Voltage Divider Calculator, Loaded and Unloaded

Voltage Divider Formula

A voltage divider is the most-used circuit in electronics and the most-misused. The formula takes one line; what it leaves out — that the answer changes the moment you connect anything — is where almost every divider problem comes from. Both halves are below.

One current, two voltages

The two resistors are in series, so there is one path and therefore one current. That is the only physical fact needed. From Ohm’s law applied to the pair:

The output is the voltage across R2, and the voltage across R2 is that same current times R2. Substituting gives the formula, with the current cancelled out:

Notice what is absent: any absolute resistance. Only the RATIO appears, so 1k over 1k and 1M over 1M both give exactly half the supply. Choosing between them is not about the voltage at all — it is about current, noise and loading, which is the subject of the sections below.

Rearranged for each unknown

Four quantities tied by one relationship, so any three fix the fourth. These are the four forms the calculator above picks between:

• Output: Vout = Vin x R2 / (R1 + R2)
• Input: Vin = Vout x (R1 + R2) / R2
• Top resistor: R1 = R2 x (Vin − Vout) / Vout
• Bottom resistor: R2 = R1 x Vout / (Vin − Vout)

The last two are the ones worth reading twice, because they are how a divider is actually designed: you know the supply you have and the voltage you want, and you are choosing parts. Both contain (Vin − Vout), which is why asking for an output equal to the input divides by zero and asking for more gives a negative resistance. A resistive divider only ever attenuates.

What a load does, exactly

Connect something to the output and it sits in parallel with R2, because both go from the output node to ground. The divider now behaves as though R2 were smaller:

Substituting R2eff for R2 in the main formula gives the loaded output, and that works. But there is a much more useful way to see it. Replace the whole divider by its Thevenin equivalent — an ideal source at the unloaded output voltage, behind a single resistance:

Now the load is simply a second divider against that output resistance, and the error it causes falls out as an exact expression — no approximation anywhere:

This is where the familiar rule of thumb comes from. A load ten times the output resistance gives −1/11, which is −9.1 %. A hundred times gives −1/101, or −0.99 %. So "make the load at least ten times the divider" is not folklore, it is that fraction — and if you need better than 1 %, ten times is nowhere near enough.

Worked example

The 3 V is the easy part. The 3.333 kΩ output resistance is the number that decides whether this divider is any use: a 33 kΩ load turns the 3 V into 2.73 V, and a 3.3 kΩ load turns it into 1.5 V. Nothing about the resistor ratio changed — the divider simply is not stiff enough to hold its own output.

Choosing the Resistor Values, Not Just the Ratio

The ratio sets the voltage and the magnitude sets everything else. Both ends of the range have real failure modes, and the right answer is a compromise between them rather than a formula.

Too low: wasted current and heat

A 1 kΩ over 1 kΩ divider on a 12 V rail draws 6 mA continuously, whatever is connected. On a battery-powered board that is more than most microcontrollers use asleep, and it never stops — a divider has no standby mode. It also dissipates 72 mW, which for two 0603 resistors is already a derating question.

Too high: noise, leakage and loading

Push the values up and three problems arrive together. The output resistance rises in proportion, so anything you connect matters more. The node becomes a good antenna for capacitively coupled noise, because there is little current available to hold it. And the input leakage of whatever is measuring it — tens of nanoamps for a typical ADC input, more when hot — starts to be a significant fraction of the divider current itself.

The working compromise

For a divider feeding a high-impedance input, total resistances of 10 kΩ to 100 kΩ suit most situations: microamps of standing current, an output resistance in the low kilohms, and leakage still a small fraction. For anything feeding a real load, size it from the output resistance instead — pick R_out at least ten times smaller than the load, and a hundred times if you care about the answer.

When a Divider Is Not Meant To Be Loaded

A large fraction of dividers are designed on the assumption that nothing draws current from them, and that assumption is often correct. Recognising which case you are in saves a lot of arithmetic.

• Feeding an ADC or comparator input. Input impedance is megohms, so the loading error is negligible — but a sampling ADC briefly draws charge from the node, so a small capacitor across R2 is standard practice.
• Setting a feedback ratio. A switching-regulator feedback divider drives an error amplifier input, which takes essentially nothing.
• Biasing a transistor base or a FET gate. A gate takes nothing; a base takes real current, and the divider has to be stiff against it.
• Providing a reference to something that then buffers it. This is the correct answer whenever the load is significant: divide, buffer with an op-amp follower, and the loading question disappears entirely.

The one case where a divider is genuinely the wrong circuit is powering a load. A divider has no regulation whatsoever: its output moves with the supply and with the load current, and it dissipates more than it delivers. If something needs a supply rather than a reference, it needs a regulator.

Common Voltage Divider Applications

Measuring a voltage higher than your ADC allows

The commonest use. A 3.3 V microcontroller reading a 12 V battery needs the battery scaled down by at least 4:1, and the scaling factor becomes part of the firmware. Two practical notes: allow headroom for a fully charged battery above nominal, and remember the divider draws current from the battery permanently unless you switch it with a FET.

Level shifting a digital signal

A divider will take a 5 V logic output down to something a 3.3 V input accepts, but the DC ratio is only one of four conditions — the shifted high has to clear the receiving part’s threshold, and the divider has to be fast enough for the signal. That case has its own tool and its own theory.

Reading a resistive sensor

A thermistor, photoresistor or potentiometer forms half of a divider, and the fixed resistor is chosen to put the interesting part of the sensor’s range in the middle of the ADC’s. For a thermistor, that generally means matching the fixed resistor to the thermistor’s resistance at the temperature you most care about.

Attenuating a signal

At audio frequencies a resistive divider is a perfectly good attenuator, provided the source can drive it and the destination does not load it. At radio frequencies the stray capacitance across each resistor makes the ratio frequency-dependent, which is why oscilloscope probes add a trimmer capacitor to compensate — the same reason a 10:1 probe has an adjustment screw.

What This Calculator Assumes

Stated plainly, because a number without its assumptions cannot be checked:

• DC, or a frequency low enough that stray capacitance does not matter. Above a few hundred kilohertz with high-value resistors, the divider is an RC network.
• Ideal resistors at their nominal values. Real ones have a tolerance, and the output error from that is usually larger than any error in your arithmetic — use the standard-values tool to see how much.
• A resistive load. A capacitive load does not change the DC answer but adds a time constant; an active load that draws a fixed current is a different calculation entirely.
• The source can supply the divider current without sagging. A divider across a high-impedance source loads it, and the input voltage you assumed is then wrong.
• Room temperature. Both resistors drift, and only the difference between their drifts affects the ratio — which is the argument for using two resistors of the same type from the same batch.

Common mistakes

• Ignoring the load. The formula gives the unloaded output; anything connected sits in parallel with R2 and pulls it down by R_out / (R_out + R_L).
• Putting R1 on top of the fraction. The output is measured across R2, so R2 is the numerator. R1 on top gives the voltage across the top resistor instead.
• Using a divider to power something. It has no regulation and dissipates more than it delivers — a divider makes references, not supplies.
• Choosing very high values to save current, then measuring with something that leaks. Tens of nanoamps into a 1 MΩ divider is millivolts of error, and it moves with temperature.
• Expecting a divider to increase a voltage. Two resistors can only attenuate; (Vin − Vout) in the design formulas goes negative and the answer is meaningless.
• Forgetting that the divider draws current from the source permanently. On a battery, a monitoring divider left connected is a slow discharge path.

Frequently asked questions

What is the voltage divider formula?

Vout = Vin × R2 / (R1 + R2), where R1 is the top resistor between the input and the output node, and R2 is the bottom resistor between the output node and ground. Only the ratio of the two matters for the voltage.

How do I calculate the resistor values for a voltage divider?

Pick one value first — usually from your current budget — then get the other from the ratio. For a given R1, R2 = R1 × Vout / (Vin − Vout). For a given R2, R1 = R2 × (Vin − Vout) / Vout. Then round both to standard values together rather than rounding one after the fact.

Why does my voltage divider output drop when I connect a load?

Because the load is in parallel with R2, which reduces the effective bottom resistance. The error is exactly −R_out / (R_out + R_L), where R_out is R1 in parallel with R2. A load ten times R_out costs 9 %, a hundred times costs 1 %.

What is the output impedance of a voltage divider?

R1 in parallel with R2, which is R1 × R2 / (R1 + R2). It is the Thevenin resistance of the divider and it is the single most useful number about one, because everything about loading follows from it.

Should I use high or low resistor values in a divider?

Low values give a stiff output and waste current; high values save current and become sensitive to loading, leakage and noise. For a divider feeding a high-impedance input, 10 kΩ to 100 kΩ total suits most cases. For a real load, choose values so that R_out is at least ten times below the load.

Can a voltage divider power a circuit?

No, not usefully. It has no regulation — the output falls as the load draws more, and moves with the supply — and it dissipates more power than it delivers. Use a divider for a reference and a regulator for a supply.

Can a voltage divider increase voltage?

No. Two resistors in series can only divide a voltage down. To raise one you need a boost converter, a charge pump or a transformer — no arrangement of resistors will do it, and the design formulas give a negative resistance if you ask.