I have a 5-meter-long wire connected to an ESP8266 input. The wire is connected to a regular wall-mounted switch and goes through my room's wall and is adjacent to mains power lines.

As you would expect there is significant noise on the input of the microcontroller. I intend to use an optocoupler to eliminate the noise, but I was wondering if this noise can also turn on the LED in the optocoupler.

Can electrical noise create enough voltage and current to turn on an LED? Say it's a regular IR LED with a 1K resistor in series.

I'm asking since even after putting the optocoupler in the circuit I still could read false LOW/HIGH triggers on the input.

(I have done my research about input signal noise and know that I should use twisted shielded wire, filters, etc. but my question here is specifically about the amount of power that I can expect from EMI noise created in a 5m wire in a household environment)

  • \$\begingroup\$ an optocoupler will filter some kind of noises, but there's no reason to assume a fast optocoupler won't couple over RF noise from source to sink side. \$\endgroup\$ Commented Feb 7, 2019 at 15:04
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    \$\begingroup\$ How do you intend to power the opto LED side? If you use the same power supply as the ESP you aren't actually isolating anything. \$\endgroup\$
    – Wesley Lee
    Commented Feb 7, 2019 at 15:13
  • \$\begingroup\$ @WesleyLee I am using the same power supply. But I was under the impression that since an LED draws current, electrical noise would not be able to turn it on. While a microcontroller input has high input impedance and is very sensitive to voltage changes in a wire. \$\endgroup\$
    – Pouria P
    Commented Feb 7, 2019 at 15:16
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    \$\begingroup\$ @PouriaP It has enough energy to turn on the LED for a brief moment, and if your optoisolator is fast enough that's enough to turn on the other side too. \$\endgroup\$
    – Hearth
    Commented Feb 7, 2019 at 15:38
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    \$\begingroup\$ Fully agreeing with @Hearth: I think you're trying to solve a very specific problem with something that's not meant (or even capable) to solve that problem. Maybe describe the problem before asking about your slightly specific-in-the-wrong-way solution to it! \$\endgroup\$ Commented Feb 7, 2019 at 15:41

2 Answers 2


Yes, because an optocoupler is as sensitive as any other piece of electronic maybe even more sensitive since the led voltage of the optocoupler is less than CMOS treshold. To solve the problem:

  • A resistor between the signal line and GND. Value between 100 and 900K.

  • A capacitor between the signal line and GND. Value between 0.1 and 0.47uF.

  • A resistor in serie with the signal input. Value +- 1K.


While assisting in performing 4000 volt (4KV) pulse testing, with 50 nanosecond edge speeds (or was it 10 nanosecond edge speeds), the free-space (air) electric fields changed fast enough, injecting enough displacement currents, to make LEDs on the PCB glow faintly.

What voltage appears across the in-the-process-of-opening wall-mounted light switch, with energy stored in that 5 meters of wire, striking an arc as the switch metal points first separate? by 1 micron? by 10 microns? by 100 microns? by 1,000microns (1 millimeter)?

We are told that air will break down at fields of 3,000 volts per millimeter.

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    \$\begingroup\$ How does this relate to "EMI noise created in a 5m wire in a household environment"? \$\endgroup\$ Commented Feb 7, 2019 at 20:41

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