splitters for two amps

stevdart wrote: jon02accord,  I hear that often but it's not correct.  You are talking about finding Power, which in this case is watts from the speakers.  The voltage you used is the car's electrical system voltage, where to find power you would need to use the voltage output of the amplifier. The job of an amplifier is to make voltage, after all. The fuses on an amp are there to limit incoming amperage, and must not be confused with the amperage that the amplifier produces.   The input to the amp is low voltage, high current.  The output is high voltage, low current.  Since the amplifier makes volts, and P = E X I, then the power would come from the amp's output voltage times the current it produces which is determined by the load. An example:  a USacoustics USB600D mono class D amp has a fuse of 20 amps.  It outputs 200 watts into 4 ohms, and 350 watts into 2 ohms.  By your reckoning, the maximum power output of this amp would be 288 watts, using your formula of 14.4 volts X 20 amps.  Your guess is way off track.  I have measured this particular amp and know its rating is accurate, but you can use any known amplifier and find the error in your reasoning. Just for info:  Class D amps are much more efficient than class A/B amps and so would need less amperage in the fuses.  Their efficiency allows them to make more voltage with less amperage input, so the fuse rating can't be used to tell much.  And a class A amp, being even less efficient than A/B, will have a larger fuse rating than either of the above...for the same power output.  This makes you realize that the amp's fuse rating has little to do with the power a speaker will ultimately get.

Stevdart, you have usually been pretty right-on with most of your posts, but I gotta say this time you are indeed OFF!!! He was not confusing the input current rating with the output current capability. An amplifier can only CONVERT energy, not CREATE it. If your fuses limit your INPUT power to 288 watts, YOU CANNOT GET MORE THAN THAT OUT OF IT!!! (...on a continuous basis, anyway - read: 1000 watts, one channel driven, 4 ohm resistive load, 1kHz tone, 10ms, 50% THD - see the manufacturers data sheet on any Pyrabossenhag amplifier.) Additionally, you are limited by physics and the laws of thermodynamics to a given efficiency - how much is blown off as wasted, unusable heat? A standard class AB amplifer can reach a theoretical maximum of 66%, meaning for the power put in, 33% WILL GO AWAY AS HEAT! A 60% efficiency rating is more than accurate today. (Pure class A amplifers fall around 25% efficient, and most digital amps land around 90-94%.) In regards to digital amplifiers, just because they are more efficient, it still does not mean they can EVER break one hundred percent (put out more than is put in) I appreciate the fact that you may have measured that amp, but if you do the math yourself, then 288/200=69.4% efficiency. Pretty low for a decent digital amp, loaded with a 4 ohm driver, but still PERFECTLY BELIEVEABLE as an output rating into 4 ohms. As far as the 350 watts into 2 ohms, it is probable that the amp can indeed do it, for short periods of time, but on a CONTINUOUS, SINE WAVE BASIS, it cannot - it is simply impossible. Likely, even if we give the same amp the benefit of the doubt, and call it 95% efficient into 2 ohms, you will still only be able to get 288*.95=273.6 watts CONTINUOUS. The 350 watt rating into 2 ohms is only a 28% over-rating in regards to the input capacity, and ANY amplifier worth it's salt (meaning, built with a respectable power supply) will be able to maintain a 28% peak/continuous output ratio - that is called headroom - in this case, about 1dB of headroom. The POWER a speaker sees delivered to it is a function of voltage times current - you know that. A power amplifier is a voltage source, meaning it will try to make current as long as it can to see that the load always gets the same voltage. Translation: Lower impedance (2 ohms vice 4 ohms) higher current output. THIS is where power is made. The voltage output of a typical amplifier will always try to remain the same, regardless of the load placed across the terminals. The power supply simply reacts by making more current. (I really forgot where I was going with this...........oh, yeah) This is why you see few amplifiers that actually double their power outputs all the way down to their MAXIMUM LOADS. 4 ohms 100 watts SHOULD BE - 2 ohms 200 watts, right? Ohm's Law says "YES". Why are so many amplifiers rated 150 or 175 watts at 2 ohms? (or if they are rated as such, they likely won't be rated 400 watts into 1 ohm) Because their power supply is not "stiff" enough to make the current necessary to PROVIDE 200 watts at that impedance. I'm on jon02accord's side - He's right. Those amps are WAAAYYY overrated.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 think I understand that now, haemphyst.  Thanks for the heads up.  The amp I mentioned claims >90% efficiency, so if I say:  14 volts (car) * 20 amp (fuse) = 280, * 90% (efficiency) = 252 watts.  That's below the rated power output of 300 watts into 2 ohms.  The fuse is the tell-all that lets me know that the amp will never be able to get to that level in RMS watts and still be within the 0.3% THD rating.   Although 0.3% is the number for "rated power" which I guess is only applicable to the 4 ohm load rating.Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.