I am working on a project where I need a relay, and bought an NO HKE CMP8-S-DC48V-A relay for the job. This is my first time working with relays though, and while I more or less understand what they are and how they work, I can't seem to figure out how to wire it, and most tutorials I've looked at don't demo this type of package.

I'm testing the relay on my breadboard. By looking at the diagram on their site, I've connected my "controlling" circuit to the pins at the bottom for the coil (the ones to the side that are close together) and I've connected the circuit that I want the relay to switch on the other two pins (the bigger ones close to the middle). Both circuits are just resistors hooked up to the same power source for the moment, with a switch on my controlling circuit. Apologies for making it look so sloppy, the package is not breadboard friendly.

Relay circuit

So what I expect is that when the switch is off for the coil of my relay, that there will be no current flowing the red wires, and through the 10K resistor. Yet, when the switch is off and the power is connected, I get ~12V across the 10K resistor measuring with my multimeter, while no current is flowing through the resistor in my controlling circuit with the coil. If I toggle the switch on, current obviously flows through both resistors.

I find it confusing that an NO relay is passing current while its coil isn't energized. What am I missing here?


Here is the schematic of the circuit


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Voltage can be present up to the open contacts, which means no current flows. If you draw the circuit then we can all see what you have... \$\endgroup\$
    – Solar Mike
    Oct 7, 2018 at 16:09
  • 1
    \$\begingroup\$ Why do you have a 47K resistor in series with the relay coil? You should apply the specified coil voltage directly to the coil, with no series resistor. \$\endgroup\$ Oct 7, 2018 at 17:17
  • \$\begingroup\$ With a 12v supply and a 47K series resistor, your schematic cannot possibly reflect a reality where the relay coil energizes enough to close it. \$\endgroup\$ Oct 7, 2018 at 17:27

1 Answer 1


It appears that the red jumper between the quick-connect terminals is shorting the relay contacts, so the contacts will appear closed whether the relay is operated or not.

If the part number you gave is correct, the DC48 indicates that the relay requires 48 volts on the coil to operate.

  • \$\begingroup\$ Ah, I've shorted the jumpers because the relay wasn't working. I've removed them now. I checked the datasheets of the relay, and the release voltage seems to be 4.8V or above. Does this not mean that it should switch just fine with a coil voltage of 12V? \$\endgroup\$ Oct 7, 2018 at 16:36
  • 1
    \$\begingroup\$ @thatonetallguy - "I checked the datasheets [...] the release voltage seems to be 4.8V or above. Does this not mean that it should switch just fine with a coil voltage of 12V?" No. The relay datasheet shows that for the relay part number you gave, with its 48V nominal coil, it's only guaranteed to operate (energise) at 33.6V. It might operate at a lower voltage, but IMHO it's unlikely to operate at 12V. Either change the voltage you are using to drive the coil, or use a relay with a 12V coil. As Peter pointed out, you have a 48V relay. \$\endgroup\$
    – SamGibson
    Oct 7, 2018 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.