Why thevenin theorem

When we examine the Thevenin equivalent circuit shown in figure one, we see a Thevenin-equivalent open circuit voltage VTH. Applying a known voltage to the two nodes causes an amount of current to flow between the nodes through a slight impedance. As a result, we look in and see the amount of current that flows with the linear network independent sources deactivated. When we remove the load and redraw the circuit, VOC equals 10 volts plus the rise across the 10 ohm resistor. We can use a loop equation to find that VOC equals When we move to the final equivalent circuit, we can reinsert the load resistance and find that the load current equals 0.

Looking at the equivalent circuit, we calculate the output voltage VOC with an open circuit condition symbolized by the dotted lines. In the open circuit condition, the lack of a load resistor RL results in infinite resistance or the Thevenin-equivalent voltage VTH. Short circuiting the output terminals gives us the output current IAB. The short circuit condition gives us a value of zero for the load resistance.

In addition, we use the Theorem to analyze single frequency AC circuits that have impedance values rather than resistance values. When you work with high speed PCB layouts, any mismatched impedance causes signals to reflect along the traces. In turn, the reflection causes ringing at the load receiver that reduces dynamic range and results in false triggering. You can eliminate reflections by matching the impedance of the source with the impedance of the trace and with the impedance of the load.

To accomplish impedance matching, you can use a Thevenin voltage divider. Although PSpice cannot directly calculate a Thevenin equivalent of a circuit, PSpice simulations capabilities assist with the rapid calculation by finding the open circuit voltage VOC and the short circuit current values ISC.

After building your circuit in PSpice, you can simulate a short circuit with a small resistance value and then replace the short circuit value with a larger value to obtain VOC. Working through circuit analysis can be a bear to wrestle.

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Dependent sources basically get grouped in with the resistors. Select any two nodes inside the circuit and bring them out to a port. We do the proof based on the principle of superposition. Each sub-circuit contains one source and an arbitrary network of resistors. All the other sources are suppressed. This is what we saw in sub-circuit 2 of the example circuit.

This is sub-circuit 3 of the example circuit. We have all the separate contributions to the port voltage. Now we superimpose add up all the contributions,.

The first two summation terms are voltage values based on the internals of the circuit, with nothing connected to the port. For every one of these sub-circuits the external current source was suppressed by making it an open circuit. We proved any circuit composed of resistors, voltage sources, and current sources can be reduced to a single voltage source and a single resistor. On the left is an arbitrary circuit made of any number of resistors, voltage sources, and current sources.

We select any two internal nodes we are interested in and define a port, brought out to the edge of the circuit. All we need is to remind ourselves how to do source transformation,. We find it the same way: Calculate the resistance looking into the port when all the internal sources are suppressed. Do this by placing a short across the port,. All the internal sources are turned on when you do this. The superposition strategy gave us three types of sub-circuit. Some had one voltage source, some had one current source, and one sub-circuit had no internal sources.

The third sub-circuit, the one with all internal sources suppressed, left us with just the original resistor network. The first publication by Helmholtz did not become widely known. His name is on plenty of great ideas in engineering and physics. They published their work the same month in The theorem may also be called the Mayer-Norton theorem.

Back then engineering discoveries did not travel widely or rapidly. Now we have the Web but these people did not. If you know other names for these theorems I would be interested to hear from you.

Johnson, D. Proceedings of the IEEE. In this article we did the proof with a circuit made of resistors and sources. Hi Willy! Stupendous article. Your chosen orientation of introducing the topic, with a specific example first followed by the general proof, is magnitudes more understandable than the traditional mathematical route of proof first: example later.

Other creators on YouTube like 3Blue1Brown echo this pedagogy as well. We and the rest of our content-creation group had long discussions about how to best convey complex topics with specific-example-first then generalization.

It seems to work. By far the simplest way to prove Norton is to transform the Thevenin equivalent. Be careful here, in the Thevenin proof I used the N subscript to count the current sources inside the black box; it does not represent the Norton equivalent current source. Probably should have used a different letter. Since all good engineers are lazy, doing the source transformation is the preferred route.

Toi - Sorry for the very response. That made-up current source is indeed puzzling. Let me try to justify it. If you look at the example circuit you see a 2k resistor on the far right. If you change the value of that resistor the current i will go up or down. As we work on the solution we want to account for that current i.

We have to account for all possible load resistors. One choice would be to connect a general valued load resistor, RL. This RL allows a voltage and a current to appear at the port. Remember, RL is not inside the Thevenin Equivalent. What are the other choices for things you could connect? Could be a voltage source, or it could be a current source. Either of these will allow a voltage to appear and a current to flow out of the port. The current source turns out to produce a simpler math.

So the real answer is: If we use a resistor load the math is too hard, so a current source works better. Really good explanation! I searched for a proof in youtube but everybody makes practical solution. It is exactly what I want. Mert - Very glad you found this useful. I did a similar search for a clear proof and it was way too hard, so I decided to gather a bunch of them and cast it into my own words, and share it with everyone.

Comments are held for moderation. There will be a delay before they appear. Comments may include Markdown. To share something privately: Contact me. Written by Willy McAllister. Review - solve a circuit with superposition To solve a circuit using superposition, Break the circuit up into multiple sub-circuits. Create one sub-circuit for each independent source. To create a sub-circuit keep one source turned on and turn off or suppress all the others.

To suppress a voltage source, replace it with a short circuit.