A voltage divider seems almost too simple to fail. Put two resistors in series, take the output from their midpoint, and the resistor ratio sets the voltage. With 10 V across two equal 10 kΩ resistors, the familiar calculation says:
Vout = Vin × R2⁄R1 + R2 = 10 V × 10 kΩ⁄10 kΩ + 10 kΩ = 5 V.
Then you connect a circuit to that output and measure 4.8 V, 3.3 V, or something even lower. Nothing is wrong with Ohm’s law. The original calculation quietly assumed that the output was unloaded—that nothing else drew current from the midpoint.
This article follows a 10 V, 10 kΩ/10 kΩ divider through LTspice operating-point simulations. The central idea is that a voltage divider is not a perfect voltage source. It has output resistance, and a load forms another divider with that resistance.
Figure 1. Adding a load does not change either divider resistor. It changes the circuit around them.
The formula contains a hidden assumption
In the unloaded circuit, the same current flows through R1 and R2:
Idivider = 10 V 10 kΩ + 10 kΩ = 0.5 mA.
The voltage across R2 is therefore 0.5 mA × 10 kΩ = 5 V. This reasoning works because there is only one current path.
Now connect a load resistance RL from the output to ground. Current can reach ground through both R2 and RL. Those two resistances are in parallel:
Rlower = R2 ∥ RL = R2RL R2 + RL .
The loaded divider equation becomes:
Vloaded = Vin R2 ∥ RL R1 + (R2 ∥ RL) .
Suppose the load is also 10 kΩ. The parallel combination (10 kΩ ∥ 10 kΩ) is 5 kΩ. The circuit is no longer a 10 kΩ-over-10 kΩ divider; electrically, it is a 10 kΩ-over-5 kΩ divider:
Vloaded = 10 V 5 kΩ 10 kΩ + 5 kΩ = 3.333 V.
That is a large change, but it is exactly what the new resistor network demands.
The Thevenin shortcut
Repeatedly calculating parallel combinations works, but there is a cleaner mental model. Viewed from its output terminal, any ideal resistive divider can be replaced by a Thevenin equivalent: one ideal voltage source in series with one resistance.
For the unloaded voltage:
VTH = Vin R2 R1 + R2 .
For the output resistance, set the ideal input voltage source to zero, which means replacing it with a short circuit. Looking back into the divider then places R1 and R2 in parallel:
RTH = R1 ∥ R2.
Our equal 10 kΩ divider is therefore equivalent to a 5 V source with 5 kΩ in series. Once a load is attached, the answer is another ordinary divider:
Vloaded = VTH RL RTH + RL .
Figure 2. The parallel view explains what changed. The Thevenin view makes the divider’s non-ideal source behavior obvious.
The Thevenin model gives several useful landmarks immediately. If RL = RTH, the load receives half the unloaded voltage. A load ten times RTH still causes about 9.1% error—not 1%. To hold loading error below 5%, the load must be at least 19 times RTH. Below 1% requires at least 99 times RTH.
For this divider, RTH = 5 kΩ. That means a 5% target requires approximately 95 kΩ or more, while a 1% target requires approximately 495 kΩ or more.or this divider, (R_{\text{TH}}=5\text{ k}\Omega). That means a 5% target requires approximately 95 kΩ or more, while a 1% target requires approximately 495 kΩ or more.
How much does the output move?
The LTspice sweep varies (R_L) from 100 Ω to 100 MΩ while solving the DC operating point at each value. Selected results are:
| Load resistance | Loaded output | Error from 5 V |
|---|---|---|
| 10 kΩ | 3.333 V | 33.33% |
| 100 kΩ | 4.762 V | 4.76% |
| 1 MΩ | 4.975 V | 0.50% |
| 10 MΩ | 4.9975 V | 0.05% |
As the load resistance increases, it draws less current and the output approaches the unloaded 5 V value. “High impedance” is therefore relative. A 100 kΩ load sounds large, but it is only twenty times this divider’s 5 kΩ output resistance, so its error is still nearly 5%.
Figure 3. Loading becomes negligible only when the load resistance is much larger than the divider’s Thevenin resistance.
The load also changes the current through R1. With no load, the source supplies 0.5 mA. With a 10 kΩ load, the output is 3.333 V, so R1 carries
10 − 3.333 10 kΩ = 0.667 mA.
At the output node, about 0.333 mA flows through R2 and another 0.333 mA flows through the load. Kirchhoff's current law still balances perfectly.
Lower-value resistors hold the voltage better
A divider’s ratio determines its unloaded voltage, but the absolute resistance values determine its output resistance. Scaling both resistors down by ten keeps the ratio unchanged while reducing (R_{\text{TH}}) by ten.
Consider three equal dividers:
| Divider pair | Unloaded current | Unloaded power | (R_{\text{TH}}) |
|---|---|---|---|
| 1 kΩ / 1 kΩ | 5 mA | 50 mW | 500 Ω |
| 10 kΩ / 10 kΩ | 0.5 mA | 5 mW | 5 kΩ |
| 100 kΩ / 100 kΩ | 50 µA | 0.5 mW | 50 kΩ |
All three produce 5 V when unloaded. Under the same load, the 1 kΩ pair stays closest to 5 V because its Thevenin resistance is smallest. That improvement costs continuous current and power, even when the output is doing no useful work. The 100 kΩ pair conserves power but is far easier to disturb.
Figure 4. Divider design is a tradeoff between load regulation and standing power.
This is why simply choosing “very large resistors to save power” can backfire. Leakage currents, PCB contamination, transistor inputs, protection networks, and measurement equipment all become more significant as divider impedance rises. Conversely, choosing extremely small resistors may waste battery energy or exceed resistor power ratings.
Even a measuring instrument is a load
A voltmeter does not observe a circuit from the outside; its input impedance becomes part of the circuit. A 1 MΩ oscilloscope input loads the baseline divider to about 4.975 V, an error of roughly 0.5%. A 10 MΩ digital multimeter or 10× oscilloscope probe produces about 4.9975 V, an error near 0.05%.
Those errors are small here, but the same 1 MΩ instrument would seriously disturb a divider made from megohm-scale resistors. Always compare the instrument input resistance with (R_{\text{TH}}), not just with either resistor alone.
An ADC input needs extra care. Its quoted DC input resistance may be very high, yet a switched-capacitor sampling input can draw brief packets of charge. The divider may then require time to settle between samples. That is dynamic loading rather than the purely resistive DC loading simulated here; follow the converter datasheet’s recommended maximum source impedance or add a buffer.
Designing so the load does not matter
Start with the maximum expected load and an acceptable voltage error. If the allowed fractional error is e, the Thevenin resistance must satisfy:
RTH ≤ e 1 − e RL.
For a 1% target, this is approximately RTH ≤ RL/99. Choose R1 and R2 to meet both the desired divider ratio and this output-resistance limit, then check divider current and resistor power.
For example, suppose the smallest load will be 100 kΩ and the output must remain within 1% of 5 V. The required Thevenin resistance is at most
RTH ≤ 0.01 0.99 (100 kΩ) ≈ 1.01 kΩ.
For an equal divider, RTH = R/2, so each divider resistor must be about 2.02 kΩ or less. Choosing two 2 kΩ resistors gives RTH = 1 kΩ. With the 100 kΩ load, the output becomes 5 × (100/101) = 4.9505 V, just inside the 1% target. The price is 2.5 mA of unloaded divider current and 25 mW drawn from the 10 V source.
That calculation addresses loading only. Real resistors also have tolerance and temperature coefficient, and the input supply may vary. Those error sources do not disappear when loading is solved; they need their own allowance in the error budget. A design targeting 1% total accuracy should not spend the entire 1% on loading.
If the load varies widely or needs appreciable current, a passive divider may be the wrong source. A voltage follower, regulator, reference, or other buffered circuit can preserve the target voltage without forcing the divider itself to supply the load.
Reproduce the LTspice results
The package includes the baseline load-sweep schematic, the impedance-comparison schematic, equivalent text netlists, and their generated logs and ASCII raw files.
The baseline directives are:
.step dec param Rload 100 100Meg 20
.op
.option plotwinsize=0
.save V(in) V(out) I(V1)
From this article’s directory, run:
& "$env:LOCALAPPDATA\Programs\ADI\LTspice\LTspice.exe" -b -ascii ltspice\divider_load_sweep.asc
& "$env:LOCALAPPDATA\Programs\ADI\LTspice\LTspice.exe" -b -ascii ltspice\divider_scale_comparison.asc
node assets\render.js
The renderer parses LTspice’s stepped operating-point data, validates every point against the loaded-divider equation, writes the measurement CSV, and regenerates all four SVG figures without external packages.