9
\$\begingroup\$

I have some experience using automotive relays and triggering them using the output from a two-way radio aux pin so I am aware of the need to add a diode in parallel with the coil to protect the radio's internal transitor when it is turned off.

However, I'm planning on doing a xmas project using a PIC MCU to turn on/off my xmas lights but I'm going to use solid state relays instead of the automotive type.

I haven't used SS relays before. From a quick look at some schematics from Google, it appears that the control signal wouldn't suffer from the same EMF that needs a flyback diode. Am I correct or do I still need one to save my PIC from destruction?

Thanks for your help.

\$\endgroup\$
9
\$\begingroup\$

The control side of solid state relays is usually just a LED, sometimes two LEDs back to back, and sometimes with integrated resistor. In any case, there is nothing inductive there, so no inductive kickback to protect against. If the relay is on the same board as whatever is driving it, then no inductive kickback diode is needed. It's no different than driving any other on-board LED.

However, there can be inductance in the circuit from other than the relay. If the relay is external to the board where the driving transistor is, I'd put a diode on the board to protect it on the assumption that you don't know what might get connected off board. Even just a solid state relay (LED) but with a few meters of wire can have enough inductance to at least think about. A reverse Schottky diode on the board cheap and easy insurance.

\$\endgroup\$
10
\$\begingroup\$

I have not personally worked with solid state relays before, but from my understanding they function similar to an opto-isolator, with an LED triggering a phototransistor (or photodiode) into order to close or open the "switch". As such, the triggering side is isolated from the output side so I don't think you need a flyback diode. However, if you are switching an inductive load , you should still have to include a flyback diode or other snubber circuitry on the output side to avoid damaging the load.

Looking at the "Standard" circuit for using one of the Sharp S108T02/S208T02 SSRs, they have included a diode across the input pins, but I believe this is just to avoid exceeding the maximum reverse voltage of the internal LED (which for that particular device is only 6V). The specific datasheet for those SSRs can be found here.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.