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Understanding Optocoupler circuit

  1. Sep 5, 2009 #1
    Hello all,

    If in this http://en.wikipedia.org/wiki/File:Optocoupler_Circuit.svg" [Broken], Emitter is not connected to the ground, what will be the voltage at Vo and and at the Emitter?

    Thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 5, 2009 #2

    vk6kro

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    The primary section of this circuit only generates light in the optocoupler, so it can be connected to whatever voltage you like. It does not have to be grounded at all.

    That is what optocouplers do. They isolate different parts of circuits from each other.

    In this case, the switch in the primary is open, so there is no light being generated in the optocoupler.
    So, the phototransistor is not conducting so Vo is the same as Vcc.
     
  4. Sep 5, 2009 #3
    I'm talking about the emitter of the phototransistor. If it is not connected to the ground, then what will be the voltage at Vo. In this case, I believe primary side of the circuit will be working.

    And further more, if transistor is in cut-off mode (i.e not conducting), than I believe Vo will be 0 volt.

    Thanks
     
  5. Sep 5, 2009 #4

    vk6kro

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    You need to measure voltage from somewhere to somewhere else.

    In this case, as shown, the primary switch is open so there is no current flowing in the primary (ie through the LED).

    Connecting the emitter of the phototransistor to ground has no effect as far as the operation of the circuit is concerned as long as there is a return circuit for VCC back to the emitter of the phototransistor.

    If the primary circuit is open circuit because the switch is open, there will be no light falling on the phototransistor.

    So the phototransistor will be a very high resistance.
    If R2 is any value a lot less than this very high resistance, there will be a voltage divider action where most of the supply voltage will be across the phototransistor and very little or none of it will be across R2. So, V0 will be very close to Vcc, relative to the emitter of the phototransistor.
     
  6. Sep 6, 2009 #5
    Oh, I mean the situation where the switch is closed, and LED is on (i.e the primary side of the circuit is conducting).

    I'm extremely sorry for the confusion.
     
  7. Sep 6, 2009 #6

    vk6kro

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    No confusion at all.

    In the circuit you referred to, the switch is open. Have a look.

    If you close the switch, the phototransistor will conduct. Vo would then drop, but would not go to zero relative to the emitter of the phototransistor. It will drop to something below one volt, though, if the primary current is great enough.
     
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