26.1 The Need for a Model

In order to connect circuits together, we need to be able to describe, in general terms, the ways that circuits fit together. When dealing with transistors and other power amplification devices, we often need to come up with a simplified model for how the input to a circuit or the output from a circuit behaves. Early on (in Chapter 7), we learned how to take multiple resistors in series and parallel and combine them into an equivalent single resistor.

When dealing with a power amplification circuit, it is often necessary to look at various parts of the circuit by themselves and figure out how they look to other parts of the circuit. The way that a partial circuit looks to other parts of the circuit is called the circuit’s Thévenin equivalent circuit .

Note that the single voltage source may be different than the voltage source that is actually connected. What you are doing is seeing what the circuit looks like to another circuit. For instance, a voltage divider circuit makes the output of the voltage divider look like it is coming from a lower voltage source.

Figure 26-1 shows a circuit and its Thévenin equivalent circuit . For purposes of thinking about and understanding the relationship between the circuit and things attached to the circuit, we can view the circuit as being the same as its Thévenin equivalent. Thus, having a Thévenin equivalent circuit greatly simplifies our modeling, calculating, and understanding of how circuits work together.

Any network of power sources and resistances can be converted into a Thévenin equivalent circuit. You can also get a Thévenin equivalent circuit for a circuit that includes capacitors and inductors, but the calculations become more difficult and the results are only valid for a specific frequency (each frequency will have a different Thévenin equivalent circuit). For simplicity, we will just focus on resistive circuits.

26.2 Calculating Thévenin Equivalent Values

To see how to calculate the voltage and resistance for a Thévenin equivalent circuit, this section will take a classic voltage divider circuit and analyze how it “looks” to other attached circuits. Figure 26-2 shows an example of a partial circuit. Like most partial circuits, this circuit has two output points—A and B. What we are wanting to know is this—if we attach another circuit up to A and B, is there a model that we can use to understand how the other circuit “sees” our circuit? The goal of making a Thévenin equivalent circuit is to understand what our circuit will look like to other attached circuits.

So, since our Thévenin equivalent circuit will have a voltage source and a single resistor, we need to calculate what the voltage and resistance of this circuit will be. To calculate the voltage, find out what the voltage of the circuit at the output is when there is nothing connected. That is, if we were to leave A and B disconnected and I were to connect my multimeter to A and B, what would the voltage be? This is your Thévenin voltage. Since this is a voltage divider, you can just use normal voltage divider calculations to find this out. In this case, we have a 12 V source, and the voltage divider divides it exactly in half (1 kΩ for each half). Therefore, the output voltage is 6 V. Therefore, our Thévenin equivalent circuit will have a 6 V source.

Now we need to find our Thévenin resistance. There are multiple tricks to do this, but the simplest one is to replace all voltage sources in your circuit with a wire (i.e., a short circuit) and simply compute the total resistance between A and B.¹

Figure 26-3 shows what this looks like. Therefore, to calculate the Thévenin resistance of this circuit, simply calculate the total resistance from A to B. In this case, there are two parallel paths from A to B—one through the first resistor and one through the second. Therefore, we add up the resistors as parallel resistances. As a result, our Thévenin resistance will be

Therefore, we would say that this partial circuit has a Thévenin voltage of 6 V and a Thévenin resistance of 500 Ω. Whenever we attach a circuit to this circuit, what that other circuit will “see” is a circuit like the one in Figure 26-4.

If you wanted to prove this to yourself, you can imagine a variety of different circuits attached to both our original circuit and to the Thévenin equivalent circuit. You will find that, in all cases, the amount of voltage and current the Thévenin equivalent circuit provides to the other circuit is the exact same as what the original circuit will provide.

That isn’t to say that the circuits themselves are exactly equivalent. Our original voltage divider uses up a lot of current stepping down the voltage of the voltage source. Not only does that waste energy from our battery but it probably also causes a lot of heat. However, any subcircuit that gets attached to A and B will see both our original circuit and the Thévenin equivalent circuit as providing the same output.

26.3 Another Way of Calculating Thévenin Resistance

There is another way of calculating Thévenin resistance. In this method, we first calculate what the current would be if you shorted A to B directly with a wire. This is known as the short-circuit current, or I_SHORT . Then, after calculating this, you can divide the Thévenin voltage by I_SHORT to obtain the Thévenin resistance.

When doing this, you have to remember that anything in parallel with our short will be essentially ignored—the current will always want to go through our short circuit.

Figure 26-5 shows what this looks like. What we want to do is to calculate the current going from A to B. Since A to B is a short circuit in parallel with our second resistor, we know that all of the current will prefer the short circuit. This means that the current going through A and B will simply be the current that is limited by the first resistor.

So, since we have a 12 V source and a 1 kΩ resistor, the short-circuit current will be

Now, to determine the Thévenin resistance, we divide the Thévenin voltage by this number:

As you can see, this is the same value that we got from the previous method.

26.4 Finding the Thévenin Equivalent of an AC Circuit with Reactive Elements

If a circuit has reactive elements (inductors and capacitors), we have to do a little more work to find the Thévenin equivalent circuit.

For DC circuits, this is relatively simple. Since capacitors block DCs and inductors are a short circuit for DCs, we can simply treat the capacitors as open circuits (i.e., unconnected) and treat the inductors as short circuits (simple wires). For AC circuits, you can get a feel for what this will be by assuming the opposite—that capacitors will be short circuits and inductors will be open circuits.

However, if you were to try to solve it explicitly, the problem is a little more difficult. The problem is that a full analysis of such circuits requires math involving complex numbers (i.e., numbers involving the imaginary unit i). While the technique is roughly equivalent to adding resistances in series and parallel as we have done before, it is much more difficult to do the math with complex numbers.

For the purposes of this book, the previous statements about DC and AC should suffice for a general understanding of how your circuit works.

26.5 Using Thévenin Equivalent Descriptions

Many circuits are described to their users using Thévenin equivalent descriptions. For instance, many circuits are described by their input or output impedance. This gives you a rough guide to imagine what will happen if you connect your own circuit to such circuits.

Imagine that you have a circuit that has a Thévenin equivalent output impedance of 500 Ω. If you connect an output circuit that only has 250 Ω of resistance, what do you think that will do to the signal? Well, since the output of the circuit is equivalent to going through a 500 Ω resistor (that’s what Thévenin equivalence means), then if I connect a 250 Ω resistor, then I will have created a voltage divider in which two-thirds of the voltage will be dropped by the circuit I am connecting to and I will only get one-third of the output voltage.

On the other hand, if the impedance of my circuit is 50, 000 Ω, then the voltage drop coming out of the circuit in question is negligible compared to the voltage drop within my circuit. This means that my circuit will essentially receive the full Thévenin equivalent voltage.

We can also use this to calculate the amount of current that our circuit will draw. Let’s say that a circuit yields a Thévenin equivalent output of 4 V with an 800 Ω impedance. If I connect a 3,000 Ω output circuit, how much current will flow? The total resistance will be 3,800 Ω, so the current will be V/R = 4/3800 ≈ 1.05 mA.

26.6 Finding Thévenin Equivalent Circuits Experimentally

In addition to using circuit schematics to determine Thévenin equivalent circuits, it is also possible to determine them experimentally. This way, if you are unsure of the input or output characteristics of your device, you can measure it yourself. The problem with measuring it yourself is that it requires attaching a load to the circuit. Some circuits will fry if a wrongly sized load is attached. You have been warned.

The easiest way to determine Thévenin equivalency experimentally is rather unsafe, but it will help us understand better why the method works. Imagine a voltage divider where the bottom resistor has an extremely large resistance—say, 100 MΩ. In such a voltage divider, the bottom resistor will have almost the entirety of the voltage drop, right? In fact, if the bottom resistor was infinite, it would in fact have all of the voltage drop.

Because of this, we can determine the Thévenin equivalent voltage by measuring the output voltage when there is nothing connected, because no connection means that there is infinite resistance between the output and ground. Measuring this value will give us the Thévenin equivalent voltage.

Therefore, our Thévenin equivalent circuit is 8 V with an impedance of 800 Ω.

The problem with this method is that you don’t normally want to short-circuit your output. Additionally, some circuits require some sort of a load to work properly. In order to adjust to such scenarios, there is a set of equations that allow us to measure the voltage drop across a large and small resistance (instead of infinite and no resistance) and come up with a Thévenin equivalent circuit.

The equations are a little complex, but you can actually derive them directly from Ohm’s law if you work at it (see Appendix D, Section D.6, “The Thévenin Formula,” to see it derived). The first one calculates the Thévenin equivalent voltage (V_T) from the voltage with a high resistance (V_H), the high resistance value (R_H), the voltage with a low resistance (V_L), and the low resistance value (R_L):

(26.1)

Then, we can calculate the Thévenin equivalent resistance:

(26.2)

Next we can find the Thévenin equivalent resistance using Equation 26.2:

Therefore, our unknown circuit has a Thévenin equivalent voltage of 8 V and a Thévenin equivalent impedance of 600 Ω.

What makes this method valuable is that it allows a way to experimentally determine the Thévenin equivalent of a partial circuit that you don’t have a schematic for or for which determining the Thévenin equivalent circuit might be difficult due to nonlinear components such as transistors.

Review

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