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A voltage divider uses two series resistors to produce a fraction of an input voltage. It is useful for sensing, scaling, biasing, and setting thresholds—not for powering a circuit. With R1 connected from VIN to the output and R2 connected from the output to ground:
VIN ─── R1 ───┬── VOUT
│
R2
│
GNDFor an unloaded resistive divider, VOUT = VIN × R2 / (R1 + R2). In a real circuit, the connected load, measuring instrument, ADC, leakage, resistor tolerance, and capacitance can all change the result.
The basic voltage-divider formula
Because the resistors are in series, the same current flows through both when the midpoint is unloaded:
IDIV = VIN / (R1 + R2)
The output is the voltage across R2:
VOUT = IDIV × R2 = VIN × R2 / (R1 + R2)
The resistor closest to ground determines the numerator. If you take the output across R1 instead, the equation is VOUT = VIN × R1 / (R1 + R2).
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Example: a 12 V divider
With VIN = 12 V, R1 = 9 kΩ, and R2 = 3 kΩ:
VOUT = 12 × 3 / (9 + 3) = 3 V
The divider current is 12 V / 12 kΩ = 1 mA. Resistor dissipation is:
PR1 = I²R1 = 9 mWPR2 = I²R2 = 3 mW
Choose resistor power ratings with suitable margin, and also check maximum voltage, pulse, temperature, and tolerance ratings.
Choosing resistor values
For a target output, rearrange the formula:
R2 = R1 × VOUT / (VIN − VOUT)
If you select the total resistance first:
R2 = RTOTAL × VOUT / VINR1 = RTOTAL − R2
Example: scaling 5 V to approximately 3.3 V
The required ratio is 3.3 / 5 = 0.66. A practical standard-value pair is R1 = 3.3 kΩ and R2 = 6.8 kΩ:
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VOUT = 5 × 6.8 / (3.3 + 6.8) ≈ 3.37 V
This is a nominal value, not a regulated 3.3 V supply. Input variation, resistor tolerance, transients, and the connected circuit can all move the output. Verify the receiving input’s absolute-maximum voltage before connecting it.
Divider current, power, and resistor size
For an unloaded divider, increasing both resistor values preserves the ratio but reduces current and power. Lower values create a lower-impedance, more load-resistant output, but waste more current:
| Lower resistance | Higher resistance |
|---|---|
| Lower loading error | Higher loading error |
| Lower Thévenin resistance | Higher Thévenin resistance |
| Higher current and power | Lower current and power |
| Less sensitive to leakage | More sensitive to leakage and noise |
| Usually easier to drive into an ADC | May require buffering |
There is no universal rule that every divider should use 10 kΩ resistors. Select values according to the input impedance, accuracy, bandwidth, current budget, noise, leakage, and ADC acquisition requirements.
The real problem: loading
The simple equation assumes that no meaningful current leaves the output node. A real load is connected in parallel with R2, so replace the lower resistor with:
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Then calculate:
VOUT = VIN × R2,eff / (R1 + R2,eff)
Example: a divider that sags
Start with VIN = 5 V, R1 = 10 kΩ, and R2 = 10 kΩ. The unloaded output is 2.5 V.
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If a 10 kΩ load is connected from the output to ground, the effective lower resistance becomes:
10 kΩ ∥ 10 kΩ = 5 kΩ
The output is now:
VOUT = 5 × 5 / (10 + 5) ≈ 1.67 V
This loading effect is why a divider that produces the expected voltage on a schematic can produce a much lower voltage in hardware. TI discusses this parallel-load calculation in its voltage-divider reference, while an NI laboratory example demonstrates how changing the load changes the measured output.
The Thévenin equivalent
Viewed from the output, an unloaded divider can be replaced by a voltage source and a series resistance:
VTH = VIN × R2 / (R1 + R2)RTH = R1 ∥ R2
With a load attached:
VOUT = VTH × RL / (RTH + RL)
This is the most useful practical model: the divider is not an ideal voltage source. It is the desired open-circuit voltage in series with an output resistance. A lower RTH drives a load more effectively, but requires more divider current.
As a rough starting point, designers often make the load much larger than the divider resistance. A load around 100 times R2 may keep loading relatively small in many arrangements, but this is only a heuristic. Calculate the actual error for the specific R1, R2, and load.
Why a multimeter can change the reading
A meter has finite input resistance. When it is connected across the output, that resistance becomes part of RL. For example, with a 10 V source, R1 = 1 MΩ, R2 = 1 MΩ, and a meter input resistance of 10 MΩ:
R2,eff = 1 MΩ ∥ 10 MΩ ≈ 0.909 MΩ
The meter therefore reads less than the ideal 5 V. High-value dividers are particularly vulnerable to meter loading, leakage, PCB contamination, and probe effects. Check the instrument’s input resistance, reduce divider resistance where practical, or buffer the output. Tektronix explains this measurement error using the source’s Thévenin resistance and the meter input resistance.
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Dividers are commonly used to measure a battery or other voltage that exceeds an ADC’s input range. For a maximum input and ADC voltage:
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R2 / (R1 + R2) ≤ VADC,max / VIN,max
Design using the highest possible input, including supply tolerance, transients, startup conditions, and fault voltages. Do not target the absolute ADC limit without margin.
An ADC input is not necessarily an infinite-impedance voltmeter. Depending on the device and operating mode, it may have leakage, a sample-and-hold capacitor, an acquisition-time requirement, and a specified source-impedance limit. The divider’s RTH can cause gain error or prevent the sampling capacitor from settling before conversion. Analog Devices describes these source-resistance, settling, and distortion effects in its ADC source-resistance article; TI also recommends checking the ADC’s required drive impedance.
- Determine the maximum input, including tolerances and transients.
- Choose a ratio that stays safely below the ADC maximum.
- Select resistor values that satisfy the ADC’s source-impedance and acquisition requirements.
- Include ADC leakage and input capacitance in the error analysis.
- Use a capacitor only after checking its filtering and settling-time effects.
- Add a buffer if the divider must be high resistance for low power but the ADC requires a low-impedance source.
- Check startup, shutdown, protection-clamp current, and fault conditions.
A capacitor from the ADC input to ground can filter noise, but it forms an RC network with the source resistance. That improves some noise and bandwidth problems while increasing settling time.
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The output depends on the ratio of the two resistors, not just their nominal values. For worst-case analysis, use:
R1high andR2low for the minimum output.R1low andR2high for the maximum output.
Also consider input-voltage accuracy, temperature coefficients, resistor tracking, self-heating, amplifier offset and bias current, ADC reference error, leakage, and PCB contamination. A matched resistor network can track temperature and ratio changes better than two unrelated resistors, although it may cost more or offer fewer available ratios. See Mouser’s voltage-divider network listings for the specifications that vary between parts.
Potentiometers as adjustable dividers
A three-terminal potentiometer becomes a variable divider when its outer terminals connect across a supply and its wiper provides the output:
VIN ─── outer terminal
│
resistive track ── wiper → VOUT
│
GND ─── outer terminalIdeally, the wiper moves from near 0 V to near VIN. In practice, the range and accuracy depend on total resistance, position, tolerance, wiper resistance, end resistance, contact noise, wear, and load. Potentiometers are suitable for user controls, calibration, volume settings, and thresholds. They are not regulated power supplies.
AC signals, probes, and capacitive loading
For purely resistive components, the same ratio applies to an ideal AC signal. For general impedances, use:
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VOUT = VIN × Z2 / (Z1 + Z2)
Here, Z may include resistors, capacitors, inductors, cables, probes, and input capacitance. Oscilloscope-probe capacitance, ADC sampling capacitance, PCB parasitics, and amplifier input capacitance can make the divider frequency-dependent. A waveform that looks correct at DC may show reduced amplitude, slower edges, or distortion at higher frequencies. NI explains how probe resistance and capacitance load the circuit; specialized compensated RC dividers can preserve a ratio over a wider bandwidth.
What a voltage divider should not do
A divider is normally a signal-scaling circuit, not a voltage regulator. Do not use an ordinary resistor divider to power:
- LEDs, motors, relays, or changing loads
- Digital circuits or sensors with significant current demand
- A circuit that needs a stable supply rail
- A load that charges a capacitor quickly
As the load current changes, the divider voltage changes. Use a linear regulator, switching converter, voltage reference, or buffered active circuit when the output must supply meaningful current or reject input variation. A divider can still create a low-current reference, comparator threshold, amplifier bias, or measurement signal.
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Negative and bipolar signals
The resistor equations still apply to negative voltages, but the receiving input may not tolerate them. A divider alone does not convert a bipolar signal into a safe 0–5 V signal. For example, scaling ±10 V for a unipolar ADC generally requires a bias/reference offset, level-shifting stage, protection, and a resistor network designed for both extremes. Verify common-mode range, absolute-maximum voltage, and clamp-current limits.
Including source resistance
If the voltage source itself has resistance RS, it adds in series with R1:
VOUT = VIN × R2 / (RS + R1 + R2)
Include wiring, switch, protection, and intentional series resistance when accurate ratios matter. This is especially important when the divider resistors are large.
Troubleshooting checklist
The measured output is lower than calculated
- Include the load, meter, ADC, amplifier, or transistor input in parallel with
R2. - Check for leakage paths, contamination, and an incorrect resistor value.
- Confirm the source voltage and output node.
- Inspect for a damaged resistor or a wiring error.
The output changes when another circuit is connected
This is classic loading. Estimate or measure the new circuit’s input resistance and include it as RL.
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The ADC reading is noisy or inaccurate
- Reduce source resistance if the ADC datasheet requires it.
- Check sampling and acquisition time.
- Shorten the high-impedance trace and improve grounding.
- Use an appropriately calculated filter capacitor.
- Buffer the divider when necessary.
- Verify the reference voltage and input protection.
The DC value is correct but fast edges are wrong
Check probe capacitance, ADC capacitance, cable capacitance, and the RC time constant. The divider may be bandwidth-limited even though its DC ratio is correct.
The battery current is too high
Increase total resistance while checking leakage, noise, ADC drive, settling time, and loading. For intermittent measurements, switch the divider on only during sampling and allow time for the output to settle.
Quick Recap
Quick-reference formulas
| Quantity | Formula |
|---|---|
| Unloaded output | VOUT = VIN × R2 / (R1 + R2) |
| Divider current | IDIV = VIN / (R1 + R2) |
| Loaded lower leg | R2,eff = R2 ∥ RL |
| Thévenin voltage | VTH = VIN × R2 / (R1 + R2) |
| Thévenin resistance | RTH = R1 ∥ R2 |
| Resistor power | P = I²R = V²/R |
| General impedance divider | VOUT = VIN × Z2 / (Z1 + Z2) |
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