posted
Whats the proper procedure to measure the output of an amplifier. I've been using P=V^2/R and ACV/1.414=Vrms to change volts to true rms. Am I right?
Thanks in advance.
-------------------- CRX Si 89 SS 1,2 nw, local street 3,4 157.1 dbs 2 DD9915 2 DD Z2 kinetic hc16 batts Iraggi 300 amp alt Posts: 324 | From: SJ Puerto Rico | Registered: Aug 2004
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posted
Try usen amp draw. x That by the voltage going into the amp. To get peak power. Then x it by .707 to get the true rms at 7% T.H.D. It works for me.
-------------------- Team Xtreme Loud Audio. Team Stetsom Usa Street B North East Stetsom Distributor & Rep. 1-413-433-1604 Posts: 1541 | From: Springfield Ma | Registered: Apr 2004
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quote:Originally posted by Old/School: Try usen amp draw. x That by the voltage going into the amp. To get peak power. Then x it by .707 to get the true rms at 7% T.H.D. It works for me.
This would give you the input power to the amp but it does not take into account the amplifier efficiency. The output power could be significantly less because amps aren't generally very efficient.
Using this equation (P=V^2/R) would give you a better estimate of power. This of course is assuming that the speakers impedance is what its rating is at the frequency that you are sending to the speaker. Since the actual speaker impedance actually varies with frequency, this equation would only work if you had the actual impedance of the speaker at the specific frequency that you were sending it during the test.
You do not have to do any further math to find the RMS power. All multimeters give you the RMS AC voltage (or at least a close approximation), so that equation would give you the RMS power.
Posts: 327 | From: Pleasanton & Sacramento Ca | Registered: Dec 2001
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quote:Originally posted by Old/School: Try usen amp draw. x That by the voltage going into the amp. To get peak power. Then x it by .707 to get the true rms at 7% T.H.D. It works for me.
This would give you the input power to the amp but it does not take into account the amplifier efficiency. The output power could be significantly less because amps aren't generally very efficient.
Using this equation (P=V^2/R) would give you a better estimate of power. This of course is assuming that the speakers impedance is what its rating is at the frequency that you are sending to the speaker. Since the actual speaker impedance actually varies with frequency, this equation would only work if you had the actual impedance of the speaker at the specific frequency that you were sending it during the test.
You do not have to do any further math to find the RMS power. All multimeters give you the RMS AC voltage (or at least a close approximation), so that equation would give you the RMS power.
Unfortunately, even this isn't 100% correct. You will get some error in this because of the fact that the true impedance of your sub/enclosure combination will depend on frequency played. Using the nominal impedance isn't actually correct. If you have the means to measure the current directly, then a good approximation on power is V*I. Even that isn't perfect because of phase issues. You can have a really high voltage and a really high current, but still have low power output because the phase of the voltage and current isn't aligned. This can cause your approximation of power to be inflated by as much as 30% or more. In most cases, however, I think it will be more like 10-15% error (guesstimated numbers...) If you are interested in the details of how this is possible, research "power factor" in AC systems.
posted
amen brother! input power converted to heat and power factor, the 2 main things nobody takes into account. the things 99% of the people refuse to believe and insist that their amplifier with a 50 amp fuse produces 2000 watts rms to the subs, because that piece of paper that came with it says so.
Posts: 540 | From: nor cal | Registered: Aug 2004
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posted
ohms law works as long as the R stays the same. in this case it doesn't. the best way i have found out while on a resistive load is just compare what each amp does Vac. if you want to get a minimum value, take the highest R you get on your dmm and the highest voltage you recieve on your dmm as well and do Vac^2/r. why the highest? because the highest voltage will come with the highest resistence.
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posted
im a retard when it comes to whatever kind of math you guys are talking about... but you have
P = Vac^2/R where R = resistance (ohms) right?
so lets say I throw 100vac on my sub, square that? that is 10,000 and lets say my impedence is 5 b/c of impedence rise in my box... that = roughly 2,000 right? Am I anywhere near correct on this? And lets add in error % and we could come down as low as 1,500 or so yes?
Please visit my post on Vac in this same forum... I may need the help
-------------------- back in the 50's... Kicker is 'Livin Loud' Marriage soon, new house in the works, new baby, and no money! Posts: 2497 | From: Gardiner Maine 00420ville | Registered: May 2000
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posted
the reisistence actually changes when your sub is moving so you need to just set peak hold on your dmm for resistence and voltage. this will give you a minimum number. pVac^2/pR
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quote:Originally posted by jarfunkz: the reisistence actually changes when your sub is moving so you need to just set peak hold on your dmm for resistence and voltage. this will give you a minimum number. pVac^2/pR
Can you measure resistance with a dmm when your sending signal to the subs?
posted
well the resistence changes, that is why i do peak hold on both. if you do Vac and amperes, both peaks are reading off different resistences. ohms law still works as long as you get both read on the same resistence level. to tst amps i have dummy loads that i comprised off 300w 4 ohm resistors i bought off ebay. it really is the only way to test an amps true potential in my opinion.
you can not measure resistence of the sub, because when the sub is playing mechanical resistence and heat of the coil will make the resistence rise, as well as the enclosure that the sub is in will effect the mechanical resistence.
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posted
You can't measure RESISTANCE while there is a signal going to the sub because of the way an ohm-meter works. Besides, resistance is not very useful anyway. You want impedance, which is like the AC version of resistance. Technically speaking, impedance will have a purely resistive component as well as an inductive or capacitive component. This inductive or capacitive component is what will prevent the voltage and current from being completely in phase. This complicates our calculations considerably. The best (easiest) way to determine the true power output of an amp is to test it with a purely resistive dummy load, as jarfunkz suggested. If you do this, then you can play a sine wave into the dummy load, measure the AC voltage, and then immediately disconnect the load and measure the DC resistance. You need to know the RMS voltage, not the peak. I'm not familiar with different multimeters, so I don't know which number is typically given, but you could check by measuring your wall voltage. If it's around 110-120 VAC, that's RMS. If it's more like 160, that's peak. Multiply peak by .707 to get RMS.
(RMS AC voltage)^2/resistance will give you the true power output of the amp into that ohm load. Remember that this means virtually nothing after you hook that amp up to a reactive load (subs). That is, to the best of my knowledge, how amps are typically tested. In other words, I believe that this method is how manufacturers usually get the numbers on the birth sheet. If I'm wrong about that, please feel free to correct me.
posted
Clamp on amp meter baby Mine was $250 well spent. Get a good one, apply some mathematical and electrical knowledge, and you'll get:
Accurate Wattage output of amplifers Accurate Amplifier current draw Amplifer efficency
And if you want to get really fancy, you can even do In-box subwoofer impedance curves
Come to think of it.... thats a big part in accurately figuring out what frequency your vehicle/enclosure system is tuned to, and where compliance is greatest
Hmmm... yes... what a nice tool indeed
-------------------- Nate Scholten Team Sounds And Motion SS 1-2 3rd place 2004 finals -THE BUILD- ~Our Drinking Team Has A Stereo Problem~ Posts: 4252 | From: Rhinelander,WI | Registered: Dec 1999
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quote:Originally posted by netlohcs: Clamp on amp meter baby Mine was $250 well spent. Get a good one, apply some mathematical and electrical knowledge, and you'll get:
Accurate Wattage output of amplifers Accurate Amplifier current draw Amplifer efficency
And if you want to get really fancy, you can even do In-box subwoofer impedance curves
Come to think of it.... thats a big part in accurately figuring out what frequency your vehicle/enclosure system is tuned to, and where compliance is greatest
Hmmm... yes... what a nice tool indeed
It's about as accurate as you'll get, but not 100% because of that pesky power factor. I'm sure there is some type of tool that allows you to measure the phase difference between current and voltage, but I've never seen one. With that type of tool, a voltmetere, and an amp clamp you could get a truly accurate measure of output power.
posted
I don't know how many times this issue has been addressed in one form or another. There are times I think some people may not believe published specs, and in almost every case put too much into interpeting specification sheets and their purpose. It's impossible to measure power while music is playing, unless you have a high speed photographic memory, or make a watt meter like some old home amplifiers had built in, so you can watch a needle swing back and forth.
-------------------- E.C. Wuz here Posts: 1057 | From: Vancouver, British Columbia, Canada 'eh | Registered: May 1999
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quote:Originally posted by Nomad84: I'm sure there is some type of tool that allows you to measure the phase difference between current and voltage, but I've never seen one.
That would be an oscilloscope with a current probe. You would measure voltage with one channel and current with another, and with that you could find the phase shift and thus the power factor correction number.
For the most part you can really ignore the power factor part for a car audio subwoofer. The wire resistance and distance is fairly low so the power factor wouldn't be that much anyway. How accurate of results do you need anyway?
Posts: 327 | From: Pleasanton & Sacramento Ca | Registered: Dec 2001
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quote:Originally posted by netlohcs: Clamp on amp meter baby Mine was $250 well spent. Get a good one, apply some mathematical and electrical knowledge, and you'll get:
Accurate Wattage output of amplifers Accurate Amplifier current draw Amplifer efficency
And if you want to get really fancy, you can even do In-box subwoofer impedance curves
Come to think of it.... thats a big part in accurately figuring out what frequency your vehicle/enclosure system is tuned to, and where compliance is greatest
Hmmm... yes... what a nice tool indeed
I couldn't agree more.
P = V * I
R = V / I However your impeadance curves will not only vary by note, but also by excursion....
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