wattage results

[jmrunz 10+ year member]

hi everyone it has been a while on the site. i have a question for the electricians. i want to know exactly what my subs are recieving in wattage. here are my numbers.

amp is 400x2 4ohm rms -- 800x2 2ohm max or 2200x1 2ohm max it is a soundstorm 2f2200 model amp.

subs are audiobahn 750 rms wattage 1400 max rating sub.

the output voltage at the speaker leads is 40 volts. what is the wattage here.

i found some stuff but the numbers don't add up in my mind.

amps x volts = watts

watts / amps = volts

watts / volts = amps.

using the theory that rms wattage / battery voltage = amps i get (and i am using a genaric 12 volt rating) 800 / 12 = 66.6 so then i take 66.6 x 40 = 2664 max power from the amp. is this correct. somebody please help.

james

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[tRidiot 10+ year member]

What's the final impedance the subs are wired at?

 

[Blunt 10+ year member]

acv * aca = watts but this reading needs to be taken at exactly the same time do not use peak hold.

 

[jmrunz 10+ year member]

the subs are wired in 2ohm load. just for info they are in a shared enclosure each with its own signal. the amp is seeing a 2 ohm stereo load.

 

[jmrunz 10+ year member]

blunt, what does a acv and aca stand for.

 

[IgnoreMe 5,000+ posts]

blunt, what does a acv and aca stand for.

AC voltage and AC amperage.

to find your real world wattage you need to have a dmm and a ammeter (clamp style).

play the test tone, take the reading from the speaker terminal using the DMM to get voltage. take the reading from the speaker wire that the amp clamp is around to get the amperage.

multiply the two values to get actual wattage

divide your voltage by your amperage to get actual impedance.

(the higher your ohm value the more inaccurate the result though, so if your getting wattage results for like 6 ohms you can basically disregard the data)

 

[tRidiot 10+ year member]

I thought if you did V^2/R, then that was a rough guesstimate? Obviously if you use the peak V, then that would be peaks or transients, not RMS.

Maybe we're talking about different things here...

 

[IgnoreMe 5,000+ posts]

I thought if you did V^2/R, then that was a rough guesstimate? Obviously if you use the peak V, then that would be peaks or transients, not RMS.
Maybe we're talking about different things here...

to me it seems like he's trying to get real readings, which is why i suggested doing it that way. i believe i have heard that equation before as an accepted way of getting a decently accurate result. so i dont see why he couldnt use it, especially as it seems the OP only has access to a dmm.

to the original poster: your doing it right, but the issue is that your not accounting for amplifier efficiency or anything else. if you want real power you have get a voltage reading from the speaker terminals (like you have done) and an amperage reading from the speaker terminals as well. your way of doing it is flawed and wont give you an accurate answer.

the problem with getting an amperage reading off an amp, is most people only have access to a dmm, which usually come with 10A fuses. the point of the amp clamp is so that you can read, much higher amperages without damaging tools.

 

[jmrunz 10+ year member]

ignorme. you are correct. i am wanting real world wattage results. and i only have a dmm. can you measure at the speakre leads with a dmm and get the num bers needed or does it absoultly have to be a clamp style ammeter. i don't have one of thoes and don't know where i can get one. what i was hoping for was if there was a formula that i could just plug some numbers into and get the answer. but thanks for the help. if anybody has anything else i would be glad to hear it.

james

 

[IgnoreMe 5,000+ posts]

ignorme. you are correct. i am wanting real world wattage results. and i only have a dmm. can you measure at the speakre leads with a dmm and get the num bers needed or does it absoultly have to be a clamp style ammeter. i don't have one of thoes and don't know where i can get one. what i was hoping for was if there was a formula that i could just plug some numbers into and get the answer. but thanks for the help. if anybody has anything else i would be glad to hear it.james

not really. your going to need a way to get both amperage and voltage on the outputs of the amp. you can get a ammeter for like 60 bucks at lowes, but IMO thats not really worth it, since you probably wont have a use for it after this.

tRdoc's method is an accepted one, that should give you a relatively accurate estimate. hell even using the clamp and the dmm you wont get a 100% accurate reading. the higher the ohm reading, the more inaccurate your result due to the real power, versus the apparent power thats being caluclated.

you can google "power factor" if you want more info on that.

 

[Trey803 5,000+ posts]

so how do manufacturers come up with such accurate numbers on their birthsheet?

 

[IgnoreMe 5,000+ posts]

so how do manufacturers come up with such accurate numbers on their birthsheet?

AC electric power system is defined as the ratio of the real power to the apparent power, and is a number between 0 and 1. Real power is the capacity of the circuit for performing work in a particular time. Apparent power is the product of the current and voltage of the circuit. Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power can be greater than the real power. Low-power-factor loads increase losses in a power distribution system and result in increased energy costs.

that shows a little bit on why you would need to use the power factor to get an accurate reading.

as for the mfg, they have a test bench that holds a precise load to get real power. we unfortunately have to deal with a moving mass (the subwoofer) to get amperage readings.

im having a tendancy not to believe birthsheets anywho. there is only so many 2960W @ 2 ohm kx2500.1 birthsheets i can see (including mine) before i start to doubt that all kx2500.1's put out the same exact amount of power..

 

[Trey803 5,000+ posts]

so is there anywhere that would be likely to have a test bench and/or how hard and expensive would it be to make one?

 

[n2audio 5,000+ posts]

so is there anywhere that would be likely to have a test bench and/or how hard and expensive would it be to make one?

the main component is an oscilloscope.

http://www.bcae1.com/measpwr.htm

I can tell you this though -- with 2x30A fuses 800w rms is about as good as it's going to do. I would bet lower.

 

[tRidiot 10+ year member]

You'd also need a massive 12V powersupply, unless you're just going to use an actual car battery for your power.