Help winding my own inductor?

20 millihenry, 1 amp, 25 KHz. Pot core, I presume. Don't know enough about these to know if I need a core with a gap or not. Willing to learn...
I have a 20 mH inductor (off the shelf) but is getting hot with the current I'm passing through it.
Where do I start? References and advice welcome.
Thanks,
--
DaveC
snipped-for-privacy@privacy.net
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

You need a high perm core. Large wire size makes for low I squared R as well.
Arnold makes some really good cores for this. Also use several smaller gauge wires in a litz configuration for even better I performance at your frequency.
Use a toroid. much easier than a pot core, and better as well. Easy to wind with Litz as well.
26Ga mag wire can do 1 amp, but I'd go a little bigger. Seven 32Ga mg wires makes a real good litz that can do over an amp easy.
Seven 30Ga wires makes for about just over single 24Ga, but performs much better.
Twist them together to form a single conductor.
Count only the turns *inside* the core center.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sat, 20 Dec 2003 12:19:17 -0800, DarkMatter wrote

I presume because I've only wound one or two in my lifetime, that I'm not familiar enough with these topics to understand your statement.
I always thought of winding a toroid much like threading a needle several hundred times over. To my mind, you must measure out the length of wire required and thread it through once; pull the entire length of wire through and tight to the toriod; thread again; pull entire length through... etc.
Unless the wire spool from which you're taking this from is *tiny*, or there is a gap of some kind in the core, you've got to "thread the needle".
Is there an easier way to wind a toroid? To my inexperienced hands, it really sounds like a bobbin in a pot core is going to be much easier. What am I missing, here?
Thanks,
--
DaveC
snipped-for-privacy@privacy.net
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
DaveC wrote:

I think you've got it. When you only have a few turns to wind, just stretch the wire out to its full length, and repetitively "thread the needle". If you have a lot of turns, you use a shuttle. I think you understand that, too, but if not this example might help: Imagine winding 100 feet of rope through a car tire that is not mounted on a rim. You could stretch out the rope and continually "thread the needle" until the tire was wrapped, pulling the full length through each time. Or, you could coil the rope into a loop and just keep passing the loop through, paying rope off the loop as you go. With a toroid, instead of making a loop of the wire, you wrap it around a shuttle - something that is , small enough to pass through the center hole of the toroid, but big enough to hold the length of wire needed. See the "drawing" below: _________________________ \ / /_______________________\ The wire is first wrapped lengthwise around the shuttle - you can make the shuttle from cardboard or plastic. You pass the shuttle through the toroid, back around the outside, and through again, paying out wire as you go. For smaller toroids, I've made shuttles (I think bobbins is the correct term in this case) from two pieces of brass tube. The smaller tube is inside the larger diameter tube, and is longer. The wire is first wrapped around the larger tube, then the whole thing is passed through the toroid in the same way as already described. The smaller diameter tube acts as an axle around which the larger, wire bearing tube can spin, paying out the wire.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sun, 21 Dec 2003 06:29:36 GMT, snipped-for-privacy@bellatlantic.net Gave us:

Excellent primer! Good job!
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Guys, guys, guys!
If you feed the trolls, they come back for more.
Let's stop playing "I'm righer". It's obvious who's got the wisdom that comes from years of experience. You've made your statements. Let the jury of your peers (and we beginners, too) retire in peace to ponder the evidence.
Peace,
--
DaveC
snipped-for-privacy@privacy.net
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Hello Dave, to see a picture of the shuttle, have a look here on page 8 http://www.quadesl.com/pdf/trans_inst.pdf
For tiny little toroid cores the size of a wedding ring the shuttle may be made from half a paddle pop stick from an ice lolly. Split longways.
For wire the size of coathanger wire, a shuttle about 12 inches by 1 inch is handy for a fist sized toroid. Just make a shuttle out of whatever material comes to hand. Scrap strips of printed circuit board material found under the guillotine are handy for making shuttles. 1/8 inch thick fibreglass material is good for the big shuttles.
If hundreds of turns of fine wire were involved, I used to lose count, so I cheated by bunching the turns into groups of ten turns or whatever number suited me and the job. The purists would argue that each turn had to be like a radial spoke.
Regards, John Crighton Sydney
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
DaveC wrote: (snip)

Not a thing.
--
John Popelish

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Well, it is, but I'll bet it would get hot too.
You make a "shuttlecock". It is a rectangular shaped object that is big enough to pass thru the core opening, and you wind the length of wire onto it before you begin on the toroid. Yes, it has to pass thru that many times. However many times that is.
I just wound four that operate at 19Khz. They were either 27uH or 27mH. I can't remeber which, but we tried several cores that were getting hot, and this one worked. This application was several amps.
I used 7 #22Ga for just over 16 gauge equivalent. It was about an inch and a quarter core, and it was like 12 turns.
Very high perm cores ended up getting used.
I didn't use a shuttle. Litz moves easily. I could post a pic, next week sometime.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
I read in sci.electronics.design that DarkMatter <DarkMatter@thebaratthe
com>) about 'Help winding my own inductor?', on Sat, 20 Dec 2003:

I LIKE it! just a factor of 32 times as many turns. Expert? Ha!
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sun, 21 Dec 2003 17:12:26 +0000, John Woodgate

--
1000 times, no, John?

Z = n through a transformer, but for an inductor, L changes linearly
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

No. Inductance increases with the square of the turns.
RL
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
I read in sci.electronics.design that John Fields <jfields@austininstrum
'Help winding my own inductor?', on Sun, 21 Dec 2003:

Oh, no, John, no, John, no, John, no! Not if the turns are close- coupled, as they are in a pot core.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sun, 21 Dec 2003 20:43:43 +0000, John Woodgate

Twice the turns equals 4 times the inductance.
It is easy. There is a square in the formula.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sun, 21 Dec 2003 20:43:43 +0000, John Woodgate

--
Aaarrrghhh!!! n = sqrt(L/AL)

Thanks for the reality check!-)
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
John Woodgate wrote:

How about if you adjust the gap to hold a constant flux for a given current as you change the number of turns?
--

John Popelish

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
inductor?', on Sun, 21 Dec 2003:

Then the inductance is *independent* of the number of turns, by the definition of inductance (induction per unit current).
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Mon, 22 Dec 2003 03:33:17 +0000, John Woodgate

How many Gilberts are we talking here...?
hehee...
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
John Woodgate wrote:

I have seen the definition of inductance as flux times area per current (1 henry = (1 tesla * meter^2)/amp)) but this gives my mind a snag. Perhaps you can straighten me out.
Lets say you have a large toroid core of infinite permeability with a small air gap sawed through it, and 1 turn through the hole. I measure the inductance and find x henries. So at 1 ampere, I have 1/2 x joules stored in the inductor and in the magnetic field in the air volume of that gap. I then saw the gap to twice as wide and wind a second turn through the hole, and get the same flux passing through this thicker but same area air gap, and still measure x henries. So by 1/2L*I^2 I still have the same energy stored in the inductor and in the air volume in that gap, but there is now twice the original volume of air stressed with the original flux. How can that be the same total amount of magnetic energy?
--

John Popelish

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
inductor?', on Mon, 22 Dec 2003:

The formula is L = N x [phi]/I, where N = number of turns, [phi] = flux = flux density x area and I = current.

There are now two turns, so L = 2x. If you hadn't widened the gap, the inductance would have been 4x.

It isn't. The inductance has doubled. From L = N[phi]/I, we get
L = NBA/I, B = induction, A = cross-sectional area
B = [mu]H, [mu] = permeability of air, H = magnetic field strength
and H = NI/l l = length of air-gap, since the rest of the circuit has infinite permeability
So L = [mu]N^2IA/lI = [mu]N^2A/l
To get L back to its original value, with 2 turns instead of 1, we need the gap to be 4 times longer.
If we then do a similar substitution for the stored energy:
LI^2/2 = (N[phi]I^2)/(2I) = N[phi]I/2 = N[mu]HAI/2 = N[mu](NI/l)AI/2 = [mu]/2 x N^2I^2A/l
N^2 has gone up 4 times and l has gone up 4 times, so the energy remains the same. H, B and [phi] doubled due to the two turns, but dropped by a factor of 4 due to the longer air-gap, giving a net halving.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.