We have an opto-isolator diode connected to a contact closure to ground. When the contact is closed, the cathode reads 0V when measured to ground. When the contact is opened, will I measure a voltage that is close to 28V? I can't remember if that node is treated as a "floating" node or not.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ The voltage drop across a diode ultimately depends on the current. With the switch open, the current is only that passing through the voltmeter. A real world voltmeter will draw current, but an ideal one will not - so first you need to known what you are measuring with. \$\endgroup\$ Dec 3, 2018 at 22:50
  • \$\begingroup\$ node is treated as a "floating" node .... which node are you referring to? .... please add a label to the node in the schematic \$\endgroup\$
    – jsotola
    Dec 4, 2018 at 2:54
  • \$\begingroup\$ What precisely does an "opened" diode mean? \$\endgroup\$
    – Andy aka
    Dec 4, 2018 at 8:19
  • \$\begingroup\$ 75 ohms? really? \$\endgroup\$ Dec 4, 2018 at 10:35
  • \$\begingroup\$ @jsotola He's referring to the voltage from cathode of D1 to ground. Look at where the leads of the voltmeter (VM1) are connected to. He is wondering if the diode acts as an "ideal" switch. In which case, the cathode would be floating with respect to ground. Or if the source potential appears at the cathode, like you would see for a passive element, like a resistor. \$\endgroup\$
    – mrbean
    Mar 14 at 19:45

2 Answers 2


Yes, the meter will draw a negligible amount of current and you will read a voltage very close to 28V. Depending on the meter, the diode may have some drop across it, but it will certainly be less than 0.7V.


you'll get "close to 28V" (probably around 27V) most optocoupler LEDs take less than 1V


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