Calculating power consumption.

[Radioflyer97 10+ year member]

Reguarding a recent fiasco with an alternator, i'm trying to figure out roughly how much power my system is consuming. Thus far I've been calculating based on percentages of the total peak draw:

The total fuse ratings on all 3 amps are 110 amps.

If i set all the amps gains to 50% i should draw no more than 55 amps at any time

If i have the volume at 50% i should draw no more than 27.5 amps at any time.

Is this the correct formula?

( I will be getting an ammeter installed on the main power wire next week to be certain)

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[ShakesAllDay 10+ year member]

Gains are correlated to pre-amp voltage, so saying "gains at 50%" doesn't mean much without knowledge of the pre-amp voltage. 50% @ 2v will be different than 50% @ 5v.

 

[Radioflyer97 10+ year member]

Gains are correlated to pre-amp voltage, so saying "gains at 50%" doesn't mean much without knowledge of the pre-amp voltage. 50% @ 2v will be different than 50% @ 5v.

I'm using 4v pre-outs

 

[starkfu 10+ year member]

amps work a little different, even with no incoming signal you are still going to **** at least a few amps of current, and the draw isn't linear. The only way to know for sure is to measure the load with a load tester.

 

[Decipha 10+ year member]

general rule of thumb

power x amperage = wattage

14.4 x 110 = 1584, but there is lots of factors that come in to play such as effeciency, wiring, etc...

 

[Radioflyer97 10+ year member]

general rule of thumb
power x amperage = wattage

14.4 x 110 = 1584, but there is lots of factors that come in to play such as effeciency, wiring, etc...

This formula would give me my peak output (assume 100% eff) . How would I calculate my regular output?

The amps being used are:

(2) Alpine MRP F250

(1) JL 500/1

 

[Flipx99 5,000+ posts]

general rule of thumb
power x amperage = wattage

14.4 x 110 = 1584, but there is lots of factors that come in to play such as effeciency, wiring, etc...

I use this as it provides a better margin of safety....

I am guarenteed to overkill.

 

[kross 10+ year member]

The total fuse ratings on all 3 amps are 110 amps. If i set all the amps gains to 50% i should draw no more than 55 amps at any time

This is incorrect. An amp can produce maximum output with the gain at almost any position, depending on the source signal. The gain control is used to match the source signal to the amplifier input stage. Because an amplifier has to be able to handle head units (and other sources) that have a maximum output of anywhere from 1v to 9v, the amp has to be able to produce maximum output from any of those source signals. That is what the gain control is for, and is why setting the gain properly is critical.

 

[brynm 10+ year member]

I use this as it provides a better margin of safety....
I am guarenteed to overkill.

I disagree an amp that is only 80% efficient will use more amps than your formula. amps= (power in watts/volts) for a 100% efficient amp.

For an 80% efficient amp it would be amps=(power in watts/volts)/ %Efficiency

 

[Radioflyer97 10+ year member]

I disagree an amp that is only 80% efficient will use more amps than your formula. amps= (power in watts/volts) for a 100% efficient amp.
For an 80% efficient amp it would be amps=(power in watts/volts)/ %Efficiency

How/where do you find this effeciency rating?

 

[Flipx99 5,000+ posts]

I disagree an amp that is only 80% efficient will use more amps than your formula. amps= (power in watts/volts) for a 100% efficient amp.
For an 80% efficient amp it would be amps=(power in watts/volts)/ %Efficiency

I did not calculate it that way. I always thought it was * % effeciency.