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#1 |
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diyAudio Member
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I wonder if you put 8 ohm resistor @10watt in parallel with a 8 ohm tweeter could the resistor handle putting 60 watts of power going to it?
The tweeter handle a 100w at 8 ohms, I'm looking to drive at 4 ohm with 60 watts. This should work right or am I crazy
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#2 |
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diyAudio Member
Join Date: Apr 2007
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That makes sense if you have something like a Motorola piezo tweeter .
i suggest you to move in the opposite direction ,which is having maximum efficiency ,and not throwing useful energy in a resistor that would transform it into heat . Also ,putting a resistor in parallel with a driver ,would mimic the effect of an Helmholtz resonator ... ![]() ![]() which is not the case of a tweeter , even if some have a back chamber ... |
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#3 |
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diyAudio Member
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I'm using Goldwood GT-400CD Bullet Piezo Horn Driver | Parts-Express.com
Which handling a 100 watts(tested), but only for about 6 hours, so if I scale it back this should be fine thank you
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#4 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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It is effectively a capacitor.
This makes it it's own crossover filter. The piezo tweeter is for up to 100W systems. The piezo cannot survive 100W of drive. Due to the filter it is handling mW while the amp is delivering W to the main drivers.
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regards Andrew T. |
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#5 |
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diyAudio Member
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Well I'm using a TDA2052, so this never be able to push that much power, but i was hoping to get close 50 or 60W @10% distortion
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#6 |
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diyAudio Member
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Check the datasheet. If you accept 10 % THD, ±18 V give you 22 W into 8 Ohm e.g. the tweeter or 40 W into 4 Ohm, so 20 W into the tweeter (9,1 % less) and 20 W into the resistor (waste). And that does not yet take supply voltage sag into account.
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If you've always done it like that, then it's probably wrong. (Henry Ford) |
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#7 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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Hi,
adding a resistor to a Capacitor tweeter changes the response curve. The resistor can be used to correct a response error/characteristic that the designer does not want/need. But equally the resistor may introduce a response error that sounds wrong.
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regards Andrew T. |
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