1
\$\begingroup\$

There is a solid state relay (AQH1213) that switches the load to 220 V using 5 V (Arduino).

The insulation voltage between the low and high voltage circuits in this relay is 5 kV.

The challenge is to provide additional isolation for this relay and controller contact.

For these purposes, I plan to use an optocoupler.

Solid State Relay LED Characteristics:

Forward Current: 50 mA Peak Forward Current: 1A Reverse voltage: 6 V Recommended input current per LED: 15-25 mA A current limiting resistor of 300 Ohms is installed between the relay and the controller.

What optocoupler output current should I choose?

Initially, there were doubts about the high peak current of the solid state relay (1A). At this maximum current, the optocouplers require a supply voltage> 5 V, which in my case is not suitable, since all control circuits have a voltage of 5 V.

A current-limiting resistor should clearly reduce these indicators, but I do not know how accurate.

Applying Ohm's law (U = 5 V, R = 300), I got a very low current value of 0.016 A, is that true? If this is true, then most likely it does not take into account the maximum forward current.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ What insulation spec do you need? Do you have a power supply for the output side of the optocoupler that is insulated to this same spec? \$\endgroup\$
    – The Photon
    Oct 23 '19 at 16:12
  • \$\begingroup\$ Yes, isolated DC/DC converter \$\endgroup\$
    – Delta
    Oct 24 '19 at 3:39
  • \$\begingroup\$ What isolation spec do you need? \$\endgroup\$
    – The Photon
    Oct 24 '19 at 3:42
  • \$\begingroup\$ @ThePhoton, If the voltage, then 1 - 1.5kV is already good. The optocoupler must save the controller output if, nevertheless, the solid-state relay fails. \$\endgroup\$
    – Delta
    Oct 24 '19 at 12:12
1
\$\begingroup\$

You don't want to use the peak forward current from the "Absolute Maximum Ratings" section of the data sheet. If you let 1 A flow for more than a tiny fraction of a second, you will blow out the LED. 50 mA is the absolute maximum sustained current allowed, and even then should be avoided.

According to the datasheet, that SSR is meant to run closer to 20 mA forward current through the LED.

While that SSR is typically operated with 20 mA, 16 mA should be more than enough to operate it.

\$\endgroup\$
1
  • \$\begingroup\$ thanks for the answer! \$\endgroup\$
    – Delta
    Oct 24 '19 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.