Total Impendance?

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Hi, i was wondering how you work out the total impedance of a loudspeaker if it uses the MTM configuration? surely in this case you would have an 8 ohm tweeter and the 2 woofers (presumably 8 ohms each) in parallel to give you 4 ohms, what i don't understand is what does the amp see if this is the case? Wouldn't there be a problem if the impendance is swaying between 4 and 8 ohms depending on which frequencies are playing. I think i'm missing something here so could someone please fill me in.
 
I had to read this again. If you wire two 8 ohm drivers in parallel giving 4 ohm, then in parallel again to an 8 ohm tweeter you get around 6 ohm, then factor in the resistance of your crossover you'll come in somewhere between 6 and 8 I would think. Whichever you come up with most amplifiers will not have a problem driving speakers in the 4 to 8 ohm range. As for swinging, the actual resistance that exists will range quite a bit farther than just a 4 ohm swing anyway.
 
ok thanks, i think i understand now. Just wasn't sure whether there was a drop in impedance when lower frequencies were being played and if so would the amp handle them. But if i follow most of the MTM designs i should be ok i think, because many i've seen include crossovers.
 
Depending on your crossover, Imagine Coils and Caps as frequency dependant resistors,

you can calculate that resistance as Xc = 1/(2PIfc) And Xl=2PIfl

Xc is at -90 degrees, Xl is at 90 degrees.

for eg. If you have a 4 uF cap and a 8 ohm tweeter in series, at 2000 hz, Xc = -19.8i <- That shows its at 3/4 PI.

Now you have you tweeter thats in series,

So Zt = 8 + -19.8i = sqrt (8^2 + 19.8^2)= 21.3 Ohms (with an angle, of course)

For those interested, the angle of the impedance is tan^-1 of 8/19.8 = 22.03 degrees, so 45 degrees - 22.03 degrees =
22.97 degrees

as here we are in the fourth quadrant, thats -22.97 degrees or 337 degrees..

For paralell components, (R1^-1+R2^-1+Rn^-1)^-1

So here, open Excel... and make something nice....
 
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