Peter Murray
Well-known member
You can get a series cable but then you are still not in an optimal impedance range with 8ohm cabs. You would connect one 2x10 to each speaker output on the amp. Then you connect another 2x10 to the open jack on the 2x10 with a series cable. So:
(2x10 (in series) 2x10) in parallel with ((2x10 (in series) 2x10))
=
((8ohm + 8ohm) + (8ohm + 8ohm)) / 4
=
8 ohm load on amp
Of course if you had four 4ohm 2x10's then you could get a 4 ohm load this way.
This looks correct to me.
But your problem is that no matter how you slice it, you can't get more than 500W out of the Little Mark II, no many how many cabs you manage to string up.
An SD800 would give you 750W, so that's one option.
Probably given what you want to do, it would be best to get four 8 ohm 2x10s and two 500W heads. That would give you 1000W total and a lot of flexibility and portability.
But it wouldn't be cheap.
You could achieve the same result with two CMD 102P combos and two TRV 102P cabs.
Even if you did want to make something work with 4 ohm cabs, they won't be available in the US until 2008.
By the way, a lot of players start conceiving their Markbass rigs before getting much practical experience with the gear--once you play a few gigs you realize how loud the stuff is for its size, and how the clarity of the sound allows you to get away with less power. You well may find you don't need a 1000W Markbass rig.
Cheers,
Peter