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    MSP430G2231+TRIAC

    This question is answered
    bala krishnan
    Posted by bala krishnan
    on Jul 14 2012 02:12 AM
    Prodigy95 points

    Hi there,

    Actualy i am firing a triac using a optocoupler thru a transistor from the msp 430 output pin.I am getting the desired outputs when i connect the output pins to LED but when i connect a 1K resistance to drive the transistor it is not getting the drive.Kindly help.

    Regards ,

    Bala

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    • old_cow_yellow
      Posted by old_cow_yellow
      on Jul 14 2012 04:05 AM
      Guru25715 points

      You did not supply enough information about the optocopler, the transistor, and how they are connected to the 1k resistor.

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    • Andy Neil
      Posted by Andy Neil
      on Jul 14 2012 05:30 AM
      Guru31975 points

      old_cow_yellow

      You did not supply enough information about the optocopler, the transistor, and how they are connected to the 1k resistor.

      Indeed!

      You really need to post a schematic - trying to describe a circuit in words (especially in a foreign language) is a prettty futile exercise.

      It would also help to give links to the datasheets of the components concerned; or, at the very least, full manufacturer's name & part number.

      If possible, use original datasheets direct from the manufacturer's own website - use 3rd-party sites only as a last resort.

       

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    • bala krishnan
      Posted by bala krishnan
      on Jul 15 2012 22:55 PM
      Verified Answer
      Verified by bala krishnan
      Prodigy95 points

      Details of the circuit are as follows.

      From the Microcontroller output there is an 1K quarter watt resistor which is connected again to the transistor BC-107 Base.The emitter of the same is grounded and there is a 330 ohm resistance connected to pin numbr 2 of optocoupler MOC-3021 and then the triac is BT136 which is connected to MOC via a 360 ohm and a 470 ohm resistance in series with pin number6 of MOC .Pin number4 of MOC is connected to pin number 1 of BT136.The load is connected through pin number 2 and a snubber circuit to the load. 

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    • Jens-Michael Gross
      Posted by Jens-Michael Gross
      on Jul 16 2012 12:38 PM
      Verified Answer
      Verified by bala krishnan
      Guru140085 points

      And what is your supply voltage that goe sinto the MOC?

      The internal LED of a typical optocoupler usually has a voltage drop of ~2V on 10mA. Now there is the transistor C-E voltage and the 330Ohm resistor in series.
      To have 10mA flowing through teh diode, you'd need ~6V supply voltage. 2V for the diode, 330Ohm*10mA=3.3V for the resistor and >0.5V or even more for the C-E.

      IF you try this with the MSPs VCC of 3.6V max, you'll likely have only a very small LED current. And an even less output current from the MOC, asthese devices usually have a transmission ration of <50%.

      It is definitely better, easier and cheaper to just use the MSP output, maybe with a small series resistor (100R or less) if you don't want the full port driving capability to go into the LED.

      _____________________________________
      Before posting bug reports or ask for help, do at least quick scan over this article. It applies to any kind of problem reporting. On any forum. And/or look here.
      If you cannot discuss your problem in the public, feel free to start a private conversation: click on my name and then 'start conversation'. But please do so only if you really cannot do it in a public thread, as I usually read all threads. And I prefer to answer where others can profit from it (or contribute to it) too.

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    • bala krishnan
      Posted by bala krishnan
      on Jul 16 2012 22:47 PM
      Prodigy95 points

      """"""And what is your supply voltage that goe sinto the MOC?""""""

      Suply voltage is 5v.

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    • Jens-Michael Gross
      Posted by Jens-Michael Gross
      on Jul 19 2012 10:29 AM
      Verified Answer
      Verified by bala krishnan
      Guru140085 points

      bala krishnan
      Suply voltage is 5v.

      Well,

      worst case, the MOC requires a trigger current of 15mA. And has a diode voltage of 1.5V. So 5V minus 1.5V minus 0.5V for the transistor leaves 3V on the resistor. 3V/330Ohm is 9mA and not enough to trigger the MOC worst case.

      Anyway, teh MOC already has an internal TRIAC. Triggering a triac by a triac is not trivial. Maybe your circuit is unable to fire the second triac at all.

      Besides this, you're torturing your transistor with VCC/1k = ~3.3mA base current. Way too much unless your transistor has a beta <5 :) You can use a 10k.

      Also, if I understood right, the AC/HV side has a BT136 with 830Ohm in series to the MOC triac and the gate. The BT136 requires a gate trigger voltage of 0.7 to 1.5V and a  gate current of (worst case) 25mA. This means the minimum operating voltage (worst case) would be 22V, since only then 25mA will flow through your resistors and the MOV to the BT136 gate. But even best case, 12V are needed.

      _____________________________________
      Before posting bug reports or ask for help, do at least quick scan over this article. It applies to any kind of problem reporting. On any forum. And/or look here.
      If you cannot discuss your problem in the public, feel free to start a private conversation: click on my name and then 'start conversation'. But please do so only if you really cannot do it in a public thread, as I usually read all threads. And I prefer to answer where others can profit from it (or contribute to it) too.

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    • bala krishnan
      Posted by bala krishnan
      on Jul 24 2012 00:50 AM
      Verified Answer
      Verified by bala krishnan
      Prodigy95 points

      I am sorry fr the late post.was out of town....any ways now my circuit is absolutely working and achieved the desired output......thanks a lot fr ur help all of u..looking ahead for more help.....

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