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KVL loop in BJT biasing

  1. Oct 2, 2015 #1
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    Screen Shot 2015-10-02 at 2.48.17 pm.png Screen Shot 2015-10-02 at 2.48.17 pm.png Screen Shot 2015-10-02 at 2.48.26 pm.png
    Please help how Ve is calculated what i think is that it should be:
    -Ve=-Vee+IeRe;
    How Ve is taken positive please someone draw equivalent circuit for same
     
  2. jcsd
  3. Oct 2, 2015 #2

    NascentOxygen

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    Staff: Mentor

    VE is defined as the emitter voltage wrt ground. Nothing about this says that VE must turn out to be a positive voltage, though. You can see that VE will be somewhere between -VEE and +VCC.

    If NPN emitter current is the current out of the transistor emitter, then the upper end of RE is the more positive. Draw arrows to show current direction, and voltage drop across resistances, and apply Kirchoff's Voltage Law.
     
  4. Oct 2, 2015 #3
    Thanks will you please help me in this question also i am confused in Vcc equations Screen Shot 2015-10-02 at 6.47.15 pm.png
     
  5. Oct 2, 2015 #4

    NascentOxygen

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    Staff: Mentor

    Explain which line in the solution you are having trouble with.
     
  6. Oct 2, 2015 #5
    Last line,how they have applied kvl to an open loop.
     
  7. Oct 2, 2015 #6

    NascentOxygen

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    Staff: Mentor

    There is no open loop.

    RB is supplying base current from the positive rail, VCC. Because the transistor is operating in the active region, its base is forward biased and its B-E potential is labelled VBE.

    The wire from the base ending in mid-air simply indicates where the amplifier's AC input will be impressed when the circuit is later completed to function as an audio amplifier, for example.
     
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