Few questions about BJT in practice

Discussion in 'General Electronics Chat' started by xxxyyyba, Feb 21, 2015.

  1. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Hi!
    I have two questions :
    1. I studied BJT in theory for some time and now I want to use it in practice. What are parameters of BJT that should be taken in consideration in order to use BJT properly in practice (withoud damaging it)?
    2. How total power dissipation for BJT is calculated? (parameter from BJT datasheet)
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    You have a number of "absolute maximum" parameters that are listed in the datasheet. Those generally mean that if you exceed those ratings that the device may be damaged. So try not to exceed those ratings. But small-signal transistors are cheap and so don't be too timid in your experimentation -- letting the magic smoke out of parts is a right-of-passage in this game.

    Total power for a BJT is calculated like total power for anything else -- the product of the current through a pair of pins and the voltage across those pairs of pins and then summed up over all the pairs of pins. Normally, you need to relate this power to the amount of power that the device can dissipate without raising the junction temperature above it's rated limit, and that depends on the thermal resistance between the junction and the ambient air (or other coolant).
     
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  3. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    So power would be P=Vbe*Ib+Vce*Ic, and this value must be lower than one given in datasheet for BJT to work properly?
     
  4. #12

    Expert

    Nov 30, 2010
    16,321
    6,818
    Theoretically, yes.
    We like to stay at or below half the maximum ratings so we don't burn our fingers.
     
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  5. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    Yes. As #12 said, give yourself some margin. A factor of two usually works well but is sometimes not practical, but make a real effort to stay below 80%.

    Because, in normal operation, Ic<<Ib and, usually, Vbe<Vce, the Vce*Ic term completely dominates and you can just use that as an approximation for the power, particularly if you aren't pushing the limits too hard. The exception can be when you are in very hard saturation.
     
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