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Amperage formula


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ty 
Copper - Posts: 137
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Posted: March 03, 2005 at 2:03 AM / IP Logged  

Ok, I want to put an end to all of the confusion about this formula, I have read that to calculate the draw you should divide watts into volts.  I read that you double the rms giving you max, and divide by your volts, which I measured mine when the car was running and got 14.45 all day w/o running any accessories.  I have also heard recently that you should divide watts into efficiency rating into volts.  Is this correct?  D amps would be 85%, and A/B would be 65% correct?  Then I have also read from a thread a while back that you divide watts into idle volts which is 12.6, and that is how this site got the amperage rating on the power and ground link to my left.  So could someone please clarify, and explain why people are saying so many different things, and which way should I calculate to see how many amps I am pulling? 

Also is there any way I can measure how many amps I am pulling without getting a cobalt ammeter or expensive equipment?  Thanks.

stevdart 
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Posted: March 03, 2005 at 7:41 AM / IP Logged  

This is to address your first questions but not the final one....

The best way to go about this is to separate known fact from unknown guesswork.  What are the things that are known as absolute fact and what are the things that are guessed at?

Facts:

  • Ohm's Law
  • Properly measured results of anything
  • Amplifier fuses
  • etc.

Guesses:

  • Power output rating of amplifier (published specs)
  • Efficiency of an amplifier
  • Car's voltage under load conditions if not properly measured
  • etc.

Figuring amperage draw is done for a variety of reasons;  some are precise when all knowns are facts, and some are educated guesses based on what is known and what is conjectured.  You rarely have all known facts...so you use what you have to work with.

If you've measured 14.4 volts steady in your car you have a fact....until you connect a load onto the system.  Once you run accessories or a power amplifier your "fact" now has become the second category - a guess.  Only by measuring under absolute load conditions, now, will give you a fact to work with.

This calculated evaluation of amperage draw if most often used for the purpose of figuring either size of power wire requirement or alternator needed.  A wire will sustain quick bursts of over-amperage without failing (as can a fuse), but an alternator is looked at a little different:  if it doesn't supply the short bursts of higher amperage that is being asked of it, you will see the results in the robbing of power going to accessories when your amp is requiring the big hits (when your amplifier is fed with a larger wire than other accessories are, it will get the lion's share of the power).  The alternator, then, requires the stricter measure.

Nothing is wrong in what you heard, but all are in the "guess" category.  Some guesses are meant to find the worst possible scenario (e.g. amperage pull using 12.6 V vs. 14.4 V due to the known fact Ohm's Law) and some guesses are a little more lenient.  The answer (formula) may be different in specific cases because some cases are almost all guesses while other cases contain more facts. Generally you will have a high ratio of guess-to-fact, because until you get the alternator and wiring in place and the amp running....you have nothing to measure.

In your examples, let's look at what would be priority over another:

  • Amplifier total fuse rating is fact.  It has priority in determing amperage pull over the other method of guessing the two unknowns - power rating of amp and efficiency.  In what you read recently on this forum, the latter (consideration of amplifer's efficiency) was used because the fuse rating was an unknown.  (In that case the amp was not yet purchased.)
  • 14.4 V is your measured voltage under low load conditions.  It is a fact at the moment but becomes an unknown when you want to predetermine wiring needs to hook up an amp load.  Idle volts or battery voltage (12.6 V) will constitute an increase in amperage needs because of Ohm's Law so it is a stricter measure and is given priority.

So, you can see that there is versatility here and although it may look like confusion it's really not.  The ways you cited are using the factors available for those given situations to make educated guesses. 

Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.
stevdart 
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Posted: March 04, 2005 at 12:07 AM / IP Logged  

ty wrote:
and which way should I calculate to see how many amps I am pulling?
 

If your situation is that you have the system complete, up and running you can make some measurements. 

  • Find proper procedure and measure amplifier output voltage
  • Measure voltage at car battery with a full load (amp running strong)

You now have some known facts:  nominal impedance load on the amp (measurement not needed) and voltage output of amplifier.  You can use Ohm's Law to find power output of the amplifier:  P = E^2/R.  This will give you power in watts.

With the other known fact of car voltage under a load, and you have already found the wattage at the amp's highest output level, you can then find how much pull the amplifier is making on the car's electrical system.  Use Ohm's Law I = P/E with power being the wattage output of the amp and voltage being the input voltage measured at the battery.

You will find that this measure usually comes out less than the "double the RMS" guess, because you will have made the measurements and are calculating measured numbers.

http://www.bcae1.com/ohmslaw.htm for reference.

Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.
ty 
Copper - Posts: 137
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Posted: March 04, 2005 at 2:41 AM / IP Logged  

thanks for the help steve, now correct me if I am wrong, but the voltage output by amp is measured in AC volts correct?  And I measure them where my wires meet the box right, not at the amp?  I don't have a 0db sound wave, so when I measured mine to see if it was clipping the voltage readouts were all over the place, but I just took a song with lots of bass and made sure it never went over the maximum voltage level I calculated, would I do the same thing in this instance, just take the highest voltage level and double it, and divide by nominal impedance?  Thanks again.

stevdart 
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Posted: March 04, 2005 at 5:28 AM / IP Logged  

Yes, it's all AC voltage at the amp.  You have everything set up and the gains set and you know the maximum level you can turn the volume of the head unit.  Have this all done using your ear and tweaked over time.  At this point you should be confident that when you play your system to the loudest you ever play it that the music is still clean.  So you are going to measure output voltage in a real-life scenario, not to find gain setting.

You must have these tools:  A DMM (you have) and a professionally-recorded CD of test tones in various frequencies at 0 db.  You will use these test tones for measuring voltage so the reading isn't jumping all over the place.  You don't want to burn up the sub by making it play a tone at a loud and steady level so you disconnect one of the speaker wires at the amp.  The amp still has signal input and will make voltage.

Read AC volts at the amplifier terminals that the subs are (or would be) connected to.  The volume of the deck is at the maximum you ever put it, the gain is set on the amp, so the tones played will show output at very close to the maximum voltage.  Find the freq reading with the highest output level.  That's the amp output voltage.

See the post above for formula to find watts using this AC volt reading and the nominal impedance of the speaker load.

(If you were to use the "doubling of RMS" value you wouldn't need to go through all this testing to find voltage.  So, no, you wouldn't double the highest reading you find, you just use that reading.  This will be pretty close.  Remember, you're asking how to find actual amperage draw of the audio system; you're not evaluating what wire size you may need.)

Then when you know voltage, use the second formula described above to find how many amps the car has to feed the amplifier to make that voltage.  Read the voltage in DC volts at the battery at this time.  With all things electrical, though, there will be loss in heat.....so there will always be a gray area and you should add a little to the amperage you find to compensate for what was lost in heat on the way from the alternator to the amplifier.  Test your results by looking at the amplifier's fuses and seeing if they are rated below what you found.  If they are less than what you figured the amperage to be, then something was done wrong.

ty 
Copper - Posts: 137
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Posted: March 05, 2005 at 2:45 AM / IP Logged  

Thanks for the help steve, now I am not trying to figure what fuse size I need or what gauge wire I need because I am running 1/0 gauge, and I have them fused at 300 amp fuse, what I am trying to do is find how much watts my system is putting out without getting my amp bench tested.  The reason being is I heard the amp that I have, hifonics bx1500d doesn't put out what it claims.  I heard a guy had his bench tested and thought it was running 1500 watts at 1 ohm, and it actually ran 1163 watts at 1 ohm, so I really want to find out what mine is doing without having to waste money to have someone tell me.

So, basically what I am going to need to do is get a cd with various frequencies at 0db, where can I find one of these?

stevdart 
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Posted: March 05, 2005 at 3:44 AM / IP Logged  

Ahh, I was wondering what the underlying reason was.  You know you asked how many amps your amplifier was pulling, but you really want to know how many watts your amp is producing. 

Sorry, you need the controls in place that a bench test provides for this type of accuracy.  The guy with 1163 watts instead of 1500 watts:  he's only getting 77% of the power he thought he was getting, right?  Well, that's not the right way to look at it when you're dealing with watts vs. decibels.  If he were getting the full 1500 watts his system would be barely ONE decibel louder! 

When you're testing in a car with minimal equipment, thus minimal control over the test, there will be some gray area and guesswork involved.  So whether you find that your amp makes more power than his through your efforts or not, it won't be a valid comparison.  And as you can see with the big 1 decibel difference we're talking about, you might find it's really not worth the effort.

The amplifier manufacturer, in the design stage, sets up a very controlled environment to test....and the results are sometimes achieved in real life but more often not.  Every amp that comes off the assembly line will be a little different than the other.  Why?  In electrical components there is variance.  Look at  this example.  This component is allowing a 10% variance, which is common.  Now think of all the components involved in the makeup of an amplifier and the variance that each is allowed.................doesn't take long to see the point, does it?

Another reason there are differences in wattage is the impedance load.  1,2 or 4 ohms is nominal impedance, not absolute.  You could have one sub driver that measures 3.2 ohms DCR at the coil, making the sub a 4ohm nominal but another driver that measures 3.4 ohms at the coil.  The nominal impedance is the same, but you know there is some variance in the actual voice coil measurements so the overall impedance will necessarily be somewhat different.  Now use Ohm's Law and look at what a difference there is from an absolute 1 ohm load to a load that actually is closer to 1.3 ohms:

Say you measure the voltage at 38.75 volts.  You use Ohm's Law P = E^2/R to find power, so 38.75 * 38.75 / 1 ohm = 1501 watts. 

Now, say the load was 1.3 ohms.  38.75 * 38.75 / 1.3 = 1155 watts!

Those explanations should help put things into perspective, eh?

http://www.bcae1.com/measpwr.htm is what you need to do for bench testing.

Here is one CD with a few tones, but if you do a search you'll find more.

The ultimate test comes to this:  is it LOUD enough for you?  And does it sound good?

Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.
ty 
Copper - Posts: 137
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Joined: September 16, 2004
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Posted: March 05, 2005 at 1:56 PM / IP Logged  
Cool thanks a lot for explaining everything Steve, I know you only get 3db for double the watts, but I wanted to make sure that the amp is doing what it claims, but now I know that it has to do with the driver as well.  But is this right for Hifonics or any amplifier manufacturer to make an amp that does not do what it claims in real life scenario's, it just does not seem right to me, I mean even with the 10% variance that is still 150 watts from 1500, should be 1350 watts, not 1163.  Anyway, how do you think their Zeus series would do, it is class A/B, would this make any difference in variability?  It says it puts out 4x85 watts rms at 4 ohms, do you think that would close?  I was thinking about getting it for my mids, but now I don't know if it will even put out 85 watts like it claims.  Thanks again.
stevdart 
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Posted: March 05, 2005 at 2:59 PM / IP Logged  

ty wrote:
it just does not seem right to me

   (sighs)

Buy equipment that you know has a reputation for quality.  Be careful and accurate in setting it up.  Put in your favorite CD and enjoy it.  'Nuff said.


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