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Old 22nd September 2003, 01:07 AM   #1
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Default mid/tweet power division ?

Hi All,

I've been looking up on the division of power between mids and tweets for a bi-amped system..

I'm designing my speakers to go to 105Db peaks, with an average of 85Db. (Mid - Tweeter x/o @ 3000Hz)

The only info I've found is on Rod Elliots site which states that at 3000Hz the power is split 85% - 15%

Therefore, if the mids are peaking at 105Db, can I assume my tweeters will peak at 105/85 * 15 = 18.5Db ?

If so, as they're 92Db sensitivity I can happily use a 2.5 watt amp to drive them..


Hopefully somebody can confirm for me,

Cheers

Rob
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Old 22nd September 2003, 01:35 AM   #2
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Seeing as Franks on the loudspeaker section right now, I'll add that I'm cosidering the Lilliput amp to drive my tweeters....

Cheers


Rob
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Old 22nd September 2003, 03:09 AM   #3
ojg is offline ojg
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Actually, your calculations are a little strange, but you somehow arrived close to the right answer!

Here's how it should be calculated (I'm sure somebody here can double-check for me). You need to do the antilog before multiplying with the percentage ratio:

You want 105dB SPL @1m and speaker efficiency is 92dB/1W/1m. So you need a total amplifier power of 10^((105-92)/10) = 20W

Tweeter needs only 20 * 0.15 = 3W
Mid needs 17W, more if it has less than 92dB sensitivity.

But remember that Elliott states on his page that the 85/15 ratio is on average! Which means that with certain music the tweeters may need much more than 3W.
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Old 22nd September 2003, 08:56 AM   #4
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Hi ojg,

Thanks for the input.

I guess this is a worse case scenario aswell, I probably listen at a 75 - 80 Db average, rather than 85.

Cheers

Rob
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