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I am making a circuit for a button to be far away from the circuit, about 50 meters round-trip. I have went decided to use an optocoupler to protect the uP from the worst of the outside environment and hopefully just get a blown optocoupler rather than a fried uP. The optoucoupler will be driven by a constant current IC to 10mA.

I have made a minimal schematic of both configuration.

enter image description here

What would be the difference in choosing high or low side configs? I am not particular in which offers better response time since a mechanical button can only be pressed very few times per second (but I am keen to know which one offers better response time). I am more concerned about which configuration is less susceptible to failure such as ESD from let's say a worse one, lightning. I went ahead and placed a TVS diode which should clamp at around 40 volts, the highest the CC IC can handle. I should probably increase it a bit more to make sure not to conduct so that the LED won't turn on.

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  • \$\begingroup\$ Have you considered using an isolation amplifier? \$\endgroup\$
    – tim
    Jul 6, 2020 at 19:53
  • \$\begingroup\$ @tim they are really expensive compared to an opto where all i need is just to know when a button is pressed \$\endgroup\$
    – Jake quin
    Jul 6, 2020 at 20:14
  • \$\begingroup\$ You also need to reject the common mode signal which will be highly significant over 25m especially with a lightning strike hazard. For the LOW SIDE circuit, a negative transient to the left of the button might turn the LED on. For the HIGH SIDE circuit, a positive transient to the right of the button might turn the LED on. So a differential input with a transorber across its inputs with an isolator on the output would be good. See Silicon Labs AN614 for a good implementation. The Si8610 is as cheap as your opto-isolator. \$\endgroup\$
    – tim
    Jul 7, 2020 at 4:39
  • \$\begingroup\$ @tim the IC you recommended has a 6 channel version which would net to be cheaper for my 8 channel application (but i would go 2x 4channel still cheaper). Would the circuit be really as straight forward as just like in the datasheet example circuit ? No more additional TVS, filtering, button debouncing needed ?? \$\endgroup\$
    – Jake quin
    Jul 7, 2020 at 14:11

2 Answers 2

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Here is a potential idea which uses a digital isolator (Silicon Labs Si8610 Data Sheet) on the output. The input has lightning strike protection diodes and a differential amplifier with high common mode rejection ratio; both required for a 25m cable exposed to hazards. You could also add filter capacitors. All parts are relatively inexpensive.

The differential amplifier has a gain of 2, and has a high common mode rejection ratio guarding against transients triggering a false button press.

When the button is not pressed, 1.65V appears across the inputs, giving an output of 3.3V.

When the button is pressed, it shorts the inputs, giving 0V from the amplifier output.

You don't need a separate LED driver chip and opto-isolator, so that will offset the cost of the digital isolator.

Texas Instruments has a great series of videos on the subject of galvanic isolation which discusses inductive, capacitive and optical isolation techniques.

Remote button using isolation amplifier

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  • \$\begingroup\$ there seem to be a lot of complimentary components, because if thats all just for common mode there is an optocoupler with common mode rejection, are they all necessary? \$\endgroup\$
    – Jake quin
    Jul 7, 2020 at 18:51
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    \$\begingroup\$ You definitely need lightning strike protection and CMR/CMTI. Whether that's implemented with an expensive one-chip OTS solution or with cheaper discrete components is up to you. What cable are you using for the button? Is it a controlled impedance such as 50 Ohms or 75 Ohms coax or twisted pair? \$\endgroup\$
    – tim
    Jul 7, 2020 at 20:22
  • \$\begingroup\$ Have a look at this Opto has a 5kV CMR/CMTI and its cheaper than the Si8x but more expensive than my opto. knowing that would the circuit then be back to the schematic i showed? No impedance controlled wire, just regular twisted pair. \$\endgroup\$
    – Jake quin
    Jul 7, 2020 at 20:29
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    \$\begingroup\$ I think that is talking about the common mode between the input and output of the chip. See page 15 of the Si81xx datasheet for the test circuit; the Si86xx has 50 kV/us immunity. However, there is also the common mode between the pairs of the 25m cable which the differential amplifier rejects. The issues with your circuit is that the input is single-ended, the opto degrades over time and consumes ~40x more power. You could also add capacitors to the differential circuit create a low pass filter of ~20 Hz. \$\endgroup\$
    – tim
    Jul 8, 2020 at 15:46
  • \$\begingroup\$ The tvs and resistor are hurting the limited space i have, 603 smds are the smallest i can solder. for the TVS i might simplify to this . but i found diff Op-Amps that has ESD protection so can the tvs be removed which greatly reduces the number of components, what do you think ? \$\endgroup\$
    – Jake quin
    Jul 9, 2020 at 15:11
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I suggest use this schematics. No need of B1. If used voltage at button side is not 15V - reduce resistors accordingly (to get 10-20mA current on optocoupler)

schematic

simulate this circuit – Schematic created using CircuitLab

R1 and R2 must be equal for better noise suppression . C1 is for RF noise reduction. D1 protects optocoupler from reversed voltage.

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  • \$\begingroup\$ I would still like B1 to be there to extend the lifespan of the opto, and with the added benifit of widening the range of the voltage i can use. \$\endgroup\$
    – Jake quin
    Jul 7, 2020 at 14:12

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