Quote:
Originally Posted by jhowell
Adding a bit doubles the number of possible voltage levels. This is a log10(2)=0.3 bel (3 dB) increase. However, power increases with the square of voltage (for a fixed load resistance) which means that dB of sound intensity are double dB of voltage, or 6 dB per bit.
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That's not quite right - dBs are normally calculated differently for powers ratios and voltage/field ratios specifically to avoid the confusion you describe.
For a voltage ratio, dBs are calculated as 20*log10(V1/V2). For power ratios, it's 10*log10(P1/P2). This means that when comparing two signals, you get the same dB value irrespective of whether you compare their voltage or power.
Hence adding a bit adds 6dB to the DR, whether you're talking about the voltage or the resultant power.
/JB