Calcometry

Calculadora de divisor de tensão

Ao usar nossas calculadoras, você concorda com nossos Termos de uso (English).

Classic two-resistor voltage divider for reference and sensor scaling.

Divisor

Tensão de saída

2.5 V

Divider ratio: 50%

Corrente do divisor
0.25 mA
Vout / Vin
0.5

Calculadoras relacionadas

Voltage dividers

Two resistors in series split input voltage proportionally. Output at the junction: Vout = Vin × R2 ÷ (R1 + R2). Load current drawn from the tap lowers Vout — this formula assumes negligible load current.

Dividers set reference voltages, scale sensors, and bias transistors. For stable output under load, use a buffer op-amp or size resistors small enough that load current is negligible (watch power dissipation).

Example: 5 V input, R1 = 10 kΩ, R2 = 10 kΩ → Vout = 2.5 V. Current through divider = 5 V ÷ 20 kΩ = 0.25 mA.

Equal resistors give half the supply at the midpoint — the classic 2.5 V reference from 5 V logic. Unequal values scale anywhere from near 0 V to near Vin depending on the R1:R2 ratio.

Power dissipation in each resistor is I²R. High-value resistors minimize current but become noise-sensitive; low values stabilize voltage under light load but waste power — pick a compromise for battery-powered sensors.

Worked example with defaults: Vin = 5 V, R1 = R2 = 10 kΩ → Vout = 2.5 V, divider current 0.25 mA. If a load draws significant current from Vout, measured voltage will sit below 2.5 V unless buffered.

ADC scaling maps a 0–5 V sensor output into a 3.3 V microcontroller range using unequal dividers. If Vin = 5 V and you need Vout ≈ 3.3 V, ratio R2/(R1+R2) = 0.66 — solve for R1 and R2 from standard E24 values, then verify divider current stays low. Unequal dividers scale 0–5 V sensors into 3.3 V ADC range — pick E24 pairs near R2/(R1+R2) ≈ 0.66 before layout.

Thevenin equivalent at the tap is R1∥R2 in parallel. A 10 kΩ load on a 10 kΩ∥10 kΩ divider (5 kΩ Thevenin) pulls Vout below 2.5 V significantly — either buffer with an op-amp or size R1/R2 an order of magnitude below expected load. 10 kΩ load on a 5 kΩ Thevenin tap sags Vout — buffer with an op-amp when load current rivals divider current.

Changing R1 from 10 kΩ to 20 kΩ while R2 stays 10 kΩ drops Vout from 2.5 V to 5 × 10/(20+10) ≈ 1.67 V. Small resistor tolerance (±1%) shifts the tap by tens of millivolts — critical when feeding a 10-bit ADC spanning 3.3 V on the default 5 V supply.

Perguntas frequentes

Why does my measured voltage differ?

Multimeter loading, supply tolerance, and resistor tolerance affect readings. Heavy load on Vout pulls voltage down.

Can I use this for AC?

Same formula for resistive AC at an instant, using RMS values. Capacitive dividers need different math.

What resistor values minimize load error?

Make divider current at least 10× the expected load current, or add a unity-gain op-amp buffer so the tap sees high impedance.