What is the voltage divider output formula?

For an unloaded divider: Vout = Vin × R2 / (R1 + R2). R1 connects Vin to the output node; R2 connects the output node to GND. The ratio R2/(R1+R2) is always between 0 and 1, so Vout is always less than Vin.

How does a load resistance change the output voltage?

A load resistance RL connected across R2 forms a parallel combination Rp = R2 × RL / (R2 + RL). The loaded output voltage is Vout = Vin × Rp / (R1 + Rp), which is always lower than the unloaded value. The calculator expresses this as a loading error percentage: (Vout_unloaded − Vout_loaded) / Vout_unloaded × 100.

How do I minimise the loading effect?

Make the divider current much larger than the load current. A rule of thumb is to size R1 + R2 so that the quiescent current is at least 10× the maximum load current. Alternatively, buffer the output with an op-amp voltage follower, which presents near-infinite impedance to the divider.

What are E-series resistors and why do they matter?

E-series (E12, E24, E48, E96) are standardised sets of preferred resistor values spaced logarithmically within each decade. A component picked at random from a reel will be one of these values — not an arbitrary ohm value. The E-series finder in this tool calculates the exact resistor needed for a target Vout, then snaps it to the nearest available standard value and shows how much that deviates from your target.

What is the difference between E12, E24, E48, and E96?

The number gives the count of values per decade. E12 has 12 values per decade (roughly 10 % tolerance), E24 has 24 (5 %), E48 has 48 (2 %), and E96 has 96 (1 %). The tighter the tolerance series, the closer the standard value is to your ideal calculation, and the smaller the resulting Vout error.

How is power dissipation calculated for each resistor?

For the unloaded case, the current through the divider is I = Vin / (R1 + R2). Power is then P_R1 = I² × R1 and P_R2 = I² × R2. When a load is present, the current splits: R1 carries the full current I_R1 = Vin / (R1 + Rp), while R2 carries I_R2 = Vout / R2 and the load takes I_load = Vout / RL.

Voltage Divider Output Calculator

A voltage divider output calculator that handles both the ideal unloaded case and the more realistic loaded scenario, where a downstream circuit pulls current from the output node and lowers the voltage. It is aimed at electronics engineers, students, and makers who need to quickly size a resistor pair, verify a divider under load, or find the nearest catalogue resistor for a target output.

How it works

The classic voltage divider places two resistors in series across a supply. The output is tapped at the junction between them:

Vout = Vin × R2 / (R1 + R2)

R1 is the “top” resistor connecting the supply to the output node; R2 is the “bottom” resistor from the output node to ground. The fraction R2 / (R1 + R2) — called the divider ratio — is always between 0 and 1, so Vout is always a fraction of Vin.

The loading effect

The formula above assumes no current flows out of the output node. Any real circuit (ADC input, microcontroller GPIO, op-amp, transistor base) draws some current, which effectively places a load resistance RL in parallel with R2. This reduces the bottom leg of the divider:

Rp = R2 ‖ RL = R2 × RL / (R2 + RL)

The loaded output voltage is then:

Vout (loaded) = Vin × Rp / (R1 + Rp)

Since Rp < R2, the loaded Vout is always lower than the unloaded value. The calculator quantifies this as the loading error:

error = (Vout_unloaded − Vout_loaded) / Vout_unloaded × 100 %

A loading error under 1 % is usually acceptable. Above 5 % you should either reduce R1 + R2 (use lower-value resistors to increase quiescent current) or buffer the output with an op-amp voltage follower.

E-series resistor finder

Resistors are only manufactured in standardised values defined by the E-series — 12, 24, 48, or 96 values per decade. When the exact R1 or R2 computed from your target Vout doesn’t exist on a reel, the tool snaps it to the nearest E-series value and re-calculates the actual Vout so you know the resulting error before soldering anything.

Worked example

Design a 3.3 V reference from a 5 V supply, driving a 10 kΩ ADC input.

Step 1 — choose R2 = 10 kΩ, find R1 (unloaded):

R1 = R2 × (Vin / Vout − 1) = 10 kΩ × (5 / 3.3 − 1) = 10 kΩ × 0.515 = 5.152 kΩ

Step 2 — snap to E24 series: nearest is 5.1 kΩ

Actual Vout (unloaded) = 5 V × 10 / (5.1 + 10) = 3.311 V (0.33 % error)

Step 3 — check loading effect with RL = 10 kΩ:

Rp = 10 kΩ ‖ 10 kΩ = 5 kΩ

Vout (loaded) = 5 V × 5 / (5.1 + 5) = 2.475 V — a loading error of 25 %!

This is why the loading check matters. The fix: choose a lower-impedance divider, for example R2 = 1 kΩ, R1 = 510 Ω. Now the quiescent current is 3.3 mA, far above the ADC pull current, and loading error drops to 0.5 %.

Vin	R1	R2	RL	Vout loaded	Loading error
5 V	5.1 kΩ	10 kΩ	10 kΩ	2.48 V	25 %
5 V	510 Ω	1 kΩ	10 kΩ	3.27 V	1.3 %
12 V	10 kΩ	10 kΩ	100 kΩ	5.74 V	4.3 %

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