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#1 |
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diyAudio Member
Join Date: Nov 2003
Location: England
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Why aren't there loudspeakers for valve amps that have super long coils of fine wire on them to negate the OPT?
Is the trade off between copper on the speaker not linear and so you'd need a disproportionately heavier mass on the speaker than with higher currents? |
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#2 |
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diyAudio Member
Join Date: Nov 2003
Location: Brighton UK
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Hi,
I think the voice coil wire would be too thin to be practical. |
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#3 |
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diyAudio Member
Join Date: Dec 2003
Location: Tulips, windmills and wooden shoes.
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Philips made some 800 ohm speakers in the past.
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#4 |
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diyAudio Member
Join Date: Nov 2003
Location: England
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Thanks for the quick replies....
there's probably a simple answer to it, it's just something I was thinking about whilst lying in bed. current wise, at 30 -> 40 awg 477 -> 43ma safe current 40 -> 50 awg 43 -> 4.39mA and... 50 -> 56 awg 4.39ma -> 1.08ma I know that guitar pickups are usually wound in the 38 to 42 awg range and that's quite easy to do. I've done it by hand myself a few times. And I mean.... by hand, winding the coil and positioning each turn. As hinted at in my first post, I'm guessing that the problem with doing this isn't so much the extra winding time it takes (some audio nutters would have made a 'one of a kind super pricey' speaker by now if it was - probably audionote, and it'd be silver wire), but that you can drive the same output volume from the speaker with less overall copper using a higher current and shorter, but thicker, wire. And that the benefits gained by doing this outweigh the losses of having an OPT in the way... i.e. it's easier to make a better OPT than speaker with a long voice coil. |
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#5 |
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diyAudio Moderator
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Off the top of my head - I think the voice coil would be too big and too heavy. That's a lot of wire, even if it is thin.
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#6 |
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diyAudio Member
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I did find a 100ohm speaker once, it was a small one from a alarm clock or something.
A speaker with that high(800 ohm) of impedence would be for a tube amp only so would cater to a much smaller market. I also suspect that the current capability of tube amps vary thereby making no specific impedence right for a all tube amps. The point above in regards to wire gauge is probably the main cause. |
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#7 |
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diyAudio Member
Join Date: Nov 2005
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Crossover inductors would get a mite largish
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#8 |
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diyAudio Member
Join Date: Nov 2003
Location: England
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If it's merely a practical issue, I would still expect to have seen at least one or two guys making them specifically for ultra expensive valve amps.
I think there's actually a mathematical, electrical, mechanical reason behind not doing it. That the weight of the coil increases disproportionately and you loose a whole load of range from the speaker due to damping, or something similar. Although I'm not sure, which is why I asked. If you had a single valve outputting a few tens of mA say, you could just wind the voice coil with something like 40awg and you'd be close to the limit of the wire's maximum safe current (so you'd keep wire to a minimum). That wouldn't be at all hard to wind. It can't be as simple as being difficult, since so many people sell audiophile gear with weird, excessive tweaks to them saying it's the best way to improve the sound. I'm thinking of the 2" thick cables, half ton turntables and speakers with 36 individual cones in them. At least with this idea you're cutting something quite substantial out of the amp, the whole OPT problem. I've had a go at some rough calculations for this but they're a bit tricky. Looks like I might have to actually get a bit of paper out and write the equations down to work it out properly. I worked out you'd need ~170g for a single EL34, which could be miles off. I can't imagine there's that much on the back of a normal 16 ohm speaker. Does anyone have some more numbers for normal speakers? Actual wire guages, rough sizes / weights of coils / DC resistances etc |
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#9 |
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diyAudio Member
Join Date: Oct 2006
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I'm pretty sure it has to do with the fact that the force exerted on the cone is proportional to the current through the current, I am sure BIl has something to do with it, where B is the magnetic flux, I is the current and l is the length of wire.
So if you had very low currents then you would have very little driver diplacement and hence low acoustic output. It seems such a long time since those uni classes! |
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#10 | |
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diyAudio Moderator
Join Date: Apr 2002
Location: Chatham, England
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Quote:
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