# Why the “20” in calculating db ratio?

In wikipedia, I read the formula to convert a power ratio to db. This formula is :

10*Math.log10(power_out / power_in);


But, for the sound pressure, I read the formula is :

20*Math.log10(soundLevel_out / soundLevel_in);


I read in some places a reason : it is because the sound pressure, like the voltage, have to be squared in the ratio and this squared becomes a 20 instead of a 10. Ok, but why ?
When I want to double the sound "feeling", I was told I need to double my power amplifier so, I understand it's a question of power (with no squaring) and not a question of voltage, but it's wrong.

Who can help me to understand where does this square comes from ?

• I assume you know that $10\log(x^2)=20\log(x)$ (?) – Matt L. Dec 19 '18 at 18:13

According to this source, the sound power received by an aperture is proportional to the pressure squared, i.e.: $$P = \frac{A p^2}{\rho c} \cos \theta,$$ where:

• $$P$$ is the received power
• $$A$$ is the surface area of the receiving aperture
• $$p$$ is the pressure
• $$\rho$$ is the density of medium
• $$\theta$$ is the angle between the propagation direction and the surface normal

When you double the voltage you quadruple the power, since power is proportional to the square of the voltage. Same for sound pressure and sound power. Hence you use 20 for linear quantities (pressure, voltage, current, velocity, volume velocity) and 10 for energy quantities (power, intensity, energy). In either case doubling of the linear quantity will result in +6 dB.

Perceived loudness is different though. The human auditory system is quite complicated and you need perceptual models to understand what's happening. For most frequencies, signals and levels, perceptual loudness will roughly double for a 10dB increase. Roughly speaking: to get twice as loud, you need 10 times the power.

• Thank you for that second explanation which is also very clear and helpful. – Dr_Click Dec 20 '18 at 6:17