[size=4][font=Arial, sans-serif]Hi all, [/font]
[font=Arial, sans-serif]So I think I've got most of this and apologies if this has been answered already but there's one piece that's not quite clear/I haven't found the answer yet on this thread/the web.[/font]
[font=Arial, sans-serif]My setup (picked up through the BC for-sale boards) is as follows:[/font]
[font=Arial, sans-serif][url="http://usa.yamaha.com/products/musical-instruments/guitars-basses/amps/bbt500h/"]Yamaha BBT500H head[/url], 500W at 2ohm, 250W at 4ohm, class D amp.[/font]
[font=Arial, sans-serif]Cab 1 is a matched Yamaha BBT210S (250W handling at 4ohm). So far so good and one happy customer.[/font]
[font=Arial, sans-serif]I'd like to add another cab for bigger venues and the obvious thing is to keep it simple and get a 250W (or greater) 4ohm cab (so 500W total at 2ohm = happy days). However let’s say I want to [hypothetically] use an 8ohm 250W cab instead...[/font]
[font=Arial, sans-serif]The overall impedance is going to be 2.667ohms (still fine for the class D head).[/font]
[font=Arial, sans-serif]I understand the power will be split unevenly between the two cabs with 2/3 going into the 4ohm and 1/3 going into the 8ohm cab.[/font]
[font=Arial, sans-serif][u]My question is[/u]: is this one thirds/two thirds power splitting based on the max head power (i.e. 500W, which would put 160W into the 8ohm and be fine but 333W into the 4ohm cab and ultimately shred it subject to “Alex’s first rule”) [u]OR[/u] is this power splitting based on the power at the overall impedance (i.e. 500 * 2/2.667 = 375W) in which case the power nominally going into each cab is essentially fine (250W into the 4ohm and 125W into the 8ohm)?[/font]
[font=Arial, sans-serif]Thanks![/font][/size]