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SOT-23 NMOS transistor power dissipation question

Discussion in 'Datasheets, Manuals and Component Identification' started by wave.jaco, Oct 9, 2015.

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  1. wave.jaco


    Aug 1, 2015
    Dear forum members

    I am working on a design of a control system that controls a few small solenoids, each drawing about 200 mA of current and operating at 15-18 V. These solenoids are activated by NMOS transistors. Due to the size constraint that I have for the PCB containing these transistors, I would ideally like to use SOT-23 NMOS transistors.

    Now, if one does the math, the power dissipated by each transistor activating a solenoid is P = (0.2 A)(18 V) = 3.6 W. However, it seems that most of the SOT-23 package NMOS transistors can only dissipate a maximum of about 1 W.

    My question now is this - is the maximum power dissipation rating primarily due to the heat build-up that could damage the device if it exceeds its rated power dissipation? Or is it the limit of the internal materials that the transistor is made of?

    Each solenoid in this system is typically never activated for longer than 20 seconds, and at the very extreme it might be activated for 1 minute, which happens almost never. The average on-time for these solenoids are between 1 second and 10 seconds. Will this relatively short on-time (dissipating 3.6 W of power) have a detrimental effect on a transistor that can only dissipate a maximum of 1 W? Or can one safely assume that the average on-time of the solenoid is short enough that the transistor will not heat up too much to cause any damage?

    I have used the 2N7002 NMOS transistor (which has a very low max. power dissipation rating) in a prototype for this system with no problems at all and no evident heat build-up by the transistors. Should I rather get a properly rated transistor (3.6 W or higher rated), or can I safely use a 1 W rated transistor for this system?

    Your comments and suggestions will be highly appreciated. Thanks!

    EDIT: I realize now my calculation is totally wrong. The above method calculates the power dissipated by the solenoid, not the transistor. Moderators, please feel free to remove this question. Thank you.
    Last edited: Oct 9, 2015
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    If one does the circuit correctly, the MOSFET is either on or off. The 2N7002 has a max. Rds_on of 13.5Ω Power dissipated in the MOSFET is P0V*I=I²*Rds.
    In the off state, I=0 A -> P=0 W
    In the on state I = 0.2A -> P = 0.54 W (max. value, typical value willl be less).
    0.54 W is less than the 1 W you mention, therefore the use of this FET in this application seems to be safe.

    However, the thermal resistance of this FET is Rthja=625 °C/W. At 0.54W this amounts to a junction temperature of tj=625 °C/W*0.54W+tamb ~ 360 °C at 25 ° C ambient temperature. This will burn your transistor.

    As the typical value for Rds_on =2.4Ω, the typical power dissipation is only 0.1W and tj=88 °C which is no problem at all.

    Your design therefore will typically work, but you shouldn't be surprised if it fails with a batch of transistors that have worst case parameters (while still being within spec).
  3. dorke


    Jun 20, 2015
  4. BobK


    Jan 5, 2010
  5. wave.jaco


    Aug 1, 2015
    Thank you for all the answers. My question is answered. Much appreciated.
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