Free Decibel (dB) Calculator
Convert a power or voltage ratio into decibels (dB) instantly. Choose power mode (10·log₁₀) for watts or voltage/amplitude mode (20·log₁₀) for volts, current, or sound pressure, and read the gain or loss live.
Pick a mode, then enter two quantities to find their ratio in decibels. Use power for watts (10·log₁₀) and voltage / amplitude for volts, current, sound pressure or any field quantity (20·log₁₀).
Power: dB = 10·log₁₀(P₂/P₁). Voltage / amplitude: dB = 20·log₁₀(V₂/V₁). The 20·log₁₀ form gives the same decibel value because power is proportional to the square of a field quantity (P ∝ V²), so the factor of 2 from the exponent moves out front. A positive result is a gain; a negative result is a loss (attenuation).
Quick answer
A decibel expresses a ratio on a logarithmic scale. For power, dB = 10·log₁₀(P₂/P₁); for a voltage or amplitude ratio, dB = 20·log₁₀(V₂/V₁). For example, a power increase from 1 W to 100 W is 10·log₁₀(100/1) = 10·2 = 20 dB, the same value you would get from a 10× voltage increase (20·log₁₀(10) = 20 dB).
Formula & method
Power ratio
dB = 10 · log₁₀(P₂ / P₁)
- P₂ — Output / measured power (same unit as P₁, e.g. watts)
- P₁ — Reference / input power
Use this when both quantities are powers (watts, milliwatts) or intensities. Each 10 dB is a 10× change in power; +3 dB ≈ double the power.
Voltage / amplitude ratio
dB = 20 · log₁₀(V₂ / V₁)
- V₂ — Output / measured field quantity (voltage, current, sound pressure)
- V₁ — Reference / input field quantity
Power is proportional to the square of a field quantity (P ∝ V²), so the factor 10 becomes 20. Here +6 dB ≈ double the voltage; each 20 dB is a 10× change.
Examples
- Input
- Mode = Power, P₂ = 100 W, P₁ = 1 W
- Result
- 20 dB
- Why
- Ratio = 100 / 1 = 100. dB = 10·log₁₀(100) = 10 × 2 = 20 dB. A 100× power increase is a 20 dB gain.
- Input
- Mode = Voltage/amplitude, V₂ = 10 V, V₁ = 1 V
- Result
- 20 dB
- Why
- Ratio = 10 / 1 = 10. dB = 20·log₁₀(10) = 20 × 1 = 20 dB. A 10× voltage rise equals a 100× power rise, so it also reads 20 dB.
- Input
- Mode = Power, P₂ = 2 W, P₁ = 1 W
- Result
- 3.0103 dB
- Why
- Ratio = 2 / 1 = 2. dB = 10·log₁₀(2) = 10 × 0.30103 = 3.0103 dB ≈ 3 dB — the familiar rule that doubling power adds about 3 dB.
- Input
- Mode = Voltage/amplitude, V₂ = 1 V, V₁ = 2 V
- Result
- −6.0206 dB
- Why
- Ratio = 1 / 2 = 0.5. dB = 20·log₁₀(0.5) = 20 × (−0.30103) = −6.0206 dB ≈ −6 dB. The negative sign marks a loss; halving voltage drops the level by about 6 dB.
When to use this tool
- Finding the gain or loss of an amplifier, attenuator, antenna, filter, or cable run from input and output levels.
- Converting an audio level change (loudness, signal-to-noise ratio) into decibels when you know the power or voltage ratio.
- Checking the quick rules of thumb — +3 dB ≈ double power, +6 dB ≈ double voltage, +10 dB ≈ 10× power, +20 dB ≈ 10× voltage.
- Comparing two measurements (e.g., two speaker outputs or two RF signals) on a logarithmic scale instead of raw ratios.
Common mistakes
- Using the 10·log₁₀ (power) formula on voltages, or 20·log₁₀ on powers. Match the factor to the quantity: 10 for power/intensity, 20 for voltage, current, amplitude, or sound pressure.
- Swapping the reference and the measured value. dB = factor·log₁₀(measured/reference); flipping P₂ and P₁ only changes the sign, turning a gain into a loss.
- Entering zero or a negative quantity. The logarithm of a ratio is undefined for non‑positive numbers, so both inputs must be greater than zero.
- Treating decibels as a linear unit. dB add (and subtract) instead of multiplying, and +3 dB is roughly double the power — not 3× and not +3 watts.
Frequently asked questions
What is the formula for decibels?
It depends on the quantity. For power, dB = 10·log₁₀(P₂/P₁). For a field quantity such as voltage, current, or sound pressure, dB = 20·log₁₀(V₂/V₁). Both describe the same physical change because power is proportional to the square of a field quantity, which is why the factor doubles from 10 to 20.
Why is it 10·log for power but 20·log for voltage?
Power is proportional to the square of voltage (P = V²/R for a fixed resistance). Substituting V² into 10·log₁₀ gives 10·log₁₀(V₂²/V₁²) = 20·log₁₀(V₂/V₁), because the exponent 2 comes out of the logarithm as a multiplier. So both formulas give the identical decibel value for the same real change.
How many decibels is double the power or double the voltage?
Doubling power is 10·log₁₀(2) ≈ 3.01 dB (the well-known '+3 dB' rule). Doubling voltage or amplitude is 20·log₁₀(2) ≈ 6.02 dB (the '+6 dB' rule). Halving each gives the same magnitudes with a negative sign: about −3 dB and −6 dB.
What does a negative decibel value mean?
A negative result means the output is smaller than the reference — a loss or attenuation. For example, dropping from 2 V to 1 V is 20·log₁₀(0.5) ≈ −6.02 dB. A positive value means a gain (output larger than reference), and 0 dB means the two quantities are equal.
Is this calculator for dB, dBm, or dBW?
This tool computes a relative gain or loss in dB from any two values in the same unit. dBm and dBW are absolute levels referenced to 1 mW and 1 W respectively. To get dBm, put your power in milliwatts as P₂ and 1 mW as P₁; for dBW, use watts with P₁ = 1 W.
Can I use this for sound levels in decibels?
Yes. Sound pressure level uses the 20·log₁₀ form because sound pressure is a field quantity: dB = 20·log₁₀(p₂/p₁), with the standard reference p₁ = 20 µPa in air. Sound intensity or acoustic power instead uses the 10·log₁₀ form. Pick the matching mode for your measurement.
Sources & references
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Provided “as is” for general information only — results may be inaccurate, so verify before you rely on them. No warranty; use at your own risk.
Built and reviewed by HIFreeTools against the formula shown above and any authoritative references cited on this page. See our methodology and editorial standards.
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