Amp draw to output conversion

[GatiBeats]

I've seen the formula voltage x amp draw = output (watts). Take a sundown NS-1 for example. If the maximum rms output is 6500, 6500/14.4v= ~451a. However, at full tilt, let's assume the voltage drops to 13v. 6500/13v= 500a. Does the amp draw increase as voltage decreases? This doesn't make sense to me. Can someone explain where I went wrong?

[SnowDrifter]

You are forgetting the efficiency of the amp. Plus the fuses don't necessarily blow the second you pull more power through them.

And the NS-1 won't do the same power at 13v as it will at 14.4v. It will do less power

[P4killer_]

well like you said A x V = total watts aka I iirc.

so if one of the variables drops, then the other will have to increase to keep both sides of the equation equal.. obvi.

normally voltage doesnt go up when your blasting, obvi. so as the voltage drops, the current draw would have to increase to make the same amount of power..

youll need a DC clamp meter though to measure real time efficiency, how many amps coming in versus the output..

[Broke_Audio_Addict]

Yes the amp will have to draw more current to make the same power at a lower voltage which is bad news for the amp. It makes the amp work harder and can cause overheating.

[GatiBeats]

well like you said A x V = total watts aka I iirc.

so if one of the variables drops, then the other will have to increase to keep both sides of the equation equal.. obvi.

normally voltage doesnt go up when your blasting, obvi. so as the voltage drops, the current draw would have to increase to make the same amount of power..

youll need a DC clamp meter though to measure real time efficiency, how many amps coming in versus the output..

it just seemed odd to me that amp draw would increase with a drop of voltage. I think the efficiency is a larger factor than I thought tho

[Wicks]

You can't calculate input current draw by dividing the output power by the input voltage.

Your combining unrelated input and output numbers.

Your amplifier doesn't feed that ~14v that you're referencing to your speakers.

An amplifier is basically a variable frequency inverter.

It changes DC to AC.

Those output power calculations are MUCH more complicated then multiplying voltage and current because of the heavily inductive load of a speaker.

If you want a general idea of how much power an amplifier is capable of then multiply it's running input voltage times its fuse rating. Then multiply by its efficiency.

As far as how much power the speakers are receiving, you need to be able to measure the instantaneous voltage and current at the amp's output terminals and your average DMM and clamp meters can't do that.

[SnowDrifter]

Yes the amp will have to draw more current to make the same power at a lower voltage which is bad news for the amp. It makes the amp work harder and can cause overheating.

Only if it's an amp with a regulated power supply

[P4killer_]

Yes the amp will have to draw more current to make the same power at a lower voltage which is bad news for the amp. It makes the amp work harder and can cause overheating.

Only if it's an amp with a regulated power supply

Unregulated meaning there is no limit high or low on voltage the amp.puts out? I remember something like that but forgot how it works.

[SnowDrifter]

Yes the amp will have to draw more current to make the same power at a lower voltage which is bad news for the amp. It makes the amp work harder and can cause overheating.

Only if it's an amp with a regulated power supply

Unregulated meaning there is no limit high or low on voltage the amp.puts out? I remember something like that but forgot how it works.

More voltage in = more power. And dropping low doesn't hurt it - the amp just puts out less power.