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