# Thévenin's theorem - how to

We proved Thévenin’s and Norton’s theorems in the previous article. This article covers the practical steps to create a Thévenin or Norton equivalent for a linear circuit.

**Thévenin’s theorem**

Any network of resistors and sources, when viewed from a port, can be simplified down to one voltage source in series with one resistor.

**Norton’s theorem**

Any network of resistors and sources, when viewed from a port, can be simplified down to one current source in parallel with one resistor.

The proof and the practical design method are separate ideas. You often find them mixed together in many texts.

Written by Willy McAllister.

### Contents

### Where we’re headed

To create a Thévenin or Norton equivalent,

- Pick two nodes to be the port of the circuit you want to transform.
- Remove the external component(s) connected to the port.
- Find any two of these three: $\bold R, v_{oc}, i_{sc}$ — whichever two are the easiest,
- $\bold R$ — Suppress the internal sources and simplify the resulting resistor network down to a single resistor.
- $v_{oc}$ — Restore internal sources, leave the port open, find the open-circuit voltage.
- $i_{sc}$ — Restore internal sources, short across the port, find the short-circuit current.

- The third variable comes from the other two,
- $\bold R = v_{oc}/i_{sc}$
- $v_{oc} = i_{sc}\,\bold R$
- $i_{sc} = v_{oc}/\bold R$

- The Thévenin equivalent is $\text R_\text T$ in series with $\text V_\text T$,
- $\text V_\text T = v_{oc}$
- $\text R_\text T = \bold R$

- The Norton equivalent is $\text R_\text N$ in parallel with $\text I_\text N$,
- $\text I_\text N = i_{sc}$
- $\text R_\text N = \bold R$

One of the coolest ideas from linear circuit theory is the concept of an *equivalent circuit*. Say you have a circuit and you “look into” a port using a voltmeter and ammeter. You observe some sort of $i$-$v$ behavior. If you have another circuit that displays the same $i$-$v$ behavior then those two circuits are *equivalent* from the viewpoint of the ports.

In an earlier article on simplifying resistor networks, we learned how to turn any resistor network into an equivalent single resistor. Thévenin’s and Norton’s theorems are the next step. They teach us how to simplify networks of resistors *and* sources. If you have a complicated linear circuit the theorems provide the instructions for how to construct a very simple equivalent circuit.

## Using Thévenin’s theorem

Thévenin’s theorem looks like this in schematic form,

## When is Thévenin’s theorem useful?

Think about using Thévenin’s theorem when you want to focus on a specific part of a circuit and push the details of the rest into the background. For example, suppose you care about what an amplifier does at its output port. Thévenin’s theorem creates a simple equivalent version of the complicated amplifier with the exact same $i$-$v$ behavior at the output.

Sometimes you can apply Thévenin’s theorem as an alternative to Kirchhoff’s Current Law or Voltage Law. It is another circuit design tool to have in your toolbox.

## Strategy

We use aspects of the proof to come up with practical steps to find an equivalent circuit.

A Thévenin equivalent has a voltage source $\text V_\text T$ in series with a resistor $\bold R$.

The Norton equivalent has a current source $\text I_\text N$ in parallel with the same resistor $\bold R$.

To create a Thévenin equivalent,

- Pick two nodes to be the port of the circuit you want to transform.
- Remove any external component(s) connected to the port.
- Transform what remains to its Thévenin equivalent.
- The resistor—you can call it $\bold R$ or $\text R_\text T$ or $\text R_\text N$—is the simplified equivalent resistance of the resistor network from the original circuit when all internal sources are suppressed.
- To suppress a voltage source, replace it with a short.
- To suppress a current source, replace it with an open.

- The Thévenin voltage $\text V_\text T$ is $v_{oc}$, the voltage at the port when all internal sources are turned on and the port is left open.
- Alternatively, you can find the Norton current $\text I_\text N$. It is $i_{sc}$, the current flowing from the port when all internal sources are turned on and a short is placed across the port.
- If you know $i_{sc}$ and $\bold R$, compute $v_{oc} = \text V_\text T = i_{sc}\,\bold R$ for the Thévenin equivalent.
- If you know $v_{oc}$ and $\bold R$, compute $i_{sc} = \text I_\text N = v_{oc}/\bold R$ for the Norton equivalent.

When you find $v_{oc}$ or $i_{sc}$ you have to do a circuit analysis. Choose either one based on how hard you think the analysis task will be. If you can’t tell, flip a coin.

## Example 1

We are asked to find the output voltage for several different values of the load resistor (the initial value of the load resistor is $2\,\text k\Omega$ shown on the right),

We could solve the whole circuit for each value of the load. Or, we can find the Thévenin equivalent of everything to the left of the port (the two little circles) and solve a much simpler circuit for each value of the load.

### Define the port

First, decide what part of the system you want to reduce to a Thévenin equivalent. Select two nodes you care about and define them as the *port*. Then remove any components external to the circuit.

We are interested in what’s happening at the $2\,\text k\Omega$ resistor on the far right, so we identify the port by drawing two little circles. Our goal is to simplify everything to the left of the port by finding its Thévenin equivalent. The first step is to take away the $2\,\text k\Omega$ resistor,

### Find the Thévenin components

The next task is to identify the Thévenin voltage and the Thévenin resistor.

#### Thévenin resistance

To find the Thévenin resistance, suppress internal sources and compute a single equivalent $\bold R$. We use this same resistance in both the Thévenin and Norton equivalent circuits, so we can call it $\text R_\text T$ or $\text R_\text N$ or just $\bold R$.

Here is the circuit with the two internal sources suppressed—the voltage source is replaced with a short, the current short is replace with an open,

With all the sources gone, what’s left? A resistor network. So let’s reduce the network to a single equivalent resistance,

$\bold R = 500 + (1000 \parallel 1000) = 500 + \dfrac{1000 \cdot 1000}{1000 + 1000}$

$\bold R = 1000\,\Omega$

## $\parallel$

Two vertical bars $\parallel$ are shorthand notation for “in parallel with.”

#### Thévenin voltage

To find $\text V_\text T$ we have to do a circuit analysis. If it looks like it’s going to be difficult, think about solving for $\text I_\text N$ instead. Pick the easiest route.

To find the Thévenin voltage restore the internal sources and leave the port *open*. Then find the open circuit voltage, $v_{oc}$. $v_{oc}$ is the voltage the circuit presents to the world when nothing is connected to its port. The Thévenin voltage is equal to $v_{oc}$.

Have a go at finding $v_{oc}$ yourself. (It is not that easy.) Hint: When a circuit has multiple sources one of your choices is to solve by superposition.

The answer is tucked away here—try yourself before peeking,

## Find $v_{oc}$

The circuit has multiple sources. When you see multiple sources, think *superposition!*.

Break the circuit into two sub-circuits, one for each source. Then work out $v_{oc}$ for each one. The final $v_{oc}$ is the superposition of the two sub-circuits, $v_{oc} = v_{oc1} + v_{oc2}$.

Sub-circuit #1 has the voltage source, with the current source suppressed,

Use the voltage divider formula to find $v_{oc1}$,

$v_{oc1} = 5\,\text V \,\dfrac{1000}{1000+1000} = 2.5\,\text V$

Sub-circuit #2 has the current source restored and the voltage source suppressed,

The $2\,\text{mA}$ current flows through the three resistors,

$v_{oc2} = 2\,\text{mA} (500 + 1000 \parallel 1000) = 2\,\text{mA} (500 + 500)$

$v_{oc2} = 2\,\text V$

Now superimpose the two sub-circuits,

$v_{oc} = v_{oc1} + v_{oc2} = 2.5 + 2$

$v_{oc} = 4.5\,\text V$

If you are not already in love with superposition, try to solve this circuit with the Node Voltage Method. The difference in effort is striking.

#### Norton current

The third variable you might want to find is the Norton current, $\text I_\text N$. You have to do a circuit analysis for this, too, but it may turn out to be easier than finding $v_{oc}$. Let’s see if that’s the case.

To find the Norton current restore all the internal sources and place a short circuit across the port. Then find the current in the shorting wire, $i_{sc}$. The Norton current is equal to $i_{sc}$. It is the current the circuit would generate if you forced the output voltage to zero. (Please be very careful if you do this to a real circuit. It might not be designed to drive a short.)

The solution for $i_{sc}$ is tucked away here—see if you can do it yourself before peeking,

## Find $i_{sc}$

This problem has multiple sources. Whenever I see multiple sources my first thought is *superposition!* Superposition turns one nasty circuit into multiple simple circuits.

Solve for $i_{sc}$ twice, and add the results, $i_{sc} = i_{sc1} + i_{sc2}$

Sub-circuit #1 includes the voltage source, with the current source suppressed,

Find $i_{sc1}$,

$5\,\text V = i_1 ( 1000 + 1000 \parallel 500)$

$(1\text K \parallel 500) = \dfrac{1000 \cdot 500}{1000 + 500} = 333\,\Omega$

$i_1 = \dfrac{5\,\text V}{1000 + 333} = 3.75\,\text{mA}$

Find $\goldC{v_x}$ with the voltage divider formula,

$v_x = 5\, \dfrac{333}{1000 + 333} = 5 \cdot 0.250 = 1.25\,\text V$

$i_{sc1}$ is the current flowing in the $500\,\Omega$ resistor,

$i_{sc1} = \dfrac{v_x}{500} = \dfrac{1.25}{500}$

$i_{sc1} = 2.5\,\text{mA}$

Sub-circuit #2 has the current source restored and the voltage source suppressed. This one is particularly easy to solve.

The wire across the port directly shorts across the current source. That means the voltage across the current source *and* the resistor network is zero. Therefore, the entire $2\,\text{mA}$ current flows in the shorting wire,

$i_{sc2} = 2\,\text{mA}$

Superimpose the two components of $i_{sc}$,

$i_{sc} = i_{sc1} + i_{sc2} = 2.5\,\text{mA} + 2\,\text{mA}$

$i_{sc} = 4.5\,\text{mA}$

For this example the effort to find $i_{sc}$ compared to $v_{oc}$ was similar, with $v_{oc}$ probably a little easier overall. No harm done, we got to see both. Hooray for superposition!

## Find $\text V_\text T$ from $i_{sc}$ and $\bold R$

We can use $i_{sc}$ and $\bold R$ to find the Thévenin voltage.

$\text V_\text T = i_{sc}\,\bold R$

$\text V_\text T = 4.5\,\text{mA}\,1000\,\Omega$

$\text V_\text T = 4.5\,\text V$

The same answer is the same as when we analyzed the circuit for $v_{oc}$.

#### Assemble the equivalent circuits

Put it together. The Thévenin equivalent is $\bold R$ in series with the open-circuit voltage source,

And pretty much for free we get the Norton equivalent. The Norton equivalent is the same $\bold R$ in parallel with the short-circuit current source,

### Example 1 simulation models

Open this simulation model of Example 1 in a new tab. There are five circuits shown. The top one is the original Example 1 circuit. The two below that are the circuits we used to find $v_{oc}$ and $i_{sc}$. Use these to check your own circuit analysis work. The two on the right are the Thévenin and Norton equivalents.

- Click on
**DC**to run a DC analysis. - Look at the current and voltage for the $2\,\text k\Omega$ load resistor of the first circuit. Compare that to the current and voltage of the Thévenin and Norton equivalents. Are they the same?
- Explore: Change the load resistor to some other value in all three by double-clicking on the resistor symbol.
- Run
**DC**analysis again. - The the current and voltage will be different, but all three should still match because all three circuits have the same $i$-$v$ equation. Pretty cool, eh?
- If you want to simulate the superposition solutions, suppress the sources one at a time in the two versions in the lower left and run
**DC**again.

## Example 2

Here’s a practical application. This circuit shows a common way to set up a bipolar junction transistor (BJT) as an amplifier. The BJT is the symbol in the center right, with reference designator $\text Q1$. The power supply for the amplifier is provided by the $15\text V$ source.

The $100\,\text k\Omega$ and $50\,\text k\Omega$ resistors set the voltage of $\text Q1$’s base terminal to an intermediate value between the power supply and ground. Together these resistors are called the *biasing network*. It is often useful to convert the biasing network into its Thévenin equivalent.

**Find the Thévenin equivalent of the biasing network.**

Identify the port with two small circles and isolate the biasing network by removing the external components,

Now we find the two Thévenin components, a voltage source and resistance. To test your understanding, give this a try on your own before peeking at the answer.

## Thévenin voltage

The Thévenin voltage is the voltage on the port when we leave it open, $v_{oc}$. The circuit is a voltage divider so we’ll use that formula to find $v_{oc}$,

$\text V_\text T = v_{oc} = 15\,\text V\,\dfrac{50\text k}{100\text k + 50\text k} = 15\,\text V \cdot \dfrac{1}{3}$

$\text V_\text T = 5\,\text V$

## Thévenin resistance

To get the Thévenin resistance we suppress the voltage source by replacing it with a short circuit. The Thévenin resistance is what’s left,

$\text R_\text T = 50\text k \parallel 100\,\text k = \dfrac{50\text k \cdot 100\text k}{50\text k + 100\text k} = \dfrac{5000\text M}{150\text k}$

$\text R_\text T = 33.3 \,\text k\Omega$

Assemble the two components to get the Thévenin equivalent,

## Thévenin equivalent

And the final step is to embed the Thévenin equivalent of the biasing network back into the amplifier circuit,

## Thévenin equivalent embedded into amplifier

## Summary

Thévenin’s theorem is another design tool to put in your toolbox. Use it is an alternative to Kirchhoff’s Current Law or Voltage Law.

**Thévenin’s theorem**

A circuit made of any combination of resistors and sources can be simplified down to a single voltage source in series with a single resistor.

**Norton’s theorem**

A circuit made of any combination of resistors and sources can be simplified down to a single current source in parallel with a single resistor.

Norton and Thévenin forms are interchangeable because of what we learned in the article on source transformation.

To create a Thévenin or Norton equivalent,

- Identify the port and remove the external component(s).
- Find any two of these three: $\bold R, v_{oc}, i_{sc}$, whichever two are the easiest,
- $\bold R$ — Suppress the internal sources and simplify the resulting resistor network.
- $v_{oc}$ — Restore internal sources, leave the port open, find the open-circuit voltage.
- $i_{sc}$ — Restore internal sources, short across the port, find the short-circuit current.

- The third variable can be derived from the other two,
- $\bold R = v_{oc}/i_{sc}$
- $v_{oc} = i_{sc}\,\bold R$
- $i_{sc} = v_{oc}/\bold R$

- The Thévenin equivalent is $\bold R$ in series with $v_{oc}$.
- The Norton equivalent is $\bold R$ in parallel with $i_{sc}$.