Remember, though, most drivers have a resonant peak around its fs, or a peak around the fb of the box (2 peaks if vented, more if horn loaded, etc.... )
Such peak(s) has an impednace of tens of Ohm or more (my woofer is approaching 200 Ohm). Beside the peak, the impedance is obviously rising at frequencies below 100 Hz or so (some comes even higher up). Find impedance sweeps of any speakers, you would see.
That is, most speaker's actual impedance at bass is FAR from the nominal value, any fix resistor (say, 24 Ohm) in parallel with the impedance peak would inevitably share most current and eat up the power in the bass range.
Such peak(s) has an impednace of tens of Ohm or more (my woofer is approaching 200 Ohm). Beside the peak, the impedance is obviously rising at frequencies below 100 Hz or so (some comes even higher up). Find impedance sweeps of any speakers, you would see.
That is, most speaker's actual impedance at bass is FAR from the nominal value, any fix resistor (say, 24 Ohm) in parallel with the impedance peak would inevitably share most current and eat up the power in the bass range.
As I said I put back the old driver in the enclosure this afternoon.
While the emclosure was open I decided to replace the crossover with a 2 way that I had just to see if the woofer still have break-out.
Mind you the crossover is not the right one for the driver I will let it run for a few days and check if the problem re-occurs. If I'm lucky the crossover was at fault if not I will be sure the woofer is faulty.
Thanks everyone for your valuable comments.
While the emclosure was open I decided to replace the crossover with a 2 way that I had just to see if the woofer still have break-out.
Mind you the crossover is not the right one for the driver I will let it run for a few days and check if the problem re-occurs. If I'm lucky the crossover was at fault if not I will be sure the woofer is faulty.
Thanks everyone for your valuable comments.
Well someone sure must have changed the EE curriculum since I got my BSEE if you don't understand what I said.
When there are two paralleled resistors being driven from a voltage source, the power to each will be divided based on their resistance values. Each resistor obviously sees the same voltage. Thus when that voltage is squared to calculate the power being dissapated by each, that power is inversely proportional to their resistances. If one resistor has twice the resistance of the other, it will be dissipating half as much power as the other. As far as paralleling a resistor with a 15-ohm driver to achieve a nominal 8-ohm "impedance" to match the 8-ohm impedance of the woofer in the other speaker as the OP said, clearly this won't accomplish what he's after; it won't likely end up with equal woofer sensitivities in the two speakers. And, the total power being delivered to the parallel combination of driver and resistor will be divided between them based on their resistance/impedance values.
Paul
When there are two paralleled resistors being driven from a voltage source, the power to each will be divided based on their resistance values. Each resistor obviously sees the same voltage. Thus when that voltage is squared to calculate the power being dissapated by each, that power is inversely proportional to their resistances. If one resistor has twice the resistance of the other, it will be dissipating half as much power as the other. As far as paralleling a resistor with a 15-ohm driver to achieve a nominal 8-ohm "impedance" to match the 8-ohm impedance of the woofer in the other speaker as the OP said, clearly this won't accomplish what he's after; it won't likely end up with equal woofer sensitivities in the two speakers. And, the total power being delivered to the parallel combination of driver and resistor will be divided between them based on their resistance/impedance values.Paul
Theorical and practical is not quite the same.
Trying it makes it real, I realized it by experimenting the subject of my topic.
Worst case scenario I will search for an identical 8 ohm woofer.
I'm sure there's a good 12" Wharfedale woofer on Internet somewhere.
Again, thanks a lot I learned one more time from this forum.
Long life to DIY Forum.
Trying it makes it real, I realized it by experimenting the subject of my topic.
Worst case scenario I will search for an identical 8 ohm woofer.
I'm sure there's a good 12" Wharfedale woofer on Internet somewhere.
Again, thanks a lot I learned one more time from this forum.
Long life to DIY Forum.
Paul the point that you seem to be glossing over is, that the power delivered to the speaker does not change with a parallel impedance. After all this is what matters to the sound ie SPLs, A draw back is the amp has to deliver extra power to the matching resistor. This is one of the trade offs to make the OP's existing replacement driver work. Rather than focus on the amps total power ie efficiency delivered, is pointless as practical modern amplifiers are concerned. The brute force paralleling a resistor across the whole freq. band is not the only approach that you seemed to miss. A parallel zobel will provide a match as needed, over a narrower frequency range limiting the wasted power. Perhaps it's just a philosophy difference at work here, because I'm an EE that considers all the options before flat out stating something won't work.
My, my, aren't you impressed with yourself! Go back and reread Post #12 where I clearly stated what you said I've been glossing over (in agreement with you, actually!). Also, the fact that I didn't pursue any other possibilities, like the Zobel network, does not equate with "flat out stating something won't work". I was addressing only what the OP proposed and did not "miss any approaches".
Paul
Paul
well Paul I'm just trying to understand were you are coming from by telling me "I missed the point". You still haven't really clearly stated what your position is concisely, anywhere that I can see in this thread. This is just handwaving over using power eff. vs dB V/m as far as I'm concerned, to be avoided. apples to apples
You missed another person's point to which you replied regarding his comments about how the power would be divided up between the driver and the paralleled resistor (Posts around 7-9, I believe). If you cannot see anything concise in what I've posted, that's on you, not me. And, I've not been "waving my hands" or trying to confuse anyone in some manner. Please leave your ego out of your posts.
Paul
Paul
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