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In Reply to: RE: Watts vs speaker ohms.. posted by earnie3 on May 17, 2016 at 18:50:27
For solid state amplifiers with very low output impedance, power is inversely proportional to speaker impedance. Therefore, an amplifier that produces 80-watts into 6-ohms will produce 60-watts into 8-ohms. In other words:
80 x 6 = 60 x 8 = 480
The constant is the voltage across the speaker, which is the square root of 480 = 21.9-Volts RMS. This is because:
Power x (Speaker Impedance) = (Voltage)2
Another way of putting this is:
Power = (Voltage)2 / (Speaker Impdeance)
Follow Ups:
Or convert everything to dB.
Amplifier Power output: 10xLOG(80)= 19dBW
But it is the peak power that is important. +3dB = 22dBW
Speaker Sensitivitry at one meter +87dB = 109dB peak levels.
adjustment for normal listening distance - 6dB = 103dB peak level
Add second loudspeaker +6dB correlated (Mono) = 109dB peak level
Adjust for room acoustics +3 for live, -3dB for dead ~0 109dB peak
Adjust for program loudness (Crest factor) -10 (MP3rock downloads) to -25 (Classical wide Dynamic range) average -15dB = 94dB average level capability. Which is pretty loud. Average amplified concert levels 5 second running average maximums at about 95-105dBC (and that's why I wear ear plugs to concerts or go to small jazz clubs.
"The hardest thing of all is to find a black cat in a dark room, especially if there is no cat" - Confucius
it would be easier to jump through the various hoops and over the various hurdles to arrive at an answer which doesn't matter anyway, since speakers aren't of constant impedance across the audio spectrum, and there's that pesky issue of reactance.
In any case, it's not just about power output, it's also about instantaneous current capability.
:)
Ya see, this is one reason why I'm not a physicist or even an EE. As my woodwinds (clarinet, flute, etc., for those in Rio Linda) teacher once said, "I remember what I need to remember, and I flush the rest".
On the other hand, ask me something important, like "How many sabines of absorption are on that wall?", and I'll reply with "At what frequency?".
:)
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