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.










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    up vote
    3
    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
      3
      down vote

      favorite









      up vote
      3
      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 Nov 23 at 23:08









      Programmer

      22218




      22218






















          1 Answer
          1






          active

          oldest

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          up vote
          7
          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
            Nov 23 at 23:48






          • 2




            As a curiosity: if you connect 5V on the Vin, it'll be injecting current back to the USB (mainly if it's powered down). Or if you accidentally short 3v3 to 5v, threshold voltage jumps over 10V. (the first one was caused by 5V voltage regulator instead of 9V. Second one by some short on shield and 9V on Vin)
            – KIIV
            Nov 24 at 9:16











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          7
          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
            Nov 23 at 23:48






          • 2




            As a curiosity: if you connect 5V on the Vin, it'll be injecting current back to the USB (mainly if it's powered down). Or if you accidentally short 3v3 to 5v, threshold voltage jumps over 10V. (the first one was caused by 5V voltage regulator instead of 9V. Second one by some short on shield and 9V on Vin)
            – KIIV
            Nov 24 at 9:16















          up vote
          7
          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
            Nov 23 at 23:48






          • 2




            As a curiosity: if you connect 5V on the Vin, it'll be injecting current back to the USB (mainly if it's powered down). Or if you accidentally short 3v3 to 5v, threshold voltage jumps over 10V. (the first one was caused by 5V voltage regulator instead of 9V. Second one by some short on shield and 9V on Vin)
            – KIIV
            Nov 24 at 9:16













          up vote
          7
          down vote



          accepted







          up vote
          7
          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 Nov 23 at 23:47

























          answered Nov 23 at 23:41









          Majenko

          65.4k42874




          65.4k42874












          • "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
            – Programmer
            Nov 23 at 23:48






          • 2




            As a curiosity: if you connect 5V on the Vin, it'll be injecting current back to the USB (mainly if it's powered down). Or if you accidentally short 3v3 to 5v, threshold voltage jumps over 10V. (the first one was caused by 5V voltage regulator instead of 9V. Second one by some short on shield and 9V on Vin)
            – KIIV
            Nov 24 at 9:16


















          • "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
            – Programmer
            Nov 23 at 23:48






          • 2




            As a curiosity: if you connect 5V on the Vin, it'll be injecting current back to the USB (mainly if it's powered down). Or if you accidentally short 3v3 to 5v, threshold voltage jumps over 10V. (the first one was caused by 5V voltage regulator instead of 9V. Second one by some short on shield and 9V on Vin)
            – KIIV
            Nov 24 at 9:16
















          "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
          – Programmer
          Nov 23 at 23:48




          "The diode imposes a voltage drop" I noticed this too and was curios on what's going on. Thanks for your clear answer
          – Programmer
          Nov 23 at 23:48




          2




          2




          As a curiosity: if you connect 5V on the Vin, it'll be injecting current back to the USB (mainly if it's powered down). Or if you accidentally short 3v3 to 5v, threshold voltage jumps over 10V. (the first one was caused by 5V voltage regulator instead of 9V. Second one by some short on shield and 9V on Vin)
          – KIIV
          Nov 24 at 9:16




          As a curiosity: if you connect 5V on the Vin, it'll be injecting current back to the USB (mainly if it's powered down). Or if you accidentally short 3v3 to 5v, threshold voltage jumps over 10V. (the first one was caused by 5V voltage regulator instead of 9V. Second one by some short on shield and 9V on Vin)
          – KIIV
          Nov 24 at 9:16


















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