transistor biasing

Discussion in 'Homework Help' started by lemon, Apr 18, 2010.

  1. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    Hi:
    Please look at the image below. I'm having trouble understanding how these formulae come about. This diagram is showing conventional current, I believe. And it is an npn BJT biasing.

    The first part says the voltage across a forward biased junction is 0.7. I think this is because it acts like an np diode.

    But it says:
    [​IMG]
    How can this be if the current is flowing from the positive terminal (conventional current)

    [​IMG]

    And what is the reason we have two voltage values?

    Any help please
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes its true
    Hmm ???
    [​IMG]
    Two voltage sources ?
     
  3. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    Hi:
    Ahh - two voltage sources - but if voltage sources are pushing current in direction of arrows, the voltage source Vcc is minused from Vbb because as it goes across the base junction it opposes Vbb? I can't understand how this works physically.

    hmm??
    I didn't give the full working for that part of question. I was gonna do in stages but I guess I confused with that.

    [​IMG]

    So this is ohms law
    and Ic is the current gain multiplied by Ib? How does this happen?
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    This two voltage source act independent.
    To open (ON) a bjt transistor you need Ib current to flow.
    So Vbb supplies Ib current to open (ON) BJT.
    +Vbb--->Rb--->base-emitter---> - Vbb

    And when BJT is open (ON) this mean that Ic current may begin to flow.
    Form +Vcc --- >Rc--->collector -emitter ---> -Vcc

    And of-course you can use single voltage source

    [​IMG]

    And Ic = β *Ib and Ie = Ib + Ic is a basic principle of a transistor "action"

    [​IMG]


    [​IMG]

    http://forum.allaboutcircuits.com/showthread.php?p=165135#post165135
     
    Last edited: Apr 18, 2010
    anhnha and kingdano like this.
  5. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    So in a forward biased npn, the base junction acts like a tap. Once the 0.7V has been reached current will start to flow (this is forward biased, right?).
    So the n collector is heavily doped with an excess of electrons but they can't jump the depletion region which is lightly doped with only a few available carrier holes. The emitter is moderately doped with electron overflow.
    When the 0.7V has been reached by the base junction supply, the depletion region is narrowed and the flow can commence from the n collector.
    How is that?

    I still don't understand what is Vbe. Is this an output voltage, and it is taken across the base emitter junction?

    Nice water pipe diagrams, by the way.
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    If base-emitter junction is forward biased the base current is flow. And Vbe is just a voltage drop in a base-emitter junction. So Vbe is a input voltages.
    However, notice that the base-to-collector junction is reverse-biased. The collector is at a much higher positive potential than the thin base. So when emitter inject electrons to the base terminal electrons will flow toward the very positive-looking depletion area created in the collector area close to the base layer.
    The end result is a much higher current flow through the collector than is flowing through the base.
    http://www.allaboutcircuits.com/vol_3/chpt_2/8.html

    The bjt act like control by base current current source
     
  7. kingdano

    Member

    Apr 14, 2010
    377
    19
    this is a great post, with awesome visual aids.

    very well done
     
  8. lemon

    Thread Starter Member

    Jan 28, 2010
    125
    2
    So - Vbe is a voltage drop across the base-emitter junction due to its resistance, the drop being in the region of about 0.7V, usually.
    Once this drop begins to become overcome the depletion region starts to narrow more because the emitter is full of free electrons which diffuse across the b-e junction to the lightly doped and already narrow base region. But there are only a small amount of holes there so only a small percentage combine to form a small trickle of current out the base. The rest diffuse into the collector region, the c-b junction, where they are pulled across by the positive attraction of the depletion layer field. The electrons now move through the n region collector attracted by the positive potential of the collector terminal.
    And the flood gates open.
    :)
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    This Vbe voltage drop exist thanks to "barrier potential or built-in potential" in P-N junction.
    http://en.wikipedia.org/wiki/P-n_junction#Equilibrium_.28zero_bias.29

    Water pressure will drop because you need a force to open a flood gate in the base.
     
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