choosing inductors for crossovers

[ark]

I need to purchase a few inductors for crossovers - two way high efficiency speakers.. TAD components.. etc..
don't want to skimp on the crossovers.. but confused about what type of inductors to buy?

I need 2- .6 mh and 2- 1 mh inductors.

I thought I read someplace to buy the lowest AWG inductors I could find? is this correct?

North Creek Music has as low as 8 AWG but those are Way expensive.. would 10 AWG be better than 12?

Oh and I'll never feed them over 30 watts...

Please help clarify. Thanks a lot!

[sreten]

http://www.partsexpress.com/webpage....TOKEN=15468895

Hi,

There is a lot of choice. Going for ultra low DCR is pointless.

/sreten.

[owdi]

The Jantzen 15 awg air core are the largest guage inductors I've used. IMO, spend your money on low DCR inductors that are in series with the least efficient driver in your speaker. The rest, especially parallel indutors, can be a small guage.

IMO(2) - low DCR inductors on padded tweeters are a waste.

Dan

[ark]

thanks. not much of an expense to upgrade to at least 14 awg or 12 for the woofer... if it can't hurt then might go for the meat and buy a few less cups of coffee this week. = p

[owdi]

Some more info:

The reason you want to minimize series resistance in the least efficient driver is because it acts as a voltage dividier. This is very easy to calculate using the following equation:

R1 - DCR of inductor
R2 - DCR of driver (ignoring inductance to keep things simple)

Loss in output equals 20 * Log[R2/(R1 + R2)] decibels.

For example, using a .5 ohms inductor with a = 6 ohm speaker.

20 * Log[6/(.5 + 6)] = -0.7 db

What if you used a 1 ohm inductor?

20 * Log[6/(1 + 6)] = -1.33 db

If you want to try this yourself, paste the following function into Excel.

=20*log(b2/(b1+b2))

Put the DCR of the inductor in B1
Put the RE of the driver in B2

If I screwed this up somehow, someone please correct me.

Dan

[ark]

Thanks I will work with this. Very helpful and help make some sense out of why the lower dcr coils sound different..

[Svante]

Originally posted by owdi
For example, using a .5 ohms inductor with a = 6 ohm speaker.

20 * Log[6/(.5 + 6)] = -0.7 db

What if you used a 1 ohm inductor?

20 * Log[6/(1 + 6)] = -1.33 db

Yes, this is true where the impedance of the driver is 6 ohms. However, since the impedance varies with frequency, so will the attenuation effect. Some thing like this:



I think one can see the -0.7 and -1.33 dBs at for example 200 Hz, where the impedance is ~6 ohms.

[owdi]

Svante - oh I agree completely. I was just trying to keep things simple to illustrate the effect of inductor DCR on overall efficiency. As a generalization the equation holds up pretty well even after you consider impedance.

Dan

[cjd]

Remember that series impedance will alter the Q as well.

Some projects I find speccing low-DCR is worth it on the woofers - squeeze the most out of 'em as far as output SPL (every little bit helps keep total up particularly once BSC is counted in) and to keep box size down while maintaining a good Q. Runs from amp to speaker also add impedance.

I noted just a bit ago that B&W suggests cabling no more than 0.1ohm on their 800 series. This is likely why.

C

[Svante]

Originally posted by cjd
Remember that series impedance will alter the Q as well.

Actually no. Not "as well".

The change in frequency response (in my graph above) can be explained in alternate ways; one would be that the Q of the driver is changed, another would be simple voltage division between the series resistance and the frequency-varying impedance ot the driver.

The two explanations are equivalent, and one has to select one of them.