So, the ‘d’ in ‘dB’ in the G and SPL representations of voltage and sound pressure still refers to the factor 10 (‘deci’) used to make power comparison easy. But in the calculations, and in our interpretation and discussions, we use a factor of 20 to match the voltage and sound pressure levels to the associated power ratios in dB. The field factor is twice as high as the power factor, which sometimes has confusing consequences...
For example, a ‘factor of two’ increase in a voltage or sound pressure is represented by roughly 6dB, whereas a ‘factor of two’ increase of power or sound intensity is represented by roughly a factor of 3dB. Or, in other words; doubling a voltage or sound pressure value gives a 6dB higher output level, but it also requires 6dB more electrical power or sound intensity - which equals a factor of four, not two.
A practical case where this may lead to confusion is the calculation of the SPL output of a loudspeaker with a given amplifier power. The loudspeaker sensitivity is always given as ‘dBSPL@1W1m’: the SPL that will be generated at one metre distance on-axis if the loudspeaker is driven by a voltage that corresponds with one watt (in case of an 8Ω speaker that’s 2.84 volts). Let’s say that our loudspeaker has a sensitivity of 95dBSPL. If we drive it with 100W, we have to add the dB ratio of 100W to 1W. The trick is that, although we are calculating SPL - for which we normally use ‘20log’ to calculate ratios - in this case we have to add the power ratio, which uses ‘10log’. In our case this boils down to a 20dB power ratio, so the output SPL is 95dBSPL + 20dBP = 115dBSPL.
Of course if we had calculated the result using signal voltage levels, the result is exactly the same. To achieve a power amplifier output of 100W to 8Ω, we need to apply a voltage of 2.84 times the square root of 100, which calculates to 28.4V. The dB increase is then 20*log(28.4/2.84) = 20*1 = 20dB. But by using the power levels we can skip calculating the voltage levels, making engineering easier and quicker.
Although I have discussions and make electro-acoustic calculations every working day using decibels, I still have to keep reminding myself that dB at the end is about power ratios and that the ‘d’ represents a factor 10. If I think of a voltage ratio, the ratio in dB refers to the electrical power ratio that the voltage could generate if it was connected to a meaningful load (e.g. a dummy resistor). If I think of an SPL ratio, then the ratio in dB refers to the sound intensity ratio needed to produce the sound pressure ratio.