Tutorial: Thévenin Equivalents Three Ways
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In this problem, we'll consider finding the Thévenin equivalent for the following circuit, in a number of different ways:
Because this circuit is comprised of only linear components, we know that its i-v constraint will have the form: v = v_{\rm TH} + iR_{\rm TH}, forming a line through the i-v plane whose v-intercept is v_{\rm TH} and whose slope is 1 / R_{\rm TH}.
Our goal here is to characterize that line for this circuit, so that we can replicate it with a circuit of the following form:
In this problem, we'll explore three different ways of characterizing the line representing this circuit's i-v constraint:
- first, by finding two points on the line;
- then, by finding a single point but finding the slope directly;
- and lastly, by holding i or v fixed and solving for the other, finding the equation directly.
1) Method 1: Find Two Points
One method we can use to characterize the line is to find any two points on it. We can do this by setting either i or v to a constant value and solving for the other. Two points that are often relatively straightforward to find are the two intercepts:
- The "v-intercept": v when i=0{\rm A}
- The "i-intercept": i when v=0{\rm V}
We can find the first of these, the v-intercept (also called the "open-circuit voltage") by connecting up a 0{\rm A} current source and measuring the voltage drop across it (remember that the terminal i-v relationship holds regardless of what element(s) we connect up there, so we have free rein to connect up whatever we want that will help us figure out the line):
This is equivalent to leaving the terminals disconnected from anything and solving for the voltage between the terminals:
v_{\rm oc} =~
We can also find the i-intercept by connecting up a 0{\rm V} voltage source and measuring the current through it:
This is equivalent to shorting the terminals together and measuring the current that flows through that short:
When v=0, i =~
Using these two points, we already have our Thévenin voltage (it's exactly that v-intercept value), and we could solve for the slope of the line to find the Thévenin resistance.
It's worth noting that finding any two points on the line would be enough to back-solve for all of the relevant information, and so it would be fine to pick any two points you prefer; we just often choose the intercepts because they allow us to view this as adding a short or an open instead of an arbitrary voltage or current source, which generally makes solving easier.
2) Method 2: Find the Slope Directly
Another approach would be to find the slope of the line directly. We can then use this in conjunction with either of the intercepts to find the complete equation for the line.
We do this by "zeroing out" all of the independent sources in the circuit (in the same way we would zero them out when using superposition).
Consider the following two circuits:
Option A
Option B
Which of these matches the original circuit with the source values set to 0?
R_{\rm eq} =~
This resistance is exactly the Thévenin resistance, which, in conjunction with either of the two points from the previous section, gives us enough information to completely represent the line.
3) Method 3: Find the Whole Equation Directly
It is also generally possible to find the equation directly, rather than finding multiple points or a point and the slope. We can do this by treating either i or v as a known value, and solving for the other (using any of the techniques we have developed for solving circuits so far). Let's see what this looks like.
One thing we could do would be to treat i as a constant and to solve for v in terms of that value i. Doing so gives us an equation that directly relates the two (i.e., the equation of the line we're interested in). Logically, we can think of things by attaching a current source (with known, fixed value i) to our circuit:
And now we can solve for v (in terms of i) using any of the methods we know and love (including superposition if we're so inclined, since there are two sources in the circuit now). Go ahead and do that; when you've got an answer, it should be in the form of v = (\text{a voltage}) + (\text{a resistance})\times i.
v =~
Alternatively, we could hold v constant and solve for i, which we can think of as connecting up a voltage source instead:
Then we can solve for i in terms of this (now known) value v. Let's try that, too:
i =~
From either of these equations, we can pick out the Thévenin voltage and resistance directly.
4) Putting It Together
Using results from any of the methods above, what are the Thévenin equivalent voltage and resistance for the network from above (reproduced here for convenience)?
is equivalent to
5) Summary
In this problem, we've explored three different methods of achieving the same goal: characterizing the linear i-v relationship enforced by the circuit above so that we can find an equivalent circuit (with fewer components) that imposes that same constraint on any external circuit connecting up to it.
It is a good idea, as you move forward with other questions that involve computing i-v curves and/or Thévenin equivalents, to try out more than one of these methods on each problem so that you can get a sense of which strategies work best in what situations! In most of the rest of the problem set solutions/explanations this week, we will only walk through one method of solving. But it is important to keep in mind that, in many cases, we could use more than one method.