Update: Thanks for the help -- working circuit below!

I am trying to turn on a 110V AC array of LEDs with a small 5V relay with some explosive results!! :-)

I am not doing something right as I have successfully blown-up two of my relays...

Relay: HK4100F-DC5V-SHG

Spec Sheet: https://img.ozdisan.com/ETicaret_Dosya/445413_4369639.pdf

The Relay

There are 6 pins, the middle two pins are the coil. When I apply 5V DC I can hear it clicking which I think is expected operation.

I attach the load (array of LEDs) on the left of the photo.

I attach 110V AC from a wall socket on the right side.

Case 1:

  • a) No voltage to Coil.
  • b) Load and Power Source attached
  • c) Result is the LEDs are quite dim.
  • When I applied DC to coil the LEDs went out.

I'm not exactly sure why my relays exploded on subsequent attempts. Could it be that I reversed wires on the load, or power source? Should this even make a difference with AC?

Hoping this makes sense!

relay pinout from datasheet

Update: Working circuit below: enter image description here

  • 1
    \$\begingroup\$ your post implies that you are trying to achieve explosive results. ... please change the wording. ... your drawing makes no sense ... what are the three grounds? ... please draw a detailed schematic, including the relay coil and relay contacts \$\endgroup\$
    – jsotola
    Commented Feb 3, 2018 at 2:05
  • \$\begingroup\$ Your 5V is connected to one side of the mains? Respectfully, you should stop before you hurt someone. \$\endgroup\$ Commented Feb 3, 2018 at 2:11
  • \$\begingroup\$ I just peeked into this one. I think you want to try something more like this: Activate LED PS via 5 V relay. \$\endgroup\$
    – jonk
    Commented Feb 3, 2018 at 3:34

2 Answers 2


As I read the datasheet, the two terminals on the left are connected together, and aare the common terminal (moving contact) of a single-pole, double throw switch. The top right is the Normally Open (NO) contact - with the relay not energized, it is not connected to anything. The bottom right terminal is the Normally Closed (NC) contact. With the relay not energized it is connected to the two left terminals.

When you energize the relay, the top right terminal will be connected to the two left terminals, and the bottom right terminal will have no connection.

Given that description, your drawing doesn't make sense - you seem to be applying 110 VAC between the NO and NC terminals.

  • \$\begingroup\$ That's exactly what I must have been doing -- explosive results. I'm still a bit lost. Your explanation of the relationship between the moving contact and NO, NC makes sense. But, now I am failing to see how it is possible to create an AC current if you can attach only one of the NO, NC? \$\endgroup\$
    – MangoHeat
    Commented Feb 3, 2018 at 1:58
  • \$\begingroup\$ You pass the AC Neutral directly from source to load, unswitched, and just switch the AC Hot/Line/Live wire. The AC Hot wire would be connected to the relay NO contact, and the relay Common connected to the load Hot terminal. \$\endgroup\$ Commented Feb 3, 2018 at 7:18
  • \$\begingroup\$ Success! I think the top right is NC and the bottom is NO. When I attach the live wire from the power source to the bottom right pin, the live wire from load on a left pin, and apply a voltage across the coil, my LEDs turn on. That makes me think bottom right is normally open, until voltage crosses the coil, and creates a closed connection. This is just naming convention now -- I understand the proper function. Thanks so much for your help Peter! \$\endgroup\$
    – MangoHeat
    Commented Feb 3, 2018 at 22:36

An AC relay coil is designed for a certain impedance at 50 or 60 HZ. Its DC resistance is much lower so it will act like a heater and burn up.

There is also a possibility that in the OFF state you are shorting out your AC power source and that will burn out the relay contacts every time. Double-check before you make a wreck!


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