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Um...Ohms??

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Drunkkidcon0r

May 18, 2006
55
1
Northern Virginia
www.myspace.com
#1
I have been playing bass for eight years and I still don't feel like I have a good understanding of how Ohms work.

Is there anyway to pair an 8 Ohm cab and a 4 Ohm cab out of the same head? Specificly, A bergantino NV610 (4 ohms) and a shroeder 112 (8 ohms) via a Markbass LMII or III?

Im pretty sure the Berg says something along the line of at least 4 ohms, does that mean I could run the whole thing at 8 ohms and be fine?

I know the watts-vs-loudness argument is a touchy subject, but can I assume that running my 6x10 at 4 ohms would actually be louder than running both cabs at 8 ohms?

Thanks in advance!

Chad
 
#5
No. and there would be no point, since the surface area of the 610 is so much greater in proportion than the 112. As far as total impedance, (one divided by 4) + (one divided by 8) = 0.375

One divided by 0.375 equals 2.67 ohms. Your markbass head should not be operated below 4 ohms.
 
#6
How Ohms work in parallel is a bit ant-common sense. When you add them together the answer is always smaller than either.

Think of water running down a pie, then the pipe splits into two. You now have twice the cross section, half the water runs down each & twice as fast.

It's a bit like plugging 2 x 8 Ohm cabs in to a head. Combined they give 4 Ohms.

The formula is 1/N= 1/n1+1/n2+.............

In the above case, 1/N=1/8+1/8 = 2/8

1/N=2/b therefore N/1=8/2=4

For an 8 & a 4 Ohm cab it would be:

1/N= 1/8 + 1/4
= 1/8 + 2/8
= 3/8

N = 8/3 = 2.6666667 Ohms.

So, unless your head can run safely at 2 Ohms, it's not advisable.
 
#7
Drunkkidcon0r said:
I have been playing bass for eight years and I still don't feel like I have a good understanding of how Ohms work.
Click to expand...

Ohms = resistance. When adding resistance, in series you simply add the values, however, unless you have a specially made cable or one of those rare amps where the outputs are in series (some vintage amps) this is not something you'll really ever need to consider.

For cabs in parallel, you add resistance as 1/R(1) + 1/R(2) = 1/R(total). For your cabs, it would be 1/4 + 1/8 = 3/8 = 1/R(total) thus R(total) = 2.66.

You can also do it as R(total) = (R1*R2)/(R1+R2), or in your case R(total) = (4*8)/(12) = 2.66.

This website has good information and more examples if you want to dig deeper.
http://www.colomar.com/Shavano/spkr_wiring.html
 
#8
Drunkkidcon0r said:
...
Is there anyway to pair an 8 Ohm cab and a 4 Ohm cab out of the same head? ...

Chad
Click to expand...

I routinely do this through a Yamaha BBT500H, using a Carvin BRX10.2Neo 4 Ohm speaker and Mirage 1X15 8 Ohm speaker for a total 2.6 Ohm load. The Yamaha's manual recommends running the amp at 2 Ohms for efficiency. Works well for me. For some reason my head has less noise at 2.6 or lower than it does with a 4 or 8 Ohm load, but I have no idea why.
 
#15
Drunkkidcon0r said:
I must be a guitarist...

Does this mean I could not use the LMII or III with the 8 ohm cab unless I pair another 8 ohm cab with it??
Click to expand...

No, you'd be just fine with one 8-ohm cab ... you just wouldn't have as much power available. With solid state amps, you can always run speakers at a higher impedance than what the amp is rated for. Just not lower. So your 4-ohm head will work fine with an 8-ohm cab, two 8-ohm cabs (equalling a 4-ohm load), or one 4-ohm cab. A combination of a 4-ohm cab and an 8-ohm cab is no bueno, because they combine for a 2.7-ohm load, which is lower than your amp is rated for. You get optimum performance with the 4-ohm load.
 

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