Finally picked up a clamp meter and tested my zx...

[Immacomputer 10+ year member]

a watt is a unit of power, which is made up of voltage and amperage
an amplifier takes DC in and goes through a conversion and switching process to produce AC

There is no other real way to measure power out of your amps, you can do the *.707 bs that other people do and that just gives you a lower number, he is testing what his amps are doing peak during a 2 second burp

You don't need to explain what power is to me; I have a pretty good understanding of electronic basics. It appears that you do not. Do you have any clue what the difference between real and imaginary impedances are?

 

[tommyk90 5,000+ posts]

I never said you were an idiot and I never said you don't know anything. What I have said is that you are wrong. I'm sorry that you act like a child when you hear that. The people who think they're getting actual power readings from measuring like this are wrong and I would tell that to anybody who things otherwise.


I have told you before but you either don't understand or you didn't ever read it.

I would love to do that on my own if I had the equipment and time.

For starters, you need a purely resistive load that can handle large amounts of power and not heat up much. I know places sell them because I have seen then at work. They are large metal bars like you would see on a furnace or heater and they dip down into a 5 gallon bucket like container. If you attach that load to your amp and set the gains with a scope, you will be able to measure the actual power of the amp while playing pink noise through it. In this case, using a nice scope and nice clamp meter will be able to give you a much, much, more accurate number.

But why would I want to test with a purely resistive (fixed resistance?) load?

That's not how a speaker acts in the least.

I'd hate to sound childish to you, but you seem to be making things more difficult than they really have to be. You say you know all about these theories by reading and studying them, but have you actually USED them to determine their validity in the car audio realm?

 

[teamsubgopoof]

Getting power using that method only applied to purely resistive circuits (DC circuits). When you're dealing with AC circuits that have inductance and capacitance (which is what you will have when dealing with speakers), you will have an imaginary part of the impedance that will affect the actual power measured in watts. By imaginary, it like the sqrt(-1) like you have heard of in algebra.
In AC circuits, power in watts is measured by:

Current(rms)*Voltage(rms)*Power Factor = Power (in watts).

If you leave off the power factor, you just have volt amps.

and btw I know about power triangles and what power factor is

and if you want to get very technical about it then everybodies testing is flawed and not really give real power, just apparent.

BUT, since apparent power is standard it still gives a comparison to other peoples testing, and is therefore an efficient measurement of power in the car audio world //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[PsychoTheRapist 5,000+ posts]

edit: I'm w/ tapout on this one..

 

[teamsubgopoof]

he just told you to look up "power factor"...
i did. from what i can tell, immacomputer is right. no disrespect to tommyk.

i remember when i got schooled by immacomputer about my 20.1... hahah //content.invisioncic.com/y282845/emoticons/crying.gif.ec0ebefe590df0251476573bc49e46d8.gif

as far as the theory goes he is right.

HOWEVER you can't have just tommy saying my amp does this much REAL power and everyone elses results are wrong because that isn't the standard for testing power out of amplifiers

 

[tapout 5,000+ posts]

where is squeak, audioholic, and helotaxi when you need them?

 

[teamsubgopoof]

You don't need to explain what power is to me; I have a pretty good understanding of electronic basics. It appears that you do not. Do you have any clue what the difference between real and imaginary impedances are?

and i also understand about real and imaginary impedences, just because you are probobly taking an electronics class and learning theories, don't assume that no one else on this site doesn't know about electrical engineering and theories of electronic circuits //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

the impedence he is seeing is the impedence of the woofers, which is a real resistance.

also, the way tommy is testing is by far the EASIEST way to get close to actual power, most people that compete in car audio have no education on electronic theories, just being able to mulitply voltage and amperage and get a comparison is definately a fast and easy way to compare one amp to another

 

[tommyk90 5,000+ posts]

Well I have read up on power factor, and see it's significance.

But I have neither the equipment or the means to test out this theory in my vehicle, along with 99% of people out there who test power the same way.

From what I recall, people used to multiply the va by .707 to get a more "real" number, but I never saw the purpose of that.

Theories can only get you so far. In the car audio world, they get you nowhere. In "theory" alot of things should happen a specific way. In the real world, thats hardly the case. Can we agree on that at least?

 

[Immacomputer 10+ year member]

But why would I want to test with a purely resistive (fixed resistance?) load?
That's not how a speaker acts in the least.

I'd hate to sound childish to you, but you seem to be making things more difficult than they really have to be. You say you know all about these theories by reading and studying them, but have you actually USED them to determine their validity in the car audio realm?

You use a purely resistive load because using any reactive load will not give you the numbers you're looking for! You want to know POWER right? Well power is measured in WATTS. You're not going to get an accurate measurement of watts if you're measuring volt amps.

It's not as difficult as you think it is but just not something anybody with an amp can do. It does take precise and reliable measuring equipment.

And yes, what applies to all forms of electronics applies to car audio as well. I don't know if you realize it, but your sub in it's enclosure is just like a circuit with resistors, capacitors, and inductors. I have played with tons of circuits and I have seen the difference between volt amps and watts every time I make something.

 

[IgnoreMe 5,000+ posts]

as far as the theory goes he is right.

HOWEVER you can't have just tommy saying my amp does this much REAL power and everyone elses results are wrong because that isn't the standard for testing power out of amplifiers

i understand and agree with that part to a point. what immacomputer was saying that the higher the load is (for example tommyk's 4 ohm load compared to someone elses 1.1 ohm load) the more skewed the power is. so i think it works when your have 2 amplifers with 2 similar loads and the end but isnt accurate when you compare say a 4 ohm load to that other amps 1.1 ohm load.

im only speculating. this is too much for my head and i prefer to go with the standard volts x amps //content.invisioncic.com/y282845/emoticons/blackeye.gif.66a1670f5aaf7f406e783a63e3387dc5.gif

 

[Immacomputer 10+ year member]

and i also understand about real and imaginary impedences, just because you are probobly taking an electronics class and learning theories, don't assume that no one else on this site doesn't know about electrical engineering and theories of electronic circuits //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif
the impedence he is seeing is the impedence of the woofers, which is a real resistance.

Where did I ever say I was the only person here with electrical knowledge? Judging from your posts, you don't seem to have a very decent grasp though.

And since when has apparent power ever been a standard for labeling amps?

 

[Immacomputer 10+ year member]

i understand and agree with that part to a point. what immacomputer was saying that the higher the load is (for example tommyk's 4 ohm load compared to someone elses 1.1 ohm load) the more skewed the power is. so i think it works when your have 2 amplifers with 2 similar loads and the end but isnt accurate when you compare say a 4 ohm load to that other amps 1.1 ohm load.
im only speculating. this is too much for my head and i prefer to go with the standard volts x amps //content.invisioncic.com/y282845/emoticons/blackeye.gif.66a1670f5aaf7f406e783a63e3387dc5.gif

The reason why the results get more skewed is because the imaginary part of the impedance gets larger which increases the phase angle between the two. The phase angle is used to determine the power factor. As the angle increases, the power factor gets smaller and smaller which means the actual power gets less and less.

 

[tommyk90 5,000+ posts]

You use a purely resistive load because using any reactive load will not give you the numbers you're looking for! You want to know POWER right? Well power is measured in WATTS. You're not going to get an accurate measurement of watts if you're measuring volt amps.
It's not as difficult as you think it is but just not something anybody with an amp can do. It does take precise and reliable measuring equipment.

And yes, what applies to all forms of electronics applies to car audio as well. I don't know if you realize it, but your sub in it's enclosure is just like a circuit with resistors, capacitors, and inductors. I have played with tons of circuits and I have seen the difference between volt amps and watts every time I make something.

I want to know actual power that my SUB is receiving, not a completely resistive load.

How do i measure ACTUAL power my SUB is getting by placing a purely resistive load on my amp? Like I said above, theories are all fine and dandy, but they really mean d*ck in the car audio world. There's far too many other factors that come into play.

 

[dbman150 5,000+ posts]

//content.invisioncic.com/y282845/emoticons/popcorn.gif.32dd9e22fd77e77bc3c907062768fcd2.gif

 

[IgnoreMe 5,000+ posts]

The reason why the results get more skewed is because the imaginary part of the impedance gets larger which increases the phase angle between the two. The phase angle is used to determine the power factor. As the angle increases, the power factor gets smaller and smaller which means the actual power gets less and less.

so would i be correct in assuming that if you have a final load of 1 ohm with amp A and a final load of 1 ohm with amp B, that the results can be considered fairly accurate? not in a sense of oh this amp puts out xxxx total watts, but this amp puts out roughly xxx watts MORE than said amp?

just so that we could have something to go on when comparing amps head to head?