Wah Inductor.- Creation time.

[QSQCaito]

Ok so i've made severals(loots) of phone calls, and from all, i was told a 500mh inductor could be done. First i finally got to a house that had ferrite, they onley sold thoroidal and told me that i'd would be hell of a job to wire a 500mh inductor with that. Kind guy gave me another phone number. And finally got to a house with lots, loots , of ferrite. These are to make inductors. But these is how far i've arrived. I knew there was a formula given the Al factor to calculate amount of turns needed.

Well i have all these to choose between Click here

I really don't know which should i choose, which type of wire. In Argentina(and excuse it's in spanish. but i think it's deducible) i can't buy inductors, nor cheap wahs to steal to them. My only choice is to make an inductor, and im determined to do it 

bye bye

hope someone points me in the right direction

DAC

PS BTW which is the best value for wah inductors, 500.550.660?

Bye bye!

[Seljer]

this here may be of help: http://www.arielfx.com/inductor_guide/guide.htm

[danymal_X]

Hey, DAC

I was just reading through RG Keen's wah article and found this, thought you might be interested:

http://geofex.com/Article_Folders/wahpedl/wahped.htm#BM_ryo

Good Luck

dan

[QSQCaito]

this here may be of help: http://www.arielfx.com/inductor_guide/guide.htm

I though of that, get the same part.. but i can't get that part :/

I'll still searching though..

BTW, does anyone knows the formula involving the AL factor of the ferrite? i can't remember it

bye bye

thanks a lot

dac


Edit:

Hey, DAC

I was just reading through RG Keen's wah article and found this, thought you might be interested:

http://geofex.com/Article_Folders/wahpedl/wahped.htm#BM_ryo

Good Luck

dan

It's almost impossible to get dead wahs in here.. so expensive, no one would let them death..
Shelf inductors, i should put 500 in series 
And i've got no reference of any audio transformer of similar inductance, nor even can get that radio shack, no radio shack here.
After market wah inductors, bring them from usa, 15USD is lots of money, add at least 15USD more for shipping and all that..

I m left with the last chance:
Roll your own. Not recommended unless you have a personal calling from some deity or other. It's lots of work, and hard to find the very specialized parts and pieces...
It's the thing im trying to see.. i think it's just a matter of keep trying

bye bye
thanks a lot

DAC

[Paul Perry (Frostwave)]

I think you should talk to your local ferrite company. Tell them what you want to do, and ask their advice. Tell them you are thinking about manufacturing!!

Alternatives that have worked in the past include the output transformers from small transistor radios, and even the 600 ohm isolation transformers inside old fashioned telephones & early modems. Good luck!!

[R.G.]

BTW, does anyone knows the formula involving the AL factor of the ferrite? i can't remember it

It's N = 1000* SQRT(L/A_l) Where L is in millihenries. So if for example you had a pot core like the FE4508210 18mm x11mm, you'd have A_l=5290 and
N = 1000* SQRT(500/5900) = 291 turns.

If you used a potcore like the 18mm x 11mm FE4508222 with A_l = 400, then
N = 1000 * SQRT(500/400) = 1118 turns.

The advantage is that the second pot core is much more linear because of the air gap and saturates much less easily. But that's not what we need.

Get yourself a high A_lpotcore. Many real wahs were made with 18 x 11 pot cores.

Get the bobbin with external pins on it.

When you get it, measure very accurately the area inside the bobbin for winding wire. Then divide that area by the number of turns you need to make, giving the space available for one turn of wire, and take the square root of that to get the maximum diameter of wire you can use. That is an upper estimate.

You need to make a more conservative estimate by then selecting the next-smaller wire diameter that you can actually get, and dividing the actual wire diameter into the width of the winding area on the bobbin. That tells you how many turns you can really get on one layer of the bobbin. Then divide the total number of turns by the turns per layer to get the number of layers.

With the number of layers, multiply the number of layers by the diameter of the wire and see if that is less than the window area height. If it is less than about 90% of the height, you're in. If not, choose the next smaller wire size.

Then just go wind your wire on.

[QSQCaito]

First of all thanks a lot everyone who helped, and RG specially 

In 11x18, the one with highest A_l factor is the one you amrked FE4508210. I could use these, or i could use one with higher A_l factor as the 30x19 FE4508470 with an A_l factor of 11500.
I will try to make calculations with both, in order to determine whic is better to use.

N= Number of turns
L= value that i want to be achieved in milihenries
A_l= A_l factor.

The given formulae is N=1000*√(500/11500)=208.514~208
With the FE4508210, RG has already done it.

When you get it, measure very accurately the area inside the bobbin for winding wire. Then divide that area by the number of turns you need to make, giving the space available for one turn of wire, and take the square root of that to get the maximum diameter of wire you can use. That is an upper estimate.

Im actually watching the graphic of the POT CORE(both are pot cores) at the bottom of This file. And if i'm not wrong, i don't wind directly to the ferrite, i should give the turns to the bobbin reel(don't know if it's spelled like these, it's labeled "Carrete" on the drawing at the .pdf), not to the ferrite. So what i would indeed be measuring was this "reel" right?

If this is correct, lets supose the that reel measures are: 18mm x 11mm.
The diameter of the reel is 18mm. Given the formulae for perimeter=(perimeter being P, radius being R, diameter being D, π being pi)

P=2xπxR, which is equal to P=πxD

P=π*18
P=56.548P
PxH=WA.(H being height, WA being winding area)
56.548x11=622.03.

WA/N=S(S=Space available for one turn of wire)

622.03/208.5=2.983

√S=Maximum diameter of wire i can use.

√2.983=

Maximum diameter of wire i can use=1.727
If my calculation are correct( i think that's a reasonable value) the upper estimate of the maximum diameter of wire i could use is 1.727.


Upto there i could understood im stuck now:P i hate language barriers 

Well, i hope im doing fine upto now..

bye bye

dac

PS Calculations are not correct, it's just done to see if im doing things ok.

[R.G.]

For the 18 x 11 pot core (which I recommend because there were several versions of real Crybaby wahs with that size), the bobbin must fit inside the pot core. That is, its inner diameter will be bigger than the d3 diameter ( or > 7.6mm); its outer diameter will be less than d2 (or < 14.9mm), and its height will be less than h2 (or 7.3mm).

We know the bobbin reel will have some thickness. Let's assume that it is a uniform 1mm (0.039") thick everywhere. So the cross sectional area of the bobbin to wind is is
(d2 - (d3+1.0)) * (h2-2.0) = 6.3*5.3 = 33.39mmSq . We do not subtract from d2, because there is no outer wall on the bobbin, we add 1 to d3 because the inner cylinder is  1mm thick, and we subtract two from h2 because it has two thicknesses, one on each end.

We calculate that we need 291 turns, so the area per turn is 33.39/291 = 0.114mmSq, or a diameter of wire of 0.3387 mm. This is probably not an exact size. Looking up some metric wire sizes, I find size 0.32mm nominal size has an actual max diameter including insulation of 0.334mm. That ... might work, but it's awfully tight. I'm guessing that we'll have to drop to 0.250mm, which has a max diameter with insulation of 0.301mm.

So if the wire is 0.334mm, and we have h2-2.0 = 5.3mm, we can get 15.86 turns in one layer on the bobbing width. Fractional turns are a bad bet to make, so we'll say 15 turns per layer. That's 291/15 = 19.4 layers, or 19 full layers and one of six turns. Twenty layers high is 6.68mm, and the available height is only 6.3mm. Oops. The wire will spill over the bobbin even if we wind it perfectly. If we were able to actually make 15.5 turns per layer and get to only 19 layers, it would stil be 6.346 mm tall, and not fit in the pot core.

We'll try 0.25mm wire. The max diameter is 0.301, so each layer gets 5.3/0.301 = 17.6 turns. 291/17 turns is 17.1 layers. Call it 18 layers. That's 18*0.301= 5.4118mm high, and the wire fits neatly inside the bobbin, and the bobbin fits inside the core.

So you can wind a 0.5H inductor on an 18 x11 potcore bobbin with 0.25mm nominal wire, about #30AWG.

The perimeter length of the wire turn has nothing to do with the winding area. It's all the cross sectional area of the bobbin and wires.

Did that help?

[QSQCaito]

   

Blessed you , you got my homework done

Ok, it's really 17 turns and 10 more rounds right?, but as we are using max value as a reference, 7 more turns won't make that difference, right?

Well, i guess it's just time to go and buy the parts. I think this post will be helpful to many others in similar situation, i'll post my results when i finish, and i bet it'll sound good!

About winding, i should start from upper position, when i arrive to lower position, i scotch it, drive the wire up again, and start the second round, if i do it up and down and up.. they will just cancel each other, right? Or i just avoid scotching it, it will make it too thick.. can i stick it with multi purpose pegament?? the ones that dry fast?

Thanks a lot RG

bye bye

dac

PS i have to remember to measure the reel before starting, to see it's around the 1 mm speculated.
once again, thanks a lot.

[Paul Perry (Frostwave)]

About winding, i should start from upper position, when i arrive to lower position, i scotch it, drive the wire up again, and start the second round, if i do it up and down and up.. they will just cancel each other, right?

So long as you keep winding in the same direction around the core, it doesn't matter whether you are going up and down or not. You can wind up and down as much as you need. Maybe super glue would hold the wire.. but nail varnish is probably easier to work with, if you have a little patience!

[R.G.]

Good reply, Paul.

The exact way you wind wire affects how many wires you can get in the bobbin. The way I computed the wire size gives you an upper limit on the wire size, or alternatively tells you how sloppy you can be.

The more orderly you wind wire, the more turns fit inside a given space. In geometry, physics and mechanical engineering, they have determined that there are two maximally-dense ways to pack round things into a square space (looking at the cross section again here). Those are (a) cubic, where the centers of the round things form a square and the edges of the round things just touch, and (b) hexagonal close packed, where the centers of each successive layer ride right in the groove formed by the next-underlying layer. Both densities are the same, it turns out. But the idea is that each layer is rigidly orderly, each circle touching four (cubic) or six (HCP) others. If you break the order by not placing each circle exactly touching its neighbors, then there is space used up that holds nothing that could have held circles - or wire in the winding case.

So absolutely orderly level layers of wire is the most compact way to wind. Any randomness leaves air holes and makes the winding "fluffier" and therefore bigger. If you're right at the edge of fitting all the wire in a winding, then you absolutely must get every wire in the correct place. If you have some extra space available, then you have some tolerance for not fitting everything perfectly. It's hard to say how high random winding gets, since there are so many ways to wind randomly ... 

In the case of the bobbin we're using as an example, you have 18 layers, and the 18th is only barely used. The bobbin height would let you put almost three more layers on top, for 21 layers (almost, which counts in random because the layers are no longer strictly units). So some randomness would be OK. Too much randomness and the wire will be too tall to fit inside the bobbin again.

How small a wire can you use? It depends on the resistance of the wire. The resistance of copper wire is well known, and most wire size tables also give you the resistance, measured as ohms per 1k feet or ohms per kilometer. NOW is the time to calculate an average length of a turn of wire, known in the winding biz as "mean turn length".

The MTL is the average between the innermost turn and the outermost turn. In this case the innermost turn has a diameter of 8.6mm and the hypothetical outer one has a diameter of 14.9. So the average turn length is L = average circumferance = pi * (d3+1.0+d2)/2 = 36.9mm. And 291 turns is 10.74m of wire.

An ideal wah inductor has a DC resistance around 50 ohms, judging only from what I read. You can get that two ways, either by the resistance of fine wire, or by adding a resistor externally in series.

#30 AWG is 103.2 ohms per 1000 feet, and 338.5 ohms per km. we need 10.74m, or 0.01074km, so the wire resistance can be estimated at 3.63 ohms.

You can use smaller wire to raise the wire DC resistance and be sure that you have enough room on the bobbin, so you can use anything smaller than #30 or 0.25mm wire.

[Paul Marossy]

You're brave QSQCaito. Let us know if it works for you! 

[swt]

Hey! i live in Argentina also. where are you?. i'm from neuquen. one way to do a simple inductor is to cannibalize the ferrite from old radios, i've got plenty of them...the only problem is that they are a little bit noisy, because they are open inductors, but with a little patience, it works. i've done it, at least working ones. You will have to use 0.1mm wire if you want the sound of the old halo inductors...and maybe .07 for the newer ones. let me know how that goes...

[Paul Perry (Frostwave)]

swt, if you make two 250mH open ferrite inductors and join them in series (but beside each other & arranged so the windings are clockwise in one & anticlockwise in the other) you would have a 500mH "humbucking open ferrite wah inductor". The first in the world probably.
Seriously, I'm sure it would work! Good luck!

[Paul Marossy]

if you make two 250mH open ferrite inductors and join them in series (but beside each other & arranged so the windings are clockwise in one & anticlockwise in the other) you would have a 500mH "humbucking open ferrite wah inductor".

That's a brilliant idea! 

I wonder if it really could minimize hum picked from things that radiate EMI.

[QSQCaito]

Thanks a lot for the help, to all.

So, there will be enough space if i scotch after every layer, right???if i have enough space for 21 layers(if done tidy..)

I think im going to go with this wire.., to get the "desired" resistance, can i add resistors, in that way i could test with different resistances to check the one i like msot

About the winding,

So long as you keep winding in the same direction around the core, it doesn't matter whether you are going up and down or not. You can wind up and down as much as you need. Maybe super glue would hold the wire.. but nail varnish is probably easier to work with, if you have a little patience!

with this you ment to always, for exmaple, wire clockwise, all rounds clock wise, can these be done: (which would you recommend)



bye bye thank you all lots for your help

dac

PS Im not brave, i've got no choice.. well indeed, maybe a bit.. cause im not buying it

[R.G.]

if you make two 250mH open ferrite inductors and join them in series (but beside each other & arranged so the windings are clockwise in one & anticlockwise in the other) you would have a 500mH "humbucking open ferrite wah inductor".

It would work. My version of that was with four of Mouser's axial lead 125mH inductors mounted side by side with opposing winding directions. However, with pot cores, you have much less to gain by that. Pot cores have the windings sitting inside a magnetically-conductive shell. There's not much room for external flux to get in. It's one of the reason that Bell Labs came up with the pot core. Still, better is better.

with this you ment to always, for exmaple, wire clockwise, all rounds clock wise, can these be done: (which would you recommend)

Only one of those makes sense - the second one. The first one leaves a hump in the winding for the return of the wire. That return wire does you no good, adds DC resistance, and makes winding later layers harder.

Do use a thin tape after ever layer to keep things neat. Randomness creeps in from the edges.

[QSQCaito]

Thanks a lot RG! i will use a thin tape just for prevention

I couldn't understand what you ment with this :

Randomness creeps in from the edges.

bye bye

thanks a lot!

dac

[R.G.]

What I meant was, it's easy to make the middle of the coil smooth, evenly wound. Most coils start out neatly layer wound. Then there's a little glitch in the "bump" from the first layer to the second layer. When you wind back to the bump on the next layer, it's even harder to make the end right without an even bigger bump. So the ends of the coil are where the smooth, even layer winding breaks down and the wires get more randomly places as you build up more layers. What you want is 100% smooth even wound layers - in fact, that's what your wire size calculation is based on. The less smooth and even the winding is, the more space it eats up and the more likely you are to blow out the top of the bobbin.

The little glitches start at the ends of each layer and make succeeding layers harder to wind smooth and even.

Randomness creeps in from the edges - of each layer.

[QSQCaito]

Ok now i could get it. I'd pay special atention then, thanks a lot. I will try to make it as neat as possible, i have no hurry, i want a nice sounding wah

What i was thinking, maybe, was, as per round enters really a bit more than 17 turns, i could put some kind of"something" to make it enter exact 17 turns, that would probably help, ill just try-

Thanks a lot RG!


Bye bye


dac

PS: Can i add a resistor to reach desired 50ohm??