I'm powering a heater with an SSR and esp8266 nodemcu module. I've found out that I need a higher voltage to effectively trigger it than what the nodemcu can put out on a digital pin (from previous post SSR ticking when controlling heater).

I have added a transister and diode combo into my network (based on https://electronics.stackexchange.com/a/220504/137437) which allows the 3.3v from the esp8266 to trigger 5v: enter image description here

This works but when the digital output from the nodemcu is off, an LED on the the SSR still maintains a very slight glow.

If I power the nodemcu from a different wall wart that is plugged into an external wall socket (see diagram below) the led on the SSR is off when the digital output is off. enter image description here

Is there some feedback or something that is making the SSR Led be on when there is no power to the P1 pin? I have tried testing the voltage across the SSR low voltage input but only have a rudimentary voltage tester and when I try it doesn't read anything and the led goes off.. Or is my wall wart used in the first diagram just dodgy?


One of two solutions might help you.

  1. Add a 10k resistor from the Port pin to ground.
  2. Add a 1k resistor directly across the SSR low voltage input pins.
  • \$\begingroup\$ Thanks, can you provide any further information on exactly what the effect of each of these would be? \$\endgroup\$ – AlexS Feb 3 '17 at 16:46
  • \$\begingroup\$ 1 - reduce potential for Q1 to turn on under reset or leakage from MCU 2 - recue sensitivity of opto to small currents \$\endgroup\$ – Jack Creasey Feb 3 '17 at 19:01
  • \$\begingroup\$ OK so I tested both. Adding a 10k resistor from Port pin to ground (which is like a pull down resistor I think) didn't do anything in my case. But adding a 1K resistor across SSR low voltage pins did. This means it is essentially in parallel with the diode. Is that OK, or does it negate having a diode there? Also, does this indicate that the transistor is leaking a tiny bit of voltage and could be faulty? \$\endgroup\$ – AlexS Feb 11 '17 at 17:55
  • \$\begingroup\$ Glad you fixed the problem. There are always leakage currents to consider and in this case I think you have in addition a very sensitive (high gain) optocoupler. I'd imagine the solution will be fine. \$\endgroup\$ – Jack Creasey Feb 11 '17 at 18:05

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.