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Beginner
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5V 2 Channel Relay being powered by Intel Edison Breakout Kit

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Hi,

I am trying to power a 5V 2 channel relay using the intel Edison's GPIO pins.

Can I power up the 5V 2 channel relay using the GPIO pins that are already there on the Edison Breakout Kit or do I need to do something else in order to power it up?

Thanks in advance!

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Valued Contributor II
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Even if you use level shifters - you still cannot drive the relay coils directly.

What I would do, is use the existing 1.8V GPIO to drive a MOSFET or even a Darlington-Pair, or even just a pair of NPN transistors, and drive the relay coil from that.

I'd tie one side of the relay to +5V and switch the other to GND using the MOSFET or Darlington. No level shifter needed :-)

Remember to protect the circuit from high breakdown currents when the relay is switched off by putting (at least) a diode in reverse across the relay coil.

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Community Manager
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Hi Ashwinkannan,

 

 

Thanks for your interest in the Intel Edison platform.

 

 

In the Intel Edison breakout board the pins are pulled from the Intel Edison Compute module directly, using 1.8V as logic "HIGH" and don't allow much current to go through. That means if you hook them up with 5V or 3.3V, there is a high risk of damaging the Edison board.

 

 

In order to power up your 5V 2 channel relay you'd need to use level shifters to convert 1.8V signals to 5V. Please take a look at this thread for level shifter suggestions: /message/276726# 276726 Suggestions for level shifters to use with mini breakout board.

 

 

Hope this information helps.

 

 

Regards,

 

-Yermi A.

 

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Valued Contributor II
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Even if you use level shifters - you still cannot drive the relay coils directly.

What I would do, is use the existing 1.8V GPIO to drive a MOSFET or even a Darlington-Pair, or even just a pair of NPN transistors, and drive the relay coil from that.

I'd tie one side of the relay to +5V and switch the other to GND using the MOSFET or Darlington. No level shifter needed :-)

Remember to protect the circuit from high breakdown currents when the relay is switched off by putting (at least) a diode in reverse across the relay coil.

View solution in original post

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