What's the function of this transistor on Arduino Mega?











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I am trying to get information about the transistor on the Arduino Mega board below:



enter image description here



It says "340P" on the transistor. I want be able to know which model transition is this so that I can check its documentation. It also want to know what it is driving or its function on the Arduino board.










share|improve this question


























    up vote
    1
    down vote

    favorite












    I am trying to get information about the transistor on the Arduino Mega board below:



    enter image description here



    It says "340P" on the transistor. I want be able to know which model transition is this so that I can check its documentation. It also want to know what it is driving or its function on the Arduino board.










    share|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am trying to get information about the transistor on the Arduino Mega board below:



      enter image description here



      It says "340P" on the transistor. I want be able to know which model transition is this so that I can check its documentation. It also want to know what it is driving or its function on the Arduino board.










      share|improve this question













      I am trying to get information about the transistor on the Arduino Mega board below:



      enter image description here



      It says "340P" on the transistor. I want be able to know which model transition is this so that I can check its documentation. It also want to know what it is driving or its function on the Arduino board.







      arduino-mega transistor documentation






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 5 hours ago









      Programmer

      1656




      1656






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          3
          down vote



          accepted










          It's a P-channel MOSFET. It's job is to act as a "dropless" diode.



          The principle is this:




          • MOSFETs have a built-in diode across them in reverse bias (an effect of the chemistry)

          • The P-channel MOSFET is connected backwards in series with the incoming USB power.

          • The internal diode conducts power when there is +5V in to the USB to give power to the board.

          • The diode imposes a voltage drop, but there is enough voltage still to run the circuitry.

          • An op-amp compares the incoming voltage from the barrel jack (divided by 2) against 3.3V. If it's less than 3.3V (6.6V incoming) or not there at all, it turns on the P-channel MOSFET

          • The MOSFET then short circuits the internal diode removing the voltage drop, giving the full 5V from the USB to the rest of the board.


          schematic





          simulate this circuit – Schematic created using CircuitLab



          Note: diode shown is internal to the MOSFET.



          The MOSFET is designated T2 in the schematic, and is a FDN340P on the reference design, although the actual model is not that critical, as long as the threshold voltage is above about -3V and it can happily handle 500mA or more.






          share|improve this answer























          • "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
            – Programmer
            4 hours ago











          Your Answer






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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          It's a P-channel MOSFET. It's job is to act as a "dropless" diode.



          The principle is this:




          • MOSFETs have a built-in diode across them in reverse bias (an effect of the chemistry)

          • The P-channel MOSFET is connected backwards in series with the incoming USB power.

          • The internal diode conducts power when there is +5V in to the USB to give power to the board.

          • The diode imposes a voltage drop, but there is enough voltage still to run the circuitry.

          • An op-amp compares the incoming voltage from the barrel jack (divided by 2) against 3.3V. If it's less than 3.3V (6.6V incoming) or not there at all, it turns on the P-channel MOSFET

          • The MOSFET then short circuits the internal diode removing the voltage drop, giving the full 5V from the USB to the rest of the board.


          schematic





          simulate this circuit – Schematic created using CircuitLab



          Note: diode shown is internal to the MOSFET.



          The MOSFET is designated T2 in the schematic, and is a FDN340P on the reference design, although the actual model is not that critical, as long as the threshold voltage is above about -3V and it can happily handle 500mA or more.






          share|improve this answer























          • "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
            – Programmer
            4 hours ago















          up vote
          3
          down vote



          accepted










          It's a P-channel MOSFET. It's job is to act as a "dropless" diode.



          The principle is this:




          • MOSFETs have a built-in diode across them in reverse bias (an effect of the chemistry)

          • The P-channel MOSFET is connected backwards in series with the incoming USB power.

          • The internal diode conducts power when there is +5V in to the USB to give power to the board.

          • The diode imposes a voltage drop, but there is enough voltage still to run the circuitry.

          • An op-amp compares the incoming voltage from the barrel jack (divided by 2) against 3.3V. If it's less than 3.3V (6.6V incoming) or not there at all, it turns on the P-channel MOSFET

          • The MOSFET then short circuits the internal diode removing the voltage drop, giving the full 5V from the USB to the rest of the board.


          schematic





          simulate this circuit – Schematic created using CircuitLab



          Note: diode shown is internal to the MOSFET.



          The MOSFET is designated T2 in the schematic, and is a FDN340P on the reference design, although the actual model is not that critical, as long as the threshold voltage is above about -3V and it can happily handle 500mA or more.






          share|improve this answer























          • "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
            – Programmer
            4 hours ago













          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          It's a P-channel MOSFET. It's job is to act as a "dropless" diode.



          The principle is this:




          • MOSFETs have a built-in diode across them in reverse bias (an effect of the chemistry)

          • The P-channel MOSFET is connected backwards in series with the incoming USB power.

          • The internal diode conducts power when there is +5V in to the USB to give power to the board.

          • The diode imposes a voltage drop, but there is enough voltage still to run the circuitry.

          • An op-amp compares the incoming voltage from the barrel jack (divided by 2) against 3.3V. If it's less than 3.3V (6.6V incoming) or not there at all, it turns on the P-channel MOSFET

          • The MOSFET then short circuits the internal diode removing the voltage drop, giving the full 5V from the USB to the rest of the board.


          schematic





          simulate this circuit – Schematic created using CircuitLab



          Note: diode shown is internal to the MOSFET.



          The MOSFET is designated T2 in the schematic, and is a FDN340P on the reference design, although the actual model is not that critical, as long as the threshold voltage is above about -3V and it can happily handle 500mA or more.






          share|improve this answer














          It's a P-channel MOSFET. It's job is to act as a "dropless" diode.



          The principle is this:




          • MOSFETs have a built-in diode across them in reverse bias (an effect of the chemistry)

          • The P-channel MOSFET is connected backwards in series with the incoming USB power.

          • The internal diode conducts power when there is +5V in to the USB to give power to the board.

          • The diode imposes a voltage drop, but there is enough voltage still to run the circuitry.

          • An op-amp compares the incoming voltage from the barrel jack (divided by 2) against 3.3V. If it's less than 3.3V (6.6V incoming) or not there at all, it turns on the P-channel MOSFET

          • The MOSFET then short circuits the internal diode removing the voltage drop, giving the full 5V from the USB to the rest of the board.


          schematic





          simulate this circuit – Schematic created using CircuitLab



          Note: diode shown is internal to the MOSFET.



          The MOSFET is designated T2 in the schematic, and is a FDN340P on the reference design, although the actual model is not that critical, as long as the threshold voltage is above about -3V and it can happily handle 500mA or more.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          Majenko

          64.9k42874




          64.9k42874












          • "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
            – Programmer
            4 hours ago


















          • "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
            – Programmer
            4 hours ago
















          "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
          – Programmer
          4 hours ago




          "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
          – Programmer
          4 hours ago


















           

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