Old SS Vox Head

[Mark7]

I have an old Vox Defiant head. Apparently they were originally sold with a three speaker 15 ohm cabinet.

Anyone care to hazard a guess at how the speakers were wired up?

Here's some info on it.

Note: Mine is the 100 watt Custom model.

[Imaduck]

Well, it's two speakers with a horn. Horns are generally used for high end, much like you would use tweeters. To separate the high end and the low end (and I should mention these are only general terms since most Celestions are already designed to accentuate the high end) there is a crossover in the cabinet. The crossover is set to a frequency and sends all signals above that frequency to the tweeter and anything below to the bassier speakers.

I hope that answers your question. If you're wondering about impedence I haven't the foggiest, though I'd imagine it'd be 4 or 8 ohms.

[Mark7]

Actually it's 15 ohms

[Outlaws]

Originally Posted by Mark7

Actually it's 15 ohms

Well, even most speakers that are labled 16ohms are actually more like 15.x +/- so there isn't really a true 4, 8, or 16 speaker in any guitar amp. But if the speakers are 4 ohms each, and wired in Parallel, then they would be providing 12 ohms of resistance to the amp. That wouldn't work all that well. But then back in the old days they might have used 6 ohm speakers that when placed in parallel would give 18 ohms of resistance to the amp that needs at least 15 ohms. That would work out fairly well. And since its solid state its less of a factor anyways.

But for your situation, two 8 ohm speakers in parallel (or two 16 ohm speakers in series) would work fine. But if you ever wanted to, you could even get three 8 ohm speakers and wire them in parallel and not have much problem.

[Imaduck]

You've got it backwards.

You'd want two 8 ohms in series or two 16's in parallell.

This is impedence, not capacitence.

(if only i could spell...)

Also, with a crossover, as I mentioned the original speakers used, the calculations are a bit different. (don't ask me the specifics, I kinda forget, and I'd have to look it up and stuff, which takes effort)

[AGCurry]

Originally Posted by Imaduck

You've got it backwards.

You'd want two 8 ohms in series or two 16's in parallell.

This is impedence, not capacitence.

(if only i could spell...)

Also, with a crossover, as I mentioned the original speakers used, the calculations are a bit different. (don't ask me the specifics, I kinda forget, and I'd have to look it up and stuff, which takes effort)

I'm afraid you're both a little confused.

Two 8 ohms drivers in series = 16
Two 16 ohm drivers in parallel = 8

If there is a true crossover, i.e., involving capacitance and inductance, the amp's power output is connected directly to it, so its impedance is what the amp will see. A variance in the drivers connected to the crossover will affect the crossover point.

[Outlaws]

Ya, I was a bit out of it when I typed that. lol