Sizing Diodes

The shunting diodes are commonly used to protect switching transistors. If small contacts are switching larger contactor coils off and on, then you probably don't need need a shunt. But if you were concerned about contact damage from frequent cycling then you might want to include one to maximize the life of the contacts.

You typically only need to select a diode that can withstand a reverse voltage that you use to power the contactor. A diode with a Forward current of an amp or more would most likely take care of it unless it is a really big contactor coil. The collapsing magnetic field will probably deliver a peak current of 10 to 100 times the forward current but only for a few microseconds. That isn't long enough to do any significant damage.

Just make sure you reverse bias it when you hook it up or you may damage your DC circuit.

From first principles, on partially-inductive DC loads the shunt diode is there to prevent the formation of a large back-EMF when the magnetic field in the load collapses urpon de-energising; the magnetic field is trying to sustain the forward current through the device and cannot do so as a contact has opened. The shunt diode is there to suppress it. Therefore the shunt diode must be rated no lower than the forward current through the partially-inductive load.

So, if the relay, contactor or whatever has a forward current of, say, 5A then a shunt diode of a minimum of 5A with a reverse bias withstand voltage in excess of the supply voltage should be selected. Correctly connected, this diode will protect the contacts on everything wired upstream of it that could switch it.

If the relay contacts are switching 'something else' DC partially inductive load, then the something else needs a shunt diode across it as well, again selected such that its forward current equals or exceeds the load current so that, when the something else is de-energised, its own back-EMF is suppressed thereby protecting the contacts of everything upstream of it including, one suspects, the contacts on this particular relay/contactor.

Therefore it is wise to shunt every partially-inductive DC load, and not just the coil on this relay/contactor. It is each load that requires the shunt diode, and not the contacts.

Many manufacturers' DC relays today already have a shunt diode included in the package so as to save on external connections. Check the datasheet on each one, as appropriate. However, the final load may 'need and not have', and it is wise to add one in as part of the design process for the circuit.

Or if you're really strapped for cash, build your own...

http://www.aaroncake.net/circuits/motorcon.asp

Yes, they cause all kinds of problems. Starter manufacturers make arc suppressors or you can use "MOV"s. The best answer is a DC Drive, Reliance (and others) have been making good ones for years. They don't cost that much an you don't have to reinvent the wheel. Good luck..

Thanks folks for all the ideas. I wanted to do the small DC drives but the only power source is 125DC, directly across the battery bank and all the drives I found were AC input. I think we will just use some 1A, 600V diodes on the coils. manufacturer finally came back and said the coils drew 0.5A. For some reason they didn't have that information available on the literature.