What is the loading effect in a voltage divider?

When you connect a load (RL) to the output of a voltage divider, it appears in parallel with R2. This reduces the effective resistance of the lower leg, pulling the output voltage below the unloaded value. The loading error is (Vout_unloaded − Vout_loaded) / Vout_unloaded × 100%.

How do I minimize loading error in a voltage divider?

Keep RL at least 10× greater than R2. For example, if R2 = 10kΩ, use RL ≥ 100kΩ for less than 10% error. Alternatively, buffer the output with a voltage follower op-amp, which presents virtually infinite input impedance to the divider.

Loaded Voltage Divider Calculator

Calculate output voltage with and without a load, and see how load resistance affects your divider's accuracy.

Divider Values

Input Voltage (Vin)

Top Resistor (R1)

Bottom Resistor (R2)

Load Resistance (RL) — optional

Results

Vout (no load)6.0000 V

Divider ratio0.5000

Quiescent current600 µA

Rule of thumb: RL ≥ 100.0 kΩ for <10% error

The Loading Effect Explained

An ideal voltage divider assumes no current is drawn at the output. In practice, any load connected to the output forms a parallel combination with R2, reducing the effective bottom resistance and pulling the output voltage down.

This "loading effect" is the difference between the unloaded and loaded output voltage, expressed as a percentage. It becomes significant when the load resistance is comparable to R2.

Unloaded Output

Vout = Vin × R2 / (R1 + R2)

Loaded Output (R2eff = R2 ∥ RL)

Vout = Vin × R2eff / (R1 + R2eff)

Key Points

RL ≥ 10× R2 keeps loading error below ~10%
RL ≥ 100× R2 keeps loading error below ~1%
Lower divider resistors reduce loading but increase quiescent current
A unity-gain op-amp buffer eliminates loading entirely
ADC input impedance acts as a load — check the datasheet

Applications

Battery voltage sensing with ADC
Reference voltage generation
Signal level shifting
Bias networks for transistors

Practical Examples

12 V → 6 V with 10 kΩ load

Simple 1:1 divider (R1=R2=10 kΩ) from 12 V with a 10 kΩ load. Shows how loading collapses the output voltage.

R2∥RL = 5 kΩ · Vout = 12 × 5/(10+5) = 4.0 V instead of expected 6 V — 33% error!

Stiff divider fix: 1 kΩ resistors

Same 12 V to 6 V divider but with R1=R2=1 kΩ and the same 10 kΩ load. Loading error drops dramatically.

R2∥RL = 909 Ω · Vout = 12 × 909/(1000+909) = 5.71 V — only 4.8% error

Did you know? A voltage divider loaded by a high-impedance ADC input (e.g. 1 MΩ) barely affects the output voltage, but a 10 kΩ load on a divider made of two 100 kΩ resistors can cause more than 8% error. This is why op-amp buffers are recommended after precision dividers.