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    laser as z probe

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    • themelleundefined
      themelle @chilli
      last edited by themelle

      @chilli said in laser as z probe:

      @dc42 I didn't quite get what you meant, sorry

      what would happen if you feed a negative voltage (-3,3V) into the Duet Z probe input?

      What David tried to make clear to you:
      While it may be perfectly possible (from a wiring and software point of view) to use this laser sensor as a z probe, it may actually be quite unusable because your setup needs a timing resolution of at least 200 femtoseconds (read: 0,000 000 000 000 200 seconds) - and must be reliably reach this resolution for every measurement taken. On this scale small variations in timing imply huge differences in the measured distance.

      You can clearly see the effect in a demo
      https://www.youtube.com/watch?v=S2jaAQEv3Yo&feature=youtu.be&t=510
      where the measured distance deviates in millimeters.

      hurzhurzundefined 1 Reply Last reply Reply Quote 0
      • hurzhurzundefined
        hurzhurz @themelle
        last edited by

        @themelle I think the sensor @chilli wants to use does not work by the Time-of-Flight principle like the VL53L0X.
        It probably uses triangulation / measureing the reflection angle by distance via a CMOS image sensor.
        I saw a video about a similar sensor a while ago: https://youtu.be/0D9nJIzNVTY?t=130

        chilliundefined 1 Reply Last reply Reply Quote 0
        • chilliundefined
          chilli @hurzhurz
          last edited by

          @hurzhurz yes that one looks kinda similar. And that principle would work?

          @themelle @dc42 Do you know what would happen if I feed negative voltage (-3.3V) into the Z probe input?

          dc42undefined hurzhurzundefined 2 Replies Last reply Reply Quote 0
          • dc42undefined
            dc42 administrators @chilli
            last edited by

            @chilli said in laser as z probe:

            Do you know what would happen if I feed negative voltage (-3.3V) into the Z probe input?

            Probably nothing, because the ESD protection diode in the MCU will protect the input. But for safety, you could use a series resistor and a Schottky diode to ground.

            Yes I was assuming that it is a time of flight sensor.

            Duet WiFi hardware designer and firmware engineer
            Please do not ask me for Duet support via PM or email, use the forum
            http://www.escher3d.com, https://miscsolutions.wordpress.com

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            • hurzhurzundefined
              hurzhurz @chilli
              last edited by hurzhurz

              @chilli Yes, I think so. Though, the needed resolution of the image sensor probably needs to be very high to achieve this crazy measurement resolution of 2-4µm... maybe they also use some clever algorithms to interpolate.
              I just found this video with an illustration: https://youtu.be/g-EJ4FDB4qQ?t=34

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              • chilliundefined
                chilli
                last edited by

                I've been trying out a bit and it seemed to kinda work. I'm getting a signal in.
                Can't work on it till probably next week, cause the z-axis clutch broke.

                Another thing is, that the laser is giving 0V if the distance is 0. if it's more than 0 the voltage rises till 5V and stays there and same with negative, falling to -5V.
                As the z probe input works with 3.3V i was thinking about using a resistor to ground and also in series a schottky diode for the negative voltage (thanks for that info @dc42).
                I'm just not sure about what resistor and what diode to use. Any suggestions?

                What current does the input draw at 3.3V?

                maybe this schottky diode:
                https://www.conrad.de/de/p/panjit-schottky-diode-gleichrichter-1n5817-do-41-20-v-einzeln-1304858.html

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                • hurzhurzundefined
                  hurzhurz
                  last edited by

                  Do you really need to deal with negative output voltage?

                  Pages 5&6 in the il1500 manual it look like you can set different output voltage range modes:

                  • 0V to 5V
                  • -5V to +5V
                  • 1V to 5V

                  So I think you can simply use the first one together with a voltage divider?

                  Well, any maybe you could simply use the digital output "LOW/HIGH judgment output".
                  Here you could use a optocoupler for isolation between it and the duet.

                  chilliundefined 1 Reply Last reply Reply Quote 0
                  • chilliundefined
                    chilli @hurzhurz
                    last edited by

                    @hurzhurz yes these modes are possible to set, but the problem is that the voltage stays after the set point is reached.
                    For example 0-5V: above 0 the voltage is 0V, the closer it gets to 0 the voltage rises (the point from which is the voltage rises can be set too) : 10mm 0V; 5mm 2,5V 0mm 5V.
                    If the distance gets negative the voltage stays 5V
                    So I thought it would be better that the voltage above and below 0 would change, so that it doesn't get too deep and still thinks it's at 0 distance

                    Not sure if I was able to explain it understandably...

                    The idea with the digital output and the octocoupler sounds possible, will try that!

                    hurzhurzundefined 1 Reply Last reply Reply Quote 0
                    • hurzhurzundefined
                      hurzhurz @chilli
                      last edited by

                      @chilli I hope I understood it correctly, but I would do it like this for the analog output way:
                      For your example values: 10mm=0V / 5mm=2,5V / 0mm=5V
                      I would define the middle (5mm) as my target trigger point and later set my probe height offset accordingly (G31 Zxxx).
                      Therefor I would adjust the probe trigger value (G31 Pxxx) so it matches the value that the duet measures at 2,5V (or 1,65V in case of an ideal voltage divider).
                      This way, you can see if you are below or above the trigger point.

                      Btw, we could also have a conversation in German if that helps...

                      chilliundefined 1 Reply Last reply Reply Quote 0
                      • JoergS5undefined
                        JoergS5
                        last edited by JoergS5

                        I checked the optoNCDT sensor, they start used at 400$ (older models than the one mentioned in the video). Does someone have cheaper solutions with the same high resolution?

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                        • chilliundefined
                          chilli @hurzhurz
                          last edited by

                          @hurzhurz yeah i think you got me right. that could actually work too.
                          In what unit would i put the trigger point value (behind the P of G32 Pxxx)?
                          Or is it just 0?
                          I think german would be easier, but english would be better for the others here 😃

                          @JoergS5 Sorry, don't have any solutions for that

                          hurzhurzundefined 1 Reply Last reply Reply Quote 0
                          • chilliundefined
                            chilli
                            last edited by

                            @dc42 What current does the duet draw at the 3.3V Z probe input?

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                            • hurzhurzundefined
                              hurzhurz @chilli
                              last edited by

                              @chilli It is a number (between 0 and 1000 if I remember correctly) that represents the voltage:
                              0 for 0V and 1000 for about 3,3V (or the other way round, depends on the "I" parameter of M558). So you would enter something like G31 P500.
                              But I would try that out / look what G31 (without parameters) exactly tells when it is in the correct position.

                              You also have to consider that these 1000 steps are the available resolution for the probe input.
                              So if you map the distance of 10mm to the sensor output voltage range, you actually get a resolution of about 0,01mm at the probe input.
                              Because of this, maybe the digital output would give you a better repeatability.
                              I think you should try both variants...

                              For the probe input: @dc42 may correct me, but I think it shouldn't draw any noticeable current because it is a high impedance input.
                              There is also a 10k resistor from a RC filter which should limit it to 0,33mA (at 3,3V).

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