This circuit is being used for mains detection. Input is being fed directly from 220V 50Hz mains and output goes to Arduino which is running on 3.3V. Theoretically the optocoupler LED should burn out during reverse polarity of 220VAC in the circuit given below:

ac det

Here is the voltage graph which appears across the LED of opto-coupler:

enter image description here

It shows a peak reverse voltage of 50V only instead of expected 220V (at least that's what I expected). However 50V alone should be able to destroy the LED.

I have used this circuit in a project and it has been working perfectly for about 6-7 months. Why does this circuit work?

Here are the Absolute max ratings from the datasheet: AMR

And these are characteristics for the device: EOC

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    \$\begingroup\$ I guess the reverse current is not high enough to destroy the LED reliably. But it is still a bad design. \$\endgroup\$ Apr 1, 2016 at 12:55
  • \$\begingroup\$ Why do you think that? The reverse breakdown current is just as limited as the forward current. \$\endgroup\$
    – pjc50
    Apr 1, 2016 at 12:57
  • \$\begingroup\$ When you say "work"; have you (physically) built this circuit? \$\endgroup\$
    – Tyler
    Apr 1, 2016 at 12:58
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    \$\begingroup\$ @Tyler - yes I have and I am using it continuously for 6-7 months. \$\endgroup\$ Apr 1, 2016 at 13:06
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    \$\begingroup\$ @pjc limits in a datasheet are the limits up two which the manufacturer garantees a certain behviour. Beyond the limits, nothing is garanteed, and that includes the possibility of normal operation. \$\endgroup\$ Apr 1, 2016 at 13:32

3 Answers 3


You may misunderstood reverse current, see http://www.renesas.eu/products/opto/technology/standard_p/index.jsp

The LED is a diode, so it is not intended to conduct in the reverse direction. However if you still force enough high reverse voltage to its pins, this very little reverse current does flow.

Scope (with proper insulation transformer) the voltage on the LED. LED - just like any other diode - does have a reverse breakdown voltage. This is the Vr in the datasheet. In reverse breakdown you can imagine the LED as a Zener, so once more than 4V applied in the reverse direction, current will flow.

Refer to this picture: http://reviseomatic.org/help/e-diodes/Led-graph.gif You can read more at wiki: https://en.wikipedia.org/wiki/LED_circuit

If you drive the led in reverse, performance of the optocoupler degrades over time, see http://www.renesas.eu/products/opto/technology/standard_p/index.jsp Vr.

Therefore it is a good idea to add a standard diode either in series (so no reverse current can flow), or to the led of the optocoupler in the reverse direction (so it shunts the reverse voltage).

Moreover, as this is a zero-crossing circuit, you can consider using a rectifier bridge then connect the led to the output of the rectifier. This results very clean zero crossing spikes in both halfwave.


There is a parameter called integration time.

With certain limitations, absolute maximum ratings given in technical data sheets may be exceeded for a short time. The mean value of current or voltage is decisive over a specified time interval termed integration time. These mean values over time interval, Ti, should not exceed the absolute maximum ratings.

This might be a hint if we can also include dynamic reverse resistance and junction capacitance.

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    \$\begingroup\$ But for diode breakdown (which isn't a thermal process) the integration time to produce damage should be quite short. A half-cycle of mains is 10ms, during most of which the diode is being driven out of spec. \$\endgroup\$
    – Chris H
    Apr 1, 2016 at 15:16

What is happening, is that when the LED is reverse biased, there is "enough" backwards current (about 1ma) to drop most of the voltage (200v) across the 200k resistor, this leaves about 20v (50v p.p) across the LED. Also, even though the maximum reverse voltage allowed (and not damage the LED) is 6v, in reality it has to be higher in order to guarantee the lower value.
Even though your particular LED is "working," you or your company will be open to lawsuits.
When the LED fails, it will fail in the mode that says "there is no voltage mains," but this is not true. Then when you want to transfer your liability to the LED manufacturer, it will not work, because you operated it outside its maximum limits.


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