8 ohm + 4 ohm cab?

[JeffKissell]

I have an AI Clarus III and a Bergantino HT112 cab, which is 8 ohms. I'm thinking about getting an AI contra EX which is 4 ohms. Can I use them together? I don't think I would use them together very often, but I would like the option.

1st question - Is it possible to run a 8 ohm speaker and a 4 ohm speaker load together?

2nd question - What about adding an 8 ohm resister (sp?) to the Bergie cab to reduce the load to 4 ohms. Is this just crazy talk?

much obliged

-J

[ricobasso]

Originally Posted by JeffKissell

1st question - Is it possible to run a 8 ohm speaker and a 4 ohm speaker load together?

2nd question - What about adding an 8 ohm resister (sp?) to the Bergie cab to reduce the load to 4 ohms. Is this just crazy talk?

much obliged

-J

Very crazy No, No No, Don't do the resistor thing. It will just waste power and get very hot or even maybe start a fire.

The Clarus is designed to give 150w/8ohms, 250w/4ohms, 350w/2ohms. The way it works is this; if you have two 8 Ohm speakers (in parallel) it looks like 4 Ohms to the amp and they share the extra power equally. If you added a 4 Ohm speaker in parallel with an 8 Ohm you get 2.6 Ohms, which is OK, the Clarus will drive it, but the 4 Ohm speaker will take twice as much power then the 8 Ohm. i.e about 200W to the 4 Ohm cabinet and 100W to the 8 Ohm cab. Assuming equal sensitivity, the 4 Ohm cabinet will always be 3dB louder at any volume setting.

[JeffKissell]

Thanks for the info. I've always wondered about unmatched ohm loads.
-J

[MT Spaces]

Originally Posted by ricobasso

The Clarus is designed to give 150w/8ohms, 250w/4ohms, 350w/2ohms.

For what it is worth, I think you may be quoting the older Clarus Series I and II numbers. The Series III has appreciably more power now: 500w @2 ohms, 400w @4 ohms, 250w @8 ohms. I believe the original poster said he had a Clarus Series III.

[mje]

Just for reference: When you combine two speakers in parallel, the impedances (Z1 and Z2) combine as follows:

Ztotal = (Z1 * Z2) / (Z1 + Z2)

For an 8 and a 4 ohm cabinet:

Ztotal = (8 * 4) / (8 + 4)

= 32 / 12

= 2.6666 ohms.

Thus the amp will be producing somewhere between 400 and 500 watts. And as noted, the 4 ohm cabinet will draw twice the current of the 8 ohm cab.

[JeffKissell]

I'll see if I can find a small 8 Ohm cabinet that works for the quiet stuff I was thinking of. Otherwise, if I get the AI (4 Ohm) cabinet I just won't run it with the Bergantino. It's less flexible but it still works for me.
-J

[drurb]

Originally Posted by mje

Just for reference: When you combine two speakers in parallel, the impedances (Z1 and Z2) combine as follows:

Ztotal = (Z1 * Z2) / (Z1 + Z2)

For an 8 and a 4 ohm cabinet:

Ztotal = (8 * 4) / (8 + 4)

= 32 / 12

= 2.6666 ohms.

Thus the amp will be producing somewhere between 400 and 500 watts. And as noted, the 4 ohm cabinet will draw twice the current of the 8 ohm cab.

For the record, and I know that mje knows this, you cannot extend this convenient form to more than two impedances. In other words, you cannot tack on a third impedance to the form mje gave. In general:

1/Ztotal = 1/Z1 + 1/Z2 +1/Z3....1/ZN where N is the number of combined impedances. When N=2, that is, when you are combining two impedances, the algebra (try it at home ) works out to give you that nice formula mje provided.

[mje]

What he said. ;-)