Speaker SPL calculator
Enter the speaker's sensitivity, the power you feed it, how many identical speakers play and how far away you sit. The calculator estimates the sound pressure level in dB at that distance.
Runs in your browser: nothing you enter is sent anywhere.
How to use it
- Enter the sensitivity from the speaker’s specification, in dB at 1 W and 1 m.
- Enter the amplifier power delivered to each speaker and the number of identical speakers playing the same signal.
- Choose metres or feet and enter the distance to the listener.
- Read the sound pressure level, with each part of the sum below it.
SPL = sensitivity + 10 log₁₀ P + 10 log₁₀ n − 20 log₁₀ d
P is the power per speaker in watts, n the number of speakers and d the distance in metres (feet are converted). This is the free-field inverse-square law; it ignores the room, the amplifier’s headroom and the speaker’s power handling.
Worked example
An 88 dB speaker driven with 50 W, heard from 3 m:
- power: 10 × log₁₀ 50 = 17.0 dB, so 88 + 17.0 = 105 dB at 1 m;
- distance: 20 × log₁₀ 3 = 9.5 dB;
- SPL = 105 − 9.5 = 95.4 dB.
Two 90 dB speakers with 100 W each at 4 m: 90 + 20 + 3.0 − 12.0 = 101 dB.
Tips
- Peaks in music are often well above the average level: leave headroom in the amplifier rather than running it at its limit.
- Check the speaker’s power rating; the formula does not stop at what the speaker can take.
- To add the levels of different sources, such as a speaker and background noise, use the decibel addition calculator.
Frequently asked questions
How do I calculate speaker SPL?
Start from the sensitivity, the level at 1 W and 1 m. Add 10 × log₁₀ of the power in watts, add 10 × log₁₀ of the number of speakers and subtract 20 × log₁₀ of the distance in metres. An 88 dB speaker with 50 W at 3 m gives 88 + 17.0 − 9.5 = 95.4 dB.
How much louder is double the power?
Doubling the power adds 10 × log₁₀ 2 ≈ 3 dB. Ten times the power adds 10 dB, which most people hear as about twice as loud. That is why sensitivity matters: a speaker 3 dB more sensitive needs half the power for the same level.
Why does the level drop 6 dB per doubling of distance?
In free space sound spreads over a sphere, so its intensity falls with the square of the distance: 20 × log₁₀ 2 ≈ 6 dB per doubling. In a room, reflections keep the level higher than this at a distance, so the result is a conservative estimate indoors.
My speaker lists sensitivity at 2.83 V. Can I use it?
For an 8 Ω speaker 2.83 V is 1 W, so it is the same number. For a 4 Ω speaker 2.83 V is 2 W, and the 1 W figure is about 3 dB lower: subtract 3 dB before entering it.