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I am new to Electronics and I am currently trying to wire up a SPDT Relay. I must admit I am totally confused. I want to have a SPDT switch that simply switches the relay over. Any pointers would be appreciated. When I currently energize the relay, I hear it cycling, which I would image that is not good. I simple want it to "switch" normally on and then when I energize the coil, switch is back.

Here is a copy of the Datasheet: enter image description here

datasheet from radioshack

Edit - Solution

Bread Board Solution

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  • \$\begingroup\$ How have you wired it up ? \$\endgroup\$ – Rocketmagnet Mar 7 '13 at 16:08
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    \$\begingroup\$ Take a picture of what you have now (i.e. your wired up breadboard). You will get more better/gooder answers that way. \$\endgroup\$ – Hair_of_the_Dog Mar 7 '13 at 16:15
  • \$\begingroup\$ Yes, but its not right. I guess for now I am looking for pointers on which pins go where. I know its vague, but like I said, I'm struggling a bit. \$\endgroup\$ – PhillyNJ Mar 7 '13 at 16:16
  • \$\begingroup\$ @PhilVallone - It would really help us to help you if we can see what you have now. Even if it's not right. \$\endgroup\$ – Rocketmagnet Mar 7 '13 at 16:19
  • \$\begingroup\$ So that we can explain why it's right that way and wrong your way. We're focused on learning rather than end products. Or at least I am. \$\endgroup\$ – user17592 Mar 7 '13 at 16:34
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I added #'s to the pins in the diagram to help describe wiring.

Using pins 3 & 4 - when a current flows between those pins it creates a magnetic field that then moves the internal contact.

When the relay is unenergized pin #1 is connected to pin #5, this is called Normally Closed (NC).

When the relay is energized (power applied) Pin#2 is connected to pin #5. In the unenergized state pin #2 and Pin #5 are not connected -i.e. they are Normally Open (NO).

This has a DC coil and the various specifications show you that when you supply at least 3.5 V at 89.3 mA then the contact closes (Pickup).

The contact will stay engaged until the voltage drops back to the 0.25V level at which time it disengages.

This difference in pickup and drop out is called hysteresis and helps prevent relay chatter if the input signal swings too much.

To wire this into a circuit, (it is rated for 1A at 120V AC) you would break the line wire, and connect the relay pins #2 and #5 inline. Neutral wire will remain connected. in #5 should go to the load, Pin#2 to the source (for safety reasons).

You will need to supply a 5 V source to Pins #3 & #4 that is capable of at least 100 mA.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Cool - I will take some some tonight to read and disgust your response. I'll post back tonight or tomorrow! Thanks \$\endgroup\$ – PhillyNJ Mar 7 '13 at 20:29
  • \$\begingroup\$ Posted edited with a link to my bread board. I am struggling with how to wire up the coils. Thanks for the help. \$\endgroup\$ – PhillyNJ Mar 8 '13 at 2:36
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This relay is about as simple as it gets. The two pins in the bottom left of the pin diagram are the coil. That's what the curly looking symbol means. If you apply 5 V accross those pins, the switch part of the relay will go one way. When the 5 V is disconnected from the coil, a spring in the relay causes it to switch back to the other state.

The bottom right and the two top pins connect to the switch part of the relay. The bottom right is the center of the "pole" of the SPDT (Single Pole Double Throw) switch. The pole is always electrically connected to one of the two top pins. It's a little unclear, but I'm guessing the top left pin is NO (Normally Open) and top right NC (normally closed). In this case "normal" refers to the coil not energized. With no voltage on the coil, the bottom right pin will be connected to the top right with the top left unconnected. When the coil is energized, the bottom right will be connected to the top left with the top right unconnectd.

There is a possibility that the two top pins are opposite of what I just said. This is easy enough to verify with a ohmmeter or continuity tester.

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  • \$\begingroup\$ Thanks for the help. I will read this tonight and respond. \$\endgroup\$ – PhillyNJ Mar 7 '13 at 20:28

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