multiple sources and resistors will be converted into a single voltage source and series resistor. If you need to iterate or vary your load, this makes it much easier to do the calculations. So you can convert a circuit that looks like this:
Into something that looks like this:
The neat part about this is that, from the viewpoint of the load, these circuits are perfectly equivalent. No matter how you change the load in either circuit, the voltage across and current through the load will be the same in both circuits.
With that, let’s go over the steps to make this conversion and then a few samples to show this in action.
- Review your circuit, identify your load and the nodes it is connected to.
- Remove the load resistor.
- Find RTh by shorting all voltage sources and by open circuiting all the current sources and then see what the resistance looks like from the point of view of the nodes where the load resistor was located.
- Find VTh by finding the voltage across the nodes the load was originally hooked to, using standard circuit analysis methods.
- Replace the load and find the current flowing through the load with these new values.
- Sanity check!
Before we get into the examples and see these steps in action, I’d like to emphasize that during the analysis, voltage sources become shorts while current sources become open circuits. The way to remember this intuitively is to think what would happen if you had a 0V voltage source - it would act like a wire, right? We assume there is no voltage across a wire no matter how much current there is, so if we want to remove a voltage source from a circuit, we replace it with a wire. But what if we have a 0A current source? No current can flow, no matter how much voltage, so it’s an open circuit when we want to remove it from the circuit. Hopefully, this helps you keep these two straight!
Alright, let’s get into the first of our two examples.
#1 Easy Example
1. Looking over this circuit, we can see the voltage source on the left and the load resistor on the right. The load resistor is attached to ground on one side and is connected to the same node as the 100, 200, and 300 ohm resistors. We also take this time to recognize that, in this case, being a DC circuit, the capacitor can be treated as an open.
2. Remove the load resistor.
3. Once the load resistor is gone, you short the voltage source, and if we had a current source, we would replace that with an open. When that’s done, we verify where the nodes of the resistors are attached and find the equivalent resistance of the remaining resistors. In this case, we find the resistance to be 100||200||300 in parallel, which gives us 54.5 ohms.
4. To find the Thevenin equivalent voltage, we put the voltage source back in and find the voltage across the still open spot where we had the resistor. We were able to simplify the circuit by finding the equivalent parallel resistance of the 200 and 300 ohm resistors, and then we noticed that the voltage across those resistors is actually the Thevenin Voltage. With some simple Ohm’s Law, we calculate it out to 5.45V. (1)
5. And here is the Thevenin equivalent circuit that you can hook the load back into and, from any load’s perspective, will provide the same voltage and current as the original circuit.
Despite there being a few steps and the details being easy to forget if you don’t go through this very often, the process is quite straightforward. Let’s go onto a slightly more complicated scenario.
#2 The Slightly Harder Example
1. Again, we start with our circuit and figure out what is going on. We have the load on the right and we have both a voltage source and a current source. We note that we can put the 300 and 200 ohm resistors in parallel but once we remove the sources, it will simplify even further. We also note that the load resistor is attached to what we will consider ground and shares nodes with the 100 ohm resistor and the 1mA current source. (2)
2. Remove the load resistor.
3. Now we are looking for the Thevenin equivalent resistance and to do so, we remove both power sources. The removal of the voltage source puts the 200 ohm resistor in parallel with the 300 and other 200 ohm resistors, but, interestingly enough, none of that matters. As we remove the 1mA current source, it isolates the 100 ohm resistor and, from the perspective of the load resistor, eliminates the others. This leaves us with the Thevenin equivalent resistance of 100 ohms.
4. Now we need to find the Thevenin equivalent voltage, which we do by finding the voltage across the nodes where the load would be connected. As seen above, the load is in parallel with the 100 ohm resistor, so we know that whatever the voltage is across the 100 ohm resistor will be our Thevenin voltage. Since the 1mA current source is located where it is, we know that the current through the 100 ohm resistor will be 1mA. Using Ohm’s Law, we see that the voltage comes out to be -.1V. This gives us our Thevenin Voltage of -.1V.
5. Now we put those values in as we create our equivalent Thevenin Circuit. As you can see, the resulting circuit is quite simple and, if I did everything right, will act exactly like the original circuit from the perspective of the load. Put that load back in and start messing around with it. Very exciting!
But what about dependent sources? If we have a dependent source, we will need to treat the circuit slightly differently. For step 2, when calculating the equivalent resistance, instead of removing all of the sources and finding the equivalent resistance of the remaining components, short circuit the load. Replace the load with a short circuit and then calculate the current through that short circuit. Once you know the short circuit current, you can use Ohm’s law, where RTh = VTh / Isc. There is another method for dealing with dependent sources and, as far as we can tell, they yield the same results, so don’t be surprised if you learned a different method than this somewhere else.
And that’s it! Thevenin’s Equivalent Circuits are very easy to create and the process is simple, you just need to remember the steps. Next, we’ll learn about Norton’s Theorem, which is related but slightly different.
(1) Ohm’s Law - https://www.britannica.com/science/Ohms-law
(2) Nodes - https://academy.binance.com/en/articles/what-are-nodes
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