I have several contactors with 24 VAC coils, but I need 240 volt coils. The coils are wound on a simple plastic form, and would be very easy to rewind. Is there a way to calculate what gauge wire to use and how many turns, so it has sufficient pull-in force but won't overheat?
Buy the right ones from either the wholesale house or an used equipment dealer.
On Mon, 2 Aug 2004 13:42:52 -0700, SQLit put forth the notion that...
If that was the answer to everything, there wouldn't be much use for this group, would there? The last time I got a quote on a coil for a contactor, they wanted $75, which was more than I paid for the whole contactor. Why would I want to do that, when the wire to wind a new one is worth about a buck? Maybe someone a little more knowledgeable has a more useful answer, but thanks anyway...
equipment
Buy the right coils.
Wire to rewind coil $1
Time spent on researching how to rewind the coil, coupled with time spent winding coil $500
Time spent fiddling around wondering why it doesn't work, and rewinding coil $1000
Explaining to insurance company why the place burned down....priceless.
Bill
On Mon, 2 Aug 2004 21:32:05 -0500, Bill Shymanski put forth the notion that...
That's how we learn things, Bill. If I wanted to just buy coils, I'd have done so. I'm looking for a knowledgeable electrical engineer who can actually contribute something besides a smart-assed answer... seen any?
[...]*plonk*
My opinion:
10 times the turns of the present winding with the largest wire size that will fit on the bobbin.The current should be 1/10 of the original and could be used to estimate what wire size you should be looking at.
On Mon, 02 Aug 2004 21:58:48 -0500, ED put forth the notion that...
It doesn't look like the bobbin would hold that much wire of the same gauge. Typically, the wire gauge is a lot smaller with a higher voltage coil. I could measure the coil resistance, but then there's the inductive load to consider. I think it would also depend on the contactor size, because bigger contactors have bigger armatures and stronger springs. What I'm hoping is there's a relatively predictable way to calculate this... perhaps something on the order of X number of turns of Y gauge wire at 12 VAC = x number of turns of y gauge wire at
240 volts. Maybe 20 times as many turns, and whatever gauge would give you the same current draw at the higher voltage, since the current (I think) is in direct proportion to the amount of magnetism produced.10 times the original turns for 24 volts AC will increase the inductive reactace for 240 volts AC and should give the same magnetic flux for both cases.
The increase in inductive reactance will limit the flow of current allowing you to use a smaller wire.
When you find out how many turns you have to have and have measured the bobbin to find out the available space for the winding, you can calculate the wire size that will fit on the bobbin. The wire size will probably be 1/10 the area of the 24 volt wire.
If you are looking to calculate the turns the formula for inductive reactance is Xinductve=2*pi*frequency*inductance
The inductance will depend on the permability of the core material, the area of the core, the number of turns as well as the length of the core.
There are a number of books and web pages that show the formulas. The hard part is filling in the variables with the correct numbers to come up with an answer.
I would count the number of the turns in the present coil and assume the enginer that designed it has sized it properly. Ratio to the new voltage and rewind.
As another has told you, your "new" coil would have 10 times the number of turns.
You MIGHT be able to determine the number of turns without taking the old coil apart.
You have to wrap as many turns of whatever magnet wire you have available around the existing coil. Energize the coil and measure the voltage. You can now determine the turns/volt on the voil.
Take the turns/volt and multiply by 120/240/24 (whatever) and you have the turns you need.
Have fun.
On Tue, 3 Aug 2004 02:59:55 -0400, John Gilmer put forth the notion that...
Sounds do-able. So you're saying the wire gauge isn't that important, as long as long as there's room for 10 times as many turns as the 24 volt coil has?
Greetings.
Shall I assume that you've already considered and, for some unknown reason, gainsaid the idea of purchasing a quality, low power 240VAC to
24VAC transformer and driving the contactors with the voltage they were designed for? Admittedly, my implied answer doesn't answer your question, but it does seem like a better solution.Cordially, Richard Kanarek
P.S. Don't forget the fuse(s) (everywhere)!
P.P.S. In reply to your most recent post, I'm not sure that offering contactor coil rewinding advice is the raison d'etre for this group, although I confess that it might be better than the ones that (apparently) are.
| That's how we learn things, Bill. If I wanted to just buy coils, I'd | have done so. I'm looking for a knowledgeable electrical engineer who | can actually contribute something besides a smart-assed answer... seen | any?
For something as simple as a contactor coil, most knowledgeable electrical engineers I am sure will just spec the voltage and force needed and "buy the right one". But if your goal is to become one of those knowledgeable electrical engineers who is in a position of specifying the manufacture of them, then I suggest a knowledgeable electrical engineering teacher. In lieu of enrolling in an EE program at a nearby university, you might be able to find the information you need online with the help of Google. I know I have seen exactly that info online. But I didn't bookmark it since I didn't need that info myself.
Personally, I'd think winding your own contactor coils would be boring. But I suppose I might ask the question much like you did when my real objective was to build a massive earth orbit launching solenoid. Obviously I wouldn't want anyone else to know my real plans so they wouldn't realize what a lunatic fringe fool I really am.
On Tue, 03 Aug 2004 10:54:24 GMT, Richard Kanarek put forth the notion that...
I've considered that option. It's the easiest solution, but space is a consideration. Also, I'd just like to know how to do it.
| Sounds do-able. So you're saying the wire gauge isn't that important, | as long as long as there's room for 10 times as many turns as the 24 | volt coil has?
Just guessing, here, but I'd say that the total _volume_ of wire would remain a constant to get the same amount of "kick" for the same amount of power at the same frequency with the same surface for dissipation of heat. Physics often has a nice way of doing things like that. You're just changing the voltage to current ratio (something a transformer does with two windings).
Oh dear. I've been "plonked" by Red Green. However will my fragile self-image survive this?
Bill
On 3 Aug 2004 16:05:58 GMT, snipped-for-privacy@ipal.net put forth the notion that...
This is what's got me temporarily fascinated. It seems like a relatively simple problem, but there are several factors that have to be taken into consideration... coil resistance, inductance, magnetic flux, physical space, etc. I don't want to do this for a living, but I'd like to know how to calculate what the physical properties of the coil need to be... partially because I have several contactors I want to convert, and partially because it's a good learning exercise. I suppose I could buy one and dissect it, since I have three contactors I want to use at
240 volts, but I'm betting someone in here knows an easier way.| On 3 Aug 2004 16:05:58 GMT, snipped-for-privacy@ipal.net put forth the | notion that... | | |> On Tue, 3 Aug 2004 01:09:49 -0700 Checkmate wrote: |> |> | Sounds do-able. So you're saying the wire gauge isn't that important, |> | as long as long as there's room for 10 times as many turns as the 24 |> | volt coil has? |> |> Just guessing, here, but I'd say that the total _volume_ of wire would |> remain a constant to get the same amount of "kick" for the same amount |> of power at the same frequency with the same surface for dissipation of |> heat. Physics often has a nice way of doing things like that. You're |> just changing the voltage to current ratio (something a transformer |> does with two windings). | | This is what's got me temporarily fascinated. It seems like a | relatively simple problem, but there are several factors that have to be | taken into consideration... coil resistance, inductance, magnetic flux, | physical space, etc. I don't want to do this for a living, but I'd like | to know how to calculate what the physical properties of the coil need | to be... partially because I have several contactors I want to convert, | and partially because it's a good learning exercise. I suppose I could | buy one and dissect it, since I have three contactors I want to use at | 240 volts, but I'm betting someone in here knows an easier way.
Sure, resistance will go up. And inductance will go up. They should since the voltage is going up and the current is going down. For 10 times the voltage, impedance needs to be 100 times as much to get 1/10 the current to keep the power level constant.
Here are some example figures for a 6 watt load at several voltages common in North America:
amps: 0.50000000 ohms: 24.0000000 volts: 12.0000000 watts: 6.00000000 amps: 0.25000000 ohms: 96.0000000 volts: 24.0000000 watts: 6.00000000 amps: 0.12500000 ohms: 384.000000 volts: 48.0000000 watts: 6.00000000 amps: 0.05000000 ohms: 2400.00000 volts: 120.000000 watts: 6.00000000 amps: 0.02886751 ohms: 7200.00000 volts: 207.846096 watts: 6.00000000 amps: 0.02500000 ohms: 9600.00000 volts: 240.000000 watts: 6.00000000 amps: 0.02165063 ohms: 12800.0000 volts: 277.128129 watts: 6.00000000 amps: 0.01732050 ohms: 20000.0000 volts: 346.410161 watts: 6.00000000 amps: 0.01443375 ohms: 28800.0000 volts: 415.692193 watts: 6.00000000 amps: 0.01250000 ohms: 38400.0000 volts: 480.000000 watts: 6.00000000 amps: 0.01000000 ohms: 60000.0000 volts: 600.000000 watts: 6.00000000
So if you have 10 times the wire length, at 1/10 the cross sectional area (1/3.16227766 the width) you'll end up with 100 times resistance. But you'll also have 100 times the inductance, so the resistance will still be about the same proportion. Model the wire as 10 resistors in parallel that you are changing over to 10 resistors in series.
Loss is proportional to I^2*R. Since I is 1/10th as much and R is 100 times as much, it comes out the same. It should get just about as hot.
The issue you could run into is that the wire insulation might have to be thicker to handle the higher voltage. That can end up making the whole coil assembly larger, requiring longer wire, introducing more resistance. You can compensate with larger wire, but now the assembly gets even larger.
Aren't we forgetting something here?
I see no mention of the temperature and voltage rating for the "new" wire insulation.
Assume the insulation has to be 10 times higher rated to accept the higher voltage. That takes up space on the bobbin that may prevent the success of the project due to size constraints.
Also assume the surrounding components need to be able to withstand 10 times the voltage. Some materials are fine at 50VAC or lower but are not well suited to voltages above 160.
Be careful with the seemingly simple answer... it is more detailed than one might think at first blush...
:)
On Tue, 10 Aug 2004 04:15:38 GMT, John Smith put forth the notion that...
Manufacturers use the same contactors for different voltages. The only thing you have to do to change them, is take out the existing coil and drop in one wound for the correct voltage. They're all wound with ordinary enameled magnet wire as far as I know. Skinny wire with lots of coils for higher voltages, and fatter wire with less coils for lower voltages. Furnas makes a dual voltage coil for some of their contactors, where you either connect two windings in series for 240 volts, or in parallel for 120 volts.
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