I have a fan on a PC case running at 100% speed making 40dB/A of noise (sound pressure). Then I add a second fan right next to it running at 100% speed making 30dB/A of noise.

  1. What is the combined noise level?

  2. What happens if add and extra 50dB/A fan?

  3. And then 10 more 30dB/A fans?

Forget about space, treat them as sources coming from the same point in 3D space.


The total signal level from sources with different strengths can be calculated as:

Lt =  10 log ((S1 + S2 ... + Sn) / Sref)  


Lt = total signal level (dB) 

S = signal (signal unit)

Sref = signal reference (signal unit)


I can't directly use this formula because it requires the sound power of fans instead of the sound pressure manufacturers such as Noctua typically provide. How can I convert it? What will I need?

  • $\begingroup$ I'm looking for a basic formula that isn't space dependent. What if they come from the same source? $\endgroup$ – Vulkan May 4 '20 at 15:24
  • $\begingroup$ You want to look at sound power level vs sound pressure level. But your new noise level won't by much higher than 40 dB. $\endgroup$ – jko May 4 '20 at 15:54
  • $\begingroup$ I know that it won't be much higher than 40db but how much exactly? The ratings are the ones provided by the fan manufacturers. An approximation is the best. $\endgroup$ – Vulkan May 4 '20 at 22:01

This is a pretty straightforward calculator: https://www.engineeringtoolbox.com/adding-decibel-d_63.html

Basically noise level is dictated by your 2 loudest sources, so 50 dB + 40 dB (assuming they are at or near the same location) would result in about 50.5 dB.

  • $\begingroup$ I've edited the main post. Any way to convert the sound pressure (dB/A) manufacturers provide to sound power? $\endgroup$ – Vulkan May 7 '20 at 11:04
  • $\begingroup$ It's easier to do that in reverse. Your sound power level (dB) = 10 * log (sound power [ in watts]/ 10^-12 [watts]). Sound "power" is incredibly weak mechanically, so it is based off a reference power of 10^-12 watts. So a 40 dB source is 4 powers of ten greater than 10^-12 watts, or 10^-8 watts. A 50 dB source is 10^-7 watts. If you combine both of those you get 11^-7 watts of power, which is only a 0.4 increase in dB. $\endgroup$ – jko May 7 '20 at 12:07
  • $\begingroup$ I'm looking to create a calculator where you add each fan and the noise it makes, then it calculates the total noise levels. I need a formula to work with this based on sound pressure ratings the manufacturers provide.It needs to be a rough estimation based on accurate readings. $\endgroup$ – Vulkan May 7 '20 at 12:32
  • $\begingroup$ Equation (2) from the engineering toolbox link is the formula you need. For each dB input you just need to deconstruct it into the sound power (watts) from my previous comment. You just sum up all those powers, divide that sum by 10^-12, take the log, and multiply by 10 to get dB. $\endgroup$ – jko May 7 '20 at 12:41
  • $\begingroup$ By deconstructing the logarithm the sound power = 10^-12 * (2^(0.1*dB)). The sum = 10^-12 * (2^(0.1*dB1) + 2^(0.1*dB2) + ... + 2^(0.1*dBn)). When you divide the sum by 10^-12 you are left with just 2^(0.1*dB1) + 2^(0.1*dB2) + ... + 2^(0.1*dBn). Finally, noise = 10 * log (2^(0.1*dB1) + 2^(0.1*dB2) + ... + 2^(0.1*dBn)). When I used the values 40db and 30db like in the example above I got 13.8db... Any ideas what's wrong? I triple checked everything. $\endgroup$ – Vulkan May 7 '20 at 13:14

The standard rule of thumb for adding decibels is that doubling the sound power increases the dB reading by +3dB. Using an on-lone calculator for adding dB's, we find that adding two sound sources that differ in strength by 10dB yields a dB total of 40.25dB.

To the human ear, a difference of 0.25dB is completely inaudible.

  • $\begingroup$ If I add 3 more fans at 30db the total noise output will be 41db? If I add an extra fan at 50db the total output will be >51db? $\endgroup$ – Vulkan May 4 '20 at 23:26
  • $\begingroup$ try the on-line dB calculator! -NN $\endgroup$ – niels nielsen May 5 '20 at 0:39

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