Wiring up two 8ohm drivers for bookshelves.

[Rufes1]

Hey all,

Well this is my first post here so a hello to everyone.

I have the 2 perfect drivers to use in my new bookshelves. Problem is I want the total impedance to be 8ohm, but the impedance of each driver is 8ohm.

So in series its 4, parallel; 16.

Now what I wanted to know is how much power I am going to be loosing if I run them at 16ohm? I suspect there is a formula in the theory of maximum power transfer to work out exactly how much I am loosing? I couldnt find much..

Secondly.. Is there any way to wire them in parallel and then add a resistor either in series or parallel to alter my total impedance/resistance? Or will this consume too much power?


Thanks again

[ThomasW]

A resistor will get 1/2 the power so that's not a good thing to do.

Actually you don't loose any power with a 16 ohm load, the amp simply won't make as much with that load.

[Karma]

HI Rufies,
I think you have the formulas for equivalent resistance in networks reversed. Equivalent resistance for parallel resistance is always less than the lowest value resister. In your case its 8 ohms in parallel with 8 ohms for a value of 4 ohms.

Series resistances add so in your case it's 8 ohms + 8 ohms equals 16 ohms.

Actually, resistors are used very often in crossovers to balance both impedance and unequal driver sensitivity. They are one of the very few tools you have available to do this. And yes, they can waste power which must be considered when specifying overall speaker sensitivity.

Sparky

[Rufes1]

Hmmmm

yeh your right.. A resistor would just be a waste whichever way you look at it.

Next question, say I my amp pumps out 80wrms per chan at 8ohm. How would I figure out how much it is going to pump out at 16ohm?