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I am very bad at electrical things, as I don't understand much of electrical engineering.

I was tasked to design a software that will take some inputs, randomize them and give some outputs to some relays. I did the software and somebody else designed the circuit. The problem is that now that somebody else is on vacation and I need to explain how to make connection to the relay and I have no idea how.

First of all, I found terms of Normally Closed, Normally Open, Active Low and Active High and I have no idea what they mean. I read about them but I do not understand.

Secondly, this is the relay module: https://www.tme.eu/ro/details/oky3012/module-pentru-relee/okystar/

https://i.postimg.cc/kg9xyGJ6/prime-presentation-pres-0004.jpg

It doesn't say which is NO, COM, or NC.

Thirdly, I understand that I can use a multi-meter to check for resistances from only-God-knows-where to I-have-no-idea-which point.

The program that I did on an ESP32 sets the relays output pin to HIGH, which means that the relay is off or not activated (for me it's output. basically the pin that will control something):

pinMode(Relay1_1,OUTPUT);  
digitalWrite(Relay1_1, HIGH);

And when I want to activate the relay or switch it on (make it click) I send a LOW signal to it.

digitalWrite(Relay1_1, LOW);

To what pins of the relay do I need to connect something to be able to control it?

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    \$\begingroup\$ What do you want to control? \$\endgroup\$ – HandyHowie Sep 4 at 9:54
  • \$\begingroup\$ We made this for a buyer. They will control some LEDs and some alarms. I don't know the exact type of those devices. \$\endgroup\$ – bleah1 Sep 4 at 9:59
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    \$\begingroup\$ When you activate the relay, it connects the COM and NO connections together. When you deactivate the relay it connects the COM and NC connections together. \$\endgroup\$ – HandyHowie Sep 4 at 10:22
  • \$\begingroup\$ Ok. Great. But which is which ? I couldn't find any datasheet. \$\endgroup\$ – bleah1 Sep 4 at 10:25
  • \$\begingroup\$ What does it say on the module? Can you add a photo? \$\endgroup\$ – HandyHowie Sep 4 at 10:51
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When you activate the relay, it connects the COM and NO connections together. When you deactivate the relay it connects the COM and NC connections together.

If this is your module -

enter image description here

Image from - https://okystar.com/wp-content/uploads/2018/05/1-5.jpg

Then the blue connector on the left is showing two switch symbols. The middle connection in each case is the COM. The one that is show as connected by default is the NC. The one open by default is the NO.

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  • \$\begingroup\$ Good answer but you should probably add where the OP needs to connect the signals driving the relay. I would guess this is the IN1 and IN2 pins on the 2,54mm strip. Vcc is probably(?) 12V and signal levels are probably(?) 5V. Looks like two optocouplers in the middle there, so the coils aren't driven directly. \$\endgroup\$ – Lundin Sep 4 at 11:02
  • \$\begingroup\$ The relays are connected. I can make them switch on and off. \$\endgroup\$ – bleah1 Sep 4 at 11:06
  • \$\begingroup\$ To be clear, are these the correct connections ? up = NO, middle = COM, down = NC : i.postimg.cc/q70sBRhX/Honsy.jpg \$\endgroup\$ – bleah1 Sep 4 at 11:12
  • \$\begingroup\$ @bleah1 You asked "To what pins of the relay do I need to connect something to be able to control it?" But ok there you go then, this should answer your question. Up = NO indeed, and so on. \$\endgroup\$ – Lundin Sep 4 at 11:13
  • \$\begingroup\$ @Lundin You are correct ! I have edited the title of the question. I hope it's less ambiguous now. \$\endgroup\$ – bleah1 Sep 4 at 11:15
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Normally Open (NO) and Normally Closed (NC) work as follows.

When there is no power applied to the relay coil, the Common (C, or COM) is connected to NC. "Normally", in relay land, means "when the coil is not being driven". When you apply voltage to the module input, the relay will be activated. At this time, COM will disconnect from NC and connect to NO.

You can see for yourself. Take your DMM and set it on "Ohms". Set it on the lowest setting, probably "200 ohms".

Now connect 5 volts DC to the input at upper right on your photo. Connect the + to "VCC" and the - to "GND". Also connect the "IN1" pin to "GND". Leave "IN2" disconnected.

Now take your meter and connect to the upper 3 pins on the left side. Starting from the top, they are NO, COM, and NC. See the little graphic to the left of the terminals? The diagonal represents a switching arm which rotates from 45 degrees down to 45 degrees up when power is applied. And no, it doesn't do that inside the relay, it's just a convention to show how the connection is made. So, with no power applied (Normal), the center connects to the lower terminal, and the upper terminal is not connected.

Check between NO and COM. You should see an overload on your meter, probably a flashing "199" or something similar. At any rate, it will be whatever you see when there is nothing attached. This is generally called "open", shorthand for "open circuit". Now check between NC and COM. You should see something like 1 ohm or less (it will depend on your test leads and DMM). This is generally called "short" or "shorted", shorthand for "short circuit".

Now connect IN2 to VCC. You may well hear a slight click, and this is a good thing.

Looking at the outputs, NO and COM should show a short, and NC to COM should show open.

How you "connect something to be able to control it" depends on exactly what you want to do.

For instance, if you want to drive a 5-volt lamp or LED, you can ground the - lead of the lamp/LED, and connect +5 to COM. Now you have a choice. If you connect NO to the lamp/LED +, the LED will turn on when you connect 5V to IN2, and turn off when you don't. On the other hand, if you connect NC to the LED, it will be on until you drive the relay, and will then go off.

If you have a 12-volt lamp, you can use a 12-volt power supply, and connect it to COM as above. However, you must continue to use 5 volts on the VCC and IN2 inputs. When you get more comfortable with the subject, you can use 12 volts on IN2, but you'll need a separate resistor, and you can ask a question about that, but that will be for another time.

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