I am experimenting with AC optocoupler (SFH628A-4) to sense AC voltage(240 VAC) in a circuit. As per the data sheet (https://www.mouser.in/datasheet/2/427/sfh628a-1767413.pdf), the absolute maximum input DC forward current for the optocoupler is 50mA, however I am unable to understand the minimum or optimal current that is needed for the optocoupler to function.

I did some tests using 100K, 200K, 300K, 400K, 500K, 600K resistors in series with the AC (240VAC) supply and they seems to work fine (The opto turns ON). The optocoupler failed to turn ON only when I used 650K or above resistor values. From the experiment I guess we can conclude that the optocoupler turns ON even with current as low as 0.40mA i.e while using 600K resistor in series with 240VAC. Also using higher resistor values like 400K or 500K seems to be ideal as it produces much less heat. However I am not sure if my understanding/assumptions are correct and whether using lower current will have some drawbacks.

Can someone please help me to understand the minimum or optimal current for the optocoupler to function?


1 Answer 1


Look at figure 8 in the data sheet to get a good general idea: -

enter image description here

As you can see it's telling you what your output transistor will conduct at for various input currents from 0.5 mA to 5 mA. So instantly, you should recognize that going below 0.5 mA is going to be a bit flaky because it's not covered in the data sheet.

But, if you are monitoring AC the waveform naturally falls through 0 volts and so there isn't a continuous sustained current into the input LED so, you have to accept this and anticipate that your opto output will be pulsating with the AC input frequency.

Back to 0.5 mA and the graph above; it's basically telling you that you cannot expect anything more than 1 mA collector current for a typical device AND this will be significantly worse for those devices with a less than typical CTR (current transfer ratio).

Hence, the current you need to apply at your input needs to be enough to cope with the output current load and the input voltage waveform passing through zero not giving you too-long a period where the output is "false".

using higher resistor values like 400K or 500K seems to be ideal as it produces much less heat

You can alternatively use a dropper capacitance. Pick a capacitor with the right impedance at 50/60 Hz and it won't generate heat but, always use some series resistance to prevent surges causing an unholy amount of current flow that blows up the LED inside the opto.

  • \$\begingroup\$ Excellent. Any reason the graph is showing only current till 5mA even though the maximum rating is 50mA? I am connecting the output of optocoupler to a microcontroller's GPIO pin using 1K resistor with a 3.3v DC supply, hence the output current requirement will be within 1mA limit. Since 300K or 400K resister gives current above 0.5mA, I hope I can skip using additional capacitors and rely on the resistors alone. \$\endgroup\$
    – Zac
    Commented Jul 10, 2021 at 17:02
  • \$\begingroup\$ Sorry @Zac I can't help you with that one. It seems bizarre that it doesn't go up to at least 20 mA (a fairly standard value of LEDs) but, if your output current is 1 mA and the CTR is 80% then you only need to feed 1.25 mA into the LED to get that current. \$\endgroup\$
    – Andy aka
    Commented Jul 10, 2021 at 17:25

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