Can someone please tell me if it's possible to configure (2) MTX 5500 8 ohm DVC and (2) MTX 5500 4 ohm svc to a Kicker IX1000.1 (2 ohm stable) amplifier?
Taking it to a shop that's done good work on previous installs but would like to be well informed before hand.
Taking it to a shop that's done good work on previous installs but would like to be well informed before hand.
Yes, if the two 4 ohm drivers are in series. If they are paralleled, the 2 ohm limit is reached and any other impedance placed in parallel with it will drop it below 2 ohms. In series (for 8 ohms) and placed parallel with the other two 8 ohms, the total impedance will be 2.7 ohms.
edit: ugh! 8 ohm DVC, that makes things trickier but it's still possible.
edit: ugh! 8 ohm DVC, that makes things trickier but it's still possible.
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um, is that an "8ohm" d4 configuration, or a d8 configuration? don't think they have a d8, so i'm going to go with the first. technically, you could wire that right up to 2 ohms, by series the subs to 8ohms, and then parallel them all, but the 4 ohm would be getting as much power as the pair of dvc subs. don't do it. find another configuration of subs.
If the "8 ohm" are dual 4 ohm voice coils, then my original configuration of 8||8||8 should work.
oops, i just saw there were (2)svc subs. in order to ballance the power, you must run the mis-matched loads in series. i have done this a few times before, and running odd loads in series is the only way to ever get a close ballance. so you would parallel a d4 to 2 ohms, then series with the 4 ohms to 6 ohms, and then repeat and series them to 3 ohms total... real world results. if it was really d8, then you could just treat them all like 4ohm subs and series-parallel to 4ohm final load. the 8/8/8 configuration is treating 2 drivers as 1, and the others 2 drivers as 2, so they would again, unbalance the load.
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