I've got a Knowles SPH1642HT5H-1 (pdf link) analog MEMS microphone that I'm planning to connect to one of the ADC inputs on an STM32. I've done what I think are the correct calculations to determine the gain and offset that I will require but would like someone to just check that my interpretation of the datasheet is correct.
The sensitivity of the microphone is -38dbV/Pa and the MCU AVDD is 3.3V. The small signal is carried on a relatively large DC offset. I calculate for the signal:
$$
Vrms = 10^{-38/20} = 12.589mV\\\\
Vpp = Vrms . \sqrt{2} = \pm 17.8mV\\\\
Gain = \frac{AVDD}{2 . Vpp} = 92\\\\
Offset = \frac{AVDD}{2} = 1650
$$
To make full use of the range of the single-ended ADC I need to design a pre-amp with a gain of 92 and an offset of 1650mV.
For reference, these are the datasheet parameters:
Edit
Based on the @JRE's comments below I can recalculate as follows.
If I want 70dB SPL to be full scale on the ADC then, using this calculator I convert dB SPL to Pa and get 0.0632Pa. The new gain calculation will be:
$$
Gain = \frac{\frac{0.5 . AVDD}{\sqrt{2}}}{10^{-38/20}. 0.0632} = 1466
$$
Based on @Andy aka's answer I can adjust further based on how things actually perform in the real world.
Best Answer
Well that calculation works fine for sine wave audio being measured by the microphone but real audio has a significant crest factor and this means that the peak voltages will be much greater for the same acoustic power signal. If you want to avoid distortion/clipping consider crest-factor: -
Picture from here.
You should also consider that your ADC will never be operating fully from rail to rail so assume 50 mV top and bottom of the 3.3 volt range is off-bounds. This is because of zero offset and gain errors.