Legion Sound 8x10 re wire

I'm not to sure how the wiring on horns works, I'm sure someone will chime in on that. You'll need to wire the speakers as two groups. The top four speakers wire in parallel, and do the same for the bottom four. This gives you two groups with resistance of 2 ohms. You'll then wire the two groups in series to make it 4 ohms.

The two 'inputs' is actually an input and output which are wired in parallel to send power to another cabinet.

 

If you wire the two jacks in parallel, than it does not matter at all, which jack you use for input.

Is there a highpass for the Tweeter / horn? There should be one. If not you are allowed to wire the horn in parallel to one of the jacks.

Ciao

Uwe

 

If its a piezo tweeter you don't need a crossover and you just wire it directly to the speaker wires.

 

That would blow your amp unless it's designed for a 1 Ohm load.

Do this for each group of four speakers and wire the two sets parallel, before connecting the + and - to the jacks.
http://www.eminence.com/wp-content/uploads/2010/12/wiringseriesparallel4.gif

This configuration makes an 8 Ohm load for each group of four speakers.

If you use an amp that only handles an 8 Ohm load and has one output jack, DO NOT connect it to all of the speakers by using a Y cord- the 4 Ohm load won't make your amp happy. If it's a tube amp, it may have a separate 4 Ohm output, so this would be OK. If the amp has two 8 Ohm output jacks, connect one output jack to each input jack and DO NOT make any connection between the jacks on the cabinet.

What amp are you using?

FYI- if one jack is for input and one if for output, it's for sending signal to another cabinet, not power. Power is just a calculated value. Only connect it that way if you know your amp can handle the additional load.

Also, that "big, thick resistor" is probably a capacitor, used to filter the mids/bass from the signal going to the horn. Leave it in place- it helps a lot.

 

No- copy the diagram in the link. Each pair of speakers in series will show 16 Ohms. Wired parallel with the second pair, it drops to 8 Ohms. The second group of four speakers will also be 8 Ohms and wired parallel to the first group of four, the result will be 4 Ohms. Wiring in series, the "impedance" of each speaker is added. This applies to any speakers in the circuit and the impedance of each doesn't matter- you would calculate this by simply adding the numbers. You would wire two speakers in series by connecting the wire from the jack's + to the first speaker's + terminal and then, connect from the first speaker's - terminal to the second speaker's + terminal. Connect the second speaker's negative terminal to the jack's - terminal.

Parallel is trickier- The easy way to calculate the impedance of two speakers is by multiplying the two numbers and dividing that by the sum of the two numbers. If they're both 8 Ohms, you use (8 x 8)/(8+8), or 64/16, which comes to 4 Ohms. If one is 8 Ohms and the other is 4 Ohms, you'll have (8 x 4)/(8+4), or 32/12=2.67 and you may see this formula stated as "product divided by the sum" or "product over sum".

If you have doubts about getting this to work by yourself, you could go to a car stereo shop- they SHOULD be able to wire this easily- they work with series/parallel speaker loads all the time.

 

Here's a way to do this- I'll label the speakers, from top to bottom, L1 (top left), L2, L3, L4 (bottom left) and R1 (top right), R2, R3 and R4 (bottom right).

Put a piece of two conductor speaker wire on each speaker that's long enough to go to the jacks and cut another piece, to be used later. Connect the (-) of L1 to the (+) of L2 and the (-) of R1 to the (+) of R2. Connect the two remaining (+) wires together and do the same with the remaining two (-) wires. Repeat this with L3/L4 & R3/R4 and their remaining wires. Then, combine the paired (+) wires with one end of the spare piece's (+) wire and combine the paired (-) wires with one end of the spare piece's (-) wire. Connect this wire to the jack.

L1/L2 = 16 Ohms, R1/R2 = 16 Ohms and these pairs wired parallel = 8 Ohms (16 x 16)/(16+16)= 256/32=8 and again, wiring the two four speaker groups parallel will result in a 4 Ohm load.

Hopefully, this image will help- my drafting program is messed up, so I had to use Paint.
http://www3.picturepush.com/photo/a/13471971/640/13471971.png

 

Imagine the cab as a pair of 8ohm 410's. Wire each group of 4 in series/parallel, then wire the 2 groups in parallel, just as if you were stacking two 410's to make one 810.

If your part is in fact a resistor, then your HF horn is a piezo. Make sure the resistor is in line with the + wire going to the horn. If it's a capacitor, then your HF horn is a regular diaphram/coil driver. Make sure the capacitor is in line with the + wire going to the horn.

Actually, it doesn't even matter if you know what that part is, just put in line with the + wire going to the horn.

 

edit: removed due to incorrect diagram

 

I'm pretty sure that this will five you 16 ohms at the jack, not 4 ohms.

 

The hand-drawn diagram is two series-wired sets of four 8 ohm speakers in parallel to each other for a total impedance of 16 ohms. What you want is two parallel-wired sets of of four speakers and the sets in series to each other.

 

Oh crap! Good catch. I don't know why I was thinking they were 2 ohm speakers. I'll re do the diagram once I'm home!