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I have two relays to convert +12 volt circuits to "ground" inputs into a fuel injection ECU. If I connect the relay coil ground to the same ground as the "source" will I need to install a diode to prevent possible fly-back to the ECU?

Here is the diagram:

enter image description here

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    \$\begingroup\$ Are you sure they're high-side (12v) switched outputs? It's more common to ground the relay coil to switch it, saves having to generate 12v in the ECU logic side. \$\endgroup\$
    – John U
    Feb 18, 2014 at 15:34

2 Answers 2

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Sometimes, relay coils are used together with switches, and no freewheeling (flyback) diodes are used. This will work, but every time you open the switch contacts, an arc will burn for a short time, which shortens the lifetime of your switches.

With a transistor output driving a relay, a freewheeling diode is absolutely necessary, because the voltage spike will destroy the transistor.

When using a switch to turn on/off a relay, omitting the diode will work, but your switch will be happier when you use a diode (one individual diode for each relay). The best place for the diode is right at the relay, Cathode at input from switch ("86"; in case you switch the + end, which is indicated in your diagram) and anode at GND ("85").

Good diodes for this purpose are those of the fast switching type, e.g. a 1 A / >=50 V Schottky barrier. A SB160 or SB1100 or similar will likely be o.k. Note that on automotive supply rails, there are often high voltage / high energy spikes caused by turning off other inductive loads. Therefore allowing some safety margin and using a 100 V diode won't hurt. Those parts are cheap anyway.

Note: You say If I connect the relay coil ground to the same ground as the "source"... This will not decide if you need the diodes or not. Actually, if you were not using the same ground for switch, battery and the relays, no current would flow at all. Including the battery, switches and return paths (GND) into your diagram makes things clearer (even though it looks a bit messy):

complete diagram

Please make sure that this diagram is correct (relay coil between 86/85, common contact at 30, NO at 87, NC at 87a). Also confirm that the GND (chassis) connections are o.k. like drawn here.

You say you are worried about damage to the ECU. Since the battery is not a perfect voltage source, the flyback pulse will lift the positive end of the battery somewhat, and there may be a spike. However, in a car, there are loads much worse than a relay, causing much higher spikes. Any good automotive gadget should tolerate these spikes.

Some background info:

You need the flyback diodes because at the instant you open the switch, a relay's inductance will try to maintain the current. Adding a diode will create an easy path for this current to circulate until the relay's inductance has lost all its energy.

Charging the inductance from the battery and via the switch: charging the relay's inductance

Discharging the inductance via the diode: discharging the relay's inductance

Note how the direction of the current in the relay itself does not change and satisfies the rule that an inductance does not allow a rapid discontinuity of its current.

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  • \$\begingroup\$ Some supporting info including appnote from Littelfuse in this thread: electronics.stackexchange.com/questions/55371/… \$\endgroup\$
    – John U
    Jan 28, 2013 at 13:45
  • \$\begingroup\$ Thank you for your help and clear explanation of the circuit. I was fairly sure I was going to need to install a diode to help protect the ECU. Here is a link to the "rest" of the diagram I came up with for this application. It includes an 8-relay circuit board which already has diodes installed. \$\endgroup\$
    – BPT Racing
    Jan 28, 2013 at 18:54
  • \$\begingroup\$ (i382.photobucket.com/albums/oo264/bptracing/…) \$\endgroup\$
    – BPT Racing
    Jan 28, 2013 at 19:11
  • \$\begingroup\$ @zebonaut It might be a dumb question but I am just curious to know. What will happen if we switch ground and not the +12V. Will it be safe to do so? \$\endgroup\$
    – user49520
    Jul 17, 2014 at 6:30
  • \$\begingroup\$ Unless there is some "hidden" connection to groud (like a metal case of the relay connected interneally), both ways will work just fine. The current path for the "on" switch (red circle in first image) will be completed no matter if you switch the high side (+12V) or the low side (GND). With the switch open, it's just a matter between the relay coil and the diode. To be sure of any hidden connection, open the switch and the ground connection (where your switch goes then) and check the continuity from the relay coil to GND (chassis) and +12V. Your Ohm-meter should read infinite resistance. \$\endgroup\$
    – zebonaut
    Jul 17, 2014 at 6:54
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This may be late coming, but might be useful for the next reader...

I'm using automotive relay TE 1432793-1, which is a SPDT Panel Mount Non-Latching Relay, 12V. You can get it from RS Components. The relay has this following diagram on it's side. You can see that the diagram even gives a polarity for connection, which is only necessary in a relay's case if there is a protection diode. Also, it's data sheet is shown below. Both clearly indicate that the relay has a protection diode built in. So in this case, my answer is no. You do not need a flyback diode with all automotive relays.

relay side relay datasheet

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