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I have this design. By a possible mistake from user, the circuit would receive 12 V AC, and may destroy the optotransistor on reverse voltage cicle.

I can put a diode D1 before the resistor, but the diodes I already know, may have voltage drop or resist just small currents (300 mA), but the AC source can have 2000 mA.

I need a Diode with Low Voltage drop.

The low voltage drop is required to can handle the logical input at right side.

The 5.0 volts is not a source, it comes from a switch connected to a device sealed. When optocoupler gets on, the logical inputs receive a signal and the device acts.

Which diode can I use?

my optocoupler would recibe ac in output

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  • \$\begingroup\$ is the load suppose to have a dc or ac voltage? \$\endgroup\$ – drtechno May 3 '18 at 16:21
  • \$\begingroup\$ It might help us to know your intended circuit function...for any switch state, "logical input" gets 5.0 v. The opto-isolator does nothing. \$\endgroup\$ – glen_geek May 3 '18 at 17:09
  • \$\begingroup\$ the 5.0 volts is not a source, it comes from a switch connected to a device sealed. When optocoupler gets on, the logical inputs receive a signal and the device acts. \$\endgroup\$ – Ando May 4 '18 at 3:40
  • \$\begingroup\$ the load doesn't matter for optocoupler, its works for other circuit. \$\endgroup\$ – Ando May 4 '18 at 3:41
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You don't need to worry about the reverse current rating of the diode. Unless the voltage is so high that it breaks down, the diode will not allow any current to flow.

Actually, a tiny amount of reverse leakage current may flow, but it will be so small you can ignore it.

That's for the reverse direction. The maximum possible current for the forward direction will - approximately - be your peak voltage (about 17V: \$12\cdot\sqrt{2}\$ V) divided by R1's resistance. You haven't specified the resistance but you should be able to calculate that.

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The current of a reverse biased diode(when the voltage goes to the negative cycle in your circuit) is negligible and will not affect the operation of your circuit as long as the reverse voltage doesn't exceed the diode breakdown voltage.

Roughly the reverse current can range from few nanoamps all the way up to micro amps.Everything depends on the diode you will be using (i.e Shottky or standard rectifier) and on how much reverse voltage is applied.

If the reverse voltage is less than the breakdown voltage and the leakage current(Current passes when the diode is supposed to block current) is negligible then the diode in your circuit will do the job.

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