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Question about biasing BJTs

  1. May 5, 2016 #1
    What happens if for argument's sake you have a voltage of 15V and a resistor of 1k connected to the base of a BJT thus biasing it with 15/1000=.015amps

    and then you apply a constant current source to the emitter/collector that is

    1. less than beta*baseCurrent
    2. more than beta*baseCurrent
    3. Equal to beta*baseCurrent
     
  2. jcsd
  3. May 5, 2016 #2
    With a grounded emitter, you'll have less than 15 mA base current due to the base-emitter voltage drop. This is also a terrible way to bias the BJT, since you're relying on knowing device parameters that vary wildly. You should use something like a stiff voltage divider instead to set the quiescent current.

    With the current source oriented correctly, it will just provide whatever current you set it to by applying the needed voltage across the BJT, which will be small if you give the BJT enough base drive.
     
  4. May 5, 2016 #3

    phyzguy

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    Is this a homework question? If so, have you seen the homework template? You need to make an attempt before we can help. You might start by thinking about what a constant current source does. For example, what happens if you have a constant current source and apply it to an open circuit?
     
  5. May 5, 2016 #4
    no this is not a homework question. I like pondering about things is all.

    Your question can be easily answered if referring to resistive circuits but semiconductors are different.

    with a varying R, a current source will supply the voltage necessary to keep the current constant.

    With a BJT though, the base determines what it 'wants' the collector to have flowing through it. If a current source is applied to the collector though, the two are kind of at odds aren't they? They both 'want' two different things. Someones not going to get what they want.
     
  6. May 5, 2016 #5
    That's not how it works. It's not a great model, but think of the BJT as more of a variable resistance. All the BJT can do is be either fully on, off or somewhere in between, i.e. it can, approximately, act as a short circuit, an open circuit or some load in between. How does that affect the current source?
     
  7. May 5, 2016 #6

    phyzguy

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    As you said, the current source will supply the voltage necessary to keep the current constant. But it has voltage limits also. If it takes 1,000,000 volts to keep the current constant, this it is probably going to hit its voltage limits before it can keep the current constant. So the question is whether the current source will be able to keep the current constant or whether it will hit a voltage limit. Also, remember that the model of the BJT having collector current = beta * base current is only true if it is in the forward active mode. If it goes into saturation, for example, this no longer holds. So in your first case, the collector voltage will just drop until the BJT is saturated and the Ic = beta * Ib relation no longer holds.
     
  8. May 5, 2016 #7
    If this current source is in "collector side" then transistor will enter saturation if Ib*β > Isource --- >15mA > Icource and Ie = 15mA + Isource.
    But if we put this current source into "emitter side" ( from emitter to ground) the Ib current will be equal to Ib = Isource/(β + 1) and transistor will be in active region as long as 15V < Ib*1kΩ
     
  9. May 5, 2016 #8
    Thanks for the explanations. Could someone also explain what happens in terms of the hole/electron physics?
     
  10. May 5, 2016 #9
    Jony, how would you put it into active mode when the current source is connected to the collector?
     
  11. May 6, 2016 #10
    I can't help you because I don't know nothing about transistor physics.
    There are many ways to achieve it. For example we can use some negative feedback from collector to base. Or we can simply set Isource around Ib*β value.
     
  12. May 7, 2016 #11
    Wouldnt hfe*B just need to be higher than Ic and that will make in active mode?
     
  13. May 7, 2016 #12

    NascentOxygen

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    You mean hFE*IB? That condition puts it in saturation.

    To be in the active region, hFE*IB = IC
     
  14. May 7, 2016 #13
    @Ry122.... We have to first see the importance of biasing. The approach which is used in electronics engineering is mostly not hole-electron pair explanations. These hole-electron pair phenomena is also not accurate. You need to learn semiconductor physics and promise yourself that Scrodinger's equation best describes the atomic world and solve it to get Fermi levels(using Fermi-Dirac Statistics), conduction bands, valence bands and so on. Then, you need to study extrinsic p and n semiconductors. Then you need to see what happens when you join both( to form a diode)....Then, you have to extend the same to transistor.
    The problem does not end with transistor alone. You need to know something like Eber-Bohr Transistor model, how to model it and so on. Using this model, you can construct for yourself various bjt biasing circuits and prove that the voltage and current values don't deviate much.
    It may look daunting but with little differential equations, and little algebra, you can manipulate things to prove for yourself
     
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