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I'm exploring new methods to find if relay contacts have been stuck together without using auxiliary contacts of the relay. while searching I find out this block circuit: enter image description here

at this Toshiba page

as far as I can tell that photorelay just behaves like an optocoupler, thus from this block circuit I don't understand how they can detect the mechanical state of a relay. in particular the low-side relay. Also, how can I maintain galvanic isolation with that setup?

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    \$\begingroup\$ What is the problem that this is supposed to solve? Using safety relays with forcibly guided contacts would be much easier, less error-prone and probably cheaper too. (Which is saying a lot, since safety relays tend to be expensive.) \$\endgroup\$
    – Lundin
    Nov 29, 2022 at 10:30
  • \$\begingroup\$ When the switch is open a current is meant to flow through the opto-coupler. The resistors ensure the current is quite small. \$\endgroup\$ Nov 29, 2022 at 18:29

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An opto-coupler gives you galvanic isolation. Even quite low spec ones will withstand thousands of volts between the input and output side.

If there is a voltage across the input to the opto-coupler, then the output side will give you an indication of that, either with a logic level, or closing a conducting path like a FET or bipolar transistor. You use that output to determine whether there's a voltage across your relay terminals when there should not be, or none when there should be.

The input to the opto-coupler need only draw a few mA, hopefully little enough to avoid compromising the relay being able to switch the load off.

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  • \$\begingroup\$ I know how the optocoupler works, but what you say is true to check if there is a voltage at the outputs of the relays, if you look at the picture, they used the optocoupler to check the mechanical state of both relays, by connecting them in parallel, thus there is no close circuit for the led inside the optocoupler \$\endgroup\$
    – David
    Dec 18, 2022 at 20:44
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Well it's pretty easy. When the relay is closed, it sends power to something, right? So they are putting a very high-impedance LED (essentially) in parallel with the relay contacts.

When the relay is closed, the voltage across its contacts is zero and the LED does nothing. When the relay is open, the voltage across its contacts is system voltage, and the LED sees power (leaked through the load to complete the circuit). The load gets a tiny amount of current - too little to do anything.

Then a sensor sees the LED light and says "therefore, relay is open".

schematic

simulate this circuit – Schematic created using CircuitLab

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