nagnuggets
New member
I have 3 3x12 speakers and was wandering how I would go about wiring them together to equal the same ohm's??? Has anyone ever just used 3 speakers in a cab????
bsp01
New member
I think the problem is that no matter how you wire it the numbers are wrong. 8 + 8 serial = 16 ohms, 8 + 8 + 8 serial = 24 ohms. 16 + 16 parallel = 8 ohms, 16 + 16 + 16 parallel = 5.3 ohms. You didn't specify what the ohm rating of the speakers in question are, I just picked 8 for an example.
Unless I've got the way that works wrong.
Unless I've got the way that works wrong.
bsp01
New member
These guys have pretty easy to follow explanation of how it works. http://www.colomar.com/Shavano/2x12wiring.html
technomancer
Well-known member
IIRC you wire 2 16 ohm speakers in parallel for 8 ohms then the pair is wired in series with a third 8 ohm speaker to yield 16 ohms.
Are these the old 3x12 Trace Elliot cabs? I've got one here I've been meaning to get re-tolexed
Are these the old 3x12 Trace Elliot cabs? I've got one here I've been meaning to get re-tolexed
bsp01
New member
technomancer":357eff8s said:
Cool, I wouldn't have even thought of doing that.
nagnuggets
New member
sorry 1- 8 ohm and 2-16 ohmbsp01":2yre58zx said:
dainbramage0
Active member
I run a 4x12 in a 3x12 config. I used 2 jacks and wire them independantly, sorta like 2 cabs in one. 8ohm EV12L to one jack, and 2x16ohm(8ohm) v30s to the other jack. Works really nicely with my Fortin modded D60 
I lifted the idea from the boys at Laboga, they offer a 4x12 in the same speaker combo with their mr hector heads.
I lifted the idea from the boys at Laboga, they offer a 4x12 in the same speaker combo with their mr hector heads.
mboogman
Active member
technomancer":wjzcmj97 said:
That was my thought too.
RedRider
New member
Or do 2 8 ohm speakers in series running parallel to a 16 ohm for a total of an 8 ohm load...
or 2 4 ohm speakers in series running parallel to an 8 ohm speaker for a 4 ohm load....(do they even make 4 ohm speakers anymore??)
or 2 4 ohm speakers in series running parallel to an 8 ohm speaker for a 4 ohm load....(do they even make 4 ohm speakers anymore??)
petejt
Active member
I'm wondering how to do this too.
petejt
Active member
I just managed to work out how to wire up a "3x12" cab for 8 ohms as well as 16 ohms.
For the 8 ohm cab, I need to wire a single 4 ohm resistor in series with speaker C.
For the 16 ohm cab, I need to wire two 4 ohm resistors (in series), that together are wired in parallel with speaker C.
Speaker C in both cabs is an ElectroVoice "12S" 200 watt 12" speaker (similar to the 12L).
The power for 'speaker C' is 70-75 watts, which combines both speakers. I guess since the actual speaker is 200 watts, then I won't need a high wattage 4 ohm load resistor? Or do I? What wattage do I need?
For the 8 ohm cab, I need to wire a single 4 ohm resistor in series with speaker C.
For the 16 ohm cab, I need to wire two 4 ohm resistors (in series), that together are wired in parallel with speaker C.
Speaker C in both cabs is an ElectroVoice "12S" 200 watt 12" speaker (similar to the 12L).
The power for 'speaker C' is 70-75 watts, which combines both speakers. I guess since the actual speaker is 200 watts, then I won't need a high wattage 4 ohm load resistor? Or do I? What wattage do I need?
thunda
New member
RedRider":q559ffbx said:
this. a resistor is not the best way as it will limit flow to the speaker...
GuitarGuyLP
Active member
petejt":3a0u3ddo said:
A resistor will not be ideal. A speaker is an active load as in the resistance changes with frequency that is why a speaker load is called an impedance, not a resistance. The higher the frequency the higher the effective resistance of the speaker.
Easiest way is either two 16 ohm in parallel for 8 ohms in series with an 8 ohm speaker. The 8 ohm speaker will get 50% of the power, and each of the 16 ohm speakers will get 25%
or
two 8 ohm speakers in series in parallel with a 16 ohm speaker. In this case the 16 ohm speaker will get 50% of the power, and the 8 ohm speakers will each get 25%
petejt
Active member
GuitarGuyLP":2yl6qmu7 said:
Ha! I've edited my post again.
At first I was a bit pessimistic, but now you've given me an idea to use a 'dummy' 16 ohm speaker as a dedicated load, rather than a resistor. I have two spare 16 ohm speakers which I think are low wattage (need to check). My main problem is that in one of the cabs, the fourth crappy speaker is stuck. So I have to get it removed somehow.
I'm trying out impedance calculations etc. to work how to wire them up. Thanks for the advice!
petejt
Active member
I'm stuck trying to work out how to get 16 ohms total impedance when using three 8 ohms speakers and a 16 ohm speaker.
I think I will have to get some resistors to either subtract or add impedance to get it to match, at least somewhat close.
I think I will have to get some resistors to either subtract or add impedance to get it to match, at least somewhat close.
GuitarGuyLP
Active member
two 8 ohm in series = 16 ohm put that in parallel with the 16 ohm speaker. that will be 8 ohms. Put that in series with the other 8 ohm speaker, and you will have 16 ohms. 50 % of the power will go to the last 8 ohm speaker, 25% to the 16 ohm, and 12.5 % to each of the 8 ohm speakers in series.
petejt
Active member
GuitarGuyLP":18v41rct said:
Thanks for that man!
Why does the last speaker get half of the power? Wouldn't the first speaker get most of it?
GuitarGuyLP
Active member
From an electrical point of view the order of the speakers does not matter. The best way to look at it is that electricity is lazy it takes the path of least resistance.
The first two 8 ohm speakers in series appear as a single 16 ohm speaker with each getting 1/2 the power. You put that in parallel with a 16 ohm speaker now you basically have what appears to be an 8 ohm speaker. If you put that in series with another 8 ohm speaker that 8 ohm speaker will get half of the power, and the other half will go to the 8 ohm load created by the three speakers.
The first two 8 ohm speakers in series appear as a single 16 ohm speaker with each getting 1/2 the power. You put that in parallel with a 16 ohm speaker now you basically have what appears to be an 8 ohm speaker. If you put that in series with another 8 ohm speaker that 8 ohm speaker will get half of the power, and the other half will go to the 8 ohm load created by the three speakers.
petejt
Active member
^^ Thanks again for all that.
I have a 200 watt speaker which I want to take most of the power.
The 16 ohm speaker I have I suspect is of low wattage (it's some MSP Radial Beam 'Duo Speaker' with a green-tinted shell), so I am wary of blowing it up if it gets too much power. So I might have to change things up so it is the path of most resistance, and thus least of the power.
I was getting confused earlier because I was using this website calculator (http://www.speakerimpedance.co.uk/?act= ... calculator), and thinking if I combined different impedance speakers it would alter the total impedance more than just halve or double it.
I have a 200 watt speaker which I want to take most of the power.
The 16 ohm speaker I have I suspect is of low wattage (it's some MSP Radial Beam 'Duo Speaker' with a green-tinted shell), so I am wary of blowing it up if it gets too much power. So I might have to change things up so it is the path of most resistance, and thus least of the power.
I was getting confused earlier because I was using this website calculator (http://www.speakerimpedance.co.uk/?act= ... calculator), and thinking if I combined different impedance speakers it would alter the total impedance more than just halve or double it.
petejt
Active member
EDIT: Scratch that (idea for a slight mismatch)
The 16 ohm speakers that I have, I got from someone who built their own 2x12 cabinet. They are "MSP Radial Beam Duo Speaker" 12PQCB" 12" speakers. I don't know what wattage they are, but I suspect they are 25 watts.
I'm concerned that if I connect them so they receive 25% of the power (50 watts, peaking to 100 watts), they'll blow up. Would this happen?
The 16 ohm speakers that I have, I got from someone who built their own 2x12 cabinet. They are "MSP Radial Beam Duo Speaker" 12PQCB" 12" speakers. I don't know what wattage they are, but I suspect they are 25 watts.
I'm concerned that if I connect them so they receive 25% of the power (50 watts, peaking to 100 watts), they'll blow up. Would this happen?