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I have these devices:

  • Sensor Keyence FS-N41P
  • Camera Flir Blackfly S
  • Controller Teknic Clearcore

The sensor triggers the camera and the controller. The controller supplies 24V to the sensor.

This is a schematic that works. I also added some technical details from the manuals: Camera-Trigger-6 Download KiCad Schematics

Some more info for the camera: flir1 flir2

My big question is: Am i missing something or is something wrong?

  • I am not sure if the pull-up/down resistors are correct?
  • The camera opto out is open drain and requires a pullup resistor. The controller input seems to have a internal 5k pullup to 5V. The camera manual recommends 1k pullup for 5V. Is this a problem?
  • Camera opto out is described as "open drain". Do i need another resistor like in the Clearcore "Digital PNP" example? Is this a pulldown? Can you have a internal pullup and an external pulldown at the same time?
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1 Answer 1

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  1. Pullup resistor is a resistor that connects an arbitrary point to VCC. If you have several pullups then all those resistors are in parallel, and so the well-known formula for parallel resistors connection applies. In particular, if you have one 5K pull up and add another one of 1.25K then you'll get an equivalent of 1K pullup in total.

  2. "Open collector" means that an output pin is directly connected to the collector of (internal) NPN transistor (like in "digital NPN sensor" picture), or PNP (the next picture). If NPN is opened the output pin gets ground potential and if NPN is closed the pin becomes floating. So you need a pullup with NPN. While with PNP it's the other way round.

"Open drain" refers to the drain of some internal MOSFET. And if it's N-MOS (and usually it is) then you need a pullup resistor like with NPN BJT.

  1. If you have both pullup and pulldown then they form a voltage divider, obviously. This is something you don't really need for a digital output.
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  • \$\begingroup\$ Thank you for your answer. It definitely sheds some light on my questions. 1. Lesson learned. I didn't know that the equivalent overall resistance is less than the smallest parallel resistance. So I could add a 1.25k to 5V and meet the 1k recommendation. In this particular case I don't have access to the 5V! So is it a problem to just leave it to the 5k pullup? Am I right that this would mean 10mA for the gpi optocoupler? \$\endgroup\$ Commented May 20 at 12:36
  • \$\begingroup\$ 2. Very good explanation. The logic is inverted. \$\endgroup\$ Commented May 20 at 12:36
  • \$\begingroup\$ 3. I'm glad I don't need this schematic because I don't understand it. Are they trying to get the voltage below 1V with the 499 ohms? 0.454V seems to be the output voltage. The resistor is 2W. Does this mean that it absorbs a lot of energy from the internal 5k resistor? So this design wouldn't be very economical? \$\endgroup\$ Commented May 20 at 12:36
  • \$\begingroup\$ @MartinHenne Yes, that's right. I'm not sure why 2W resistor is needed. Maybe that part is supposed to work with 24V voltage level, I guess. \$\endgroup\$
    – Matt
    Commented May 20 at 13:03

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