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I want to use home automation, so I am thinking of using an input to an ESP8266 from a PC814 bi-directional optocoupler for a 230 VAC mains physical switch.

Using this scheme I don't have to modify my house wiring, and I can just connect a physical wall switch output, which was going to a device, to go to a PC814 optocoupler using a 100 kΩ resistor. The PC814 output NPN switch will be connected to a NODEMCU's GPIO.

I am concerned that at 50 Hz my input will be toggling every 20 ms. How can I avoid this toggling, and using a mimimum number of components get a continuous signal until power is present on the input of the PC814?

enter image description here

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    \$\begingroup\$ Possible duplicate of Opto-coupler output \$\endgroup\$
    – Long Pham
    Commented Jul 31, 2018 at 12:57
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    \$\begingroup\$ Is Nodemcu not capable of processing an interrupt every 20ms? \$\endgroup\$ Commented Jul 31, 2018 at 13:08
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    \$\begingroup\$ What does "continuous pulse signal until Power is present on input of PC814" mean? It's not clear what you want. \$\endgroup\$
    – Andy aka
    Commented Jul 31, 2018 at 13:30
  • \$\begingroup\$ have seen one simple circuit to detect low frequency pulsing in this form by Vince Patron. I have not tested it. Check your luck - electronics.stackexchange.com/questions/325118/… \$\endgroup\$
    – user19579
    Commented Jul 31, 2018 at 18:38
  • \$\begingroup\$ Sorry typo error. Its not pulse signal. Simple if 230 VAC is present input of Nodemcu should be 1. As soon as power is off input to GPiO is 0 \$\endgroup\$
    – Kalpesh
    Commented Aug 1, 2018 at 0:45

2 Answers 2

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schematic

simulate this circuit – Schematic created using CircuitLab

Capacitor C1 prevents the output from returning immediately high, during the intervals between mains peaks. Mains input will have to disappear for a few cycles, to give C1 time to charge, and finally bring the output high.

You should probably also share the 350V (peak) mains voltage (and power dissipation) across a few resistors in series, as shown, so that you don't exceed their voltage rating, and so that they can be smaller ¼W or ½W models. Total average power in a single 100kΩ resistor would be \$\frac{(250V)^2}{100k\Omega}= 630mW\$.

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50Hz input gives 100Hz positive pulses at zero crossings. ( power on = logic “0” out)

With 22k pullup the collector voltage needs to be below 1/3Vdd for marginal “0”. Since RC=T is 63% towards Vdd the Cap load on collector needs to be 2T min. to cut slew rate voltage rise < Vdd/3 or 2RC=10ms or C>=0.47uF.

Making C too big may delay the detection of power due to current limiting and cap pull down rate from input Current Transfer Ratio (CTR). You could use C>=0.047 and Rc=220k

A power switch rated for >2A cannot be used for <10% of rated current due oxidation of contacts so if you are planning to do this, use a gold plated switch rate for >=2A.

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