A voltage divider’s familiar output formula assumes nothing is drawing current from its output. Connect a device—or even a voltmeter—and that device becomes part of the circuit. For a conventional divider, a resistive load across the output sits in parallel with the lower resistor, changing the resistance ratio and usually lowering the output voltage.
Why the output voltage changes
In a conventional divider, R1 connects the input voltage to the output node, and R2 connects that node to ground. With no load attached, the output is Vout = Vin × R2 / (R1 + R2). This assumes the output draws no current.
When a resistive load RL is connected between the output and ground, current through R1 supplies both R2 and the load. The effective lower resistance becomes R2 || RL, where || means parallel. The loaded output is therefore Vout = Vin × (R2 || RL) / (R1 + (R2 || RL)). For a finite resistive load to ground, this is below the unloaded output.
The load does not simply “use up” a portion of the voltage; it changes the resistance network that sets the output. A transistor base, for example, draws base current from the divider and can shift the intended bias voltage.
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Calculate the loaded output
- Calculate the effective lower-leg resistance: R2,eff = (R2 × RL) / (R2 + RL).
- Substitute that value into the divider equation: Vout,loaded = Vin × R2,eff / (R1 + R2,eff).
- Compare the result with the unloaded value, Vin × R2 / (R1 + R2). A load much larger than the divider’s output resistance has little effect; a smaller load causes more droop.
For a concrete example, Electronics Tutorials describes a potentiometer with a 6 V unloaded wiper output and a 1 kΩ lower section. Connecting a 3 kΩ load makes the lower equivalent resistance 750 Ω and the output 5.4 V (Electronics Tutorials’ voltage-divider explanation).
Use the Thevenin model to understand sensitivity
Viewed from the output, an ideal-source divider can be replaced by a Thevenin source: its voltage is the unloaded divider output, and its source resistance is R1 || R2. A load then forms a second divider with that source resistance. This makes the key comparison clear: the load’s input resistance should be large relative to the divider’s output resistance if you want little voltage change.
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Increasing both divider resistor values raises the output resistance, making the divider more sensitive to a given load. Lower values make the output stiffer, but they also increase current continuously through the divider.
Your voltmeter may be the load
A voltmeter connects across the points being measured. Since a real meter has finite input resistance, it adds a parallel path just like a load. Higher meter input resistance means less current drawn from the circuit and a smaller measurement effect; the effect is especially noticeable in high-resistance circuits. All About Circuits explains the ideal limit: “An ideal voltmeter has infinite input impedance, drawing zero current from the circuit under test.” (All About Circuits’ voltmeter-loading lab).
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The size of the effect can be dramatic when the divider resistors are very large. In a published All About Circuits example, two 250 MΩ resistors divide 24 V to an unloaded 12 V midpoint. A 10 MΩ voltmeter across the lower 250 MΩ resistor makes that leg about 9.615 MΩ in parallel and the meter reads 0.8889 V. Those values describe that specific worked example, not a typical meter-and-divider combination (All About Circuits’ worked example).
Choose a fix that matches the load
| Approach | When it helps | Trade-off |
|---|---|---|
| Use lower-value divider resistors | The load is known and modest, and a little droop is acceptable. | More continuous current and resistor power dissipation; check resistor ratings and tolerances. |
| Use a higher-input-resistance load or meter | The downstream device has input-impedance options, or measurement loading is the issue. | Confirm the actual input resistance and include it in the calculation. |
| Add a buffer | The following stage should not draw meaningful current from the divider. | The buffer adds design complexity; choose it for the required voltage range, current, stability, and accuracy. |
| Use a regulated supply | The load needs substantial or varying current, or the voltage must remain stable as the load changes. | A divider is not a general-purpose power supply; load-related power is dissipated in its resistors. |
There is no universally best fix. Compare expected voltage droop, continuous divider current and resistor dissipation, measurement loading, accuracy, and implementation complexity. Very high resistor values also make a meter’s internal resistance more likely to affect the measured divider, as noted in the All About Circuits voltage-divider lab.
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Check what the connected device actually looks like electrically
The parallel-resistor calculation is appropriate when the load can be modeled as a fixed resistance from output to ground. Not every electronic input behaves that way: it may be nonlinear, frequency-dependent, or active. For a precision result, use the device datasheet or model its input under the relevant operating conditions rather than assuming a fixed RL.
Reproduce the effect on a breadboard
A simple experiment makes the cause visible: build a divider, measure its unloaded output, then connect a known resistor across the output and measure again. Use the meter’s input resistance in your predicted calculation as well. A breadboard and resistor assortment are enough to explore how changing divider and load values affects droop; the All About Circuits lab provides a divider-building exercise.
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