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    Converted to Split 12V/24V - No Communication HELP

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    Duet Hardware and wiring
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    • dc42undefined
      dc42 administrators
      last edited by

      There is an LED and resistor connected to the heater output, that's why you can measure VIN less the voltage drop of the LED at the heater output pin.

      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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      • oleeundefined
        olee
        last edited by

        The bed heat LED is off on your first photo but lights on on second one. Maybe this is a cue?

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        • wwmotorsportsundefined
          wwmotorsports
          last edited by

          That was my thought as well. I am unsure why connecting the grounds of the bed heater control and the VIN should allow current to flow between the two.

          I have no bed heater connected in these photos, so in my mind there should be no reason for the voltage to get from the bed heat + to the bed heat - unless something has gone wrong.

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          • elmoretundefined
            elmoret
            last edited by

            @wwmotorsports:

            That was my thought as well. I am unsure why connecting the grounds of the bed heater control and the VIN should allow current to flow between the two.

            Because your multimeter doesn't have infinite resistance.

            @wwmotorsports:

            I have no bed heater connected in these photos, so in my mind there should be no reason for the voltage to get from the bed heat + to the bed heat - unless something has gone wrong.

            As David says, with the bed heater off you should see the ground of the bed heater at (Vin - LED voltage drop), because the LED is connected between the + and - terminals of the bed heater terminal strip.

            With the bed heater off, the positive terminal will show Vin and the negative terminal will be floating, pulled up to Vin (less the forward drop of the LED) by the bed heat LED and its resistor.

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            • oleeundefined
              olee
              last edited by

              If the multimeter is in good shape it would have input resistance in order of mega-ohms so it can't provide an essential current to light LED and to make such a big voltage drop across LED. I suspect that heater's nMOS switch is partly fried.

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              • elmoretundefined
                elmoret
                last edited by

                @zov:

                If the multimeter is in good shape it would have input resistance in order of mega-ohms so it can't provide an essential current to light LED and to make such a big voltage drop across LED. I suspect that heater's nMOS switch is partly fried.

                Nope, I have a perfectly functioning Fluke 179, it provides enough current to light the LED enough to be visible.

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                • oleeundefined
                  olee
                  last edited by

                  Look at the spec's of my multimeter (Fluke 289), input impedance 10MOhm < 100pF so the current through LED is about 12 uA in our case. Don't think this current is able to fire up any perceivable glow in LED. To speak frankly I don't check it by myself so can be mistaken and contemporary LEDs might be such effective.

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                  • resamundefined
                    resam
                    last edited by

                    zov: I think what elmoret is talking about is the "diode test" mode, available on most multimeters. In this mode the multimeter provides typically 3V on the probes, and limits the current to 2-3mA. If you put your probes onto an LED in this mode, the LED will light up dimly - but it will definitely emit light! This mode can be used to measure the forward & reverse bias voltage of diodes (and light-emitting diodes of course).

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                    • oleeundefined
                      olee
                      last edited by

                      If you look at photos of OP you see he is measuring voltage between 2 terminals not diode test.
                      ADD: And I have mistaken in the post above: the current through LED should be about 1.2 uA not 12 uA. So it makes the multimeter much more unlike as a shunt to lights LED on.

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                      • elmoretundefined
                        elmoret
                        last edited by

                        @resam:

                        zov: I think what elmoret is talking about is the "diode test" mode, available on most multimeters. In this mode the multimeter provides typically 3V on the probes, and limits the current to 2-3mA. If you put your probes onto an LED in this mode, the LED will light up dimly - but it will definitely emit light! This mode can be used to measure the forward & reverse bias voltage of diodes (and light-emitting diodes of course).

                        No, wasn't talking about diode test mode. Regular DC voltage mode.

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                        • elmoretundefined
                          elmoret
                          last edited by

                          @zov:

                          If you look at photos of OP you see he is measuring voltage between 2 terminals not diode test.
                          ADD: And I have mistaken in the post above: the current through LED should be about 1.2 uA not 12 uA. So it makes the multimeter much more unlike as a shunt to lights LED on.

                          My multimeter also has a 10Mohm input resistance, and it is enough to light the LED, faintly. You can see an LED's light output at as little as one microamp.

                          http://forum.arduino.cc/index.php?topic=183526.0

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                          • oleeundefined
                            olee
                            last edited by

                            Won't argue with you as I don't have any practical experience of using LEDs with such low currents. Anyway thanks for a new knowledge you gave me.

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