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I wanted to convert volts into dB SPL. Thankfully, I found this article

convert Volts into dB SPL

While trying to understand why he was adding the mic sensitivity along with 94 dB SPL, I found another article

Volts into dB SPL explained

In the first article, during the dB SPL calculation the reference Voltage used is the microphone reference voltage, 0.005012 V RMS (microphone sensitivity of -46dBV/Pa , this gives 0.005012 V RMS) , which I thought is correct and made sense.

However in the second article, he uses a relative voltage of 1V during the dB calculation. He should have used 0.0063096V (microphone sensitivity is -44 db which translates to 0.0063096 V/Pa.)

I still think the first article is right. Can someone clarify if which article is correct or if I've misunderstood something ? Thank you.

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Both articles are correct. There are so many gains and reference levels flying around, and options in the order that you use them, that it's easy to get confused. So let's go back to fundamentals, so we know where we are.

dB, by itself, is a ratio between two levels.

dB(reference), also spelt dBreference, and \$dB_{REFERENCE}\$, and unfortunately all too often also spelt dB with a reference that should be obvious in context but isn't mentioned explicitly, is an absolute level, which has the quoted dB ratio with respect to the reference level.

So you have to be careful if you see a bare dB that you understand which of these it is. Professional scientific publications usually get it right, popular and amateur publications, especially when talking about acoustics, sometimes get it wrong.

Where does this magic 94dB come from?

The reference level for Sound Pressure Level (SPL) is \$20\mu Pascal\space rms\$. This is taken to be the threshold of hearing, the 0dB reference of Sound Pressure Level, roughly the noise of a mosquito at 3m. This is 0dB(20\$\mu\$Pa), often written 0dBSPL.

So 1Pa rms will be +94dBSPL, as 20\$log_{10}(\frac{1}{20\mu}\$) is +94dB. If we work with respect to a 1Pa rms reference, we add 94dB to get dBSPL.

Let's track what happens to a sound as it's received and converted. If the microphone sensitivity is given in volts/Pa, then we need the sound level referenced to Pa, not dBSPL. In this case, microphone gain is -46dBV/Pa, let's call that M, and the preamp has some gain G.

Sound at xdBSPL, is x-94 dB(1Pa). This becomes x-94+M dBV (or x-94-46 dBv in this case) at the output of the microphone. At the output of the amplifier, it's x-94+M+G dBV (x-94-46+G dBV).

If you have a measured voltage of Y dBV, then your original SPL must have been Y-G-M+94 dBSPL (Y-G-(-46)+94 dBV) (M is a negative number, take care of the sign when subtracting negative numbers!)

If you have a measured voltage of Z volts pp, then you have two extra steps. Divide by 2.828 to get from pp to rms. Then turn your volts rms into dBVrms.

The difference between the two articles is the order in which these gains are applied.

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  • \$\begingroup\$ Thank you for that detailed explanation. Just one other thing to clarify. If you compare your answer for calculation, which is, If you have a measured voltage of Y dBV, then your original SPL must have been Y-G-M+94 dBSPL (Y-G-(-46)+94 dBV) to the first article (electronics.stackexchange.com/questions/96205/…), where he does (-46) + 53.96 = 7.95 + 94 = 101.95 Db SPL so rather than (Y-G-M+94 dBV) or (Y-G-(-46)+94 dBV), he does (Y-G+M+94 dBV). which one is right ? \$\endgroup\$ – whoknowsmerida May 23 '18 at 5:29
  • \$\begingroup\$ So should 'M' be subtracted or added ? \$\endgroup\$ – whoknowsmerida May 23 '18 at 5:36
  • \$\begingroup\$ Using my definitions, M is always added, when going sound to voltage. M is always subtracted, when going voltage to sound. However, note that M (usually) has a negative sign, and subtracting a negative number is the same as adding a positive number. It always confuses me, which is why I have to go back to first principles and work through. Which is clearer I wonder, changing the sign and changing the operation, or always adding and trusting the Cartesian convention to keep it right. I don't know, which is why I presented it from first principles. \$\endgroup\$ – Neil_UK May 23 '18 at 6:13
  • \$\begingroup\$ yes it is a bit confusing. But since I'm calculating the Vpp which comes from a sound. I think I'll have to 'add' a negative M. Thank you. that was helpful :) \$\endgroup\$ – whoknowsmerida May 23 '18 at 6:30

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