# Clarity on dB SPL calculation

Based on this post, let me assume that I have a input source of 94 dB SPL / 1 Pa that a microphone is picking sounds from.

The microphone has a sensitivity of -46dBV/Pa , this gives 0.005012 V RMS/ Pa.

Let us assume the ADC has unity preamp gain and no additional gain. Let us also assume an ideal case where there is no degradation in signal due to noise before ADC.

Now, I'm guessing the signal measured at ADC would be the same as input, which is 0.005012 Vrms.

20×log 0.005012/005012 = 0 dB

so the dB SPL will be (-46) + 0 = -46 + 94 = 48 dB SPL.

Why is the output not 94 dB SPL which is what I thought the output should be because we are inputting a 1 Pa sound?

• The microphone converts the sound pressure level (SPL) signal into a voltage signal. So after that it doesn't make sense to talk about dB SPL. You could convert it back into SPL using an amplifier and a speaker if you want. – mkeith Feb 19 at 10:13
• @mkeith I'm clear now with the explanation that 'justme' provided. But just to take you up on your answer,even if I do not have an amplifier, can i not work the dB SPL value with the measured voltage referenced to the input voltage ? – whoknowsmerida Feb 20 at 1:24