Odd impedance?

Bronze Member

Username: Sevin7

Post Number: 87
Registered: Apr-05

Posted on Tuesday, May 03, 2005 - 05:01 GMT

if you wire 3 dual 4ohm subs you get a 2.67ohm load:
http://www.crutchfieldadvisor.com/learningcenter/car/subwoofers_wiring.html?subs =3&impedance=DVC4x2
I dont have 3 or want 3, but im curious, how can you tell how much power your subs are getting? No amp tells you how much it puts out @ 2.67ohms, do you just estimate between the 2 ohm and 4 ohm rating?

Fishy

Gold Member

Username: Fishy

Tamarac Ft.Laud, FL USA

Post Number: 1121
Registered: Sep-04

Posted on Tuesday, May 03, 2005 - 09:27 GMT

If you have a regulated power supply and are running the amp within its limits then power should double every time you halve the load. Its a little more complicated when you have a crazy impedance like 2.67 ohms. An amplifier simply provides a specific maximum voltage at its highest output and you can use this formula to figure out how that relates to the load your driving:

P = V^2/R

V^2 is a constant if the amplifier is running within its design parameters so once you figure out what V^2 is you can just plug any resistance into that formula to figure out how much power your amp "should" be developing.

Example:

Your amplifier provides 200 watts into a 4 ohm load.

1) figure out V^2

V^2 = P*R = 200*4 = 800

2) plug in a resistance into the first formula to find the aproximate maximum power output.

P = V^2/R

Power at 2 ohms = P(2) = 800/2 = 400 watts

makes sense right?

Power at 1 ohm = P(1) = 800/1 = 800 watts

again it makes sense and is easy to figure out w/o the formula.

Power a 2.67 ohms = P(2.67) = 800/2.67 = 299 ~ 300 watts

Bingo!

You can use this for any load.

Say you had 3 4 ohm drivers

You can use RF's site or another formula to figure out the equivalent resistance.

Series is easy. You just add em up.

For parallel drivers its:

Req = 1/(1/R1 + 1/R2 + 1/R3.... + 1/Rn)

= 1/(1/4 + 1/4 + 1/4) = 1/(3/4)

= 4/3 or 1.34 ohms just like it says on RF's site.

If you have the same amp then

P(1.34) = 800/1.34 = 597 ~ 600 watts

If you use 4/3 instead of the 1.34 aproximation you'll get exactly 600 watts.

If you have a fair grasp on basic algebra its not as difficult as it looks. :-)

If you want to figure out how much power your amp should deliver just calculate the equivalent resistance by using the Req formula or a site like RF's and then do steps 1) and 2) and you're golden.

-Fishy

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Odd impedance?

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