A good while back, I recall Bill saying there's a 6dB gain for every doubling of voltage. Up 'til that point, I'd always done the "doubling/halving wattage is 3dB". It got my brain turning a little bit in new ways. Opened new thought processes. Thanks for that (and many other bits of knowledge and insights over the years) Bill.
Anyway, one of my initial thoughts was, what if we want to figure the gain for a voltage amount other than double and/or what's an easy way to double, then double, then double 2.83 in my head or even manage keeping track of the number of times I doubled on a calculator? That just kinda rattled around in my head since then as an unsolved mystery.
Well, here we are, years later and I experienced a moment of clarity and inspiration to try and figure it out. Tinkering in an excel spreadsheet for an hour or so, I finally just worked out a new-to-me formula for calculating volts required to gain a specific dB and thought I'd share it with you guys. I've never seen this calc anywhere (wish I had), but I can't imagine I'm the first to come up with it. It calcs a final voltage from dB gain instead of my initial curiosity of calculating dB gain from voltage.
I know, for a lot of you this will be really boring stuff. Sorry about that. Don't bother reading any further. But just know... It's pretty damn cool for numbers guys (well... I'm impressed with myself anyway). If any of you guys
DO have an audio math reference spreadsheet, you might want to add this equation in there...
The product of 1.122 ^ dB increase, times initial voltage equals final voltage.
(1.122 ^ 20dB) x 2.83V = 28.29V
(1.122 ^ 21dB) x 2.83V = 31.74V
(1.122 ^ 22dB) x 2.83V = 35.61V
(1.122 ^ 23dB) x 2.83V = 39.95V
Any dB number and any initial voltage gives the final voltage required to make that gain.
Say we're already at 35 volts and want to know how many volts is required to gain another 1, 2, or 3dB (or any number dB)...
(1.122^
1dB) x
35V = 39.27V to achieve a 1dB gain
(1.122^
2dB) x
35V = 44.06V to achieve a 2dB gain
(1.122^
3dB) x
35V = 49.44V to achieve a 3dB gain
Since we know 3dB is essentially double the power we can double check and prove the math...
Volts x Volts ÷ Ohms = Watts
35V x 35V ÷ 8 Ohms = 153W
49.44 x 49.44 ÷ 8 Ohms = 306W... twice the initial wattage
Checks out.
Pretty sure I shared this last one already, but I'll mention it again...
1.2589 ^ dB increase over base sensitivity equals watts needed to achieve the specific gain
1.2589 ^ 20dB = 100W
1.2589 ^ 21dB = 126W
1.2589 ^ 22dB = 158W
1.2589 ^ 23dB = 199W
Any dB number gives the watts needed to achieve the gain.
That calc assumes 1 watt is the initial figure. But, just this moment, I realized we can modify the equation to the format of the Voltage equation if we want to know the final power needed to make a specific gain above a number of Watts other than 1. Nice little ah-hah moment of incidental realization I had there. Cool. Anyway, here's how that looks...
If we're running 250W (or any number of Watts) and want to know how many total watts we'll need to realize a specific gain of 1, 2, 3db (or any number of dB)...
(1.2589 ^ 1dB) x 250W = 315W
(1.2589 ^ 2dB) x 250W = 396W
(1.2589 ^ 3dB) x 250W = 499W
Any dB increase over any starting wattage get's you the final wattage.
Anyway, I know I get a little more geeked out in the numbers than most of you. But, figured I'd share for anyone interested and for the sake of posterity. I might need to come back here and find it again sometime. I
am creeping up on 50 and all.