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schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

I have an inductive sensor (LJ18A3-8-Z/BX,) normally open 6V to 36V 300mA. It is connected to a 12V 80kHz single channel opto-coupler. At the moment it is connected to a 12V DC power supply.

The positive wires (brown) from the power supply and sensor are screwed into one terminal on the opto-coupler board, the negative (blue) wires between the power supply and sensor are directly connected and the final wire from the sensor (signal?) connects to the second terminal on the opto-coupler board. According to the specification sheet for the sensor, it operates at 12V 8mA or 24V 15mA.

Can I change the supply voltage to 24V and if so, would I need to change or add a resistor to the opto-coupler board?

Between the terminals on the board and the opto-coupler chip is a 1Kohm resistor and a red LED on the other side. My thinking is I've essentially increased the power in the circuit by a factor of 4, assuming I should also increase the resistor by a factor of 4 as well?

optocoupler link data on the optocoupler link to supplier

Octocoupler board

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    \$\begingroup\$ please replace the lengthy description with a schematic diagram of the circuit \$\endgroup\$ – jsotola Jan 12 at 23:50
  • \$\begingroup\$ We also need to have a link to the manufacturer's datasheet for the optocoupler. \$\endgroup\$ – Elliot Alderson Jan 12 at 23:59
  • \$\begingroup\$ You most likely just need to change the resistor to maintain the same current into the optocoupler LED. But without datasheet and circuit, that's just a guess. \$\endgroup\$ – user_1818839 Jan 13 at 0:02
  • \$\begingroup\$ There is a schematic editing tool available when you edit your question. The tool button looks like a schematic. \$\endgroup\$ – Math Keeps Me Busy Jan 13 at 0:34
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Inductive sensor parameter looks like quiscient current of sensor itself, nothing to do with output current. Output of sensor probably open collector. Optocoulpler has a LED inside, usually infrared. LEE! feed with curent, not voltage. With 12V supply you may have approximately 10 mA. For sure it could be measured with multimeter. 24V gives 12V in addition. So keep the same current resistor 1.2kOhm in series should be added. Or datasheet mention 50 mA max, the additional resistor not needed. But that info is not trustable.

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Loosely:

  1. Measure(or look up, or guess) the coupler input current, using the original 12V supply voltage (example: 20mA)
  2. At 24V there is an extra 12V to get rid of.
  3. To do this at the measured current (again say it's 20mA) takes an extra resistor of R = U/I = 12V/20mA = 600 ohm
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  • \$\begingroup\$ Ok, so the probe at 12v is 8mA at 24v its 15 mA, so i'd require R= 12/0.007 roughly an additional 1.8kohm resistance over the 1Kohm already on board? \$\endgroup\$ – Oz Sco Jan 13 at 1:08
  • \$\begingroup\$ No need. If the coupler draws only 15mA you are fine without the extra resistor. (It seems able to handle 20mA input current) \$\endgroup\$ – Peter Jan 13 at 1:16
  • \$\begingroup\$ Thanks very much for your help \$\endgroup\$ – Oz Sco Jan 13 at 1:28

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