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amp efficiency


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theetimurban 
Copper - Posts: 126
Copper spacespace
Joined: December 30, 2007
Location: South Carolina, United States
Posted: February 10, 2008 at 10:52 PM / IP Logged  
A while ago i was reading a post that gave the formula to determine an amps efficiency, i've been trying to find it again, to no avail. all i remember is that it has something to do with the fuses, and the wiring, or something like that. can anyone give me the formula??
stevdart 
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Joined: January 24, 2004
Location: Pennsylvania, United States
Posted: February 10, 2008 at 11:25 PM / IP Logged  

There's no formula to determine an amplifier's efficiency.  That's largely a supposed value based on the class of amplifier and / or manufacturer's data.  But what you might have seen on this forum was a formula that will tell you how many amps might be pulled depending on amplifer efficiency.  A class D mono amp with a supposed efficiency of 80% would be entered into the calculation as .8:

e = efficiency of amplifier

total RMS = watts demanded from source

(RMS X 2) / (e X 2) = total RMS

Then use Ohm's Law I = P/E to find demand in amperes.

Let's say you use 600 watts as the amplifier's RMS output.  This tells you that it will pull enough amperage to make 750 watts from the car's electrical source to get that 600 watts because the amplifier is only 80% efficient.  The other 20% accounts for the heat that the amplifier generates.

Are you trying to figure the amount of pull on your car's electrical system, or is your question dealing with something else?

Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.
theetimurban 
Copper - Posts: 126
Copper spacespace
Joined: December 30, 2007
Location: South Carolina, United States
Posted: February 10, 2008 at 11:55 PM / IP Logged  

yeah, i'm trying to figure out how many amps my setup is drawing, as well as power it's putting out.  right now i'm running a profile ap1200, it's a 2-channel class ab amp that i've bridged. these are some specs from the manual:

Bridged RMS output @ 4 ohms 1x600W

THD <.05%

Signal to Noise >103db

Damping Factor 300

<"Comic Sans MS" color=#31312e >

Fuse Value (ATC/ATO Type) 3 x 25A

stevdart 
Platinum - Posts: 5,816
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Joined: January 24, 2004
Location: Pennsylvania, United States
Posted: February 11, 2008 at 12:38 AM / IP Logged  

That amplifer is about 60% efficient  (normal for class A/B) but its published rating is more than you'll get in real life.  That has to do with class of manufacturer.  But I have a Profile amp, too.  It's in the attic right now but I can see using it again for something.  It works.

If you want to determine amperage pull based on the published output rating, use .60 in the formula.  The Ohm's Law equation that gives you amperage pull is read as:  I = amperage,  P= power in watts after allowing for amplifier efficiency,  E = voltage metered at the amplifier input (relates to car's applied voltage).

You should be using 4 ohms impedance bridged across the channels.  If you use 600 watts as rated output into that 4 ohm load, and .6 as efficiency factor, the pull on your car is enough to actually make 1000 watts.  To find out what that is in amperage (which is what the car provides to the amplifier), plug into the calculation:  P = 1000 and E = probably 12.6 to 13 volts.  At 13 volts the assumption (because you don't know true power output and can only assume the rating) is that the amplifier might need to pull 77 amperes.

A quick and sure tell is that if the amplifier has two 30 amp fuses installed it is not designed to handle more than 60 amps (where you have calculated that it might need to handle 77 amps).  That tells you that the amp is over rated.  On the other hand, if it has three 30 amp fuses it is more than covering its power bragging and is ready to do business.

Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.
theetimurban 
Copper - Posts: 126
Copper spacespace
Joined: December 30, 2007
Location: South Carolina, United States
Posted: February 11, 2008 at 1:17 AM / IP Logged  

okay, so this is what i've come up with: 

At idle, my car produces at 13.8v, soo...

I = 1000/13.8

I = 72.464

so the amp draws approx 72.5 amps (right?)

The amp has 3, 25 amp fuses.

So would you say it's "ready to do business"? I'm not trying to sound condescending, i'm really trying to understand it.

n2audio 
Copper - Posts: 95
Copper spacespace
Joined: March 02, 2005
Location: United States
Posted: February 11, 2008 at 9:52 AM / IP Logged  

The AP1200 is rated at 960w rms (bridged into 4 ohms)

An a/b amp will be around 60% efficient at full power so 960/.6 = 1600w

1600/13 = 123A

The most important factor to consider is the crest factor of music - crest factor is the peak-to-average ratio of the signal.  Meaning you may have peaks in a given music signal that require close to 960w, but the majority of the signal is far less than that.  So even with high demand music you end up with a crest factor of no more than 1/3, in many cases 1/4 or less.  So the actual rms current draw of your amp is not likely to exceed 30-40A.  At 50-75% of full volume it's quite possible it will be 20A or so.

theetimurban 
Copper - Posts: 126
Copper spacespace
Joined: December 30, 2007
Location: South Carolina, United States
Posted: February 13, 2008 at 12:08 AM / IP Logged  

ap1200 is 600rms

the ap2000 is 960rms

n2audio 
Copper - Posts: 95
Copper spacespace
Joined: March 02, 2005
Location: United States
Posted: February 13, 2008 at 2:48 PM / IP Logged  

yeah, my bad.

your ~70 would be about right on the 1200.  So considering crest factor and something less than full volume most of the time you're probably looking at 15-20A.


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