Voltage divider calculator
The output above assumes nothing is connected to it. The moment a load draws current, it sits in parallel with R2 and the output sags. The rule of thumb is that the divider current should be at least ten times the load current, so a divider feeding a 1 mA load wants at least 10 mA flowing through it, which for a 12 V supply means a total resistance of about 1.2 kΩ and 120 mW burnt continuously. For anything drawing real current, use a regulator.
A voltage divider outputs Vin × R2 ÷ (R1 + R2). Two equal 10 kΩ resistors on 12 V give 6 V at 0.6 mA. The output only holds if the load draws far less current than the divider itself: a factor of ten is the usual minimum.
How to design a voltage divider
Dividers are for sensing, not for supplying. Scaling a battery voltage down into an ADC input is the textbook use, and it works because an ADC input draws almost nothing. Choosing the values is a trade: high resistances waste less power but are more susceptible to noise and to the input impedance of whatever is reading them; low resistances are stiffer but burn current continuously, which matters enormously in a battery device. For an ADC divider, tens of kilohms is the usual compromise.
The figure at the top of the panel is the unloaded output, and the ten-to-one rule of thumb is looser than it sounds. On the values this page opens with, the ladder carries 0.6 mA. A load drawing a tenth of that, 60 µA, is 100 kΩ, which sits in parallel with R2 and pulls the bottom leg down to 9.09 kΩ. The output falls from 6.000 V to 5.714 V: a 4.8 per cent error, not a rounding one. Getting inside half a per cent needs the load at a hundred times the ladder resistance, a full megohm here. And a load equal to R2 drops the output by a third, to 4 V, which is what a divider used as a power supply actually does.
All of it is decided by the divider’s output impedance, R1 in parallel with R2 rather than either one alone: 5 kΩ for the defaults, regardless of the 12 V going in. Anything reading the output sees a source with that resistance in front of it, and a microcontroller ADC in particular has to charge its internal sampling capacitor through it within a fixed window. A 5 kΩ source is comfortable for almost every part; the same 2:1 ratio built from two 1 MΩ resistors presents 500 kΩ, and the reading comes back low because the sample never finished.
Tolerance is the third error and the one people size wrongly. The worst case is roughly the division you are doing multiplied by twice the tolerance: a divider that barely drops the voltage is barely affected, and one that drops it a long way is affected almost fully. Two 5 per cent resistors on this page’s 2:1 default put the output between 5.70 and 6.30 V, a 5 per cent spread. The same parts making 12 V into 1 V spread from 0.91 to 1.10 V, nearly 10 per cent. One per cent parts bring the 2:1 case to 5.94 to 6.06 V, which is usually the cheapest fix available.
What people use it for
- Scaling a voltage into a microcontroller ADC
- Setting a reference or threshold voltage
- Reading a potentiometer or sensor
- Biasing a transistor base
- Checking what a sense divider costs a battery over a year
Questions
Vout = Vin × R2 ÷ (R1 + R2), where R2 is the resistor between the output and ground.
The load sits in parallel with R2, lowering the effective bottom resistance. Keep divider current at least ten times load current, and expect a few per cent even then.
About 4.8 per cent on this page’s defaults: 6.000 V unloaded becomes 5.714 V with a 100 kΩ load. The rule of thumb buys you a small error, not none.
A hundred times the ladder resistance. On the defaults that is a 1 MΩ load, which gives 5.970 V against 6.000 unloaded.
Only microamp loads. For anything real, use a regulator. A divider has no regulation and wastes power constantly.
R1 in parallel with R2, which is 5 kΩ for two 10 kΩ resistors. The input voltage does not come into it, and it is the number to compare against whatever is reading the output.
Almost always source impedance. The ADC charges a small internal capacitor through your divider, and a high-value ladder cannot fill it inside the sampling window. Drop the resistor values, lengthen the sampling time, or buffer the output with an op-amp.
For an ADC input, tens of kilohms balances noise against current. For a low-impedance reference, go lower and accept the current. For a battery device, calculate the ladder current here and multiply it by the hours it will run.
Yes, and how much depends on how far you are dividing. Worst case is roughly the fraction of the input you are throwing away, multiplied by twice the tolerance.
Much. Two 5 per cent parts turning 12 V into 6 V give 5.70 to 6.30 V, a 5 per cent spread. The same parts turning 12 V into 1 V give 0.91 to 1.10 V, close to 10 per cent.
Round to the nearest stock value and read the output back. For a 3.3 V target from 12 V it suggests 3,793 Ω; the nearest 5 per cent value, 3.9 kΩ, gives 3.367 V, and the nearest 1 per cent value, 3.83 kΩ, gives 3.323 V.
They follow the IEC 60063 preferred-number series: E24 for 5 per cent parts, E96 for 1 per cent, spaced so the tolerance bands just meet. The standard itself is a paid IEC publication, though every distributor lists the values.
R1 connects to the input and R2 to ground, with the output taken between them. Swapping them inverts the ratio, so the output becomes the part you were throwing away.
Very little at these values: 3.6 mW each on the defaults, against the 250 mW an ordinary quarter-watt resistor is rated for. The current draw usually matters far more than the heat.
On a high-voltage rail, yes. Small chip resistors have a maximum working voltage in their datasheet that is easily exceeded by a divider across a few hundred volts, so mains and high-voltage dividers are built from several resistors in series.
For slow signals, yes, and it is common. For anything fast it fails, because the divider’s output impedance and the receiver’s input capacitance form a low-pass filter that rounds the edges off. Use a proper level shifter above a few hundred kilohertz.
Only as a rough threshold. A divider tracks its input exactly, so it drifts with the supply, with temperature, and with the tolerance spread above. A reference IC costs very little and does none of that.
It filters noise, which is often welcome on an ADC input, but it also forms an RC with the divider’s output impedance and slows the response. Work out the time constant from R1 parallel R2 before choosing the value.
For adjusting a threshold by hand, yes: a pot is a divider you can turn. It is worse for anything permanent, because the wiper drifts with vibration and age where two fixed resistors do not.