I am using the MAX4466 microphone amplifier to calculate the db SPL level of sound. I have obtained the peak to peak amplitude in volts and I know that the microphone sensitivity is -44 db which translates to 0.0063096 V/Pa. 1 Pa is 94 db SPL.
so, say my voltage reading is 1.7 volts ; in db it = 20 log (1.7/0.0063096) = 48.6 db
Now acc. to a formula I found on this thread How to convert Volts in dB SPL
to calculate db SPL, I must add the mic sensitivity + 94 db SPL i.e 48.6 + 94 - 44 = 98.6 db SPL.
I don't completely understand why we do this?
Secondly, this formula is for unity gain. I don't know the gain of my device to subtract from the db SPL level and I can't seem to find this anywhere. Please help!
Answer
The MAX4466 doesn't have a 'default' gain. Gain is set by the ratio of feedback and input resistors connected to the inverting input (IN-). For example if the feedback resistor is 220k and the input resistor is 22k then the gain will be 220/22 = 10, or 20dB. To determine the gain of your circuit you need to know the values of those resistors.
SPL is rms sound pressure, so you must convert your 1.7V peak-to-peak measurement to rms by dividing it by 2.828 (assuming a sine waveform) which works out to 0.6Vrms. dB is a relative measurement, representing a power gain or loss. To convert 0.6V to dB you must first decide on a reference power or voltage level. dBV is voltage gain or loss relative to 1V. 20(log 0.6/1) = -4.4dbV.
If your amplifier gain is (say) 20dB then the mic must have generated -4.4 - 20 = -24.4dBV. The mic has a sensitivity of -44dBV, which means that it generates -44dBV at a sound pressure of 94dBSPL. In your test it generated -24.4dBV which is 19.6dB higher than -44dBV, so the sound pressure level must have been 19.6dB higher than 94dBSPL, ie. 113.6dBSPL.
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