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I'm quite new to the electronics, and I'm currently reading a book to learn it. In the chapter I'm in, the author is talking about relays, so I tried to make a simple circuit using a relay IM46.

Here is the circuit I made:

Breadboard

I understand that the LED is lighting, I don't press the switch, so the current does not flow through the relay (if I might say), so this is logical for me so far. The not-so-logical thing (for me), is that when I press the button, the LED goes off and the other stays off. I don't understand why. From what I understood, as the current flows though the relay, the coil should react and then activate the "other side of the relay".

Well, apparently I'm wrong (or I'm using a wrong relay type). I checked the voltage drop on the resistance, it is about 12V.

The input voltage is 12V DC (as written on the relay).

Does anyone see the mistake I made?

EDIT

As asked, I tried to make the schematic version of this circuit:

enter image description here

When the switch is not pressed, D1 is on and D2 is off. When I press the switch, D1 is off and D2 is on. At least, that's what I want.

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    \$\begingroup\$ Please post a circuit. It is not fair to expect someone to figure out what you are doing from a photograph. \$\endgroup\$ – Barry Sep 30 '14 at 20:23
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    \$\begingroup\$ Can you post the schematic, its difficult to see whats going on from the top of a bread-board. \$\endgroup\$ – Jaxxs Sep 30 '14 at 20:23
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    \$\begingroup\$ If you edit your question, pressing Ctrl-M will bring up a schematic editor. Please attempt to draw a schematic for what you've wired up. \$\endgroup\$ – Adam Lawrence Sep 30 '14 at 20:26
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    \$\begingroup\$ Pin 8 of the relay should go directly to the negative terminal of the power supply, not tho the junction of the resistor and LEDs. \$\endgroup\$ – Peter Bennett Sep 30 '14 at 20:28
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    \$\begingroup\$ Looks like you're powering the relay through the LED dropping resistor. Use the full 12V (move the black wire to the other side of the resistor). \$\endgroup\$ – Spehro Pefhany Sep 30 '14 at 20:29
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If you really using the relay you linked to: that is a bistable relay, which means that current through the coil in one direction switches it to one position, and you need current through the coil in the other direction to get it in the other position.

enter image description here

The behavior that you seem to expect is that of a normal, monostable relay.

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    \$\begingroup\$ Ah... Didn't know about that particularity but it is good to know. I'll order a mono stable and try my circuit again. If it works, I'll mark your answer as correct. Thanks \$\endgroup\$ – ssougnez Sep 30 '14 at 21:06
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Apart from it possibly being bistable, what luck you have, accidentally finding the bistable fork (not quite a needle, but still) in the haystack of cheaper mono-stable relays:

Your schematic does not jive with your breadboard. One end of the relay coil on your breadboard goes to the switch, as drawn, but the other end you connected between the resistor and the LEDs. Apart from that the centre contact of the relay and the switch go to the same rail, which is oposite from the one the resistor comes from, so I'm going to make a small guess here that your resistor is in the V+ rail, not ground and that the switch and centre-contact are in GND. Now when you push the button the relay gets this:

schematic

simulate this circuit – Schematic created using CircuitLab

Now refer to "where you say you measure" and check that if there's 12V across the resistor there must be 0V across the relay (and any LEDs that become put parallel to the relay by the push of a button). So the relay has too high a coil current requirement to work with the 510 ohm resistor. This is quite predictacble.

Move the black wire from the resistor's LED side to the resistor's power-rail and you have a slightly improved chance of at least once seeing something happen.

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