Speaker Cabinets

  • Thread starter Thread starter DrResponsible
  • Start date Start date
D

DrResponsible

New member
So I'm building speaker cabinets for my guitar amp. My ideal set up is two 10" speakers and one 15" speaker. The idea was to make them into two cabinets (one 2x10 and one 1x15) that I can stack onto each other, or for a smaller gig I could just bring one or the other. What I have are two 16 ohm 10" speakers, which I basically got for free, and that cab is almost done. Still need to get my 15" speaker, but I've got the box built for it, and I still need to get the jacks and wire for both cabs.
The idea is that I would wire my two 10" in parallel, making that an 8 ohm cab, and then get an 8 ohm speaker for the other cab. Btw, my amp has settings for 8 or 16 ohm. From there I wanted to wire it so that either cab on it's own is 8 ohm and have them connect to one another in series so when stacked the impedance is 16 ohm.
I've tried finding some info how how to do this, but haven't had any luck.
Is this possible?

Thanks.
 
It's possible, but without some kind of fancy 1/4" jack that somehow completes the circuit by itself when nothing is in it, but can be interrupted to run a connected cab in series when a cable's plugged into it, you may have to wire 3 jacks into one of them. Basically wire 2 independent circuits into one of the cabs, one that gives the 8 ohm load, and one that doesn't give a complete circuit until the 3rd jack is connected to the other cab. You could then wire one of the leads into a switch so it wouldn't be possible to mess up and connect both circuits at once. Look at how insert points in mixers work to see an example of a circuit that is complete by itself but can be interrupted to put something in series just by inserting a plug into a jack. That would be nicer, and only require 2 jacks, if you can find such a piece of hardware somewhere.

Good luck :D
 
DrResponsible said:
I wanted to wire it so that either cab on it's own is 8 ohm and have them connect to one another in series so when stacked the impedance is 16 ohm.
.
Click to expand...

My understanding is that if you run two 8 ohm cabs in series, you end up with a 4 ohm load at the amp, which yours won't do. If you have two 8 ohm outputs on the back of the amp, then you can make both cabs 8 ohm and run a cable to each cab. If I were you and I knew what I was doing, I'd make each cab 16 ohm. This way you can easily use each one alone as a 16 ohm mono cab with your amp set to 16 ohms, or run them together as an 8 ohm load on the amp. Easy as pie.

I've got a 16 ohm 2x12 with a dual jackplate on the back, and a 16 ohm 4x12. When using either by themselves, I run the amp at 16 ohms. When I run them together, I set the amp to 8 ohm, and run a cable from the amp > 2x12 > 4x12.

16 ohm cab + 16 ohm cab = 8 ohm load.
 
Two 8 ohm loads in series yields 16 ohms. You're thinking of parallel, Greg - that would yield a 4 ohm load. Also, there's no way to wire the two 16 ohm speakers that he has to make a 16 ohm load. If he puts them in parallel, they'll yield an 8 ohm load, and if he put them in series, it would be 32.

16 ohm cab + 16 ohm cab (wired in parallel) = 8 ohm load.
The way your cabs are wired is normal. Usually the secondary jack is wired in parallel, and that's not tricky to do inside. What he wants to do is not normal, though. The tricky part is that the circuit has to be able to be interrupted and the other cab's circuit sort of... inserted into the middle (in series), rather than just tacked on (in parallel) like most cabinets provide the jacks for.
 
Greg_L said:
My understanding is that if you run two 8 ohm cabs in series, you end up with a 4 ohm load at the amp
Click to expand...

Minor correction: You end up with a 16ohm load if the speakers are in series. In parallel, you get a 4 ohm load.

To get a series connection you would make this circuit:

Amp + to CabA +
CabA - to CabB +
CabB - Amp -

I think you could do this in a break-out box. Use connectors of your choice - maybe SpeakONs. Connector A receives amplifier + and -. Connectors B and C connect externally to one cab each. Internally, Wire Connector A + to Connector B +. Wire Connector B - to Connector C +. Wire Connector C - to Connector A -. If the connectors are insulated you could use a metal box for durability and maybe some shielding. Obviously connect a cab to Connector B and a cab to Connector C. You will need 3 cables.

That way, the cabs need no further modification.

Check this over to see if it makes sense. I have not made this, and accept no responsibility for house fires, singed eyebrows. etc.

Paul
 
I like PRHunt's idea better than either of mine. I didn't like the idea of using some flaky interruptible jack like an insert point from the getgo (that thing's going to wear out and stop making contact when there's no jack in it eventually), nor do I like the idea of providing both a parallel and a series circuit (through the 3 jack-setup I suggested), because it's inevitable that somebody will hook those up wrong and something crazy will happen. With a little box like he described, though, you'd only need 1 normal jack on each cab, and the circuit would be done in series inside that box.

Nice one, PRHunt :)
 
Ha! Maybe one of my Guinness ("pure genius") moments :drunk:

I think the main thing would be to use an appropriate wire gauge to make the internal connection and ensure that solder joints are clean and well made.

The box should be reasonably fool-proof as long as the connections are clearly labeled.

Paul
 
I agree with using Speakons but you can do this in the cab. Use the 2 pole Speakons (or just 1/4" if you must) for the regular 8 ohm connection, but use 4 pole Speakons for the series connections. 4 poles are not that common so the odds of somebody effing up that connection is low, and the 4 poles won't mate with the 2 poles.

4 pole cab A:

1 speak +
2 speak -
3 NC
4 NC

cab B:

1 NC
2 speak +
3 speak -
4 NC

Amp:

1 amp +
2 NC
3 amp -
4 NC

Then you get 2 4-pole speaker cables wired up and it doesn't matter the order you connect the cabs.

Now if you use the parallel connection at the same time it's your own stupid fault :p
 
Hey thanks for the advice guys. I had a thought of using two 1/4" jacks on the first cab and a switch.
Not really sure if it would work, but I came up with this. (sorry for the crude diagram, I'm pretty new to this)

cabdiagram.webp

Would this do what I want it to? Basically I'm thinking have the switch clearly labeled to avoid mistakes. It seems to me that if the switch was in the wrong position in single cab set up the circuit would not complete so it just wouldn't work, and if it was in the wrong position in the two cab set up the second cab would have an incomplete circuit and it wouldn't work, but the first still would. Is this correct? Then I would just have to make sure my amp is set to the correct impedance, which I would need to do anyway.

If this wouldn't work I will probably go with the break-out box idea. I just wanted to keep it as simple as possible and avoid getting extra materials and having another box to carry around with me.

Thanks again!
 
You must log in or register to reply here.
Back
Top