Wiring a relay with a TVS diode for protection.

Forum 2005-2010 (read only) Hardware Interfacing
1

Hello,
I am attempting to wire up a 6 pin relay to my arduino. I am confused at this point and looking for some help.

Page 2 has the schematic. The left most pins are obviously what gets switched (A or B get the power). But what are the other 4 pins? Why do two of the pins have a resistor between them (or is that the coil).

I want to control a small lightbulb with my arduino, so how would this be wired up? Do you suggest using the microcontroller to drain (pin at 0) or source (pin at 1) in order to turn on the relay? (also see note on TVS below)

Next I understand I would need a TVS diode to protect my arduino from the spike that the relay can generate when switching back to off. I am confused because this spec sheet says:

Surge Voltage Between Coil and Contacts: 1,500V FCC Part 68
(10/160[ch956]s).

So, ok 1,500V. However the TVS diodes i find online are all rated in Watts, and I have no means to convert from V to watts because I do not know the current, or is the current constant based on what voltage/current I have been applying to the relay?

Also how would I wire up the TVS diode? How would this change depending on whether I use source (1) or drain (0) on my microcontrolelr to turn on the relay?

Thanks for the help!

Lucas

2

Why do two of the pins have a resistor between them (or is that the coil)

Yes that's the coil. The other two pins at the far end are both the same, they are the common line or what gets switched.

You need a transistor to drive this relay from the Arduino. Please look it up in the playground.

The diode is placed across the coil with the cathode to the +ve end of the coil. Any rectifier diode like the 1N4004 will do. The 1500V it talks about is the isolation between the coil and the contacts, it's nothing to do with driving the relay.

3

got it, thanks