Chapter 8

Applying Thévenin’s and Norton’s Theorems

In This Chapter

Simplifying source circuits with Thévenin’s and Norton’s theorems

Using the Thévenin and Norton approach with superposition

Delivering maximum power transfer

Thévenin and Norton equivalent circuits are valuable tools when you’re connecting and analyzing two different parts of a circuit. In this chapter, one part of the circuit, called the source circuit, delivers signals and interacts with another part, dubbed a load circuit. The interaction between the source and load circuits offers a major challenge when analyzing circuits.

Fortunately, Thévenin’s theorem and Norton’s theorem simplify the analysis. Each theorem allows you to replace a complicated array of independent sources and resistors, turning the source circuit into a single independent source connected with a single resistor. As a result, you don’t have to reanalyze the entire circuit when you want to try different loads — you can just use the same source circuit.

This chapter reveals just how Thévenin’s and Norton’s theorems work. It also shows you how to use the superposition technique (which I present in Chapter 7) to find equivalent Thévenin and Norton circuits when a circuit has multiple sources. Last but not least, I explain how to apply the Thévenin or Norton equivalent to show how to deliver maximum power to a load circuit.

Showing What You Can Do with Thévenin’s and Norton’s Theorems

Trying out and replacing devices in a circuit can be dull and dreary work. But you can minimize the toil by replacing part of the circuit with a simpler but equivalent circuit using Thévenin’s or Norton’s theorem.

These theorems come in handy when you’re faced with circuits like the one in Figure 8-1. Note the following parts of the circuit:

The circuit to the left of Terminals A and B is the source circuit. It’s a linear circuit with an array of voltage sources and resistors.

The source circuit delivers a signal to the load circuits, which are to the right of Terminals A and B.

The terminals at A and B make up the interface between the source circuit and the load circuits.

To find the voltage across each device in the load circuit, you’d usually have to connect Devices 1, 2, and 3 one at a time to get three different answers for three different loads. Talk about tedious!

Illustration by Wiley, Composition Services Graphics

Figure 8-1: A source circuit with multiple load circuits.

Here’s where Thévenin’s theorem comes to the rescue. Thévenin’s theorem lets you replace the source circuit — a linear array of devices having multiple independent sources and resistors — with a single voltage source connected in series with a single resistor. Figure 8-2 replaces the source circuit from Figure 8-1 with a simplified circuit, which has a Thévenin voltage source, vT, connected in series with a Thévenin resistor, RT. The Thévenin equivalent is useful when the devices in the load circuits are connected in series.

Illustration by Wiley, Composition Services Graphics

Figure 8-2: Transform-ing the source circuit into a Thévenin equivalent.

Norton’s theorem likewise allows you to simplify the source circuit. Specifically, Norton’s theorem says you can replace a linear array of devices having multiple independent sources and resistors with a single current source in parallel with a single resistor. Check out Figure 8-3 for an example. It shows the linear source circuit being replaced with a simplified circuit that has one current source, iN, and one resistor, RN, connected in parallel. The Norton equivalent is useful when you want to try loads that have devices connected in parallel.

Illustration by Wiley, Composition Services Graphics

Figure 8-3: Replacing the source circuit with a Norton equivalent.

Finding the Norton and Thévenin Equivalents for Complex Source Circuits

Norton’s theorem and Thévenin’s theorem say essentially the same thing. The Norton equivalent is the Thévenin equivalent with a source transformation (I cover source transformation in Chapter 4).

To find the Thévenin or Norton equivalent of a linear source circuit, calculate the following variables at interface Terminals A and B:

Thévenin voltage source, vT: This equals open-circuit voltage:

vT = voc

Norton current source, iN: This equals short-circuit current:

iN = isc

Thévenin resistance, RT, or Norton resistance, RN: This resistance equals open-circuit voltage divided by short-circuit current:

So where do these equivalents come from? Resistor loads can have a wide variety of resistor values, ranging from a short circuit having zero resistance to an open circuit having infinite resistance. These extreme ends of the resistance spectrum are convenient when you’re analyzing circuits because you can easily find the Thévenin voltage vT by having an open-circuit load, and you can get the Norton current iN by having a short-circuit load.

Figure 8-4 shows a source circuit and its Thévenin equivalent. The top diagram shows the open circuit load you use to find the open-circuit voltage, voc, across Terminals A and B. The bottom diagram shows the short-circuit load you use to find the short-circuit current, isc, through Terminals A and B. With the open-circuit voltage, voc, and the short-circuit current, isc, you can find the Thévenin or Norton resistance (RT = RN).

Illustration by Wiley, Composition Services Graphics

Figure 8-4: Finding the Thévenin equivalent using open-circuit loads (top) and short-circuit loads (bottom).

In the following sections, I show you how to apply Thévenin’s and Norton’s theorems, and I show you how to use source transformation to go from one equivalent to another. I also offer an alternate way of finding RT or RN: finding the total resistance between Terminals A and B by removing all the independent sources of the source circuit.

Applying Thévenin’s theorem

To simplify your analysis when interfacing between source and load circuits, the Thévenin method replaces a complex source circuit with a single voltage source in series with a single resistor. To obtain the Thévenin equivalent, you need to calculate the open-circuit voltage voc and the short-circuit current isc.

Finding the Thévenin equivalent of a circuit with a single independent voltage source

Circuit A in Figure 8-5 is a source circuit with an independent voltage source connected to a load circuit. Circuit B shows the same circuit, except I’ve replaced the load circuit with an open-circuit load. You use the open-circuit load to get the Thévenin voltage, vT, across Terminals A and B. The Thévenin voltage equals the open-circuit voltage, voc.

Illustration by Wiley, Composition Services Graphics

Figure 8-5: Using the Thévenin equivalent.

The voltage is driven by a voltage source for this series circuit, so use the voltage divider technique (from Chapter 4) to get voc:

Solving for voc gives you the Thévenin voltage, vT.

Circuit C shows the same source circuit as a short-circuit load. You use the short-circuit load to get the Norton current, iN, through Terminals A and B. And you find the Norton current by finding the short-circuit current, isc.

In Circuit C, the short circuit is in parallel with resistor R2. This means that all the current coming out from resistor R1 will flow through the short because the short has zero resistance. In other words, the short bypasses R2. You can find the current through Terminals A and B using Ohm’s law, producing the short-circuit current:

This short-circuit current, isc, gives you the Norton current, iN.

Finally, to get the Thévenin resistance, RT, you divide the open-circuit voltage by the short-circuit current. You then wind up with the following expression for RT:

Simplify that equation to get the Thévenin resistance:

Circuit D shows the Thévenin equivalent for the source circuit in Circuit A.

The preceding equation looks like the total resistance for the parallel connection between resistors R1 and R2 when you short (or remove) the voltage source and look back from Terminals A and B.

When looking to the left from the Terminals A and B, you can find the Thévenin resistance RT by removing all independent sources by shorting voltage sources and replacing current sources with open circuits. After getting rid of the independent sources, you can find the total resistance between Terminals A and B, shown in Circuit E of Figure 8-5. (Note that this tactic only works when there are no dependent sources.)

Applying Norton’s theorem

To see how to use the Norton approach for circuits with multiple sources, consider Circuit A in Figure 8-6. Because it doesn’t matter whether you find the short-circuit current or the open-circuit voltage first, you can begin by determining the open-circuit voltage. Putting an open load at Terminals A and B results in Circuit B. The following analysis shows you how to obtain is1 and RN in Circuit B.

Illustration by Wiley, Composition Services Graphics

Figure 8-6: Applying the Norton equivalent.

Applying Kirchhoff’s voltage law (KVL) in Circuit A lets you determine the open-circuit voltage, voc. KVL says that the sum of the voltage rises and drops around the loop is zero. Assuming an open circuit load for Circuit A, you get the following KVL equation (where the load is an open circuit, v = voc):

Algebraically solve for voc to get the open-circuit voltage:

The current supplied by the voltage source vs goes through resistors R1 and R2 because the current going through an open circuit load is zero. In Circuit B, you can view the current source is as a device having an infinite resistance (that is, as an open circuit). However, all the current provided by the current source is will go through R1 and R2, and none of the current from is will go through the open-circuit load. Applying Ohm’s law (v = iR), you have the following voltages across resistors R1 and R2:

The minus sign appears in these equations because the current from is flows opposite in direction to the assigned voltage polarities across the resistors.

Substitute v1 and v2 into the expression for voc, and you wind up with the following open-circuit voltage:

The open-circuit voltage is equal to the Thévenin equivalent voltage, voc = vT.

Next, find the short-circuit current in Circuit C of Figure 8-6. The current is1 supplied by the voltage source will flow only through resistors R1 and R2, not through the current source is, which has infinite resistance. Because of the short circuit, the resistors R1 and R2 are connected in series, resulting in an equivalent resistance of R1 + R2. Applying Ohm’s law to this series combination gives you the following expression for is1 provided by the voltage source vs1:

Kirchhoff’s current law (KCL) says that the sum of the incoming currents is equal to the sum of the outgoing currents at a node. Applying KCL at Node A, you get

Substituting the expression for is1 into the preceding KCL equation gives you the short-circuit current, isc:

The Norton current iN is equal to the short-circuit current: iN = isc.

Finally, divide the open-circuit voltage by the short-circuit current to get the Norton resistance, RN:

Plugging in the expressions for voc and isc gives you the Norton resistance:

Adding the terms in the denominator requires adding fractions, so rewrite the terms so they have a common denominator. Algebraically, the equation simplifies as follows:

When you look left from the right of Terminals A and B, the Norton resistance is equal to the total resistance while removing all the independent sources. You see the Norton equivalent in Circuit D of Figure 8-6, where RT = RN.

Using source transformation to find Thévenin or Norton

In this section, I show you how to apply Thévenin’s and Norton’s theorems to analyze complex circuits using source transformation.

You commonly use the Thévenin equivalent when you want circuit devices to be connected in series and the Norton equivalent when you want devices to be connected in parallel with the load. You can then use the voltage divider technique for a series circuit to obtain the load voltage, or you can use the current divider technique for a parallel circuit to obtain the load current.

A shortcut: Finding Thévenin or Norton equivalents with source transformation

To transform a circuit using the Thévenin or Norton approach, you need to know both the Thévenin voltage (open-circuit voltage) and the Norton current (short-circuit current). But you don’t need to find the Thévenin and Norton equivalent circuits separately. After you figure out the Thévenin equivalent for a circuit, you can find the Norton equivalent using source transformation. Or if you figure out the Norton equivalent first, source transformation lets you find the Thévenin equivalent. (I cover source transformations in Chapter 4.)

For example, if you already have the Thévenin equivalent circuit, then obtaining the Norton equivalent is a piece of cake. You perform the source transformation to convert the Thévenin voltage source connected in series with the Thévenin resistance into a current source connected in parallel with the Thévenin resistance. The result is the Norton equivalent: The current source is the Norton current source, and the Thévenin resistance is the Norton resistance.

Finding the Thévenin equivalent of a circuit with multiple independent sources

You can use the Thévenin approach for circuits that have multiple independent sources. In some cases, you can use source transformation techniques to find the Thévenin resistor RT without actually computing voc and isc.

For example, consider Circuit A in Figure 8-7. In this circuit, the voltage source vs and resistors R1 and R2 are connected in series. When you remove independent sources vS1 and is in Circuit A, this series combination of resistors produces the following total Thévenin resistance:

Illustration by Wiley, Composition Services Graphics

Figure 8-7: Application of a Thévenin equivalent circuit with multiple sources.

You can then use source transformation to convert the Thévenin voltage source, which is connected in series to Thévenin resistance RT, into a current source that’s connected in parallel with RT. Here’s your current source:

Circuit B shows the transformed circuit with two independent current sources.

Because independent current sources are in parallel and point in the same direction, you can add up the two source currents, which produces the equivalent Norton current, iN:

Circuit B shows the combination of the two current sources. When you combine the two current sources into one single current source connected in parallel with one resistor, you have the Norton equivalent.

You can convert the current source iN in parallel with RT to a voltage source in series with RT using the following source transformation equation:

Circuit C is the Thévenin equivalent consisting of one voltage source connected in series with a single equivalent resistor, RT.

Finding Thévenin or Norton with superposition

When a complex circuit has multiple sources, you can use superposition to obtain either the Thévenin or Norton equivalent. As I explain in Chapter 7, superposition involves determining the contribution of each independent source while turning off the other sources. After determining the contribution of each source, you add up the contributions of all the sources.

This section shows you how to use superposition to find the Thévenin equivalent, but the process for finding the Norton equivalent is essentially the same. You simply find the Thévenin equivalent — consisting of one voltage source connected in series with a resistor — and get the Norton equivalent through a source transformation.

To see how superposition can help you obtain the Thévenin equivalent, consider Circuit A of Figure 8-8. To find the open-circuit voltage due to only the voltage source vs, you turn off the current source is by removing it from Circuit A. Circuit B is the resulting circuit.

Because of the open-circuit load in Circuit B, no current will flow through resistors R1 and R2. And because there’s no current flow, the voltage drop across each of these two resistors is equal to zero, according to Ohm’s law (v = iR). The open-circuit voltage due to vs, denoted as voc1, is therefore equal to vs. Mathematically, you can write

To find the open-circuit voltage contribution due to the current source is, you turn off the voltage source vs by replacing it with a short circuit, which has zero resistance. You see the resulting circuit in Circuit C.

Because no current flows through the open-circuit load, is flows through resistors R1 and R2. The open-circuit voltage across Terminals A and B, denoted as voc2, is equal to the voltage drop across the two resistors. Using Ohm’s law, you have the following expression for voc2:

Illustration by Wiley, Composition Services Graphics

Figure 8-8: Using the superposition method to get the Thévenin equivalent.

Adding up voc1 and voc2 gives you the total open-circuit output contribution due to vs and is:

The open-circuit voltage equals the Thévenin voltage source (voc = vT).

To find the short-circuit current due to only voltage source vs, you turn off the current source is by removing it from Circuit A. The result is Circuit D.

Current flows through resistors R1 and R2 because the short connects the resistors in series. The series combination gives you an equivalent resistance of R1 + R2. Applying Ohm’s law, you get the following expression for isc1, which is the short-circuit current due to vs:

To find the short-circuit current contribution due to only current source is, you turn off the voltage source vs by replacing it with a short circuit, which has zero resistance. The result is Circuit E. Because of the short circuit, all the current provided by is flows through Terminals A and B. In other words, isc2, which is the short-circuit current resulting from is, equals the source current is:

Adding up isc1 and isc2 gives you the total contribution due to vs and is:

This short-circuit current equals the Norton current source (iN = isc).

Divide the expression for voc by the expression for isc, and you get the Thévenin resistance:

This equation simplifies as follows:

The Thévenin resistance is equal to the total resistance of the series combination when you’re looking left from the right of Terminals A and B while removing all the independent sources.

You can see the Thévenin equivalent in Circuit F of Figure 8-8. The Thévenin equivalent reduces the source circuit of Circuit A to one voltage source in series with one resistor.

With the superposition method, I get the same expressions for voc, isc, and RN as I get using source transformation (see the preceding section). You can use whichever method works best for you.

Gauging Maximum Power Transfer: A Practical Application of Both Theorems

The power p coming from the source circuit to be delivered to the load depends on both the current i flowing through the load circuit and the voltage v across the load circuit at the interface between the two circuits.

The maximum power theorem states that for a given source with a fixed Thévenin resistance RT, the maximum power delivered to a load resistor RL occurs when the RL is matched or equal to RT:

pmax when RL = RT

Mathematically, the power is given by the following expression:

p = iv

The source circuit delivers maximum voltage when you have an open-circuit load. Because zero current flows through the open-circuit load, zero power is delivered to the load. Mathematically, the power poc delivered to the open-circuit load is

poc = iv = 0 · v = 0

On the other hand, the source circuit delivers maximum current when you have a short-circuit load. Because zero voltage occurs across the short-circuit load, zero power is delivered to the load. Mathematically, the power psc delivered to the short-circuit load is

psc = iv = i · 0 = 0

So what’s the maximum power punch delivered for a given load resistance? Using either the Thévenin or Norton approach allows you to find the maximum power delivered to the load circuit.

To see how to determine the maximum power, look at the resistor arrangements for both the source and load circuits in Figure 8-9. In this figure, the source circuit is the Thévenin equivalent, and the load resistor is a simple but adjustable resistor.

Figure 8-9: Determining maximum power delivery.

Illustration by Wiley, Composition Services Graphics

Intuitively, you know that maximum power is delivered when both the current and voltage are maximized at the interface Terminals A and B.

Using voltage division (see Chapter 4), the voltage v across the interface at A and B is

In Figure 8-9, the connected circuit between the source and load is a series circuit. The current i flows through each of the resistors, so

Substituting the values of v and i into the power equation, you wind up with the following power equation:

Determining the maximum power delivered to the load means taking the derivative of the preceding equation with respect to RL and setting the derivative equal to zero. Here’s the result:

This equation equals zero when the numerator is zero. This occurs when RL = RT. Therefore, maximum power occurs when the source and load resistances are equal or matched.