I'm wondering, what is the maximum peak amplitude, in decibels, that will register in the first 8 bits of a 24 bit wav and leave the upper 16 as 0's? Would something like -99.85dB be low enough or would it have to be below -100dB (typical software volume controllers)
PCM samples in a wav file are coded in 2's complement.
For a N bits sample size, it means that the maximum positive sample value is 2^(N-1) - 1 and the minimum negative value is -2^(N-1).
For N = 24, it leads to a maximum positive value of 2^23 - 1.
The maximum positive value of a sample composed of 8 lower bits set to 1 is 2^8 - 1.
So the dB FS value of such an 8 bit sample would be 20 * log10((2^8 - 1) / (2^23 - 1)), that is approximately -90.3 dB FS.
If you are measuring level with an RMS meter and your signal is a sinusoidal tone, the RMS meter should display about 3 dB less.
(Using 2^23instead of 2^23 - 1 as denominator has a negligible effect on the numerical result, although it is an interesting question to decide which one should be used).
