Strikes me that what you’re looking for is a decent mixer – most engineer mixers (as opposed to DJ ones) will have at the very minimum a set of auxilliary bus outputs that you can use to feed signal to amplifiers, and probably a set of subgroup faders with corresponding subgroup outs – giving you control over the levels being sent to the amps…
You can also generally daisychain amps together – most amps have 2 sockets marked as inputs on each channel – basically giving you the ability to parallel feed them all from one input (so your signal feed goes into the first pair of sockets, and the second pair of sockets are connected to the next amps inputs, and so on down the line…) Because they are connected in parallel, the voltage across the whole chain of inputs doesn’t change – the full voltage is available to all of them.
As for the load impedance – thats a bit more complex, and involves having the values for both the source Z (impedance on the output of the device providing the signal to the amplifiers which can be anything from 50 Ohms to 2K Ohms), and the input impedance on the amplifiers (typically 20K Ohms). In order to transfer the signal without introducing major deviations in level and frequency response, the Source Z to Load Z should have a ratio of at least 10:1, so the combined input impedance of all the amplifiers has to be equal to or greater than 10X the impedance of the source device…
So SourceZ X 10 <= loadZ (the total impedance of all amp inputs).
If your amplifiers have differing input impedances you need to find the average for them – so add the impedance values and then divide by the number of amps (so if you have 1 amp @ 18,000ohm, 1 @ 15,000ohm and 2 at 20,000 you would do (18000+15000+20000+20000)/4 = 18,250, and 18,250/4 = 4562.5 – this would be your loadZ and the value that the sourceZ X 10 mustn’t exceed. Most amps are 20Kohm though, so usually you won’t need to bother doing this, and would just use 20,000/number of amps as your loadZ.
So, to find the number of amps that can be paralleled from a signal, you divide the average amp inputZ (or loadZ X number of amps) by the sourceZ and then divide by 10
Example: Your mixer output has an impedance of 330ohms (the sourceZ), and your average amplifier input impedance is calculated at 20,000ohms.
So 20,000/330 = 60.6060 and then 60.6060/10 = 6.06 (which rounding down to the nearest whole because you can’t get .06 of a power amp… In this situation you could parallel 6 amplifiers to that source without running into problems – in practice it’s better to err on the side of caution, because this is a simplified calculation, so you would take 1 off to be safe and not connect more than 5 amplifiers).
Obviously this means that the lower the impedance on the signal source device, the more amplifiers can generally be driven from it…
Hope that makes some sort of sense… If not, Peavey US have a very good tech paper on impedances in audio systems which can be found here: Peavey.com :: Impedance in Audio Technology (A quite fascinating read if you’re into this sort of thing :crazy_fre: )
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