Robotics · Electronics basics · Grade 7

Voltage divider and potentiometer

When you turn the knob of a desk lamp it dims — how is the voltage divided?

The situation

When you turn the knob of a desk lamp the light dims smoothly; the volume of a radio is set with a knob like that too. Inside is a potentiometer: a resistor that you change by turning. How does it “divide” the voltage? On a breadboard we first build a divider from two resistors and measure it with a multimeter, then control an LED with a 10 kΩ potentiometer — and see why an ordinary potentiometer is a poor dimmer for an LED and why the voltage drops when a load is connected.

Topic

Voltage divider and potentiometer. Controlling a lamp’s brightness by turning a knob: how a potentiometer changes the voltage

Key ideas

Voltage divider
Potentiometer
Load

Formulas

U_{chiq} = U \cdot \frac{R_{2}}{R_{1} + R_{2}}

Voltage divider: the same current flows through two resistors in series, and the voltage is divided in proportion to the resistances.

U_{chiq} = k \cdot U, k = \frac{α}{270^{\circ}}

Potentiometer: a sliding contact (the knob) divides the resistor in two; in a linear potentiometer the output is proportional to the angle.

R_{2}^{'} = \frac{R_{2} R_{yuk}}{R_{2} + R_{yuk}}

The load is connected in parallel with the lower resistor — $R_{2}$ gets smaller and the output voltage falls.

I_{L E D} = \frac{U_{chiq} - U_{L E D}}{R_{ichki} + R_{s}}

LED current: it is limited by the divider’s internal resistance $R_{ichki} = k (1 - k) R_{pot}$ (in the middle $R_{pot} / 4$ ) and the series resistor $R_{s}$ ; if $U_{chiq} < U_{L E D}$ the LED does not light.

Theory

Two resistors — a divider

We connect two resistors in series and attach them to a source. The current is the same everywhere: $I = U / (R_{1} + R_{2})$ . By Ohm’s law each resistor gets $U_{1} = I R_{1}$ and $U_{2} = I R_{2}$ — the voltage is divided in proportion to the resistances: 1 kΩ and 1 kΩ, 4.5 V → 2.25 V across each; 1 kΩ and 330 Ω → only 1.12 V across the lower one. The voltage across the lower resistor is the output: $U_{chiq} = U R_{2} / (R_{1} + R_{2})$ . The multimeter in V⎓ mode is connected across it (in parallel), and its 10 MΩ input resistance hardly affects the circuit.

The potentiometer — a resistor that divides by turning

A potentiometer is a long resistive track (carbon or wire) and a sliding contact that moves along it. It has three terminals: the two ends (the whole 10 kΩ) and the middle one (the knob). When you turn the knob it divides the track in two: $k R_{pot}$ and $(1 - k) R_{pot}$ — that is, a divider that you change by turning. In a linear potentiometer $k = α / 270 °$ : the knob in the middle gives half of the source, at the end the full voltage. If you leave one end unconnected and use only the other end and the middle, you get a rheostat: a variable resistor.

Controlling an LED by turning

We connect an LED to the potentiometer’s output through a resistor. There are two obstacles. The first is the threshold: the LED needs ≈ 1.9 V, and below that voltage it does not light at all (at 4.5 V it is dark up to 42% of the knob’s turn). The second is the potentiometer’s own resistance: seen from the output, the divider looks like a source with a resistance of $k (1 - k) R_{pot}$ — in the middle of a 10 kΩ potentiometer that is 2.5 kΩ! This resistance chokes the current: only 0.1 mA in the middle, and the LED brightens only in the last few degrees. With the rheostat connection the resistance drops to zero at the end — an LED with no resistor burns out. That is why real dimmers use a potentiometer + transistor or PWM (Arduino `analogWrite`).

The load drags the divider down

If a consumer — a load $R_{yuk}$ — is connected to the divider’s output, it sits in parallel with the lower resistor: instead of $R_{2}$ we get $R_{2} R_{yuk} / (R_{2} + R_{yuk})$ — smaller, so the output falls. 1 kΩ / 1 kΩ, 4.5 V: 2.25 V with no load, 1.09 V with a 470 Ω load. The rule: the divider resistors should be much smaller than the load (the divider current about 10 times the load current) — with 100 Ω / 100 Ω the output is 2.03 V. But small resistors draw a lot of current and get hot: $P = U^{2} / R$ — it must not exceed 0.25 W.

With the multimeter

V⎓ — voltage, the probes go in parallel with the place you measure: red to the output, black to the minus; the range is a little above the value (4.5 V → 20 V). A⎓ — current, the multimeter breaks the circuit and is connected in series (in the return path from the LED’s cathode to the minus); the small resistance of the shunt reduces the current slightly. Mistakes: connect it in series in V mode — the circuit is open and the LED goes out; connect it in parallel in A mode — a short circuit and the fuse blows. Ω — only with the battery disconnected: between the middle terminal and the end of the potentiometer you will see $k R_{pot}$ .

About the model

The circuit is calculated by nodal analysis; the potentiometer is two resistors $k R$ and $(1 - k) R$ (at least 1 Ω at the ends), linear. LED: no current up to 1.9 V, then 2 Ω; current above 30 mA builds up over time — 2 times the normal current burns it out in a few seconds, 50 times instantly. The battery’s internal resistance is taken into account. Multimeter: V⎓ 10 MΩ, A⎓ shunt (1 Ω at 200 mA, 10 Ω at 20 mA), 1999 counts; the mA fuse is 200 mA.

Did you know?

The volume knob in radios and amplifiers is a logarithmic potentiometer: the resistance does not change evenly with the angle, because the ear senses loudness logarithmically; with an ordinary linear potentiometer the volume would jump at the start and then hardly change.
Arduino and other microcontrollers read a potentiometer knob exactly as a voltage divider: the voltage on the middle terminal at the analogue input (A0) becomes a number from 0 to 1023 — this is used in the “Analog input” experiment.

Inside the experiment

A real board in a hand-inked isometric drawing: turn it, place the parts, wire the circuit, write and upload the code — current and voltage are computed at once, and a wrong connection burns a part. Inside are missions, theory, a quiz and a guided tour.

More experiments in this section

The electric circuit: a torch — Grades 5–6
Breadboard: rows — Grades 5–6
LED and resistor: Ohm’s law — Grade 7
Resistors: the colour code — Grade 7