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I tried to detect 220V AC in my microcontroller using SFH620A-3 with series resistance of 440K (2 x 220K, 1/4 watts SMD type), a 1uF at the output to smoothen the pulse and every works as expected. However the resistors produces good amount of heat specifically when placed in a closed enclosure. The heat is not high to start a fire (I can still touch), however the heat slowly propagates across the PCB and other components too get hot. I did the power calculations and the voltage across the opto is less that 1mA and the power dissipation is 130mW on each resistor. I may have multiple AC detection circuit on a single board and hence the heat would multiply.

  1. Is it a good practice to increase the resistance further? I tried some 800K and the output wasn't reliable.
  2. Are there other ways to reduce the heat ?
  3. Should I try another AC optocouplers ? If yes, can someone suggest some AC optocouplers (DIP-4 casing).

Schematics

schematic

simulate this circuit – Schematic created using CircuitLab

Note : Transformers wont be feasible due to the size.

Update

I tried increasing the resistance by 100K, 50K etc and found that the opto fails to work at around 900K. The output seems to be stable at 770K, hence can I safely use 660K for my project (i.e 330K x 2) ?

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  • \$\begingroup\$ What is the power rating of your resistors? You could use a 1/2 or 1 Watt device and elevate it above the board. Still the same thermal load, but not as concentrated as with a smaller power device, so the temperature of the resistor won't rise as much. \$\endgroup\$
    – AlmostDone
    Commented Apr 8, 2018 at 20:42
  • \$\begingroup\$ I am using 1/4 watts SMD resistors. \$\endgroup\$
    – Zac
    Commented Apr 8, 2018 at 20:45
  • \$\begingroup\$ With SMD resistors, you can lower the temp by increasing the copper trace width attached to the pads. In effect, the copper acts as a heat sink, spreading the thermal energy over a wider area which reduces hot spots. \$\endgroup\$
    – AlmostDone
    Commented Apr 8, 2018 at 20:53
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    \$\begingroup\$ Use a capacitive dropper. It uses the reactance of the capacitor to drop voltage without generating lots of heat. Lots of articles on the internet. \$\endgroup\$
    – Steve G
    Commented Apr 8, 2018 at 21:09
  • \$\begingroup\$ @SteveG Could you recommend the values for the capacitor ? \$\endgroup\$
    – Zac
    Commented Apr 8, 2018 at 21:13

2 Answers 2

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You could add a capacitor in series with the resistors (a capacitor on each side, for safety reasons). If its capacity is adequate, it's impedance will drop a significant voltage without dissipating power (except a tiny amount due to its ESR).

Be sure to use mains-rated AC capacitors "X-type capacitors". Those have a "self-healing" structure that will prevent nasty failures in the case a line spike exceeds the capacitor voltage rating.

If most of the voltage is dropped on the caps you can lower the resistors by several order of magnitude (better to keep them anyway for safety, in case one of those spikes occur).

More details can be found in this other EE.SE thread.

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  • \$\begingroup\$ As per the thread a 100nF capacitor seems to be fine, however 220V AC rated capacitors are huge and very expensive. \$\endgroup\$
    – Zac
    Commented Apr 8, 2018 at 21:51
  • \$\begingroup\$ @Zacson - what do you call very expensive - a dollar or two? \$\endgroup\$
    – Simon B
    Commented Apr 8, 2018 at 22:24
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    \$\begingroup\$ @SimonB comparing the price of these capacitors against the resistors, they are quite expensive. However this may be the solution which generates no heat. \$\endgroup\$
    – Zac
    Commented Apr 8, 2018 at 22:33
  • \$\begingroup\$ @Zacson What are your cost constraints? Again you didn't specify an important information. Is this is for an hobby project or a professional solution. Keep in mind that when dealing with mains AC safety measures are paramount. Do something wrong and you may kill yourself or someone else! \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    Commented Apr 9, 2018 at 6:49
  • \$\begingroup\$ @Zacson X-rated capacitor are more expensive and bulky because they are certified and rated for very safe mains operations. In your case if you used a normal capacitor rated for 240Vac you could be in trouble if a 2000V spike appeared on mains. That could make the capacitor fails short and this would mean that 240V will appear across your low value resistor and optocoupler input. That could provoke a catastrophic explosion and even arcing over, with possible fire in the equipment. \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    Commented Apr 9, 2018 at 6:52
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You could add a small step-down transformer at the input side, bringing down the mains voltage to, say, 5V, then use a single much smaller limiting resistor to power the input side of the optocoupler.

You will stop having so much heat to get rid of.

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  • \$\begingroup\$ Transformer does not seems to be feasible due to the size. \$\endgroup\$
    – Zac
    Commented Apr 8, 2018 at 20:43
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    \$\begingroup\$ @Zacson you didn't mention any size constraints in your question. You should add any relevant information and constraints, otherwise we could keep giving you good advices than turn out to be infeasible due to some constraint that you haven't specified in advance. \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    Commented Apr 8, 2018 at 20:51
  • \$\begingroup\$ Sorry for that. I have added the size constraint now. \$\endgroup\$
    – Zac
    Commented Apr 8, 2018 at 20:59

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