saturation vs active mode energy dissipation question

Discussion in 'Homework Help' started by circuitashes, May 13, 2008.

  1. circuitashes

    Thread Starter Active Member

    May 13, 2008
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    0
    I read somewhere on this site that a transistor in saturation mode dissipates less energy than a transistor in active mode. I would like to know why. Please help as soon as possible.
     
  2. AlexK

    Active Member

    May 23, 2007
    34
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    The dissipated power for BJT (assuming DC operation) is approximately

    P = Vce * Ic

    And for a saturated transistor Vce is always lower compared to one in the active mode.
     
  3. Papabravo

    Expert

    Feb 24, 2006
    10,179
    1,800
    There are actually two things to consider. As the previous poster has indicated, power dissipation is the product of current and voltage. In the linear region with current and voltage in the middle of their range on the load line, power dissipation is greater than in either saturation or cutoff.

    To prove this to yourself, draw a straight line with any negative slope from the vertical axis to the horizontal axis. Now as you move along the line multiply the x-coordinate by the y-coordinate and see where the maximum occurs along the line.

    If you know any calculus, then ask yourself where the maximum of the function
    Code ( (Unknown Language)):
    1.  
    2.  y = x*( m*x + b) = m*x^2 + b*x
    3.  
    4. for  m < 0, b > 0
    5.  
    6. where m*x + b is the equation of the load line in a CE configuration.
    7.  
    will occur
     
    Last edited: May 13, 2008
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    To elaborate on AlexK's comments, Vce for a BJT is generally approximated at around 0.2 volts. A BJT biased into the active region might have at least an order of magnitude larger Vce. Between the active mode and the saturated mode Vc can change dramatically but the current will often only change by a factor of 2 for a BJT biased to place its Vc halfway between the collector supply and ground for example.

    Using the formula that AlexK has already provided, that means that in the active mode the BJT will dissipate say 2*i Watts and at saturation it will dissipate 0.2*2*i or 0.4*i Watts.


    hgmjr
     
  5. circuitashes

    Thread Starter Active Member

    May 13, 2008
    37
    0
    Thanks guys - Alexk,Papabravo, and hgmjr. Comments were really helpful. Out of curiousity, i would like to know: if Vbe = 0.7 and Vce = 0.2, would it be right to say
    Vcb = Vce - Vbe = -0.5 volts?
     
  6. Caveman

    Active Member

    Apr 15, 2008
    471
    0
    That's right.
     
  7. Papabravo

    Expert

    Feb 24, 2006
    10,179
    1,800
    You must also remember that if voltage drops are positive then a votage rise is negative. So the base is 0.5 volts above the collector and that by symmetry Vbc is the additive inverse or 0.5V
     
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