I am building a cheap line array, and am trying to figure out how to wire 8, 8 ohm drivers so I get 8 ohms. I know the difference between parallel and series, but I don't know the best way to get what I want. With 4 drivers, I know I could bridge 2 sets of series connections and get 8 ohms, but how do I do this with 8 drivers? Thanks!
This link has helped me in the past with big projects. http://www.colomar.com/Shavano/impedance_info.html Hope that helps!! Shawn Keeler
You cannot get 8 ohms. You have 2 choices. 1) add a ninth driver. Wire 3 sets of 3 in series, then wire these 3 sets in parallel. That would look like this in the calculator link : 8-8-8 8-8-8 8-8-8 I know from your earlier post that you are already pinching on driver spacing with 8, but you can have the ninth driver fire up or fire to one side or even to the rear. 2) You can stay with 8 drivers by wiring 2 sets of three in series and the last 2 in series. Then wire these 3 sets in parallel. You will end up with a 6.8 ohm load which almost any amp could handle. 8-8-8 8-8 8-8-8 The drivers in the pair will each get about 30% more power, which would make them a weak link (only relevant if your goal is that they can play very loud). With this configuration, I would place the the 2 odd dirivers in the center with the sets of 3 above and below. Its hypothetically possible to get eight 8 ohm drivers to a net 8 ohm load including resisters, but you don't want to do that- just waste power.
Brian - 9 would work! However, I am scaling back my plans and am considering doing 4. Thanks everyone!!!
