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Electronics

Thevenin's theorem explained

Thevenin's theorem says that any network of batteries and resistors, seen from two terminals, acts exactly like one voltage source, Vth, in series with one resistor, Rth. Find those two numbers once and you can predict what any load will get, without solving the whole circuit again.

Finding Vth and Rth

  1. Vth is the open-circuit voltage: what the terminals read with nothing connected.
  2. Rth is the resistance between the terminals with every voltage source replaced by a plain wire (and every current source by a gap).

Why a wire? An ideal battery holds its voltage whatever current flows, so for changes it behaves like a short.

A worked example

Take a voltage divider: 10 kΩ on top, 10 kΩ on the bottom, on 12 V. With nothing connected, the output sits at half the supply:

Vth = 12 V × 10 over 10 + 10 = 6 V

Replace the battery with a wire, and both resistors run from the output to the same point, so they're in parallel:

Rth = 10 kΩ ∥ 10 kΩ = 5 kΩ

So the divider is a 6 V battery behind 5 kΩ. Connect a load RL and it forms a second divider with Rth:

VL = Vth × RL over Rth + RL

A 100 kΩ load gets 6 × 100 ÷ 105 ≈ 5.71 V. A 1 kΩ load gets 6 × 1 ÷ 6 = 1 V. One pair of numbers answers both.

Live experiment

Load the divider

A 10 kΩ track across a 9 V battery, with a sliding tap. Set it to the middle, 4.5 V, then plug a 10 kΩ gadget into the tap. Thevenin predicts the drop: 4.5 V behind 2.5 kΩ leaves the gadget 3.6 V.

Try it yourself: free lesson

“Voltage dividers” is lesson 7 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.

Rules of thumb

The output sags by roughly Rth ÷ RL: 5 kΩ into 100 kΩ loses about 5%. To keep a signal stiff, make the load at least ten to a hundred times Rth. That's why a divider can feed a sensor input but can't power a motor, and why an op-amp buffer is used when it must.

The load gets the most power when RL equals Rth, half the voltage at the most current the source can share. That's the idea behind impedance matching in radio.

Norton's version

The same network can also be drawn as a current source, IN = Vth ÷ Rth, in parallel with Rth. For the divider, that's 1.2 mA beside 5 kΩ. Thevenin and Norton give identical answers for every load; pick whichever makes the sum easier.

Free lesson

Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Voltage dividers” is lesson 7 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.

Or start from the very first lesson, no account needed

Work it out

Practice questions

Three quick questions from the lessons. Pick an answer to check it.

  1. Question 1 of 3Why can't a plain voltage divider power a 270 Ω load at a steady voltage?
    Show the answer

    Answer: The load pulls its output down. A divider acts like a battery behind Rth. A heavy load forms a second divider with Rth, so the output sags. A Zener holds its voltage instead.

    • Why not “Dividers only work on AC”? A resistor divider works the same on DC as on AC. The problem is its Rth, which sits in series with whatever you connect.
    • Why not “It makes too much voltage”? A divider's output can never be more than its supply, and a load only ever pulls it down.
  2. Question 2 of 3With the button pressed, the pull-up and the button form…
    Show the answer

    Answer: a voltage divider whose bottom part is 0 Ω. The pull-up is the top resistor and the closed button the bottom one, about 0 Ω. Vout = 5 × 0 over 35k + 0 = 0 V, so the pin reads LOW.

    • Why not “a low-pass filter”? A low-pass filter needs a capacitor. A resistor and a closed switch are a divider with a 0 Ω bottom part.
    • Why not “a series LED circuit”? There's no LED in the path. The pull-up resistor and the closed button simply split the 5 V between them.
  3. Question 3 of 3Around any closed loop, the voltages…
    Show the answer

    Answer: add up so that the rises equal the drops. That's Kirchhoff's voltage law. The energy charge gains in the battery is all given up in the drops, so you end where you started.

    • Why not “are the full battery voltage across each part”? Parts in series share the battery's voltage; only parts straight across the battery get all of it.
    • Why not “depend on how long the wires are”? Ideal wires drop no voltage however long they are. Kirchhoff's voltage law holds for any loop.
Keep practising: free lesson

“Voltage dividers” is lesson 7 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.

Questions people ask

What does Thevenin's theorem say?

Any linear network of voltage sources, current sources and resistors, seen from two terminals, behaves exactly like one voltage source V_th in series with one resistor R_th. Whatever you connect to those terminals can't tell the difference.

How do you find the Thevenin resistance?

Replace every voltage source with a plain wire and every current source with a gap, then work out the resistance between the two terminals. For a voltage divider that leaves R1 and R2 in parallel, so R_th = R1 ∥ R2.

What is the difference between Thevenin and Norton?

They describe the same network two ways. Thevenin is a voltage V_th in series with R_th; Norton is a current I_N = V_th ÷ R_th in parallel with the same R_th. You can convert either into the other.

Can I measure a Thevenin equivalent?

Yes. Measure the open-circuit voltage at the terminals with nothing connected: that's V_th. Then measure the current into a short, or into a known load, and work out R_th. With a short, R_th = V_open ÷ I_short. Only short a source that can safely take it.

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