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dumb ohm question,......again

wiring.jpg
 
nubs, you can't make a 4 ohm cab using four 8-ohm drivers. It can't be done. You would be violating the laws of math!

If you have four 8-ohm drivers, there are only eight ways you can wire them:

(1) --8--8--8--8--; this yields a 32-ohm cabinet

(2) --8--
--| |--8--8--; this yields a 20-ohm cabinet
--8--

(3) --8--
| |
--|-8-|--8--; this yields a 10.68-ohm cabinet
| |
--8--

(4) --8--8--
--| |--; this yields an 8-ohm cabinet
--8--8--

(5) --8-- --8--
--| |--| |--; this also yields an 8-ohm cabinet
--8-- --8--

(6) --8--8--8--
--| |--; this yields a 6-ohm cabinet
-----8-----

(7) --8--
---| |---
--| --8-- |--; this yields a 3.2-ohm cabinet
| |
---8---8---

(8) --8--
| |
|--8--|
--| |--; this yield a 2-ohm cabinet
|--8--|
| |
--8--

By the way, wiring as in (2), (3), (6) or (7) results in unbalanced speaker loads.
 
OK, sorry, this font is proportiional and the lines don't line up.

(1) is all four drivers in series; 32 ohms
(2) is (two drivers in parallel) in series with (two drivers in series); 20 ohms
(3) is (three drivers in parallel) in series with one driver; 10.68 ohms
(4) is (two drivers in series) in parallel with (two drivers in series); 8 ohms
(5) is (two drivers in parallel) in series with (two drivers in parallel); 8 ohms - electrically identical to (4)
(6) is (three drivers in series) in parallel with one driver; 6 ohms
(7) is (two drivers in series) in parallel with (two drivers in parallel); 3.2 ohms
(8) is all four drivers in parallel; 2 ohms
 

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