I want to light a bulb connected to the socket by using an 1 Channel Optocoupler Relay Module which accepts an input of 5V.

From my understanding, when the modul receives DC current (5V), it's going to turn on the other circuit and turn on the bulb.

I'm not sure how to connect the module in this context.

It looks like this:

    | 10A 250VAC 10 125VAC |
DC+ – 10A  30VDC 10  28VAC – NC
DC- –                      – COM
IN  –                      – NO
    | SRD-05VDC-SL-C       |

It didn't came with any documentation, so I'm not sure what these notations mean.

I assumed I have to connect the positive wire from the 5V circuit to DC+ and the negative/ground to the DC-. I'm not sure if this is correct.

However, I don't understand how to connect the wires at the other end.

I have no ideas what IN, NC, COM and NO mean.

Is there any place where I can find the documentation for this kind of relay or are these universal notations? What do they mean?

How should I make the connections?

The module looks like this:

So, I connected the bulb wires to COM and NC and the bulb is turned on when I connect it to the socket. Then I tried connecting a battery to the DC+ and DC- and the power led of the relay component is turned on, but the bulb is not turned off.

  • \$\begingroup\$ google relay contacts \$\endgroup\$ Apr 26, 2017 at 17:52
  • \$\begingroup\$ @JImDearden That's what I'm doing for 40 minutes now... \$\endgroup\$ Apr 26, 2017 at 17:53
  • \$\begingroup\$ Read The Fantastic Datasheet. \$\endgroup\$ Apr 26, 2017 at 17:54
  • \$\begingroup\$ @EnricBlanco I find it quite difficult to understand. \$\endgroup\$ Apr 26, 2017 at 17:55
  • \$\begingroup\$ The relay itself doesn't have an IN pin according to the datasheet... \$\endgroup\$ Apr 26, 2017 at 17:56

4 Answers 4


The signals on the left are the input side and the signals on the right are the output side.

Ok so now it is clear that you have a module rather than just a relay. The module has additional circuitry on it for driving the relay. Based on this the input pins are as I put in my original answer:

  • DC+ - Positive logic supply voltage
  • DC- - Negative logic supply voltage (a.k.a ground)
  • IN - Input control signal - the thing that tells the relay whether to be on or off.

On the output side a bog standard switch naming is used:

  • NC - Normally Closed (Connected to COM when relay coil off)
  • NO - Normally Open (Connected to COM when relay coil on)
  • COM - Common pin of switch

Your relay operates as a DPST switch as shown in the diagram below.

Relay Switch

You need to apply power to the DC+/DC- pins (such as a battery), and then use the IN pin to control whether the relay is on or off. Connecting the IN pin to DC+ will turn the relay on (connect NO and COM), and connecting the IN pin to DC- will turn the relay off (connect NC to COM).

You may need to add a resistor in series with the IN pin - i.e. don't connect it directly to DC+ or DC-. A 1k resistor will suffice. It is not clear from the description whether or not this is actually necessary as the information on the listing is a very bad translation.

  • \$\begingroup\$ Thanks! This is helpful! So, should I use the IN at all? \$\endgroup\$ Apr 26, 2017 at 17:55
  • \$\begingroup\$ I've never seen a relay before that needed a logic supply voltage. I looked up the part number, and this is an electromechanical relay, not a solid-state one, so I don't know if this is entirely correct. \$\endgroup\$
    – Hearth
    Apr 26, 2017 at 17:57
  • \$\begingroup\$ @Felthry the description is perfectly correct for a solid state relay. \$\endgroup\$ Apr 26, 2017 at 17:57
  • \$\begingroup\$ The relay itself doesn't have an IN pin according to the datasheet... \$\endgroup\$ Apr 26, 2017 at 17:58
  • 1
    \$\begingroup\$ @TomCarpenter Phew, it works. By connecting the IN to DC-, it turns off the bulb when the output is connected to COM and NC, like you said! Thank you so much! \$\endgroup\$ Apr 26, 2017 at 18:36

Did you bother to scroll down on that page where you bought it...

enter image description here

That the last two lines mean is gibberish to me. Try it and see. I'd guess it means you can drive it either with a high or a low signal depending on that jumper.

In a nutshell, DC+ and DC- is the power the coil will need, IN drives an opto-coupler to deliver that DC to the coil somehow.

What High is, it does not say though, but since it's feeding an opto-coupler, do not exceed the 5mA.

  • \$\begingroup\$ Oh, somehow I totally missed that! Thanks! So, do I need a third input wire? It's still unclear how the input wires are connected. Currently I only have two of them (negative and positive). \$\endgroup\$ Apr 26, 2017 at 18:22
  • \$\begingroup\$ @IonicăBizău that's the issue with buying crap from E-Bay. It's often cheap Chinese knockoffs with very limited details and often badly toleranced. \$\endgroup\$
    – Trevor_G
    Apr 26, 2017 at 18:24
  • \$\begingroup\$ Actually, somebody in a previous question recommended me to buy such a thing. I waited for it 6 months and finally I can work on my project. :-) Thanks to Tom, we figured out the missing piece in the puzzle. :-) \$\endgroup\$ Apr 26, 2017 at 18:37
  • 1
    \$\begingroup\$ Lol. That auction page has photos of two completely different versions of the board. I was trying to follow the traces to see what goes where and suddenly the silk screen flipped and extra vias appeared. \$\endgroup\$ Apr 26, 2017 at 18:53
  • 2
    \$\begingroup\$ The most important rule to follow is no datasheet = no sale. \$\endgroup\$ Apr 26, 2017 at 18:55

Your relay module has a relay plus some interfacing electronics in it.

You have to supply the 5V from your batteries to the relay through the DC+ DC- pins. The IN signal is the input control signal. You can connect this pin to an output pin from a microcontroller like an Arduino, or to a mechanical switch connected to the 5V supply.

As a plus, you also have a jumper to select whether your control signal is active LOW (i.e., a logic ZERO makes the relay switch) or active HIGH (i.e., a logic ONE makes the relay switch).

The relay internally connects the COM (common) pin to the NC (normally closed) pin when the IN signal is driven LOW (or when it's driven HIGH, if you set your jumper to "active LOW").

Then, when you drive the IN signal HIGH (or when it's driven LOW, if you set your jumper to "active LOW") the relay switches and internally connects the COM (common) pin to the NO (normally open) pin.


NC is Normally Closed, and NO is Normally Open. COM is COMmon.

What this means is that, when no signal is applied to the coil, there is a short circuit between NC and COM, and an open circuit between NO and COM. When you energize the relay coil, there will be an open circuit between NC and COM, and a short circuit between NO and COM.

I'm not sure what IN is. You're right on the DC+/- terminals, I believe.


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