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- 8 ohm to 16 ohm driver change

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- Thread starter mr ed
- Start date

If one driver really is 8 ohm at the frequency of x-over and the other really is 16 ohm at double the frequency, and the impedance traces looks otherwise similar - the effect will be double the x-over frequency. So e.g. 2kHz -> 4kHz etc...

In your case you need either an 2,5 uF cap or two 5's in series for 2,5uF... to retain the same XO as with the 5uF cap.

Try this :

https://www.v-cap.com/speaker-crossover-calculator.php

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In your case you need either an 2,5 uF cap or two 5's in series for 2,5uF... to retain the same XO as with the 5uF cap.

Try this :

https://www.v-cap.com/speaker-crossover-calculator.php

//

Last edited:

Altec 802 8g drivers, I bought a pair with out diaphragms and Radian only has 16 ohms currently.Thanks. Whatever your xover might be at, halving the uF is a safe bet anyways. I would try that and if things sounds off, do an investigation.

Why change impedance version btw?

Can you point to the actual driver model?

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Have been using 299 8a 8 ohm drivers.

Trying a different driver / horn combo.

Higher impedance will pass more, filter less, regardless high-pass or low-pass. So parallel the 5uF with another for ~10uF.

Simplified formula 1st order filters:

HPF hz=(160khz/ohm)/uF

LPF hz=(160hz*ohm)/mH

(2nd order filters: in above, calculate uF, mH for XO frequency f=5khz/sqrt(uF*mH); then respectively multiply/divide uF, mH by the same 1<q<2 to get different curves Linkwitz-Riley, Butterworth etc.)

Simplified formula 1st order filters:

HPF hz=(160khz/ohm)/uF

LPF hz=(160hz*ohm)/mH

(2nd order filters: in above, calculate uF, mH for XO frequency f=5khz/sqrt(uF*mH); then respectively multiply/divide uF, mH by the same 1<q<2 to get different curves Linkwitz-Riley, Butterworth etc.)

Last edited:

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