We are working on a project that requires a temperature sensor to tell the Arduino Due to turn a digital pin to HIGH (3.3V) to active a relay so a heating pad can be powered on.

We have diagnosed all possibilities and the only explanation as to why the switch will not active is the digital pins are only reading 1.8 V on the relay.

We are using a JQC-3F(T73) relay. We are using two digital pins to give it plenty of voltage and amps to switch on and off when needed.

so any explanation as to why it isnt working? thanks.

  • 2
    It sounds like you are directly wiring the relay to the due. Bye bye due...
    – Majenko
    Jul 12, 2017 at 1:24
  • Also posted at forum.arduino.cc/index.php?topic=488837. And indeed they had the relay directly connected to the Due's pin.
    – per1234
    Jul 12, 2017 at 20:54
  • volts go down as you run out of omph
    – dandavis
    Jul 13, 2017 at 17:48

1 Answer 1


The output impedance for the MCU used in the Due is about 46ohm. Along with the relay's impedance this forms a voltage divider that results in a final voltage that is lower than the unloaded voltage from the output.

In any case, do not connect an inductive load directly to a MCU; use a driver such as a ULN2803 instead, being certain to connect the flyback connection of the driver chip properly.

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