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I am trying to figure out how to use this relay module. There is no datasheet that I could find for the whole module, there however are for individual box.

Here is the picture below, there are three jumper pins which read VCC, VCC, GND. And there are four activator pins (I guess) they read GND, IN1, IN2, VCC

At the output end, I got current coming in connected to Middle terminal and receiving wire connected to the one showing Isolated line.

To activate it, I am applying 3V DC -ve on the GND (on the activator pins set) and +ve on the IN1.

Am I doing it wrong ? Here is the URL for further info: http://www.miniinthebox.com/2-channel-5v-high-level-trigger-relay-module-for-arduino-works-with-official-arduino-boards_p727426.html

enter image description here

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Typical relay module clone, they are all similar. Below schematic is per channel.

Keep in mind that RY-VCC (JD-VCC in the schematic) is for the Relay section, while VCC on the 4 pin connector is for the opto-isolated side.

enter image description here

https://arduinoinfo.mywikis.net/wiki/ArduinoPower for a full wiki on using these.

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    \$\begingroup\$ According to blog.wikispaces.com the wikispaces will be shutting down this year, so this answer might need to be updated to provide not just the link but overview of the contents perhaps \$\endgroup\$ – Sergiy Kolodyazhnyy Mar 23 '18 at 3:33
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    \$\begingroup\$ fixed it, Sergiy. \$\endgroup\$ – tedder42 Apr 30 '18 at 19:45
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See passerby his answer for the whole schematic of your module per channel. For the data sheet of just the relay check sbell's answer. However, to activate it you should provide the VCC with +5V. The GND connected to the GND and give a +5V signal to the IN1 or IN2 to activate it. IN1 and IN2 should be connected to one of your output pins, they will be able to switch your relay. But when I look at the schematic passerby send you it might also be possible that you'll have to put your output pin to LOW in order to activate it.

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  • \$\begingroup\$ Yea, these are active low. And the opto-isolated part can be triggered at less than 5v, if you remove the jumper and provide 5v for the relays. \$\endgroup\$ – Passerby Jun 1 '14 at 22:00
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I just wired up one of these modules a few minutes ago to confirm the info I'm posting below.

From your controller (eg Arduino or Raspberry Pi) connect Ground (0V) to the relay module's GND and Power to the module's Vcc. In most Arduinos the power will be 5V, in a Raspberry Pi it will be 3.3V - but these relay modules will work OK with 3.3V even though they are officially rated at 5V. Connect one of your I/O pins (in output mode) to IN1 and the value on that I/O pin will control Relay #1. Connect a second I/O pin to IN2 to control Relay #2.

These modules have the circuitry described above built-in so you don't need to add it - that's only needed if you are driving a stand-alone relay directly. All that's usually needed is a few Dupont jumper wires.

The other three pins (GND/Vcc/JD-Vcc on a separate connector) are normally not used and the jumper is left in place. They are used to power the relay switching and pick up the Vcc and Gnd off the data connector. By removing the jumper you can power that other Vcc pin separately.

If you need to do this or need the optical isolation feature to keep high voltages away from your controller, you need more expert advice than we are giving here; in fact if you don't already know what to do then you shouldn't be doing it as there is a good chance of damaging both people and devices.

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I just came across this as I am using the same modules, and they don't work for me. I need a relay that comes 'ON' with a 'high' input (I am using one of the contacts to power off the pi running the show) I have modified the module to work this way. To do this :-

  1. Get rid of the jumper (between JD-VCC and the left VCC)
  2. Cut the track on the back of the PCB connecting the two VCC's together
  3. Solder a link across 4 contacts; GND, GND, IN1 & IN2 (they are adjacent)
  4. Connect power (3.3 to 5v) at JD-VCC
  5. and 0v to anywhere on the joined contacts (from 3.)
  6. Use the left VCC as IN1 (+ve input trigger)
  7. Use the right VCC as IN2 (+ve input trigger)

Just tested this using the Pi 5v rail for JD-VCC and GPIO outputs (3.3v) to drive the triggers and it works fine. The inputs sink about 0.47-0.48mA Cheers Muttley

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You can activate the relay by using npn transistor(bc547) connect the output of the microcontroller to the base of the transistor via 10k resistor, collector to ground ( - negative of your supply ) and emitter to the relay input. This relay are different from the normal relays they operate by ground connection.

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    \$\begingroup\$ Your answer provides not much more information to the already written answers. Please elaborate your answer by e.g. supplying a schematic and/or more explanation \$\endgroup\$ – Ariser Apr 1 '18 at 16:19
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Based on the identifying marks in your picture, you should be powering the relays with 5Vdc to activate them (per the data sheet)

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