bizarre broblem i cant solve...

recently i acquired a reed switch! so i tried to make a simple relay with it

so basically i wrapped the reed switch with 0.3mm width wire. when i connect the wire with a battery. it works fine. i tried 1.5 v aaa battery and a small button battery. both worked

however when i connect the reed relay i made to arduino it doesnt work...

so i thought it was power problem. the button cell provided 1.4v 45ma

i am guessing the reed switch requires less than 20ma to operate

arduino provides 5v 65ma. so it is not the lack of power thats causing this problem.

i used the blink program on pin 13. when i connect - and + of the electomagnet wrapped around the reed switch, to pin 13 and gnd. the led on pin 13 stops turning on. and the reed switch doesnt operate.

is there something missing that i am not seeing...?

your help is Gratly appreciated!!

Do you have a multimeter to measure the resistance of the coil ? I mean the ohms for a DC current.

arduino provides 5v 65ma.

I hope that's not an (ex-)I/O pin you're describing there.

resistance virtually none i think about 4 ohms
i didnt wind it that much

pin 13 output 4.5v 65mah i am getting that figure

Hi, OK.. you need MANY more turns of very fine wire. DC resistance should be 100 ohms or more...

these r some pics of it i attached

as u can see i created two versions

the one with single layer coil operates well below 100ma

the one with 3-4 layers of coil operates well below 45ma

my arduino output on pin 13 provides 5v 65ma

however it doesnt aeem to provide any current to it when i connect + to pin 13 and - to gnd

also i have observed when i connect my relay mechanism, the led light on pin 13 doesnt turn on.

am i short circuiting this? becoz it doesnt seem to have any problem, it doesnt fry my board even after 30min connected.. im confused :o

and i have an arduino thats been in operation for 4months now
it is drawing more than 20mah on a pin at given moment(two reed relays at once).
it works fine

my arduino output on pin 13 provides 5v 65mah

Only until it is destroyed, possibly taking the rest of the Arduino with it.

Never attempt to draw more than 20 mA from a pin.

This is a reed relay for an Arduino : Arduino Your Home & Environment: Working with Reed Relays.
That relay uses 10mA and that is below the 20mA.

At 5V, your relay would draw 1.25A. That is 125 times more amps than that reed relay in the link.

You know that your relay is working. That's was a nice experiment. But don't try to use it with an Arduino.

40mA from any IO pin is the absolute maximum limit spec'ed in the datasheet. Exceeding that spec is likely to damage the part. You may have already damaged that IO pin.

The reason the LED goes out is that you're dragging the output voltage of the pin way down. A 4-ohm coil is practically shorted to ground as far as the microcontroller is concerned.

You're using the wrong units in your post.

"mAh" is milliamp hours, it is a measurement of charge (in practical terms, battery capacity). It has units of current multiplied by time.

"mA" is milliamp, it is a measurement of current.

So the current drawn by a coil (or anything else) would have units of mA. If you connected something that draws 20mA for 2 hours, you would use 40mAh; if you were powering from a 100mAh battery, the battery would be at 60% charge (60mAh left). (In theory at least - batteries never meet capacity spec in real world conditions).

(the A is capitalized - Ampere comes from "Andre-Marie Ampere", a pioneer in the field)

1- You have not posted any electromechanical design criteria for your DIY reed relay.
2- are you an electrical engineer ?
3- where are your magnetic calculations for the flux density needed for a reed relay ?
4- where are the voltage and current measurements for the I/O pin driving the relay ?
(voltage on the pin in ON state and OFF state/current drawn from the pin/voltage on arduino 5V pin during ON state)
5- Where are your calculations for inductance of relay ?
6-where did you get the idea to do this in the first place ?

Post a photo of the relay

Sounds like you need a dropping resistor.

Sorry for the late response and no pic, ipad problems, had to use my 6 year old galaxy tab

I am not an expert. I am just trying to get it to work

I have coto 9007 series reed relay that works fine.

The reed switch i got need rougly the same amount of power to operate.

I am guessing it needs around 10~20ma to operate. I am just trying to get the tiny electromagnet to operate directly from arduino pins

It shouldnt need too much power, it wont damage the board

Here are the things i found out so far

  1. The reed switch turns on with less than 50 turns of 0.3mm enamal copper wire, with a button cell providing 1.3v 45ma
    (the operating power could be reduced if igive it more turns of fine wire?)

(i could be wrong) 2. Arduino doesnt seem to work, when electromagnet is directly connected to its pins

3.aside from this relay problem, i would like to successfully operate a small electromagnet DIRECTLY off arduino pins , as it is the core of this problem

Well, you can find out how much current a coil will take at a given voltage - just measure the resistance across it, and use Ohm's Law. If that current is higher than 20mA, you can't drive that off an Arduino pin directly. If as you say it works with lower voltages (hence lower current), you could limit the current through it with a series resistor (provided you sized it correctly, and that the electromagnet works well enough with that current). Again, this is an ohm's law calculation.

You can't say "it shouldn't need much power" - you don't have the experience to say what a coil "should" need by thinking about it (as evidenced by your previous posts in the thread, you do not yet have a feel for how electronics work, hence you can't rely on intuitive statements like that). Do the leg work and calculate it.

Also - you're driving a coil, so you need to put a diode across the coil, with opposite polarity to the applied voltage (ie, band towards the side you'll be applying the positive voltage to). Coils make a voltage spike when you turn them off (opposite the voltage you had been applying), and you need a diode to clamp this so the spike doesn't damage stuff.

use a DROPPING RESISTOR (Google it)

thank you so much guys!
after much research and thought
i decided to use external power thats powering arduino and control the sigal with transistors.

That's not necessary ifvyou use a dropping resistor to limit the current and drop voltage.

The reed switch turns on with less than 50 turns of 0.3mm enamal copper wire, with a button cell providing 1.3v 45ma

45 mA is WAY more than any button cell was ever intended to draw. You should be using a AAA or AA battery for that.

I am guessing it needs around 10~20ma to operate

WHY are you GUESSING ?

Do you , or do you not have a meter to measure current. If so , why aren't you using it ?

Also - you're driving a coil, so you need to put a diode across the coil, with opposite polarity to the applied voltage (ie, band towards the side you'll be applying the positive voltage to). Coils make a voltage spike when you turn them off (opposite the voltage you had been applying), and you need a diode to clamp this so the spike doesn't damage stuff.

This is very important.

How do you know that the button cell is putting 45mA through the coil if you haven't measured it?

I bet it won't put out 45mA for long - with just 235mAH capacity for a large one such as CR2032
http://www.dipmicro.com/store/BAT-CR2032-L

it's gonna be dead pretty quick:
Standard Discharge Current 0.4 mA
Max. Cont. Discharge Current 3.0 mA <<< Well under 45!

Max. Cont. Discharge Current 3.0 mA <<< Well under 45!

Exactly.

If 1.3V puts 45mA through it then R is only 29 Ohms, 5V will result in 172 mA, that will blow an output pin in a few short seconds. Don't do it no more!

If 1.3V puts 45mA through it then R is only 29 Ohms, 5V will result in 172 mA, that will blow an output pin in a few short seconds.

This remains to be confirmed. We still don't know if the OP has a meter or knows how to use one to measure current.