testing amp wattage

So I couldn't find this in the forums and I'm surprised, sorry if I sound dumb... I am curious how much power my amps are putting out so should I just measure the current going into the speaker wire (for a mono) using a DMM and multiply that by my battery/alternator voltage or is there a better way? Thanks

This is a crap shoot, but easier. Amp fuse times voltage, so 40amp x14 volts. 40A X 14V=560 Watts at 100% efficiency, then multiply that by an estimated efficiency of your amp, say 85% efficient, so take 480 Watts x 85%, or .85=476 Watts And RMS is 70.7% of the peak wattage so 476 Watts x .707= ~336 Watts RMS.2003 Chevy Avalanche,Eclipse CD7000,Morel Elate 5,Adire Extremis,Alpine PDX-4.150, 15" TC-3000, 2 Alpine PDX-1.1000, 470Amp HO Alt.

I know how to find the theoretical wattage of my amp, but I'm looking to see exactly what it's putting out. Just like you could say your battery puts out 12V because it says 12V on it, but maybe it actually puts out 11.6V due to imperfections in the manufacturing process. Sorry if I wasn't clear earlier.

Without an oscilloscope to actually SEE the onset of clipping in the waveform, you can't. You must know the EXACT output rail voltages to determine actual power, as well as the EXACT impedance the amplifier is seeing at the clipped frequency. As you know, loudspeakers are dynamic impedance devices, they will have minimum and maximum impedance peaks throughout their usable bandwidth. This will affect power output at each frequency. I suppose you COULD use a shunt resistor, (this would be able to give you the current being delivered to the speaker) and a second (storage) 'scope, and plug the numbers in that way, but FAR more expensive. If you know two of the three specifications, (voltage, resistance, and/or current) you can always figure POWER. As a VERY CLOSE approximation, you could measure the DC rails WITHIN the amplifier, but this will involve opening the amplifier, and voiding any warranty you might still have. Using Ohm's Law you can plug this number in, and determine how much power the amplifier CAN make into a RESISTOR, not a loudspeaker. If you do it this way, your numbers will be a LITTLE bit high, because the transistors will have a small drop across them, affecting the REAL output, but it'll be close enough. Why the need to know so exactly?It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."

I'm just curious because it's an Alpine PDX 1.600 I got off ebay and I'm wondering if there's a reason ebay is so much cheaper for the "same" stuff (i.e. broken/defective, counterfeit, maybe didn't quite meet alpine's standards to sell in stores). The reason I don't want to go by what fuses I have and what Alpine says the efficiency should be is that if someone slapped a 100A fuse in a $10 counterfeit amp and said it puts out 100A*12V*(50% efficiency)=600W rms, while it probably doesn't do that in real life. I was just thinking if I could measure the current coming out of the amp into the speaker I could figure out the power using the voltage the amp should be running at (battery/alternator voltage). It sounds like that won't work from your post haemphyst; why is that?

Because the numbers regarding input vs. output calculations are suggestions. They are in NO way concrete, as EVERY manufacturer's efficiency will be a little bit different, based on topology, output devices chosen, power supply driver transistors, quality of the traces or bussbars internally, how MUCH Class A (in percentage) will the amplifier run... ALL of those things will have an effect on it's output - the "theoretical" output. If mfr A says this is a 500 watt amplifier, but it is only 50% efficient, it will have +/- X rail voltage, right? This rail voltage will be limited by the input fuse. If mfr B also makes a 500 watt amplifier, but it it 60% efficient, the output rails will be the same, (they HAVE to be, right) but the input fuse can be smaller because more power is tranferred to the output buss. It's just that mfr A chose a topology slightly less efficient than mfr B - maybe a different switching frequency, maybe a smaller toroid, larger filter caps... MANY things can affect efficiency, and since nobody advertises efficiency numbers, you'll have to use a number "that's close enough for government work", right? Since REAL output is based ON THE RAIL VOLTAGE AVAILABLE, you have to know that voltage, right? Meters won't get it, because you can be into a 10 or 15% clipped situation, before the meter even registers a difference in the voltage.It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."

sedate wrote: As haemphyst so polysyllabically illuminated for us, the actual useable power the amplifier is producing is highly dependent on all of these things that go into the design of the amplifier so there really isn't a way of calculating some reference efficiency with any real certainty from those numbers either.

Polysyllabically? {ROFL} I never knew I was a polysyllabic!It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."