Looking For Cheap Crossover w/o any coils
Re: Looking For Cheap Crossover w/o any coils
I'm confused with what you told me. I have never made a coil before, except for small 5-turn ones that said in the book to do 5-turns. If I wanted a crossover for a sub, which I do, I would use the equation N=(SR[a*u*u0/I])*20pi, right. (Its the one off of that website you sent me to. What are the values for each of the variables? or,if yor know, how many turns will I need. It doesn't really matter what frequency, as long as it will work for a sub. Thanks
Re: Looking For Cheap Crossover w/o any coils
A better question might be:
I would like an op-amp low pass filter like a sallen-key with a 120 Hz cutoff.<p>http://www.circuitsage.com/filter.html<p>[ March 18, 2003: Message edited by: 1206DX ]</p>
I would like an op-amp low pass filter like a sallen-key with a 120 Hz cutoff.<p>http://www.circuitsage.com/filter.html<p>[ March 18, 2003: Message edited by: 1206DX ]</p>
Re: Looking For Cheap Crossover w/o any coils
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by mikea1962375:
I'm confused with what you told me. I have never made a coil before, except for small 5-turn ones that said in the book to do 5-turns. If I wanted a crossover for a sub, which I do, I would use the equation N=(SR[a*u*u0/I])*20pi, right. (Its the one off of that website you sent me to. What are the values for each of the variables? or,if yor know, how many turns will I need. It doesn't really matter what frequency, as long as it will work for a sub. Thanks<hr></blockquote>
The frequency matters greatly. Let's say your tweeter is rated for 5000 to 25000 hz, and your woofer is rated for 15 to 7000 hz. A good transition frequency might be 6000 hz for the woofer, 5000 hz for the tweeter. If your woofer is 8 ohms, then using X(l) = 2 · ¶ · f · L where f=6000, X(l)=8, and ¶=3.1415927, we find L is .000212 henries. (I did of course mess up in the previous post --- if you multiply by 1000, that's milli-henries; multiply by 1,000,000 for microhenries; this is 212 microhenries.)<p>From that webpage, any of the formulas should get you close; the first is: L=(D*N^2)/(l/D+0,43)<p>Where <p> D is diameter in cm
l i length in cm
L is inductance in uH
N is nuber of turns <p>You already have L (assuming the 8 ohms and 6000 frequency) as being 212µH; you would select D, and then wind whatever number of turns to get there. There will be a somewhat itterative (trial-and-error) process determining length "l"; you are unlikely to know exactly in advance the length of the coil because that is a function of the number of turns!<p>Remember that one INCH is 2.54 cm. So you can convert the cm to inches to build the coil.<p>The formula for multi-layer is slightly different, but if the difference between radii of INNER and OUTER is small, the single-layer should be close...<p>If your woofer is rated up to 4000, and your tweeter is rated down to 8000, obviously you'll need a midrange speaker (and ITS crossover will be BOTH an inductor and capacitor). And there's a whole science to building enclosures; the size and depth of the port, etc. Somewhere I have an old Radio Shack book, "building speaker enclosures".<p>RE the ratings of 15-5,000 hz --- we know that Humans can only hear down to 20hz, and up to 20,000 hz (at BEST); but audio is rated at the 3-dB half-power-point. So if you buy headphones rated 20-20,000 hz, (the ODDS being that they never TESTED them but are GUESSING/LYING!), that rating means that at those two points, 20 cycles, and 20,000 cycles, HALF YOUR POWER is already GONE. If your headphones or speakers are rated to 15hz, and up to 22,000 or 25,000, then AT the limit points your sound is still strong. Make sense?<p>A simple inductor-base and capacitor-treble crossover exhibits only one pole (single order, not a steep slope on the filter); you choosing 6000 for the woofer and 5000 for the tweeter means that 5500 should be covered, BOTH contributing at reduced amplitude...
I'm confused with what you told me. I have never made a coil before, except for small 5-turn ones that said in the book to do 5-turns. If I wanted a crossover for a sub, which I do, I would use the equation N=(SR[a*u*u0/I])*20pi, right. (Its the one off of that website you sent me to. What are the values for each of the variables? or,if yor know, how many turns will I need. It doesn't really matter what frequency, as long as it will work for a sub. Thanks<hr></blockquote>
The frequency matters greatly. Let's say your tweeter is rated for 5000 to 25000 hz, and your woofer is rated for 15 to 7000 hz. A good transition frequency might be 6000 hz for the woofer, 5000 hz for the tweeter. If your woofer is 8 ohms, then using X(l) = 2 · ¶ · f · L where f=6000, X(l)=8, and ¶=3.1415927, we find L is .000212 henries. (I did of course mess up in the previous post --- if you multiply by 1000, that's milli-henries; multiply by 1,000,000 for microhenries; this is 212 microhenries.)<p>From that webpage, any of the formulas should get you close; the first is: L=(D*N^2)/(l/D+0,43)<p>Where <p> D is diameter in cm
l i length in cm
L is inductance in uH
N is nuber of turns <p>You already have L (assuming the 8 ohms and 6000 frequency) as being 212µH; you would select D, and then wind whatever number of turns to get there. There will be a somewhat itterative (trial-and-error) process determining length "l"; you are unlikely to know exactly in advance the length of the coil because that is a function of the number of turns!<p>Remember that one INCH is 2.54 cm. So you can convert the cm to inches to build the coil.<p>The formula for multi-layer is slightly different, but if the difference between radii of INNER and OUTER is small, the single-layer should be close...<p>If your woofer is rated up to 4000, and your tweeter is rated down to 8000, obviously you'll need a midrange speaker (and ITS crossover will be BOTH an inductor and capacitor). And there's a whole science to building enclosures; the size and depth of the port, etc. Somewhere I have an old Radio Shack book, "building speaker enclosures".<p>RE the ratings of 15-5,000 hz --- we know that Humans can only hear down to 20hz, and up to 20,000 hz (at BEST); but audio is rated at the 3-dB half-power-point. So if you buy headphones rated 20-20,000 hz, (the ODDS being that they never TESTED them but are GUESSING/LYING!), that rating means that at those two points, 20 cycles, and 20,000 cycles, HALF YOUR POWER is already GONE. If your headphones or speakers are rated to 15hz, and up to 22,000 or 25,000, then AT the limit points your sound is still strong. Make sense?<p>A simple inductor-base and capacitor-treble crossover exhibits only one pole (single order, not a steep slope on the filter); you choosing 6000 for the woofer and 5000 for the tweeter means that 5500 should be covered, BOTH contributing at reduced amplitude...
Re: Looking For Cheap Crossover w/o any coils
BTW, the symbol "^2" is generally SQUARED. So:
L=(D*N^2)/(l/D+0,43)
(l/d + 0.43)(L) = (D)(N^2)
(l/d + 0.43)(L/D) = N^2<p>Thus N is the quare root of the quantity [(l/d = 0.43)(L/D)]<p>You might start off GUESSING "l" at an inch, and see if you can wind it tighter; then recalculate with the REAL value and see what your inductance is. A couple of tries changing N and l, you should be able to get close to your desired inductance...
L=(D*N^2)/(l/D+0,43)
(l/d + 0.43)(L) = (D)(N^2)
(l/d + 0.43)(L/D) = N^2<p>Thus N is the quare root of the quantity [(l/d = 0.43)(L/D)]<p>You might start off GUESSING "l" at an inch, and see if you can wind it tighter; then recalculate with the REAL value and see what your inductance is. A couple of tries changing N and l, you should be able to get close to your desired inductance...
Re: Looking For Cheap Crossover w/o any coils
First, what does the comma mean between 0,43 in one of your equasions. Also, how do I solve it? there are four variables, and I only know two of them. Thanks
Re: Looking For Cheap Crossover w/o any coils
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by mikea1962375:
First, what does the comma mean between 0,43 in one of your equasions. <hr></blockquote><p>I would think the comma was a mistype, however having said that in continental Europe the comma is used instead of the decimal point - mainly in the Germanic speaking countries - including Scandinavia in that.<p>Colin
First, what does the comma mean between 0,43 in one of your equasions. <hr></blockquote><p>I would think the comma was a mistype, however having said that in continental Europe the comma is used instead of the decimal point - mainly in the Germanic speaking countries - including Scandinavia in that.<p>Colin
On a clear disk you can seek forever.
Re: Looking For Cheap Crossover w/o any coils
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by mikea1962375:
First, what does the comma mean between 0,43 in one of your equasions. Also, how do I solve it? there are four variables, and I only know two of them. Thanks<hr></blockquote><p>You might start off GUESSING "l" at an inch, and see if you can wind it tighter; then recalculate with the REAL value and see what your inductance is. A couple of tries changing N and l, you should be able to get close to your desired inductance...
First, what does the comma mean between 0,43 in one of your equasions. Also, how do I solve it? there are four variables, and I only know two of them. Thanks<hr></blockquote><p>You might start off GUESSING "l" at an inch, and see if you can wind it tighter; then recalculate with the REAL value and see what your inductance is. A couple of tries changing N and l, you should be able to get close to your desired inductance...
Re: Looking For Cheap Crossover w/o any coils
OK, I am lost. If th coil is 0.000212 henries, how much will the diameter, turns and wire length be? Does anybody know? Also, what gauge wire muct I use? Or what would be the best?
Re: Looking For Cheap Crossover w/o any coils
Have you found this calculator?
http://home.new.rr.com/trumpetb/audio/chokejs.html
I can't vouch for its accuracy. Have you done a Google search? I found this by searching for "crossover inductance". Try various combinations of crossover, coil, inductance, calculator, etc.<p>Ron
http://home.new.rr.com/trumpetb/audio/chokejs.html
I can't vouch for its accuracy. Have you done a Google search? I found this by searching for "crossover inductance". Try various combinations of crossover, coil, inductance, calculator, etc.<p>Ron
Re: Looking For Cheap Crossover w/o any coils
Hunt down the Babani book (that's the publisher) number 103 coil winding, everything from RF to AF. Depending on the vintage of the book it will either wobble on about valve AF transformers and chokes or not. My edition cost me 30p back in 1976, well worth it.<p>However it has the wire gauges etc. Also the RSGB book has this info though not in such great detail, and you may not be able to purchase that in the US.<p>Colin
On a clear disk you can seek forever.
Re: Looking For Cheap Crossover w/o any coils
<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by mikea1962375:
OK, I am lost. If th coil is 0.000212 henries, how much will the diameter, turns and wire length be? Does anybody know? Also, what gauge wire muct I use? Or what would be the best?<hr></blockquote><p>You would obviously use wire thick enough to handle the power. House wire (solid) will work, but spacing's much better with enameled. A coil diameter of ~1.5 inches should be fine.<p>You're not interested in the length of the WIRE, but the length of the COIL---hence, the "iterative method" (also known as trial-and-error). Select length to be an inch or two, and calculate your turns; now wind the coil WITH that number of turns, and measure the REAL "l". Recalculate inductance with the real l, and see if you need to add or subtract turns...
OK, I am lost. If th coil is 0.000212 henries, how much will the diameter, turns and wire length be? Does anybody know? Also, what gauge wire muct I use? Or what would be the best?<hr></blockquote><p>You would obviously use wire thick enough to handle the power. House wire (solid) will work, but spacing's much better with enameled. A coil diameter of ~1.5 inches should be fine.<p>You're not interested in the length of the WIRE, but the length of the COIL---hence, the "iterative method" (also known as trial-and-error). Select length to be an inch or two, and calculate your turns; now wind the coil WITH that number of turns, and measure the REAL "l". Recalculate inductance with the real l, and see if you need to add or subtract turns...
Who is online
Users browsing this forum: No registered users and 101 guests