Understanding Optocoupler circuit

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Discussion Overview

The discussion revolves around the behavior of an optocoupler circuit, specifically focusing on the voltage at the output (Vo) and the emitter of the phototransistor under different conditions, such as when the emitter is not grounded and when the primary switch is open or closed. The scope includes technical explanations and conceptual clarifications related to circuit operation.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the voltage at Vo and the emitter if the emitter is not connected to ground.
  • Another participant explains that the primary section generates light and does not need to be grounded, stating that with the switch open, Vo equals Vcc since the phototransistor is not conducting.
  • A different participant emphasizes the emitter's role, suggesting that if the phototransistor is in cut-off mode, Vo will be 0 volts.
  • Another reply clarifies that measuring voltage requires a reference point and notes that with the switch open, there is no current, leading to high resistance in the phototransistor and Vo being close to Vcc.
  • One participant later corrects their previous statement, indicating they meant the scenario with the switch closed, where the LED is on and the primary circuit is conducting.
  • A final reply confirms that with the switch closed, the phototransistor will conduct, causing Vo to drop but not to zero, suggesting it will be below one volt depending on the primary current.

Areas of Agreement / Disagreement

Participants express differing views on the voltage behavior under various conditions, particularly regarding the role of the emitter connection and the state of the switch. The discussion remains unresolved with multiple competing perspectives on the circuit's operation.

Contextual Notes

Participants mention the need for a reference point when measuring voltage and discuss the implications of the switch being open or closed, but do not resolve the assumptions regarding the circuit's behavior in these scenarios.

shahper
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Hello all,

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

Thanks
 
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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.
 
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
 
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.
 
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.
 
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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