A voltage divider, and the load that breaks it
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A voltage divider is two resistors in series across a supply, with a wire taken off the joint between them. The same current goes through both, so the supply splits between them in proportion to their resistance, and the joint sits at a voltage somewhere between the two rails. It is the most useful two-part circuit there is, and also the one most often used for a job it cannot do.
The circuit above is two 10 k resistors from 5 V to ground with a meter on the middle, which reads 2.499 V: half the supply, less a millivolt that the meter itself takes by loading the joint through its own 10 megohms. To the right, a pushbutton can connect a 1 k resistor from the joint to ground. That 1 k is the load: the thing you were hoping the divider would power.
Press Run, read the meter, then hold the button. The 2.5 V you had a moment ago falls to 0.42 V. Nothing is broken. That is simply what a divider does when something draws current from it, and by the end of this page you will be able to predict the 0.42 V before you press the button.
Work the divider out by hand
Start with Ohm’s law on the whole chain. Two 10 k resistors in series are 20 k, so the current from the supply is 5 V / 20 k = 250 microamps. That same current goes through each resistor, and 250 uA through 10 k drops 2.5 V. The bottom resistor has 2.5 V across it, so the joint is 2.5 V above ground.
Doing that every time is slow, so it is usually written as one line: Vout = Vin x R2 / (R1 + R2), where R1 is the resistor to the supply and R2 is the one to ground. Try it on a pair people really use, 10 k over 4.7 k: 5 V x 4.7 / 14.7 = 1.60 V. Only the ratio matters for the unloaded voltage, which is why 1 k over 1 k gives exactly the same 2.5 V as 10 k over 10 k.
The number the formula hides is the one that matters next. Seen from the joint, a divider behaves like a perfect 2.5 V source with a resistor in series with it, and that resistor is R1 and R2 in parallel: 10 k with 10 k is 5 k. Every load you connect has to draw its current through that 5 k, and the voltage falls by whatever that current drops across it.
What the load does, in numbers
Holding the button puts the 1 k load in parallel with the bottom 10 k. Two resistors in parallel combine as R1 x R2 / (R1 + R2), so 10 k with 1 k is 10 000 / 11 = 909 ohms. The divider is now 10 k over 909 ohms, and Vout = 5 V x 909 / 10 909 = 0.42 V. The meter reads 0.417, which is that.
A load a hundred times the divider’s own resistance is a different story. Change the 1 k to 100 k in the editor and hold the button again: the joint only falls to 2.38 V. The rule of thumb that falls out of this is that a load should be at least ten times the divider’s resistance, R1 and R2 in parallel, before you can ignore it, and a hundred times if the last few percent matter.
You can push the other way and make the divider stiffer by making both resistors smaller. At 1 k over 1 k the same 1 k load only pulls the joint down to 1.67 V. But the divider now draws 2.5 mA from the supply all the time, whether anything is connected or not, and it still sags. On a battery that is a real cost for nothing.
Why a divider makes a bad power supply
A power supply is supposed to hold its voltage whatever the load does. A bench supply or a regulator manages that because its output resistance is a fraction of an ohm, and it actively corrects for the load. A divider’s output resistance is thousands of ohms and nothing corrects anything, so its voltage depends on the load, and loads change: a module that draws 10 mA idling and 80 mA while it transmits will pull a divider’s output all over the place and reset itself doing it.
The classic version of this mistake is two resistors to get 3.3 V from 5 V for a 3.3 V board or sensor. It seems to work with nothing plugged in, because the meter draws nothing, and then fails the moment the module starts drawing current. For power the answer is a regulator: the 3.3 V pin on your board is one, and a small linear regulator costs pennies. A Zener diode and a resistor is the simplest thing that behaves more like a supply than a divider does, and the button below opens one.
Where a divider is exactly the right answer
Everywhere the load is tiny. An analog input on an Uno draws almost nothing, so a divider is the standard way to bring a 12 V battery voltage down into the 0 to 5 V an ADC can read, and a potentiometer is just an adjustable divider. A light-dependent resistor or a thermistor with a fixed resistor is a divider whose ratio the world changes for you. A comparator’s reference voltage is a divider. So is the level shift from a 5 V board’s TX pin down to a 3.3 V board’s RX: 1 k over 2 k gives 3.33 V, and an input pin is a load of megohms.
The ATmega328P’s datasheet asks for a source resistance of 10 k or less on an analog input, because the ADC charges a small sampling capacitor from the pin on every reading. Two 10 k resistors are a 5 k source, comfortably inside that. Two 1 M resistors would draw almost no current from a battery, but the reading would come out low and slow, and would be pulled around by whichever channel was read before it.
Common mistakes on a real bench
Powering something from a divider is the big one, covered above. The others are smaller. Swapping R1 and R2 gives you Vin minus the voltage you wanted, which looks right until you check it. Measuring a high-resistance divider with an instrument that loads it gives a wrong answer that looks like a wrong resistor: a scope probe is 1 M, and on a divider of two 1 M resistors it pulls the middle from 2.5 V down to 1.67 V. And forgetting that resistors have a tolerance: two 5 percent resistors can be 5 percent apart in opposite directions, so a divider meant to give 2.50 V can give anything from 2.38 V to 2.63 V.
The model on this page has no tolerance, so every 10 k is exactly 10 000 ohms and the numbers come out clean. That is the right thing for learning the idea and the wrong thing to expect from a bag of resistors.
What to try next
In the editor, change R2 to 4.7 k and predict the meter before you run it. Replace the load with a red LED and a 220 ohm resistor and see how little light a divider can give. Then look at the pages that use a divider on purpose: a potentiometer feeding analogRead, and an op-amp comparator whose threshold is a divider.
Questions
What is the voltage divider formula?
Vout = Vin x R2 / (R1 + R2), where R1 goes to the supply and R2 goes to ground. Two 10 k resistors on 5 V give 2.5 V; 10 k over 4.7 k gives 1.60 V. It holds only while the load on the output is much larger than R1 and R2 in parallel.
Can I use a voltage divider to get 3.3 V from 5 V?
For a signal going into an input pin, yes: 1 k over 2 k gives 3.33 V and an input draws almost nothing. For powering a board, a sensor or a module, no. Its voltage will fall as soon as the load draws current. Use a regulator.
What resistor values should I use for a divider?
Low enough that the load is at least ten times R1 and R2 in parallel, high enough that the current the divider wastes does not matter. For feeding an Arduino analog input, a total of 10 k to 100 k is typical. The ratio sets the voltage; the size sets how stiff it is and how much it costs.
Why does my divider read lower than I calculated?
Almost always because something is loading it: the circuit it feeds, or the instrument measuring it. Work out R1 and R2 in parallel and compare it with the load. Resistor tolerance accounts for a few percent; anything more than that is a load.
Build this for real
Open the editor, change a value and watch the number move with it. Nothing to install, and no account needed.