Voltage Divider Calculator

V_out = V_in × R2 / (R1 + R2)

V_in

The voltage across the whole divider, in volts, measured at the top of the chain before any current has been taken out of the middle. It is the supply the two resistors share, and it sets the scale of everything else: double it and every other number on the page doubles. A divider attenuates, so the output is always a fraction of this and can never be larger than it — if you need a voltage bigger than the supply, this is the wrong circuit

R1

The resistor on the supply side, between V_in and the output node, in ohms, kiloohms or megohms. It carries the full divider current and drops the rest of the supply voltage: whatever is not across R2 is across this one, and the two always add up to V_in. With R1 at zero the output is the input, because a zero ohm resistor is a piece of wire and there is nothing left to drop any volts

R2

The resistor on the ground side, between the output node and ground, in ohms, kiloohms or megohms, and it is the one the output is measured across. This is the detail that catches people out: swapping the two numbers gives a plausible-looking answer that is simply the complement of the right one, so a divider meant to produce 3.8 V will quietly produce 8.2 V. If your schematic numbers the resistors the other way round, enter them the other way round here

V_out

The output, the voltage across R2 and the fraction of the input that the ratio produces. It is a proportion rather than an absolute quantity, so the same pair of resistors divides any supply in the same ratio: 10 kΩ and 4.7 kΩ turn 12 V into 3.837 V and 5 V into 1.599 V, keeping the same 0.32 as the divider ratio. That is why the ratio is usually the thing you design and the supply is the thing you check afterwards

I

The divider current, the current that flows from the supply through both resistors to ground. It depends on the sum of the two resistances and not on their ratio, which is the useful thing to know: you choose the ratio to set the output and the total to set how much current the divider wastes. A pair of 100 kΩ resistors divides exactly as well as a pair of 100 Ω ones, and burns a thousandth of the power doing it

Use it to work out the output of a divider you are designing, or to find the resistor pair that produces a voltage you need. It comes up whenever a circuit has to measure or monitor a voltage larger than an ADC can accept, to scale a sensor output into a readable range, to bias a transistor into its conducting region, or to make a reference voltage for a comparator. It is the right tool when the thing reading the output draws almost no current, and the wrong tool when it draws a real one — the equation assumes nothing is connected to the middle of the chain.

1. The defaults: 12 V through 10 kΩ and 4.7 kΩ

   1. Input 12 V, R1 = 10 kΩ on the supply side, R2 = 4.7 kΩ to ground
   2. The two share the input in proportion to their resistance, so R2 takes 4.7 / (10 + 4.7) of it
   3. V_out = 12 × 4700 / 14700 = 3.837 V
   4. R1 drops the remaining 12 − 3.837 = 8.163 V, the current is 12 / 14700 = 0.816 mA, and the pair burns 12 × 12 / 14700 = 9.796 mW

   Add the two voltage rows together and you get the input back, which is the check worth doing every time: 8.163 plus 3.837 is 12, exactly. Notice also which resistor got the larger share of the voltage. R1 is the bigger resistance, so it drops the bigger voltage, and the output — taken across the smaller one — is the smaller number. Reversing the two resistances is the single most common way to get this calculation wrong, and it produces 8.163 V, a perfectly reasonable-looking answer for a divider whose schematic said 3.8 V.

2. Two equal resistors: an exact half

   1. Input 5 V, both resistors 1 kΩ
   2. Equal resistances share the voltage equally, so each takes half
   3. V_out = 5 × 1000 / 2000 = 2.5 V, and R1 drops the same 2.5 V
   4. The current is 5 / 2000 = 2.5 mA and the pair burns 5 × 5 / 2000 = 12.5 mW

   Equal resistors are the case where the ratio is obvious enough to sanity-check the arithmetic without a calculator, which makes them a good first move on any divider problem: if your answer is not half of the input, something is wrong with the setup rather than with the sums. Halving a supply this way is also a common way to make a mid-rail reference for a single-supply amplifier, and it is the case where a small change in either resistor moves the output the most.

3. Stepping 12 V down to 8.25 V

   1. Input 12 V, R1 = 10 kΩ, R2 = 22 kΩ
   2. R2 is now the larger resistor, so it takes the larger share: 22000 / 32000
   3. V_out = 12 × 22000 / 32000 = 8.25 V
   4. R1 drops only 3.75 V, the current falls to 0.375 mA because the total resistance rose, and the pair burns 4.5 mW

   The same 10 kΩ on top now produces 8.25 V instead of 3.837 V, because the resistor below it changed. This is the pair that makes the point about which resistor does what: R1 did not change at all, and the output more than doubled. It also shows the trade the divider offers — the total resistance went from 14.7 kΩ to 32 kΩ, so the current and the wasted power both fell by more than half, while the ratio moved the output up.

4. R2 at zero: the output is grounded

   1. Input 12 V, R1 = 10 kΩ, R2 = 0 ohms
   2. With the ground side shorted, the output node is connected straight to ground
   3. V_out = 12 × 0 / 10000 = 0 V, and all 12 V appears across R1
   4. The current is 12 / 10000 = 1.2 mA and R1 alone burns 14.4 mW

   R2 at zero is accepted rather than rejected, because a zero ohm resistor is a wire and the arithmetic handles it: the output node is grounded, so the output is 0 V and the whole supply sits across R1. The current is at its highest, since the total resistance is now just R1, and this is the worst case for the power rating if you are choosing parts. It is a legitimate limit rather than an error, in the same way that R1 at zero is a legitimate limit that gives an output equal to the input; only both of them at zero, a short across the supply, is rejected.

The equation assumes nothing is connected to the output node, so the figures here are the unloaded values and the real output will be lower as soon as a load draws current. A practical rule is to keep the load resistance at least ten times the parallel combination of the two divider resistors, which in practice means low resistances and a divider that wastes real current. The output is also unregulated: it follows any change in the input, and it has no ability to supply current without sagging.

Which resistor is the output taken across?

The one on the ground side, called R2 on this page. The output is the voltage at the point between the two resistors, measured with respect to ground, and that is exactly the voltage across the lower resistor. If your schematic labels them the other way round, either swap the two numbers before entering them or read the output from the other row on the result panel, which reports the voltage across the supply-side resistor as a separate line. Getting this backwards is the most common mistake with a resistor divider, and it produces a plausible wrong answer rather than an obvious one.

What is the voltage divider equation?

V_out = V_in × R2 / (R1 + R2), where R2 is the resistor to ground. It follows directly from Ohm's law: the two resistors carry the same current, so the voltage across each is proportional to its resistance, and the share that lands on R2 is its resistance divided by the total. The same formula read as a ratio gives the design form — to get a fraction f of the input, make R2 equal to f / (1 − f) times R1. That is how a 4.7 kΩ and a 10 kΩ were chosen for a ratio of about a third.

Why is my measured output lower than this says?

Because something is drawing current from the output node, and the formula assumes nothing is. Whatever is connected there sits in parallel with R2, lowering the effective resistance to ground and pulling the output down; the more it draws, the further the output falls. The usual rule of thumb is that the load resistance should be at least ten times the parallel combination of the two divider resistors, which keeps the error under about 10%. If your reading is much lower than the calculation, either the load is too heavy for the divider or one of the resistors is not the value you think it is.

How do I choose the resistor values?

Two decisions, and the equation only makes one of them. The ratio sets the output and comes straight from the formula; the total resistance sets the divider current and comes from what you can afford to waste. Lower values give a stiffer output that a load disturbs less, at the cost of more current burned continuously; higher values waste less but sag as soon as anything reads them. Somewhere between 1 kΩ and 100 kΩ total covers most signal-level work, and a divider that feeds an ADC input can usually sit at the high end.

Can I use a divider to power something?

Not usually, and this is the limitation that matters most. A divider is a reference, not a supply: its output collapses as soon as the load draws a meaningful current, because the current has to come through R1 and that raises the drop across it. It also wastes power continuously whether or not anything is connected. For a real supply use a regulator or a converter. The one case where a divider is genuinely used to feed something is a high-impedance input such as an ADC or a comparator, where the current drawn is negligible.

Does the divider ratio depend on the input voltage?

No, and that is what makes it useful. The ratio is set entirely by the two resistances, so the same pair divides any supply in the same proportion: the 10 kΩ and 4.7 kΩ on this page turn 12 V into 3.837 V and would turn 5 V into 1.599 V. What does change with the supply is the current and the power dissipated, both of which rise with it. The output does follow the input if the input moves, which is either the point of the circuit — measuring a voltage too large for an ADC — or a problem, if the input is not stable.

• Voltage divider — the derivation of the ratio from Ohm's law, the loaded-divider case, and the effect of a load on the output — Wikipedia
• OpenStax College Physics 2e, §21.1 Resistors in Series and Parallel — why two resistors in series share the supply voltage in proportion to their resistance — OpenStax
• All About Circuits, Volume I, Chapter 6: Divider Circuits and Kirchhoff's Laws — the loaded divider, the current-draw trade-off, and the ten-to-one rule of thumb for keeping a load from disturbing the output — All About Circuits