Speaker impedance with two terminals? [Archive] - Audio & Video Forums


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01-11-2008, 08:40 PM
I have what I thought was a simple question, but I can't find an answer:

I've got 4 in-ceiling speakers in a relatively small area. I only have one speaker wire going to each speaker. And my amp has 2 channels, with a right and a left for each. The amp says run 8 ohm speakers with one channel, and 16 ohm for two channels. Perfect for 4 speakers, I thought.

So I ordered speakers that have a switch allowing for 8 ohm or 16 ohm to cover my bases.

My problem is, now I have the speakers and they have two terminals on them, one for the cone, and one for the tweeter, and I only have one wire running to them.

Reading up on impedance, I know that running to each terminal on the speaker in series formation will double the impedance of each speaker. So is the impedance of the tweeter different from the cone? Are they each 8? Or is the speaker as a whole (both cone and tweeter) 8, making the tweeter and cone different?

That compounding with the fact that the speaker has the option of being 8 or 16, and the amp says to change to 16 ohm speakers when using both channels has got me perplexed.

Or is it possible to just connect the positives from the tweeter, cone and amp together, and the negatives all together? If so, what should the speaker ohms be?

Anyone have any ideas?

N. Abstentia
01-12-2008, 09:31 PM
Here I'll further confuse you:

Each driver is 8 ohms. The woofer and tweeter wired together in parallel is..8 ohms.

Why? Because each driver produces a different frequency range. If you had two woofers together in parallel it would be 4 ohms..but since it's a woofer and tweeter producing different frequencies. That is a coaxial speaker, very common in car audio.