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Help with transistor

  1. Jul 31, 2016 #1
    Hey. I can't understand something. It's about NPN transistor. If it's given a constant base voltage VB, emitter voltage VE and base emitter voltage VBE, then VB = VBE +VE. I can't understand why phisicaly an increase in VE while VB is const will cause a decrease in VBE. Can VBE even be reduced? And how is reducing VBE is reducind the conduction level of the transistor?
     
  2. jcsd
  3. Aug 1, 2016 #2

    LvW

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    Rewriting your equation we have
    VBE=VB-VE.
    As you wrote - in case of constant VB (resistive voltage divider) an increase of VE (larger emitter current IE) will reduce the difference (VB-VE).
    This reduction in VBE is the desired negative feedback effect.
    The problem is that the IC=f(VBE) exponential relation is very sensitive to temperature changes (-2mV/K). Hence, for a constant VBE the current IC would change its value - more than allowed. For this reason we are using such an emitter resistor RE which produces an increasing voltage drop VE=IE*RE for the unwanted current increase. As a result, the voltage VBE again will decrease and, thus, bring back the current IE (and withit: IC) to nearly the previous value.
     
    Last edited: Aug 1, 2016
  4. Aug 1, 2016 #3
    Oooh, I get it now. VBE is the potential difference between the emitter and the base. Decreasing it will reduce the base current. I was viewing that VBE as the 0,7V potential barrier but it was just the difference between VB and VE :D Thank you, kind sir!
     
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