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Is this the correct way to put a diode for flyback voltage protection to protect the infrared heater? And do I need to put a flyback protection for the Arduino pin or Vcc?

  • 1
    \$\begingroup\$ If you are directly using an Arduino pin to power a relay, that can't work well. \$\endgroup\$
    – Justme
    Mar 25, 2021 at 9:49
  • \$\begingroup\$ Also the coil voltage of the relay matters. If you drive it directly from a MCU pin, it must use the same voltage as the MCU. \$\endgroup\$
    – Lundin
    Mar 25, 2021 at 10:31

2 Answers 2


Flyback diode is there to avoid voltage spikes when an inductive load is turned off.

In your case, only the one on the left side is necessary as you want to protect the arduino against the coil spikes of the relay.

The one on the right side should not be there as it's AC, you will have a short.

You may want to check the Arduino pin can provide enough current to drive your relay, usually, it is better to use a BJT or MOSFET.

  • \$\begingroup\$ There is no voltage spike in AC when the load is turned off? Yes, the Infrared Emitter is rated 220-240VAC @ 25Watts; I want a simpler and basic design for the heater because making it BJT + MOSFET combination would need to create a PID controller to regulate the temperature. \$\endgroup\$ Mar 25, 2021 at 9:16
  • \$\begingroup\$ There is no voltage spike in AC when the load is turned off? <- Yes and no, there are if the load is inductive, however, since it's AC the spike will only last for half a cycle. Usually, relay contacts are designed to handle that. In AC, you cannot use flyback diodes because it's, well, AC. Voltage spikes are particularly dangerous for electronics and a coil turned off by a transistor can easily reach hundreds of volts if there is no flyback diode, destroying the transistor. \$\endgroup\$
    – Damien
    Mar 25, 2021 at 9:18
  • \$\begingroup\$ @Tnek There is no voltage spike in infrared heaters usually. And if there is, you can't solve it this way. Fortunately, switches are harder to break than Arduinos. \$\endgroup\$
    – user253751
    Mar 25, 2021 at 17:58

The diode on the left is appropriate. The one on the right isn’t -it will half-wave rectify when the contact is open, which probably isn’t what you want, and will do nothing when it’s closed.


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