About Voltage Divider Calculator — Output Voltage & Loaded Divider
The voltage divider is the most-built circuit in electronics: two resistors turn a voltage you have into a voltage you need. This calculator solves the textbook relation Vout = Vin·R₂/(R₁+R₂) and then keeps going where textbook answers usually stop — it computes the loaded output when whatever you connect (an ADC input, a transistor base, the next stage) puts its own resistance RL in parallel with R₂, and shows the droop as a percentage of the unloaded output.
That loaded number is the one that matters, because the classic divider failure is designing for the unloaded formula and then wondering why the output sags: a 10 kΩ/5 kΩ divider from 12 V reads a clean 4.0 V on a 10 MΩ multimeter, and drops to 3.0 V the moment a 10 kΩ load connects. The tool also reports the standing current and the power in each resistor — the divider burns Vin²/(R₁+R₂) continuously whether or not anything uses the output, which is what makes divider impedance a battery-life decision, not just an accuracy one.
How It Works
- Enter the input voltage and the two resistors in kilohms. R₁ is the upper resistor (from Vin to the output node), R₂ the lower (output node to ground) — swap them mentally if your schematic is drawn the other way; only the ratio matters.
- Read the unloaded output: Vin × R₂/(R₁+R₂). This is what a high-impedance meter or an op-amp buffer sees. The edge cases work: R₁ = 0 gives Vout = Vin, R₂ = 0 gives 0 V; only R₁ + R₂ = 0 is rejected (that would short the supply).
- Enter the load resistance to get the loaded output: the tool substitutes R₂∥RL = R₂·RL/(R₂+RL) into the divider and reports both the voltage and the droop percentage. If you do not know RL, look up the input resistance of what you are driving — ADC inputs are often 10–100 kΩ during sampling, classic analog panel meters can be far lower.
- Check the current and power lines: I = Vin/(R₁+R₂) flows through the divider continuously. At 12 V and 15 kΩ total that is 0.8 mA and 9.6 mW — nothing for a wall-powered board, but 7 mAh a day from a coin cell that stores maybe 220.
- The design rule of thumb falls out of the droop number: keep the divider's output impedance (R₁∥R₂) at or below a tenth of the load impedance for under ~10% droop — or buffer the output with an op-amp follower and stop caring.
Worked Example
Divide 12 V down with R₁ = 10 kΩ and R₂ = 5 kΩ. Unloaded: Vout = 12 × 5/(10+5) = 4.0 V, with I = 12 V / 15 kΩ = 0.8 mA flowing continuously — P₁ = I²R₁ = 6.4 mW in the upper resistor, P₂ = 3.2 mW in the lower, 9.6 mW total. Now connect a 10 kΩ load: it lands in parallel with R₂, making R₂∥RL = 5×10/(5+10) = 3.33 kΩ, so the divider becomes 10 kΩ over 3.33 kΩ and the output falls to 12 × 3.33/13.33 = 3.0 V — a full 25% droop. That is not a component fault; it is the divider's ~3.3 kΩ output impedance meeting a comparable load. Raising R₂∥R₁ tenfold lighter than the load, or buffering with an op-amp, restores the 4.0 V.
Formulas
- Unloaded output
Vout = Vin × R₂ / (R₁ + R₂)- Loaded output
R₂∥RL = R₂·RL/(R₂+RL); Vout′ = Vin × R₂∥RL / (R₁ + R₂∥RL)- Current and power
I = Vin/(R₁+R₂); P₁ = I²R₁; P₂ = I²R₂; P = Vin·I
Standards & References
- Direct application of Ohm's law and Kirchhoff's voltage law — the series-resistor voltage division relation of any circuits text (e.g. Horowitz & Hill, The Art of Electronics)
- Design rule of thumb: divider output impedance R₁∥R₂ ≤ RL/10 keeps loading droop under ~10%; buffer with an op-amp follower when the load is unknown or varying
Frequently Asked Questions
Which resistor is R₁ and which is R₂?
R₁ connects the input voltage to the output node; R₂ connects the output node to ground. The output is taken across R₂, so a bigger R₂ share means more output: Vout = Vin·R₂/(R₁+R₂). If your measured output is the complement of what you expected (8 V instead of 4 V from a 12 V source), the two resistors are simply swapped in your head or on your breadboard — the divider itself has no idea which resistor you meant to be which.
Why does my divider output drop when I connect the circuit it feeds?
Because the load's input resistance lands in parallel with R₂ and shrinks it. A 10 k/5 k divider from 12 V makes 4.0 V unloaded, but a 10 kΩ load turns the lower leg into 3.33 kΩ and the output into 3.0 V — 25% droop, computed exactly by this tool's loaded mode. The multimeter did not lie: its ~10 MΩ input barely loads the divider. Fix it with lower divider resistances (more standing current), or buffer the node with an op-amp follower.
What resistor values should I pick if only the ratio matters?
The ratio sets the voltage; the absolute values set two competing costs. Low values (say 1 kΩ total) make a stiff divider that ignores loads but burns milliamps continuously — bad for batteries. High values (1 MΩ total) sip current but droop under any load and pick up noise. The usual engineering middle is 10–100 kΩ total for signal dividers, then check this tool's droop line against your actual load, and go lower (or buffer) if it exceeds your error budget.
Can I use a voltage divider as a power supply for a 5 V device?
Almost never. A divider only holds its ratio while the load is negligible or constant: a device drawing varying current is a varying RL, so your "5 V" moves with every load change, and making the divider stiff enough to hide that means burning far more power in the resistors than the device uses. Dividers are for references, sensing, and signal scaling. For powering things, use a regulator — linear for simplicity, switching for efficiency.
Does the divider work with AC or a negative supply?
The resistive division ratio is polarity- and frequency-agnostic: −12 V in with 10 k/5 k gives −4.0 V out, and an AC signal divides by the same ratio (this calculator accepts negative rails directly). The caveat at AC is parasitic and load capacitance: with tens-of-kilohm resistors, a few picofarads forms a low-pass filter that rolls off high frequencies — the reason oscilloscope 10× probes add a compensating capacitor across the top resistor.