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I have a Bosch type 5-pin relay with a built suppression resistor on the coil.

I've read that a resistor is not as efficient as a flyback diode at suppressing voltage spikes when the coil de-energizes.

Can I keep the resistor in place, and add a flyback diode across pins 86 & 85 thereby having both the resistor to dissipate the voltage and the diode to reroute the voltage? Is there any harm to this method?


Thank you everyone for chiming in. I know very little about electrical wiring and I appreciate everyone with more knowledge than me helping me with what I'm trying to do. I'm installing a water methanol injection kit into a turbo charged track car. In the attached diagram, the AEM main controller drives the pump and solenoid that attaches to the injection nozzle to control flow and prevent leakage.

The AEM Failsafe is a computer with a flow sensor that monitors fluid flow. Too high a flow and I could hydro lock the engine, and too low and I could have detonation. Under too high or too flow (alarm conditions) the AEM Failsafe computer will trigger two output signals. The Yellow wire output is +12V output trigger, and the Blue wire is Ground trigger. These triggers can be used to trigger a buzzer, LED light, or to the Engine management to cut turbo boost, etc...

In my application, I intend to use the Yellow trigger wire (+12V) to trigger relay #1 to cut power to the fluid pump. Yellow wire to relay pin 86, pin 85 to vehicle chassis Ground. The relay will run as NC (using pin 87A), only cutting power to the pump when energized (alarm condition).

The Blue trigger wire will trigger Relay #2 to control the solenoid. Since the Blue wire is a Ground wire, I will connect to pin 85 on the relay, pin 86 to +12V. The solenoid positive wire will connect to pin 87A, 30 pin to +12V.

The AEM Failsafe is prone to damage, many user assuming from the yellow and blue trigger wires. That's why I'm trying to use a diode across the relay coil. But the relays I purchased already have a resistor on the coil.

Anyhow, Is my overall diagram kosher???

Thank you everyone :)

enter image description here

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closed as unclear what you're asking by Matt Young, laptop2d, Sparky256, winny, PeterJ Mar 14 '18 at 13:47

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ I don't think you can add the diode across the pins 86 & 85. You should try pins 13 and 14 instead. \$\endgroup\$ – Harry Svensson Mar 10 '18 at 19:16
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    \$\begingroup\$ It's also probably worth noting that it might be a good idea to explain the problem you're solving! \$\endgroup\$ – Marcus Müller Mar 10 '18 at 19:40
  • \$\begingroup\$ Clear as mud to me... \$\endgroup\$ – Sparky256 Mar 14 '18 at 1:43
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I assume you have looked at the Bosch data on wiring the relay.

You will notice from that data that the relay coil (Pin 85-86) is NOT polarity sensitive. This is probably the reason they used a resistor and not a diode in the unit.

I also assume you have some reason for needing a diode suppressor on the coil ...possibly because you are driving the coil from a BJT or FET rather than a simple switch.

The data (on page 7) actually shows one schematic where they added a diode exactly as you might require:

enter image description here

You can put the diode across the resistor without concern, but you have to be conscious of the polarity you used in your wiring (which pin 85 or 86 is wired to 12 V positive). The diode cathode must ALWAYS go to the positive coil terminal you used.

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One drawback to adding such a diode is that the relay will stay energized a bit longer as the current "freewheels" through it and and back through the coil inductance, whereas a resistor will dissipate that energy more quickly. If the relay is switching very quickly, the diode may get too hot and fail. The advantage of such a diode is that it will clamp down voltage spikes and protect the transistor from collector-emitter breakdown. You can put a resistor (of lower value than the built-in, that is) in series with that diode to trade faster L/R time constant against max voltage spikes, but the appropriate trade-off depends, of course, on how fast your switching needs to be and how much voltage spike you can tolerate.

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