What is the current 'Ic' in the embedded circuit diagram? Thanks in advance..

Discussion in 'The Projects Forum' started by veeresh540, Feb 9, 2015.

  1. veeresh540

    Thread Starter New Member

    Dec 17, 2014
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    Hi,
     
    Last edited: Feb 20, 2015
  2. bertus

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    Apr 5, 2008
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    Hello,

    The collector can not be connected to the 5V directly.
    You will need a current limiting resistor.
    The current in the collector is also dependent on the current through the led and the transfer characteristics of the optocoupler.

    Bertus
     
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  3. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    When you use a 10K resistor on the collector, the collector current can be 5 Volts / 10K Ohms = 0.5 mA max.
    The real value will depend on the saturation voltage of the transistor.
    Look in the datasheet of the optocoupler to find this saturation voltage.

    Bertus
     
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  4. bertus

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    Hello,

    If the powersupply is 5 Volts and the Vce(sat) is 0.2 Volts, the voltage accross the resistor will be 5 - 0.2 = 4.8 Volts.
    The current will be 4.8 Volts / 10K Ohms = 0.48 mA.

    Bertus
     
  5. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    I said that the Ic is dependent of the led current, but you should look-up the transfer characteristics of the optocoupler in the datasheet.
    When there is more current through the led of Ic times the transfer factor (most times given in a percentage), the transistor can be considered to be in saturation.

    Here is the transfer characteristics graph from the datasheet:

    [​IMG]

    You have a led current of about 1.6 mA.
    Lower the 4k7 to 1k5 to have a led current of about 5 mA, to ensure saturation.

    Bertus
     
    Last edited: Feb 9, 2015
  6. bertus

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    Apr 5, 2008
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    Hello,

    In your schematic is given a powersupply of 15 volts at the led side.
    There is a 6.2 Volts zener in series.
    The led forward voltage is about 1.2 Volts.
    The voltage accross the 4K7 resistor is 15 - 6.2 - 1.2 = 7.6 Volts.
    The current through the 4K7 is 7.6 Volts / 4K7 Ohm = 1.617 mA.

    If you can not change the resistor, can you remove the zener?
    In that case the current would be (15 - 1.2) / 4K7 = 2.936 mA.

    Bertus
     
  7. John P

    AAC Fanatic!

    Oct 14, 2008
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    Nobody else seems to be saying it, so I will--this thread is in the wrong section.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    I tend to agree.
     
  9. bertus

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    Hello,

    I moved the thread from the embedded forum to here.

    Bertus
     
  10. bertus

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    Hello,

    Have a look at page 3 of the datasheet.
    There is the CTR value given and how it is used.

    Bertus
     
  11. bertus

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    Hello,

    Using the formula from page 3 of the datasheet ( CTR = (Ic/If) X 100% ) will give you ( 0.48 / 1.1617 ) X 100 = 29.68
    This would give you the A or B version of the optocoupler.

    Bertus
     
  12. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    This sounds like homework to me,

    why dont you use an ammeter then you will know the current, if you cant understand the datasheet why are are you bothering.
     
  13. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    I moved your post to your existing thread about the same issue (see post#14).

    The 844A version is a selection of the 844 version.

    Bertus
     
  14. veeresh540

    Thread Starter New Member

    Dec 17, 2014
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    Hi Mod/Admin,
    Can you please remove my two threads(including duplicate). Thank you very much.
     
  15. veeresh540

    Thread Starter New Member

    Dec 17, 2014
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    Hi bertus,
    I need my threads to be deleted. Can you help me in this regard. Thanks!!
     
    Last edited: Feb 20, 2015
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