Formulas to approximate wattage?

[hatedonmostly 5,000+ posts]

Trying to figure out what my amp can do. I know W = V x A, where voltage = 14.4 for my amp's power ratings and an 80 amp fuse in the amp.

14.4 x 80 = 1152

That is close to my amp's rated 1200 @ 1 ohm. But there's where the problem lies, I'm running the amp at 2 ohm.

http://wiki.answers.com/Q/How_many_Watts_are_in_1.6_Amps

Given E and R:
W = E^2 / R = 14400 / 144 = 100W and,

I = E / R = 120 / 144 = 0.8333A

14.4^2 / 2 = 103.68 watts?

14.4 / 2 = 7.2 amps?

I also want to calculate for V = 13.4, as that is what my voltage drops to.

Click to read more...

 

[dragon.breath 10+ year member]

14.4 volts x 80 amp fuse = 1152 watts. but you did not calculate the amplifiers efficency in that. a d class amplifier is normally around 70% so 14.4 x 80 = 1152 x .7 = 806.4 watts

the voltage squared divided ohm load is measured on the output ( speaker ) side in ac volts not off the battery side.

 

[groundpound4200 5,000+ posts]

you need a clamp to get the numbers to plug into your formulas

 

[hatedonmostly 5,000+ posts]

How can I calculate this then lol

 

[hatedonmostly 5,000+ posts]

you need a clamp to get the numbers to plug into your formulas

I just want rough estimates. I'm not trying to factor in box rise or anything either.

 

[groundpound4200 5,000+ posts]

Post #2 gets you close for 1 ohm output. You can be conservative and cut that number in half for 2 ohms. You have the right formulas at the bottom of the OP you're just using them wrong.

 

[grassroots 10+ year member]

you need real world numbers... can't just take your fuse rating and 14.4v...

find out exactly how many amps you're drawing (with a clamp meter) and exactly what your voltage drops to while you're playing a test tone. then multiply to find wattage.

 

[hatedonmostly 5,000+ posts]

Post #2 gets you close for 1 ohm output. You can be conservative and cut that number in half for 2 ohms. You have the right formulas at the bottom of the OP you're just using them wrong.

Elaborate. I understand I forgot to factor in the efficiency, my mistake.

 

[groundpound4200 5,000+ posts]

To use the formulas at the bottom of the second post you are going to know the current and voltage being provided by the amplifier under a load. To obtain these numbers you need a DMM and an AC clamp

 

[tommyk90 5,000+ posts]

Elaborate. I understand I forgot to factor in the efficiency, my mistake.

Volts and Amps in that formula are ALTERNATING CURRENT, aka AC amps and AC volts.

You are using DC volts (your 14.4) and AC amps (the fuse rating) to try and figure out your wattage, which is how you are using the formula incorrectly.

Doing it the way you did it gives you a terribly rough idea. In order to accurately see how much power you have, you need to get a clamp meter.

 

[groundpound4200 5,000+ posts]

Volts/Amps=Ohms

Amps*Volts=wattage

These take the fuse rating and the input supply voltage out of the equation and gives you a much more realistic number. Multiplying the fuse rating on the amp by the input voltage multiplied by efficiency will give you a rough calculation that you're looking for

*edit, Tommy got it

 

[hatedonmostly 5,000+ posts]

Volts/Amps=OhmsAmps*Volts=wattage

These take the fuse rating and the input supply voltage out of the equation and gives you a much more realistic number. Multiplying the fuse rating on the amp by the input voltage multiplied by efficiency will give you a rough calculation that you're looking for

*edit, Tommy got it

I know, I'm not looking for anything more than a rough estimate. If I had the money for a clamp meter I'd put it toward something else, like the big three or a better amp.

Even with the efficiency factored in, that doesn't give me an estimate for it at 2 ohms. 1/2 of it isn't that good of an estimate.

Oh well.

 

[groundpound4200 5,000+ posts]

Without knowing how the power supply, output stage, etc. were put together in the amp its about the best you can do without a clamp and DMM

 

[hatedonmostly 5,000+ posts]

Yes, but either way, if 14.4 x 80 = 1152 and 1152 x 0.7 = 806.4, there is no way my amp is putting out anywhere near rated, especially after box rise. Wow, those numbers on my amp were inflated as ****.

 

[boogeyman 10+ year member]

If you want to try to determine the amplifier's power output, the efficiency of the amplifier class won't be a factor. Efficiency is a factor when you are trying to find out how much power is being pulled from the source (if my amplifier does X watts then how many amps do I need from the vehicle?).

Truth be told, you can't determine your amp's power output unless you put it through a controlled bench test. But if it's a respected brand you should use the manufacturer's published RMS output data. Specs should be reported at both 14.4 volts and at 12.6 volts from the source. But even if there is only one source voltage showing in the specs you can extrapolate the wattage to conform to your vehicle's actual battery voltage measurement under full load conditions (as when you are playing the system). If you take readings consistently at 13.1 volts at the battery under load conditions, find the value per volt and then you can conform the published data to fit your actual vehicle.

That is using voltage at the source, the battery. But there is also voltage (AC) at the output of your amplifier. You have to make sure you understand which voltage, DC at the source or AC at the amplifier, you need to know for any of your calculations.

With Ohm's Law, if you know any two of the given variables then you will be able to use the calculation to find the third. You are trying to find wattage, For instance, if you know the impedance of the subs and the voltage output at the amplifier's speaker output, then you can figure power output based on those numbers using P = E^2 / R. For your "known" factors, you can get close with the ohms but not entirely accurate, for example by using the number 2 as impedance. (You don't know if it's exactly 2 ohms load at the amp or if it's 2.4 or 1.8 at any given frequency....) And you can use an AC voltage measurement taken at the amp's speaker output (with sub disconnected) while playing a test tone in the sub range. But even when you are careful to use a tone recorded at 0db and have taken a steady measurement with gain set perfectly, you still can't be sure that the voltage you read is entirely accurate as it relates to the music sources that you normally play. So your calculation can only be close, as any deviation in a variable must change the calculated output.

An example: impedance is 2; measured voltage is 28 (voltage is measured at one output channel...measure both L and R but use one of them, don't add them). Ohm's Law says "power equals volts squared divided by ohms". 28 * 28 = 784, and 784 / 2 = 392. Your estimated power in this example is 392 watts.

Now at this point you could use these findings to help determine how much power is needed from the car to provide the calculated 392 subsonic watts. Efficiency of the amp comes into play at this time, and you can use this calculation:

(RMS X 2) / (e X 2) = total RMS. Then use Ohm's Law I = P/E to find demand in amperes.

e = efficiency of amplifier

total RMS = watts demanded from source

65%, if used in the calculation, is written as 0.65