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I'm currently learning electronics and right now, I'm trying to use a relays... without success. Here is a picture of my circuit:

enter image description here

First of all, don't mind the 2 cables on the top right and the relay on the bottom right, they are not used.

So, from what I understood about relays, here are my thoughts about this circuit.

When the button is not pressed, the relay does not receive any tension, so the pins connected to the lightning LED are active, which is why the LED on the top is lightning... That I get...

Now, when I press the button, a 12V DC tension is going inside the relay, so I was kind of hoping that it would make it switch and then the second LED would be lightning but... no...

The relay is an AXICOM IM26, you can find the datasheet of this one in here: http://www.te.com/commerce/DocumentDelivery/DDEController?Action=showdoc&DocId=Specification+Or+Standard%7F108-98001%7FV%7Fpdf%7FEnglish%7FENG_SS_108-98001_V_IM_0614_v1.pdf%7F4-1462039-1

However, a friend made me notice that this datasheet "talks about" a switching current of 2/5A but I have trouble to imagine that this small relay really requires that amount of current to switch, especially because it is writing "12 V DC" on it, which, for me, tend to indicate that it switches at 12V. But it would explain why it is not switching when I press the button.

Can someone explain me what is wrong with my circuit ?

Thanks

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    \$\begingroup\$ It would be better if you draw a circuit and didn't expect from us to understand it from following your protoboard connections. Anyway, many relays have polarity so (your also), did you make sure that you connected + to + and - to -? - And what type do you have, simple or set/reset? \$\endgroup\$ – Claudio Avi Chami Sep 3 '16 at 0:17
  • \$\begingroup\$ The "switching current" seems to be the contact current rating. The IM26 is coil is rated for 12 volt operation, and should require a minimum of 10.2 volts to operate, and should release when the coil voltage drops below 1.2 volts. The coil should only draw 4 mA at 12 volts. Pin 1 is the positive terminal of the coil - it may (probably?) won't work with reversed polarity. \$\endgroup\$ – Peter Bennett Sep 3 '16 at 0:29
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Your problem is very simple. You are using your breadboard incorrectly. The top pins (pins 1 and 8) are being shorted out by the contact strip. In this case, the breadboard is incompatible with the relay. The spacing between rows of contacts is 0.3 inches, which is a standard DIP IC pin spacing. Your relay, however, has a lateral contact spacing of 0.2 inches, and so cannot straddle the row separation. Since it didn't fit that way, you plugged the relay in to a set of holes that did fit, and that is the source of your trouble.

To confirm this, set up an LED and resistor to monitor your +12 volts. Unplug the relay and you will see that the LED lights up. Now plug the relay in and the LED will go out.

With this breadboard you really have no good way to mount the relay. If you have access to a soldering iron you can make extenders for the relay pins with short wires and plug them into the breadboard, but other than that I'm afraid you're out of luck.

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  • \$\begingroup\$ This actually makes perfect sense. I'll try to find a relay more adapted to a breadboard and then I'll do the same circuit again to see if this was the problem. Thanks for your help. \$\endgroup\$ – ssougnez Sep 3 '16 at 10:32
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    \$\begingroup\$ Ah awesome ! Didn't want to wait to have a new relay so folded a bit the pin of my relay to make it cross my breadboard and it works :-D Thanks a lot! \$\endgroup\$ – ssougnez Sep 3 '16 at 10:39
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Your relay has an 'ultra high sensitive coil' which suggests it is biased with a permanent magnet. If this is the case then you must apply voltage to the coil with the correct polarity, otherwise the coil will cancel out the bias magnet's force instead of adding to it, and the relay won't operate.

If the white stripe on your power supply cable is positive then it appears you have the polarity reversed (positive should go to pin 1 of the relay, and negative to pin 8).

this datasheet "talks about" a switching current of 2/5A but I have trouble to imagine that this small relay really requires that amount of current to switch

That is the rating of the contacts, ie. how much current it can switch. The coil requires 10.2V DC and has a resistance of 2880Ω, so it needs at least 10.2/2880 = 3.5mA to operate. At 12V the coil should draw 12/2880 = 4.2mA.

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AM26 is rated for 12V at 50mW

It comes in 3 ranges of sensitivity, this being most sensitive. If coil Power = 50mW @ 12V then current is 50mW/12 ~ 4mA

This is pretty easy to drive and you might be able to make it click with the 9V battery across the coil with + on pin 1. Although due to hystereis, the guaranteed or MUST SWITCH threshold is 10.20V and the MUST release threshold is 1.20V but it MAY switch in between.

FYI The lowest contact rating is 2A resistive. So this relay has a net switch gain at max load of 2A/4mA = 500 or you could say a a coil sensitivity of 0.2% or rated current for a 12V coil.

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  • \$\begingroup\$ The datasheet indicates that the minimum operate voltage is 10.2 volts, so it won't work with a 9 volt battery. \$\endgroup\$ – Peter Bennett Sep 3 '16 at 1:22
  • \$\begingroup\$ Relay specs did not give minimum "may switch" Voltage , thus it is the gaurantee level to function. Faster operation occurs at rated nominal voltage. However some may operate slower at lower voltages, some may not, but ALL parts will switch at Min. It was not a suggestion for design, just a quick test. \$\endgroup\$ – Tony Stewart EE75 Sep 3 '16 at 2:17
  • \$\begingroup\$ The switching pin are connected to a 12V tension source. \$\endgroup\$ – ssougnez Sep 3 '16 at 10:31
  • \$\begingroup\$ Did you measure +12v between pin 1 (+) and pin 8 (-)? It should work. If not , something is faulty. \$\endgroup\$ – Tony Stewart EE75 Sep 3 '16 at 15:24

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