0
\$\begingroup\$

I feel a bit silly asking this..

I have a relay, which has a switch that turns it on/off. Next to the switch I have an indicator light which needs to turn on when the switch is turned on.

I tried putting the LED in parallel (seen below) and in line, but no luck.

enter image description here What am I missing here?

(by the way, the 22k resistor is a good value for this LED at this voltage)

\$\endgroup\$
5
  • \$\begingroup\$ Your LED and 22k resistor look shorted. \$\endgroup\$
    – vrleboss
    Apr 25, 2017 at 18:10
  • \$\begingroup\$ In your diagram, the LED is providing a very high resistance path in parallel to a section of unbroken wire. How is this supposed to cause current to flow through the LED? \$\endgroup\$
    – Robherc KV5ROB
    Apr 25, 2017 at 18:10
  • \$\begingroup\$ It might work if you connect the LED and resistor parallel to the relay's switching coil (I.E. LED attached before coil, resistor attaches after it.) \$\endgroup\$
    – Robherc KV5ROB
    Apr 25, 2017 at 18:11
  • \$\begingroup\$ @Robher KV5ROB; Yes, that did the trick. Although I'm not fully understanding why. \$\endgroup\$
    – svenema
    Apr 25, 2017 at 19:36
  • \$\begingroup\$ That's doing the trick because the back-emf of the relay switching coil is a voltage potential, while the negative resistance of the wire you connected to has a virtually 0 voltage potential. Because of that, you have now created a parallel circuit between the LED+resistor combo & the relay's switchign circuit (both see ~24V, where before the switching coil was seeing 24V & the LED+resistor were seeing ~0.0001v at most. \$\endgroup\$
    – Robherc KV5ROB
    Apr 25, 2017 at 19:42

2 Answers 2

Reset to default
1
\$\begingroup\$

I'd suggest the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

R1 will allow approximately 20 mA through the LED. D2 protects the LED from the transient when the relay is turned off. D2 could go directly across the relay coil but it does not have to.

\$\endgroup\$
4
  • \$\begingroup\$ This works, so in the end my change was to move the positive from after the coil to straight from the source before the coil. The diode is a good idea though. \$\endgroup\$
    – svenema
    Apr 25, 2017 at 19:36
  • \$\begingroup\$ However; I still am not really understanding why the original circuit didn't work. Apparently there is not enough current (?) left after the coil, not even when I lower the resistor value for the LED. -- Or does this has something to do with the mentioned impedence of the power supply? (I don't really know what that means) -- Thing is, having to power the LED straight from the power supply is another wire in an already very tight enclosure. I was hoping to power the LED straight from the switch. \$\endgroup\$
    – svenema
    Apr 25, 2017 at 19:36
  • \$\begingroup\$ The "Power Supply" in the field will be two big 12V batteries / 90A generator.. Not sure if that changes anything... \$\endgroup\$
    – svenema
    Apr 25, 2017 at 19:38
  • \$\begingroup\$ @svenema Your original schematic has a dead short across the resistor & LED. The current will take the (essentially) 0 ohm path instead of going through the LED. \$\endgroup\$
    – DoxyLover
    Apr 25, 2017 at 22:00
2
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Youll want to double check your resistor value

You may also want to add a back emf protection diode, as seen here:

schematic

simulate this circuit

\$\endgroup\$
12
  • \$\begingroup\$ as @RobhercKV5ROB hinted, this will depend on how "good" your power supply is (i.e. how low it's output resistance is). \$\endgroup\$
    – BeB00
    Apr 25, 2017 at 18:13
  • \$\begingroup\$ but if everything works without the led there, then the led should light up correctly when you put it in. \$\endgroup\$
    – BeB00
    Apr 25, 2017 at 18:14
  • \$\begingroup\$ out of interest, have you measured the resistance of your relay? \$\endgroup\$
    – BeB00
    Apr 25, 2017 at 18:15
  • \$\begingroup\$ ok, it's a bit weird.. I would expect this circuit to work as you draw it, but it's not. at just +24V -> LED -> 22k resistor -> NEG, this works fine when I throw in the relay, the LED won't turn on anymore I think the resistance might be too high.. perhaps at this voltage transistors are required? \$\endgroup\$
    – svenema
    Apr 25, 2017 at 18:15
  • \$\begingroup\$ resistance of the relay is 330 ohm \$\endgroup\$
    – svenema
    Apr 25, 2017 at 18:16

Your Answer

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

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