diy relay coil

October 28, 2008 5:22 PM

DIY relay coil

I want to make a few coils for some reed switches I have to make my own relays.

I am using relays to isolate two systems. My control/relay trigger signal is 5v up to 20mA, but the lower the mA the better. The signal I am switching is 5v low mA.Both systems are battery operated, but the control system/relays can be operated from 9v wall adapter, control sig is still 5v.

I have a spool of awg37 enamel covered magnet wire for the coil. I built a small rig to wind the coil. I am not sure how much resistance I need to make the relay operate efficiently without overheating the coil, another component, or quickly depleting the batteries. The duration of closing the switch is short, but the frequency of closing is often.

The glass body of the reed switch is 1.5" long and slips inside of a straw, which is what I am winding the coil onto. I am not sure if I should make the coil over more of the entire switch body (1.5") and thinner. Or shorter (0.5 - 1.0"), thicker, and centered over switch contacts.

So far I have made one coil, of the longer thinner style, which took 500 wraps to get 21 ohms. Purchased relays I've used before were 9v 500 and 1k ohm. So if I need that kind of resistance these would be huge.

Can you help with a resistance range to shoot for and style of winding (long/thin or short/thick)? I found info on pickups and voice coils on google.

posted by sailormouth to technology (8 comments total)

You've got to balance a few things here. Generally speaking, more turns = greater magnetic flux = greater "pulling" power. But it also means higher resistance. This is good in that it reduces your current consumption, but bad in that (pure, inherent) resistance will reduce your magnetic flux. Magnetic flux can be concentrated, in your case, by increasing the number of winding layers (i.e. making the coil "fatter").

You've also got to consider the mechanical action of the reed; mechanical advantage and all that. Very roughly speaking, a little movement at the base of the reed takes a lot of energy but produces maximum movement at the contact end; at the contact end, a little movement takes less energy (but you've got to balance this with how fast / cleanly you want the contacts to operate).

Generalised version: make your windings as long as the moveable area of the reeds (i.e. ~ the internal length of the glass envelope), and as fat as you can. Current drain can be worked out with good ol' I=V/R (in amps, volts, and ohms), and that'll govern the length of wire required I'm not overtly familiar with US/Imperial gauges, so I can't give an indication of how long that'll be - consult a set of winding tables. Hell, a good set of winding tables will give you optimum dimensions for maximising flux over a given area, which can be scaled to suit your power requirements. Be aware that you'll run into diminishing magnetic returns as the current decreases (because the maximum magnetic flux / "pull" is a function of power, i.e. V*I).

Alternately, you could just go buy reed relay coils to suit. They're available in a variety of dimensions, voltages (I've seen 2.5v to 50v), & current ratings (5 ~ 200mA). I guess Farnell or Mouser would be the places to start looking.
posted by Pinback at 5:59 PM on October 28, 2008

(A quick calculation suggests that 37awg has a resistance of 1.716 ohms/metre. For 20ma @ 5v, you want 250 ohms, which is equal to ~ 146 metres.

You probably want a higher gauge / smaller diameter wire. Good luck winding that by hand... ;-)
posted by Pinback at 6:13 PM on October 28, 2008

First, I've gotta ask, why do you need to build your own relays when there are literally thousands of types available off the shelf for much less than a dollar.

Second, you would probably be better off with 42 AWG if you can find it in small quantities.

Third, just put a resistor in series to limit the current to what you need. You design the relay coil for a certain desired strength of field, then select the current and number of turns. Finally you tune the resistance with an external resistor to limit the current depending on the desired input voltage.
posted by JackFlash at 6:15 PM on October 28, 2008

Why are you trying to use a relay at all? Why not use an opto-isolator? If you were switching significant power, you'd need a relay, but you aren't.
posted by Class Goat at 6:33 PM on October 28, 2008

Thanks for the feed back so far.

I am wanting to build my own coil instead of buying because I have the reed switches and wire. I have had these for quite awhile and had not used them yet. If I can't make these work I will buy some.

JackFlash - I had tried a res in series but it was not working. I will try again.

Pinback - Thank you for your very clear description.
posted by sailormouth at 6:43 PM on October 28, 2008

Well, I misread the hard to read label on my spool. It is not awg 37 it is an 8 lbs spool of awg 43.
posted by sailormouth at 6:56 PM on October 28, 2008

If you know the part number for the reed switch you should be able to look up the actuation specs, usually given in milliTeslas or amp-turns at a specified distance. Then you can compute the strength of the magnetic field inside the solenoid coil.

Lets say you need 10 amp-turns to operate the switch. You have 500 turns so you need 20 mA of current. From ohms law 20 mA at 5 V requires 250 ohms. You have 21 ohms in the coil so you need to add a 220 ohm series resistor.

Well, if your experiment doesn't work, since your application has low voltage / low current, you probably want to use an opto-coupler switch like the 4N25. These cost about 30 cents and are much smaller than a reed switch.
posted by JackFlash at 7:39 PM on October 28, 2008

Thank you for the responses.
I have had to change the design so making my own coil is now longer needed. Making my own was really just to see if I could do it.
For those who recommended opto-coupler the relays are connected in parallel to a 4016 bi-lateral switch (4 switches in a 16 pin package).
posted by sailormouth at 10:31 AM on November 5, 2008