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  • Power loss in Bipolar Junction Transistor

    freak16

    Member

    Updated: Oct 19, 2024
    Views: 1.9K
    Is power loss in bjt Vce*Ic? :smile: It is..but i dont know Why:smile:??
    there must be some loss due to Ib(Base current also) though i completely agree it is a contolling current. but how this relation come..plz help me out.. 😁
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  • abs_vicky

    Member • Dec 18, 2010

    See buddy, In a N-P-N Bjt, The emitter is heavily doped and the base is very lightly doped and made extremely thin. So charge carriers in the base are very less, thus the recombination with incoming electrons from the emitter isnt much and due to the small width of the base most electrons cross over to the collector. Yaa there will be a loss due to the base current out of recombination but it is negligible in comparision and often neglected for simplicity in calculations.
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  • freak16

    Member • Dec 18, 2010

    Dear vicky , you are right but my question is how the relation Vce * Ic comes?? 😁
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  • Jishnu Gupta

    Member • Jan 28, 2011

    Itz very simple maniac as per simple relation P=V*I so in a bjt Ie= Ib+Ic where the value of Ib is negligible hence we get Ie = Ic now as power loss takes place at collector junction which has a battery as voltage source i.e Vce and current flowing throuugh is Ic hence the power loss is P = Vce*Ic
    Hope it is clear now any odr doubts u have lemme knw ...........😀
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  • Arp

    Member • Jan 30, 2011

    jishnu is right.
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