Alright, I plan to make a 4x12 guitar cabinet. Before I start I want to lay out all my plans. What I need to know is how to wire the cabinet so i have: a mono (8 ohm) input, individual left and right inputs (at 4ohms each) and an output.
To get a better Idea of what I want, look at this MESA jackplate.
To get a better Idea of what I want, look at this MESA jackplate.
An externally hosted image should be here but it was not working when we last tested it.
I am pretty sure this would be correct, any one want to back me up on this.
check out this link it should help also
http://colomar.com/Shavano/spkr_wiring.html
good luck
check out this link it should help also
http://colomar.com/Shavano/spkr_wiring.html
good luck
Attachments
Hey CG; One caution. If I'm reading that diagram correctly, and if you use grounded jacks, and if use a metal mounting plate . . . you'll have a dead short from hot to ground.
Let me study it some more, but I wanted to warn you before you fire anything up. Literally. 😀
Let me study it some more, but I wanted to warn you before you fire anything up. Literally. 😀
Note the path I've highlighted in green. If those two jacks share a chassis ground, it could be a problem.
An externally hosted image should be here but it was not working when we last tested it.
Hi,
that green path is a problem. The diagram is not correct.
What you need to do is assume that they are switching jacks.
That is each jack has four connections, with no plug inserted + joins +,
- joins -, with a plug inserted only the input + and - connect to the lead.
So you do not wire it as a normal parallel/series 4x12 for the 8ohm
jack, as shown. The daisy chaining output socket is fairly obvious.
The two 4 ohm jacks are wired into each series leg, and with no plugs
inserted they provide the series connection represented by the green
cross connection. This path should be broken by either 4 ohm plug.
Make sure you get all your phasing correct.
The switched tip of one 4 ohm socket connects to the switched ring of the other.
The other two switched connections remove the connection to the 8 ohm socket.
🙂/sreten.
that green path is a problem. The diagram is not correct.
What you need to do is assume that they are switching jacks.
That is each jack has four connections, with no plug inserted + joins +,
- joins -, with a plug inserted only the input + and - connect to the lead.
So you do not wire it as a normal parallel/series 4x12 for the 8ohm
jack, as shown. The daisy chaining output socket is fairly obvious.
The two 4 ohm jacks are wired into each series leg, and with no plugs
inserted they provide the series connection represented by the green
cross connection. This path should be broken by either 4 ohm plug.
Make sure you get all your phasing correct.
The switched tip of one 4 ohm socket connects to the switched ring of the other.
The other two switched connections remove the connection to the 8 ohm socket.
🙂/sreten.
An externally hosted image should be here but it was not working when we last tested it.
a little confused
Thanks for the help, are you sure it won't work? I'm not really understanding what you're trying to tell me. Do you think you could draw up a simple diagram for me to follow!😕
Thanks for the help, are you sure it won't work? I'm not really understanding what you're trying to tell me. Do you think you could draw up a simple diagram for me to follow!😕
Hi,
nearly there except your 8 ohm input is 4ohms and one side.
Consider :
sockets are labelled :
AB EF IJ
CD GH KL
The last socket is turned upside down so :
B and D are switched, not connected with plug inserted.
F and H are switched, not used, could be a different type of socket.
I and K are switched, not connected with plug inserted.
A= left pair + (tip)
C= left pair -
J= right pair + (tip)
L= right pair -
B and E are connected together.
D and I are connected together.
G and K are connected together.
Plug in 8 ohm gives both sides in series and in phase and same
phase as 4 ohm. Plug into either 4 ohm disconnects 8 ohm.
Its not all common ground, but is amplifier safer if leads are wired wrong.
🙂/sreten.
Due to multiple cabinets always make sure absolute phasing is
correct. Tip is always positive. + voltage on tip = driver forward.
A battery connected to a jack plug make this test easy.
nearly there except your 8 ohm input is 4ohms and one side.
Consider :
sockets are labelled :
AB EF IJ
CD GH KL
The last socket is turned upside down so :
B and D are switched, not connected with plug inserted.
F and H are switched, not used, could be a different type of socket.
I and K are switched, not connected with plug inserted.
A= left pair + (tip)
C= left pair -
J= right pair + (tip)
L= right pair -
B and E are connected together.
D and I are connected together.
G and K are connected together.
Plug in 8 ohm gives both sides in series and in phase and same
phase as 4 ohm. Plug into either 4 ohm disconnects 8 ohm.
Its not all common ground, but is amplifier safer if leads are wired wrong.
🙂/sreten.
Due to multiple cabinets always make sure absolute phasing is
correct. Tip is always positive. + voltage on tip = driver forward.
A battery connected to a jack plug make this test easy.
disregard that last post, i think i understand know.
So basically this wiring will give me 4 ohms left and 4 ohms right. OR 8 ohms combined? Is this correct?
So basically this wiring will give me 4 ohms left and 4 ohms right. OR 8 ohms combined? Is this correct?
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