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A question about VCEsat

  1. Oct 25, 2016 #1
    Hi! The VCEsat which occurs between the emitter and the collector of a saturated bjt is duo to the inevitable loss of energy caused by the resistance between the emitter and the collector, right?
    If so, why this voltage drop is bigger at higher currents and lower at smaller ones? For example at Ib to Ic ratio of 1 to 10, at Ic=250mA Vcesat=,2V. But at the same base to collector current ratio, at Ic=500mA Vcesat=1V.
    Also I always wanted to ask someone, as an engineer, how deep do you need to go with physical explanations?
     
  2. jcsd
  3. Oct 25, 2016 #2
    Are you talking about the same transistor at different currents, or two different transistors?Different transistors can have different characteristics.

    But if the same transistor, note that you have doubled the base current, so couldn't that push the transistor further into saturation? Also, doubling the base current would probably push the curve a bit, and the exact 1:10 ratio would have moved slightly, so I'm not sure your description is completly accurate.

    I should add - I'm not sure that Vce is totally a product of bulk resistance, some of the physics of electrons, holes, band-gaps, etc probably play into it, but that is beyond me, or I've forgotten it if I ever did know.
     
  4. Oct 25, 2016 #3
    Nope, its the same transistor. Some transistors in their datasheet have values of Vcesat for several different currents. And to ensure the transistor is saturated, the base current is 1 10th of the collector one. And for example at 250mA Ic you got 0,2V Vcesat, but for Ic=500mA Vcesat = 0,7V(These values arent 100% sure since I may have forgotten them). Oh and so, if there are some other quantum-like reasons for Vcesat you dont necessarily need to know about them, right?
     
  5. Oct 25, 2016 #4
  6. Oct 26, 2016 #5

    CWatters

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    If it was due to resistance why are you surprised?

    V= I*R
     
  7. Oct 27, 2016 #6
    Yep, you're right. I don't know why I struggled about it so much. Since the silicon itself has some resistance you can never get away with 0V drop across CE. And so you got around 0,2V Vcesat. If the ratio between Ic and Ib is still 1/10, if the currents are bigger, duo to Ohms law Vcesat will increase. But if you increase the base current even more, you turn on the transistor more and reduce it's depletion layer even further, decreasing the resistance and Vcesat will be less.
    Thanks!
     
  8. Oct 27, 2016 #7
    Thank you guys for the replies
     
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