how do you figure out the rms watts of an amp that doesnt tell you what it is
i have my sub amp its 800w max
and my 4 channel amp its 600w max
i need to set the gains properly on them
iirc to set the gains correctly you need to use a multimeter
the product should list the RMS power in the specs. If it doesnt, they are usually trying to mislead you with the MAX power, which is usually 2-3X that of RMS.
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yeah i knew that, but the thing is i want to get the most out of the amps so i was following what was in the sticky thread about tuning amps.
it said that i needed to get the rms of it so i can figure out what the volts would be
ok my four channel is 60x4 rms
so would you take 60 and get the square root or 60*4=blah blah and get the square root of that
and do you also have to get the rms of the sub amp or just go by the max power?
well the square root of that would be 15.49 so im guessing thats what it has to read on the mulitmeter
take the max power at 12v and multiply it by .687 i believe (atleast thats the number i remember from school)
Injection is nice but id rather be BLOWN!
That's a GENERAL rule, I was told that in HS too, but it all depends on the efficiency of the amp.
The only way to TRULY do it is with an oscilloscope, which are like 1000s of $$$.
You can rig up a good free PC o-scope and point a mic at the speaker it's driving and get close to the same effect.
well its done
i got 15.49 for the 4 channel
and 18.9 for the sub amp
i got them both to where they need to be
but i think it sounds worse then when i just said "hey this sounds good" lol
To find the actual wattage your amp is producing, you need to measure the amp draw at the amp, and the voltage at the same time. Then using ohm's law, you can know the actual power output.
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wysiwyg wrote:To find the actual wattage your amp is producing, you need to measure the amp draw at the amp, and the voltage at the same time. Then using ohm's law, you can know the actual power output.
You also need to know the amp's efficiency to do it this way.
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you can get a used o scope on ebay for usually under $100 shipped.
I got one about a year and a half ago. Its a Leader LBO. like 90 years old, but works great.
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20sunfire05 wrote:wysiwyg wrote:To find the actual wattage your amp is producing, you need to measure the amp draw at the amp, and the voltage at the same time. Then using ohm's law, you can know the actual power output.
You also need to know the amp's efficiency to do it this way.
P = E x I gives you ACTUAL power output. The effeciency rating of the amp doesn't matter.
So for example, if you measure the current draw to be 50 amps, while the voltage is measured to be 13.2v. that's 660 watts ACTUAL power output.
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BradSk88 wrote:wysiwyg wrote:20sunfire05 wrote:wysiwyg wrote:To find the actual wattage your amp is producing, you need to measure the amp draw at the amp, and the voltage at the same time. Then using ohm's law, you can know the actual power output.
You also need to know the amp's efficiency to do it this way.
P = E x I gives you ACTUAL power output. The effeciency rating of the amp doesn't matter.
So for example, if you measure the current draw to be 50 amps, while the voltage is measured to be 13.2v. that's 660 watts ACTUAL power output.
At what frequency?
at whatever freq you so desire to use. Burp it at 50hz, measure the amperage and current both, and there you have it. You now know how many watts your amp will put out at that freq.
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wysiwyg wrote:BradSk88 wrote:wysiwyg wrote:20sunfire05 wrote:wysiwyg wrote:To find the actual wattage your amp is producing, you need to measure the amp draw at the amp, and the voltage at the same time. Then using ohm's law, you can know the actual power output.
You also need to know the amp's efficiency to do it this way.
P = E x I gives you ACTUAL power output. The effeciency rating of the amp doesn't matter.
So for example, if you measure the current draw to be 50 amps, while the voltage is measured to be 13.2v. that's 660 watts ACTUAL power output.
At what frequency?
at whatever freq you so desire to use. Burp it at 50hz, measure the amperage and current both, and there you have it. You now know how many watts your amp will put out at that freq.
I know you're right because, hell, you're wysiwyg
But I still keep picking up flaws.
At what IMPEDANCE would this give you power reading?
BradSk88 wrote:wysiwyg wrote:BradSk88 wrote:wysiwyg wrote:20sunfire05 wrote:wysiwyg wrote:To find the actual wattage your amp is producing, you need to measure the amp draw at the amp, and the voltage at the same time. Then using ohm's law, you can know the actual power output.
You also need to know the amp's efficiency to do it this way.
P = E x I gives you ACTUAL power output. The effeciency rating of the amp doesn't matter.
So for example, if you measure the current draw to be 50 amps, while the voltage is measured to be 13.2v. that's 660 watts ACTUAL power output.
At what frequency?
at whatever freq you so desire to use. Burp it at 50hz, measure the amperage and current both, and there you have it. You now know how many watts your amp will put out at that freq.
I know you're right because, hell, you're wysiwyg
But I still keep picking up flaws.
At what IMPEDANCE would this give you power reading?
why does impendance matter? We're not measureing impendance to find power. You can't go buy impendance because it is a changing variable.
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he's asking at what impendance is it 660 watts? 2 ohms? 4 ohms? 6 ohms? I'm sure 13.2v is different output at 4 ohm then 2 maing.
but if you measure it that way that doesn't account for power loss from the 80% or lower efficiency of the amp right?
Corsica Dude wrote:he's asking at what impendance is it 660 watts? 2 ohms? 4 ohms? 6 ohms? I'm sure 13.2v is different output at 4 ohm then 2 maing.
depends on the amp, what i gave was just an example.
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Skyler Prahl wrote:but if you measure it that way that doesn't account for power loss from the 80% or lower efficiency of the amp right?
you are measuring ACTUAL current draw. not guessing at it. If you were guessing or estimating, then you could account for the amp's effeciency.
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wysiwyg wrote:Corsica Dude wrote:he's asking at what impendance is it 660 watts? 2 ohms? 4 ohms? 6 ohms? I'm sure 13.2v is different output at 4 ohm then 2 maing.
depends on the amp, what i gave was just an example.
Even so, I don't understand how the general formula would give any useful info.
P = I x E
@ 50A draw and 13.2V
P = 50 x 13.2
Okay, so that's 660watts. At 1 Ohm? or at whatever the DMM's impedance rating is? or the amps lowest available impedance?
cmon
BradSk88 wrote:wysiwyg wrote:Corsica Dude wrote:he's asking at what impendance is it 660 watts? 2 ohms? 4 ohms? 6 ohms? I'm sure 13.2v is different output at 4 ohm then 2 maing.
depends on the amp, what i gave was just an example.
Even so, I don't understand how the general formula would give any useful info.
P = I x E
@ 50A draw and 13.2V
P = 50 x 13.2
Okay, so that's 660watts. At 1 Ohm? or at whatever the DMM's impedance rating is? or the amps lowest available impedance?
cmon
you COULD hook the amp up to some known value resistors and measure it that way. useing a sine wave, but your speakers are not resistors, and they change impendance depending on frequency. ( you already knew that).
you could measure the resistance of the speaker at that certain freq if you wanted, then you could say "this amp did this power at this load".
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also note, if you hook up a dummy load "resistors" you need to use the formula P = E 2 / R
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