Decibel Calculator
Add sound levels the right way, take away background noise, and in Advanced mode turn ratios into decibels and back, convert dBm, dBV, dBu and pascals, and see how a level drops with distance.
Type the two values (any unit, as long as both use the same one) or type the decibels. Whichever you typed last is used. Leave "in" as 1 to work with a plain ratio.
Pick any unit: the box converts when you change it, and every other unit is shown below.
Free-field estimate: open space, no walls or ground reflections, wind or air absorption.
How your decibel sum is worked out
Your numbers: decibels are logs, so turn each level back into energy, add, then take the log again.
- 1
Undo the decibels
80 dB → 1 × 10⁸80 dB → 1 × 10⁸10^(level ÷ 10)
- 2
Add the energy
1 × 10⁸ + 1 × 10⁸= 2 × 10⁸
- 3
Back to decibels
10·log₁₀(2 × 10⁸)= 83.01 dB
- 4
Compare
83.01 dB − 80 dBthe loudest alone= +3.01 dB
Read more: the formulas, worked examples and notes
A decibel is a way to compare two amounts on a log scale: 10 × log₁₀ of a power ratio. Every +10 dB is ten times the power, every +3 dB about double. Because decibels are logs, you can't add them as plain numbers. Turn each level back into energy, do the sum there, then take the log again.
The formula
total level = 10 × log₁₀(10^(L₁ ÷ 10) + 10^(L₂ ÷ 10) + …)
source alone = 10 × log₁₀(10^(total ÷ 10) − 10^(background ÷ 10))
energy average = 10 × log₁₀((10^(L₁ ÷ 10) + … + 10^(Lₙ ÷ 10)) ÷ n)
power ratio in dB = 10 × log₁₀(P₂ ÷ P₁), and back: P₂ ÷ P₁ = 10^(dB ÷ 10)
voltage, current or pressure ratio in dB = 20 × log₁₀(V₂ ÷ V₁), and back: 10^(dB ÷ 20)
dBm = 10 × log₁₀(power ÷ 1 mW), dBW = dBm − 30
dBV = 20 × log₁₀(volts ÷ 1 V), dBu = 20 × log₁₀(volts ÷ 0.774597 V)
sound pressure level = 20 × log₁₀(pressure ÷ 20 µPa)
sound intensity level = 10 × log₁₀(intensity ÷ 10⁻¹² W/m²)
point source: new level = level − 20 × log₁₀(new distance ÷ old distance)
line source: new level = level − 10 × log₁₀(new distance ÷ old distance)
Worked example
Adding levels (the default): two machines at 80 dB each. Energy of each: 10^(80 ÷ 10) = 10⁸. Add them: 10⁸ + 10⁸ = 2 × 10⁸. Back to decibels: 10 × log₁₀(2 × 10⁸) = 10 × 8.30103 = 83.01 dB, which is 3.01 dB more than one machine alone.
Subtracting a background: the total is 85 dB and the background is 80 dB. 10^8.5 = 316,227,766 and 10^8 = 100,000,000, so the source's own energy is 216,227,766. 10 × log₁₀(216,227,766) = 83.35 dB, a correction of 1.65 dB.
Average: 70 dB and 80 dB. (10⁷ + 10⁸) ÷ 2 = 5.5 × 10⁷, and 10 × log₁₀(5.5 × 10⁷) = 77.40 dB. The plain average would say 75 dB, which is too low because the louder sound carries most of the energy.
Ratios: double the power is 10 × log₁₀(2) = 3.01 dB; double the voltage is 20 × log₁₀(2) = 6.02 dB. Going back, 30 dB is a power ratio of 10³ = 1,000.
dBm and dBu: 20 dBm = 10^(20 ÷ 10) × 1 mW = 100 mW = −10 dBW. Across 50 Ω that is √(0.1 × 50) = 2.236 V rms. A voltage of 1 V is 0 dBV and 20 × log₁₀(1 ÷ 0.774597) = 2.22 dBu.
Pascals: 1 Pa is 20 × log₁₀(1 ÷ 0.00002) = 20 × log₁₀(50,000) = 93.98 dB SPL, and 94 dB SPL is 0.00002 × 10^(94 ÷ 20) = 1.0024 Pa.
Distance: 90 dB at 1 m from a small source, heard at 4 m. 20 × log₁₀(4 ÷ 1) = 12.04, so it is about 90 − 12.04 = 77.96 dB. For a line source such as a road, 10 × log₁₀(4) = 6.02, giving 83.98 dB.
Handy rules
| Change | Power or energy | Voltage or pressure |
|---|---|---|
| +1 dB | × 1.259 | × 1.122 |
| +3 dB | × 1.995 (about double) | × 1.413 |
| +6 dB | × 3.981 | × 1.995 (about double) |
| +10 dB | × 10 | × 3.162 |
| +20 dB | × 100 | × 10 |
| −3 dB | × 0.501 (about half) | × 0.708 |
Adding two levels: how much to add to the louder one
| Difference between the two | Add to the louder |
|---|---|
| 0 dB | +3.01 dB |
| 1 dB | +2.54 dB |
| 2 dB | +2.12 dB |
| 3 dB | +1.76 dB |
| 6 dB | +0.97 dB |
| 10 dB | +0.41 dB |
| 20 dB | +0.04 dB |
Notes
- Adding and averaging here is for separate, unrelated (incoherent) sources, such as two machines or two cars. Two copies of the same signal in step can add up to +6 dB, or cancel out when out of step.
- Sound level meters often apply a weighting (A, C or Z, shown as dB(A), dB(C)). Add or subtract levels with the same weighting only.
- dBu and dBV are rms voltages. 0 dBu = √0.6 V ≈ 0.774597 V, so dBu = dBV + 2.2185.
- dB SPL in air uses 20 µPa, about the quietest sound a young ear can hear at 1 kHz. Underwater acoustics use 1 µPa, so the same pressure reads 26.02 dB higher there.
- The distance rules are free-field approximations. Indoors, reflections stop the level falling as fast, and outdoors the ground, barriers, wind, temperature and air absorption all change it.
- A rough guide often quoted is that +10 dB sounds about twice as loud to most people, though hearing is not a simple log scale.
- Loud noise over time can damage hearing; follow your local workplace limits.
Common questions
How do you add decibels?
You cannot add decibels like ordinary numbers, because they are logarithms. Turn each level back into energy with 10^(level ÷ 10), add those, then take 10 × log₁₀ of the total. Two 80 dB sources give 10⁸ + 10⁸ = 2 × 10⁸, and 10 × log₁₀(2 × 10⁸) = 83.01 dB, not 160 dB.
Why do two equal sounds only add 3 dB?
Doubling the sound energy multiplies it by 2, and 10 × log₁₀(2) = 3.01 dB. So two equal, separate sources are about 3 dB louder than one, four are about 6 dB louder and ten are exactly 10 dB louder. This is for separate sources such as two machines. Two copies of the same signal in step (coherent) can add up to 6 dB.
When do I use 10 × log and when 20 × log?
Use 10 × log₁₀ for power-like quantities: watts, energy and sound intensity. Use 20 × log₁₀ for field (root-power) quantities: voltage, current, amplitude and sound pressure, because power goes with their square. So double the power is +3.01 dB, but double the voltage is +6.02 dB.
What do dBm, dBW, dBV and dBu mean?
They are decibels compared with a fixed reference. dBm is compared with 1 milliwatt (0 dBm = 1 mW, 30 dBm = 1 W), dBW with 1 watt, dBV with 1 volt rms, and dBu with about 0.7746 volt rms (the voltage that puts 1 mW into 600 Ω). dBu is 2.2185 dB higher than dBV for the same voltage.
How much quieter is a sound further away?
In open space with no reflections, a small (point) source drops by 20 × log₁₀(new distance ÷ old distance), about 6 dB each time the distance doubles. A long line source such as a busy road drops by 10 × log₁₀ of the ratio, about 3 dB per doubling. Walls, ground, wind and air absorption change this in real life, so treat it as an estimate.
How do I remove background noise from a reading?
Measure the total with the source on and the background with it off, turn both into energy, subtract, and take the log again. 85 dB total with an 80 dB background leaves 83.35 dB for the source on its own. If the total is less than about 3 dB above the background, the answer is very sensitive to small measurement errors.
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