5V DC supply for relay

Hi guys, I am using this relay for my project, its equipped with optocoupler, so is it safe for me to use Arduino 5V to connect to the VCC of this relay? FYI, i would like to control the on/off of a 12V DC component by using this relay.

Can't tell anything from the picture. Would you have a link to the schematic or spec sheet?

Those relays consume ~70 ma each. To properly answer your question, more information is needed:

  1. Which Arduino
  2. How it’s being powered and at what input voltage
  3. Duty cycle of the relays

FWIW, the opto’s are just baggage, they provide no benefit when powering the relay coils from VCC of your Arduino.

You have a 12V supply, why not us 12V relays?

Do not supply your MCU (Arduino for example) with the same supply that you feed the relays.
for example, if you connect the MCU to the +5V, then do not connect this directly to the relays

WattsThat:
Those relays consume ~70 ma each. To properly answer your question, more information is needed:

  1. Which Arduino
  2. How it’s being powered and at what input voltage
  3. Duty cycle of the relays

FWIW, the opto’s are just baggage, they provide no benefit when powering the relay coils from VCC of your Arduino.

Its Arduino Uno R3, and i plan to use a 9V, 1A adapter to supply power to the Arduino.

Perehama:
Can't tell anything from the picture. Would you have a link to the schematic or spec sheet?

5V 2 channel relay interface board
From our tests, the board is 3.3V compatible, so you can use Raspberry Pi's GPIO to activate or deactivate the relay.
Maximum Current Rating: 10A
Maximum Voltage Rating: AC 250V / DC 30V
Can control various appliances with large current and high voltage
Can be controlled directly by microcontroller such as Arduino, 8051, AVR, PIC, DSP and ARM
Build in 2 units of Opto-Isolator IC 817C.

This is all i can get

HowToCODE:
Its Arduino Uno R3, and i plan to use a 9V, 1A adapter to supply power to the Arduino.

Really bad idea! The Arduino operates on 5 V which should be regulated.

Here's the circuit of your relays:

The Arduino connects to the IN terminals and Vcc. Not ground.

The relay supply connects to JD-VCC and ground. If you have a 5 V regulated supply with which to power the Arduino (because you need a 5 V supply to power the Arduino) as well as the relay board, then you need separate paired 5 V and ground wires from the power supply to the relay board, and from the power supply to the Arduino. That is, the Vcc and ground wire travel as a pair. And you should put a 470 µF electrolytic capacitor between JD-VCC and ground.

If you use a 3.3 V device such as a Raspberry Pi or ESP8266, you still connect the I/O pins to the IN pins on the relay board, and Vcc to 5 V, not 3.3, otherwise you will not have enough voltage to actuate the relays through the optocouplers.

Remember that this relay module is "active LOW" to actuate the relays, and you must digitalWrite the pins to HIGH before you set the pinMode to OUTPUT in setup.

The short answer is :
"Why buy a relay board with the isolation of an opto coupler and then defeat that by powering both sides
of the isolation barrier (MCU side & relay side) with the SAME supply ? You might as well punch a hole in your condom... (meaning why have protection if you are going to intentionally defeat it)

Actually, there would be many commercial devices which include both logic circuits and relays and which use a single supply for all. :roll_eyes:

Clearly it is a matter of design principles, knowing how to implement lead dress as I have described above. The problem only occurs when someone imagines it can be done with a rats nest of "Dupont" jumpers and considers any ground point or supply point as good as any other - because they are connected, you see. :astonished:

My explanation above utilises the isolation of the opto-coupler in the context of powering from a single power supply. There is also the problem of physically isolating the circuit switched by the relay and keeping its wiring neat, and the common ignorant misunderstanding that the "kickback" diode must be close to the inductor rather than correctly at the switching transistor.