Ampeg 810 8ohm Wiring?

metro_area · #1 04-20-2011, 12:26 PM

I've had an old Ampeg Squareback 810 cab sitting around for a good year that I pickup up for $50. Yesterday, a friend of mine who works at a guitar shop clued me into 8 brand new 10" speakers for $90, which I jumped on. I'd like to throw these into the empty 810, regardless of tone. I've got other cabs that I play out with, and thought for $140 this would at least make a decent (albeit huge) practice cab.

The speakers are 8ohm, with a max of 250W. 40 oz magnets, 2.0" voice coil. The vintage 810 just has one single 1/4" jack in the back. Can someone get me a schematic on how to wire this thing up? My SVTs run at either 2 or 4 ohms, so I'm hoping 8 8ohm speakers can be wired to a total of 2 or 4 ohms (I don't know how that stuff works).

Is this possible? Any help would be amazing!

*Side note: This will just be a practice cab for now. Not worried about how much better the tone would be with re-coned CTS speakers, etc, etc, haha!

Downunderwonder · #2 04-20-2011, 12:41 PM

series two and parallel them with another two
do same with the other set of four
run the two sets parallel,

4 ohms total.

Downunderwonder · #3 04-20-2011, 12:45 PM

actually more sense to say parallel 4 series pairs, each series being 16 ohm

coreyfyfe · #4 04-20-2011, 12:55 PM

Exactly. Or to put it in the simplest terms possible, from the jack connect positive to positive down the left column, negative to negative down the right column, and connect the left side negative to the right side positive in each row.

metro_area · #5 04-20-2011, 01:42 PM

Thanks so much guys! coreyfyfe, when you say left/right; is that looking at the back of the cones, as if looking through the back of the cab? Or is that looking at the front of the cab and cones?

Any chance one of you life-changing, helpful individuals could draw this out for me?

Downunderwonder · #6 04-20-2011, 11:11 PM

+ TIP SOCKET -SLEAVE
I__ +O- ____+O-__I
I I
I__ +O- ____+O-__I
I I
I__ +O- ____+O-__I
I I
I__ +O- ____+O-__I

IIRC it's tip to +, if not I've got you wiring it backwards which doesn't matter until the next guy tries to run two cabs and they are out of phase.

btw each O is a speaker with a + and a -

Downunderwonder · #7 04-20-2011, 11:16 PM

Draw yourself a line throught the left column of "I" and imagine that the right ones all ine up, my keyboard drawing skills not great.

walterw · #8 04-20-2011, 11:23 PM

point being you can only get 4Ω (or 16Ω) out of 8 8Ω speakers. (yeah, i know, but there isn't much call for a 64Ω cabinet

)

you're basically making two 8Ω 4x10 cabs, then wiring them in parallel for 4Ω.

Plstrns · #9 04-20-2011, 11:50 PM

The easy way

for me anyway.

Name your 8 speakers 1-8 first. Right side 4, or left side 4, upper, lower it doesn't matter. Pairs are 1-2, 3-4, etc...

+ Jack to speakers 1357 +
- Jack to speakers 2468 -
Then, wire each pairs' odd speaker's negative to each pairs' even speaker's positive. IE- 1- to 2+, 3- to 4+ etc...

That makes 4 sets of 16 ohm pairs, in parallel for 4 ohms.

metro_area · #10 05-30-2011, 01:57 PM

Quick update: This worked out perfectly! Thanks so much for all your help!

PhiDeck · #11 05-30-2011, 03:28 PM

There are four topologies that will yield 4 Ohms, one of which is asymmetrical and makes no sense. When all eight speakers are working all topologies will behave identically.
However, differences arise when a speaker fails. Which of the three useful topologies is optimal depends on:
- probabilities that a speaker will fail open versus shorted
- minimum and maximum load impedance that your amp can handle
- how much output (SPL) you are willing to lose when a speaker fails
- amount of additional power that each speaker can safely take