Need help - wiring line array to keep 8 ohms (1 Viewer)

[Kenneth Harden]

I just bought eight of these drivers:

http://www.partsexpress.com/pe/showd...number=269-570

While my first game will be to just try out one per channel for fun, I eventually want to play Dr. Amar Bose and see what I can do. These drivers are 8 ohms. I know with an array of drivers, you can wire them to either increase or decrease in impedance for the set. Would I make 2, 16 ohm pairs, then 'merge' them into a 8 ohm pair? How would I go about accomplishing this?

Thanks! I just want to have some fun and play Bose

 

[JustinSC]

Resistance in series:

R = R1 + R2 + ...

Resistance in parallel:

1/R = 1/R1 + 1/R2 + ...

So you can wire two speakers you get a 16 ohm load. Do this with another set you have the other 16 ohm load. Then you parallel this configuration to get an 8 ohm load.

I hope that helps. You should be able to find enough pictures out there that explain how to wire in series and parallel

 

[Scott Sabin]

Kenneth,

If you want to make a pair of speakers utilizing those drivers, with each speaker rated at 8 ohms, you need to wire each speaker in the following manner:
numbers refer to driver number, R(red, or plus) and B(black, or minus) refer to polarity of voice coil:


Speaker R----------1R----------3R
Speaker B----------2B----------4B
1B-----------2R
3B-----------4R

This is equivalent to:

(+)-----------
| |
D1 D3
| |
D2 D4
| |
(-)-----------

 

[Kenneth Harden]

I THINK I get it.

Lets say I take a speaker and put a volt-ohm meter on it, and lets say it measures...6.3 ohms (I know it is never 8)

After I get everything wired, can I stick the meter on there and expect to see the same impedance as I did with the single driver if it is wired correctly? Checking my work, if you will.

Thanks!

 

[Scott Sabin]

Yes, although you will find that if the drivers vary, the overall will probably not be exactly the same value as any one of them.

Basically, wire 2 drivers in series - the (-) of one drive to the (+) of the other. Now you'll have an unconnected (+) on one driver and an unconnected (-) on the driver. This makes a series pair, and the resistance of the pair is the sum the of the resistance of each individually. Repeast this series connection for two different drivers.

Now you have two "sets" of drivers, with each set being a series pair. Each set has a (+) connection and a (-) connection. Connect the two sets together by connecting the (+) to (+) and (-) to (-). This completes the speaker. The amplifier (+) will connect to either driver's (+) (since they are tied together anyways, it doesn't matter which one), and the same goes with the minus. The overall speaker resistance should be the same as 1 driver if all the drivers measure the same. Measure the overall resistance by measuring between (+) and (-) as the amplifier would see it.

If the drivers measure differently, individually, you can still check your conections. Say R1, R2, R3, and R4 represent the resistances of the 4 drivers, and R1 and R2 are used to make one series set, and R3 and R4 make up the other set. The total resistance seen by the amp is:
(R1+R2)*(R3+R4)/(R1+R2+R3+R4). The math isn't too tough if you measure each driver individually. You can see that if each measures 6.3, that the total R value is (6.3+6.3)*(6.3+6.3)/(6.3*4) = 12.6*12.6/(12.6+12.6) = 6.3.

Note that to your amplifier, it sees more than the dc resistance as it also sees the inductance. The inductive load seen by the amp if a function of frequency, but probably averages somewhere near 1.7 Ohms, making the driver an 8 ohm (ish) total. For checking wiring, you only need to worry about the dc resistance. But I wanted to explain why a true 8 Ohm speaker might measure as 6.3 Ohms on a simple ohmmeter.