Its been a while since ive last posted here, but here goes. I have a Mesa 400 (runs at 4 and 8 ohms) playing through an ampeg 4x10 (4 ohm). I want to add a 15 for bigger gigs, but something tells me that running an 8 ohm 15" and a 4 ohm 4x10 wont be very good for my gear, or work very well. Im no expert with wiring and electronics, but by scanning through the "Why would anyone buy a 4 ohm cab?" thread i noticed someone saying via switches and wiring people can split the load. I guess the question is, is it possible for me to add an 8 ohm something to my rig? Another thought ive been having is to sell my 4x10 and get some money for some used EA speakers. Would i be better off checking something else out for more power? Thanks for the help.

If your cab is 4 ohms, and your amp goes no lower than 4 ohms, you unfortunately won't be able to add anything (wired in parallel). Any second cab that you would add would drop the total load below 4 ohms, which the amp would find unpleasant. Your only real option would be to get a 4-ohm second cab, and run the two in series. Just for reference, the total load equation for two cabs in parallel is: R1 x R2 ------- R1 + R2 So, if you added an 8-ohm cab, you'd have (4x8) / (4+8) = 32/12 = 2.67 ohms. For a series connection, it's simply addition - two 4-ohm cabs in series gives 8 ohms.

Thanks for the response, although pardon my ignorance. How would I go about putting the two cabs in series? Would I get much of a volume increase or just a fullness increase?

You can split the load. I'd go with a 4ohm 15 though, for more even power sharing, and you'd save $ on the required inductor coil. Do not use two cabs in series, that can really foul up things with different drivers. Get back to me after you find a 15 and I can give you the particulars; you will have to know how to use a soldering iron. One of your cabs will require a pair of jacks.

I'm not too good with math or anything, or speak impendances for that matter but I was just wondering if my set up is getting the most out of what I have running. Hughes and Kettner BassBase 400 400w at 4 ohms into 2 H&K BR410's (same company not the matching cabs however) at 8 ohms each run in series (there is only one speaker out on my amp). With this set up am I running them properly (or could I rewire the cabs to a different ohm-age to get the most out of them) or even add more cabs (although not that practical)

The simplest installation would have a series inductor mounted inside the 15 cab, with a high pass filtered output going to the second jack on the box to go out to the 4x10. You could add either a switch or a third jack to plug into should you want to have the ability to bypass the inductor, which would be a good idea should you ever want to use the 15 on its own.

This equation only applies when the two impedances are equal, i.e., 8 ohms and 8 ohms, etc. For unequal impedances it's 1/Rtotal=1/R1+1/R2+...+1/Rn Although the speakers are chained together, they are still probably connected in parallel - I don't believe most amps or speakers allow for connecting in series (see Bill's post #4).

There is no such thing as ohm-age. The term is impedance, which is measured in ohms. If I read you correctly you're running two 8 ohm cabs in series, that's a 16 ohm load and too high for efficient operation. Put them in parallel for a 4 ohm load and you'll be fine. But I question whether you have them in series. If you're running out of the amp into the first cab and out of the first cab to the second you're probably in parallel already, as that's how dual jacks on cabs are usually wired. If your setup is parallel wired and still isn't loud enough I'd say the amp is too small. 400w@4 ohms is only 280 watts at 8 ohms, and 35 watts per driver isn't much. I'd consider 50 watts per driver to be the minimum figure for good results.

2 entries found for ohmage. ohm·age ( P ) Pronunciation Key (mj) n. Electrical resistance expressed in ohms. NOT posting to start a fight...it's just to clear up a misconception with the word.

Nope. The equation I posted applies for any two impedances in parallel, as the example with an 8-ohm and a 4-ohm cab shows. The equation you posted is handy when you have three or more impedances in parallel. Run the numbers for an 8- and 4-ohm cab in your equation.... see if you get the same answer.....

Easiest way is to have a special cable made up to put the loads in series. You would likely see a bit of both. I'm not sure what Bill is referring to by "fouling things up"; if wired correctly, it won't harm either speakers or amp, although the results can be unpredictable (if the drivers are of dramatically different sensitivities, for example).

Where'd you find that anyway? There are some pretty horrible "dictionaries" in print. There are many more bad ones online.

Unpredictable to say the least. If the drivers are identical there's no issue, but if they aren't a higher voice coil inductance on one driver will act as a low pass filter on the other, and that's just what you'd expect to result mixing a fiteen and a ten.

I would guess that Bill was referring to the fact that when you put a cabinet in series before another cabinet, the first one acts as a very complex crossover component to the second one. So, the second one gets an interesting and mostly unpredictable subset of the full signal. He was suggesting splitting the signal to send a low-pass signal to the 15 and a full range signal to the 10s. Edit: beat me to it!

He needs a 1st order HP on the tens as well, otherwise he'll have too low an impedance load below 100 Hz. Improving the power/frequency distribution is another benefit, of course, with a very reasonable component cost.

Understood. Just wanted to clarify that "fouling things up" wasn't interpreted as "damage will result". Agree that the results may be not as expected.

You are correct sir. If I had bothered to check your math first... My correction is corrected. Hmm, there is some equation that only applies when they are equal, now I'm going to have to dig that one up.

There is; I think that's the old "divide by two". Two sixteens in parallel give 8 ohms, two eights in parallel give 4 ohms, two fours in parallel give 2 ohms, etc.