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The following simple circuit turns a LED ON/OFF. It includes:

  • DC power supply 12V (EDR-120-12)
  • SSR (rated for DC input and output, DR10D12, datasheet)
  • LED, 400mA, 2.85V forward (L2-MLN1-S, datasheet)
  • resistor 50 Ohm
  • pushbutton (PVA6LRE21241, datasheet) enter image description here

My problem is that the SSR output remains ON and can only be turned OFF with turning the power supply OFF.

When the pushbutton is ON:

  • the small LED of the SSR is ON
  • the LED is ON and the measured current flowing through it is 198 mA (supposed to be 195 mA = (12 V - 2.85 V)/47 Ohm)
  • when there is no load at the SSR output, the resistance across the SSR outputs (1/L1, 2/T1) is 0 Ohm (and my multimeter says it is the same potential)

When the pushbutton is OFF:

  • the small LED of the SSR is off
  • the LED is ON and the measured current flowing through it is 185 mA
  • when there is no load at the SSR output, the resistance across the SSR outputs (1/L1, 2/T1) is 3 MOhm

With the SSR LED turning correctly ON and OFF I believe that the issue is on the output side.

I do not understand where these 185 mA come from when the pushbutton is OFF. (I think the 13 mA difference (198 mA - 185 mA) is the small LED of the SSR turning ON and OFF).

I've read about possible solutions (use a bleeder resistor in parallel to the load to remove a leaking current?) but I am still not sure I understand what is happening here, so I thought I would ask a clear question here for the record.

Thanks for any help

Edit: for clarity, removed the circuit for the LED illumination of the pushbutton

Edit: answering questions from comments

  • "Is this SSR designed for DC?" from @user253751. Yes it is: enter image description here

enter image description here

Diagram of the SSR with DC control input and DC output.

enter image description here

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  • \$\begingroup\$ The datasheet notes that in the off state, there is a maximum leakage current of 0.1mA. That is enough to very slightly light an LED (but not as brightly as the ~20mA that would run through it with the SSR conducting fully). Could you be seeing this faint glow as the LED being on? \$\endgroup\$
    – SomeoneSomewhereSupportsMonica
    Dec 15, 2020 at 5:14
  • \$\begingroup\$ Also, it looks like your terminals are backwards - swap 3/4 and swap 1/2. I think this must be a drawing issue only though, as otherwise nothing should work properly. \$\endgroup\$
    – SomeoneSomewhereSupportsMonica
    Dec 15, 2020 at 5:17
  • \$\begingroup\$ Thanks @SomeoneSomewhereSupportsMonica. It is not a faint glow, I've measured it to be 185 mA, I've edited the post. I've edited the drawing also, thanks, that was only a drawing issue indeed. \$\endgroup\$
    – thomast
    Dec 15, 2020 at 13:24
  • \$\begingroup\$ That looks you have tyristor output. Check carefully the part number or may be label wrong. \$\endgroup\$
    – user263983
    Dec 15, 2020 at 14:20
  • \$\begingroup\$ Thanks @user263983. Part number is correct: it is DR10D12, ordered from Digikey, part number CC1767-ND. The printed label on the SSR is unmistakably correct. \$\endgroup\$
    – thomast
    Dec 15, 2020 at 16:55

1 Answer 1

Reset to default
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I'm puzzled: I entirely re-wired the circuit from scratch, and it is now working as expected. I do not understand what was happening. Sorry for the false alarm, hopefully this question still provides a clean record of a working configuration.

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  • \$\begingroup\$ There is still something wrong about your diagram, which appears to have the LED reverse biased. \$\endgroup\$
    – Math Keeps Me Busy
    Dec 18, 2020 at 14:29
  • \$\begingroup\$ Thanks @MathKeepsMeBusy, edited. \$\endgroup\$
    – thomast
    Dec 18, 2020 at 20:06

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