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Transistor question-biasing

  1. Oct 17, 2011 #1
    Consider this picture, a) and b).
    [PLAIN]http://pokit.org/get/f0af323a103c1d29539836143a7784a1.jpg [Broken]

    Now I am getting confused a bit here.

    a) I know that the first picture represents a bias for current source. And we change the value on the potentiometer the current will stay the same. (considering we won't go too far and put transistor into saturation)

    From the computations I got, and simulations in national instruments, I can say that the base current is adjusting accordingly in order to keep the current constant. I can handle that idea, I think.

    b) Will the same happen here? If I change the potentiometer. Consider that resistor connected to base to be thevenin equivalent of those 2 voltage dividers and that voltage source at base to be also be within thevins theorem. And every other parameter stays the same.

    But I have the feeling that the base current here is determined only by that voltage source at base and that resistor, hence it won't change when I change the potentiometer, hence the current in collector must fall.

    Are those to schemes one and the same? Will I have the same effect?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 17, 2011 #2
    As in the diagram, you have a resistor at the emitter to 0V. That and the base voltage is really what determine the collector current. b) is the thev. eq. of a), they both set up the voltage on the base.

    The reason we always use a resistor at the emitter because the current gain of the transistor is not stable, the emitter resistor really serve as the negative feedback. Say the voltage at the base give the base current, which in turn create the emitter current. as emitter current increase the voltage across the emitter resistor increase. At the point where Vbe is about 0.7V, any further increase in emitter will cause the Vbe to drop and decrease the emitter current. It is this negative feedback that keep the current constant at the collector. This kind of circuit eliminate the dependency of base current and become more a voltage dependent.

    I don't think I explain this very well, look up a basic book of bipolar transistor and they explain much better than me the role of emitter degenerating resistor and how it take away the variable of the current gain of a transistor.
     
  4. Oct 17, 2011 #3
    But can you confirm that the base current has to rise, in order to keep the current in collector constant, whilst changing the resistance of collector resistor high up?
     
  5. Oct 17, 2011 #4

    es1

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    Base current will be the same for constant collector current.
    Ib = Ic / Bf.

    Now technically the voltage at the collector will change and the transistor will experience a change in Bf due to this which will in turn alter Ib (this is known as the early effect). But I think you can probably ignore this non-ideality if this is for homework and if you were not yet aware of it.

    http://en.wikipedia.org/wiki/Early_effect
     
  6. Oct 17, 2011 #5
    Nah this is my trying to understand things. Homework was long done. Now look, If i change the collector resistance, and the current through collector stays the SAME, something HAS TO GIVE. I feel like the base current will increase, in order to keep the collector current constant for all resistances. Assume that you can change the collector resistance so much, that you don't get transistor into saturation.
     
  7. Oct 17, 2011 #6
    If you look at the data sheet of the specific transistor, you can find a graph that show given a particular biasing condition ( like your circuit), how the collector current change with change in Vce. You'll find the change is very very little and in the first approx, you take it as constant.

    Of cause it really vary very little and yes you need to change the base bias ever so little to keep the collector constant. But really nobody worry about this with BJT. When you talk about FETs, then it's a different story, because change of drain current vs Vds is a lot more prominent and cannot be ignored.

    Unless you are getting into semi conductor physics, just assume the collector current is constant. Take our words, I design bipolar ICs and all different transistor circuits, never once I have to worry about this.
     
  8. Oct 17, 2011 #7

    es1

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    In real transistors something does give. If Rc gets larger (without saturating the transistor) and the current stays the same then the voltage drop on the resistor goes up and the voltage on the collector goes down. The CB depletion width changes in response to the changing Vc. This changes the gain of the transistor which in turn changes the base current. This is known as the early effect and was linked to above.

    However, the transistor is designed to have high gain and high immunity to this non-ideality. Say for example Ic=10mA and Bf dropped from 200 to 150 due to decreasing Vc. Then Ib changes from 50uA to 66uA. If the biasing resistors are small enough then this change in current can be ignored.

    This is why there is the design rule "make divider current 10x the base current" in the app below, for example.
    http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npncc.html#c3
     
  9. Oct 17, 2011 #8
    Yes, in semi conductor physics, it's the early voltage and it is very big for BJT.....not so big for FETs.

    For the op, if you look at the collector curve, the slope on the straight portion is the output resistance. And if you extrapolate the straight portion ( the slope is the output resistance) of different collector curves( different collector currents), they all come to a same point and the voltage is the early voltage. For NPN, the early voltage Va is always negative.
     
    Last edited: Oct 17, 2011
  10. Oct 18, 2011 #9

    jim hardy

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    Mr Bassalisk ,
    i think sometimes in teaching we forget to stress the very basics

    contemplate for a moment that resistor between the emitter and ckt common.
    it makes your circuit into a closed loop with negative feedback.


    current through the base-emitter junction follows that formula i = e^(qv/kt) ,
    where i is current, e is 2.718, q and k are constants, v is base-emitter voltage, and t is temperature...
    now look where voltage is in that term - it's up in the exponent - so a miniscule change in that voltage really affects base current.

    Sooo -- if collector current changes ever so slightly so does emitter current
    --- which changes voltage at emitter --
    ---which changes voltage across emitter-base junction which substantially changes base current,,,

    and that base current change is in direction to push collector current back toward where it was.... which is negative feedback. (negative feeedback opposes change)
    In circuit design that's the purpose of emitter resistor, to set collector current by feedback through base.
    It reduces gain because that's what negative feedback does

    but you have correctly picked up that something is missing from everyday explanations of this.

    It's not that a transistor inherently has constant collector current
    but we can use its current gain to make it do so by adding an emitter resistor and setting voltage across that resistor and base junction...
    realizing that the base junction's share of that voltage will be small ,
    and that's how we bias a transistor when we use it in a circuit..

    in ac amplifiers you'll see emitter resistor bypassed by a capacitor which removes the negative feedback for AC , allowing higher gain.

    Have you studied feeedback yet? It's a fascinating world. Take a course in automatic controls.

    i hope i'm not beating a comatose horse here. If enough explanations are tossed out here one of them will click with your internal thinking processes and ah AHA! moment will result. I hope this one helps.
     
    Last edited: Oct 18, 2011
  11. Oct 18, 2011 #10
    THANK YOU GUYS!!! I just got it. This slight change of current in base is due to early effect right? That slight incline in BJT characteristic ? But in IDEAL BJT, current would be constant through base, until it reaches saturation.

    Transistor will reach saturation when Uce is below certain value. And that certain value is determined by voltage drop across Ropt. When it takes too much voltage drop, transistor gets bottomed(saturated) and THEN base is not the function of beta anymore.

    Did I understand this?
     
  12. Oct 18, 2011 #11
    We just had a lesson about feedbacks. I didn't get the point of it as much. I might as well go study feedback a little more, since I see you guys are talking a lot about it. :)
     
  13. Oct 18, 2011 #12

    jim hardy

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    i guess we were typing at same time...

    i dont know what you mean by 'early effect'

    but sounds like you're getting it!

    we're pullin for ya -

    old jim
     
  14. Oct 18, 2011 #13
    Thank you!! gotta go now. I have another lab. With oscilloscopes!!
     
  15. Oct 18, 2011 #14
    Oh the sick irony! I went to my third lab today, exercise about oscilloscopes and amplifiers. When wouldn't you know it... power went down. We couldn't do that lab, because we didn't have electricity.
    Apparently power company was working on some transformer or something.

    I laughed so hard, because at electrical engineering faculty, we didn't have electricity :D
     
  16. Oct 18, 2011 #15
    :bugeye::rofl:
     
  17. Oct 18, 2011 #16
    No, I just read this!!! Slight change of COLLECTOR current due to early voltage effect, not the base current. I thought you ask whether you need to change the base voltage to keep collector current constant when you vary the collector resistor.
     
  18. Oct 18, 2011 #17
    Yea that was a long shot with the Early effect :D. But yes that was the question:
    "I thought you ask whether you need to change the base voltage to keep collector current constant when you vary the collector resistor."

    And I understood, that in IDEAL transistor you don't have to. But jim told me that it has something to do with feedback, which is why the current in base is changing slightly in real transistor.

    You pointed out this too, just I don't understand fully the concept of feedback.

    I just realised that everything I knew about transistors is slowly drowning. I have a feeling that I will have to invest a lot of energy to fully understand this small yet complicated component.
     
    Last edited: Oct 18, 2011
  19. Oct 18, 2011 #18
    I tried to explain the feedback in post#2 but I don't think I did a good job and don't know how to explain it better. Remember it is the voltage between the base and the emitter that govern the collector current, not just the base voltage. The emitter resistor serve as the feedback to stabilize the transistor so it does not drift that much. Hope someone else can write it better to explain to you.

    BTW, nobody claim this 3 legged animal is easy!!! Take your time, keep asking question here and people will try to answer. One book that I studied this subject is by Malvino. It is the easiest and the best book on transistor I've seen.
     
  20. Oct 18, 2011 #19
    You really made me laugh with this 3 legged animal hahahahah.

    Don't worry. I truly appreciate yours, and any other expert help that I can get here. Because, to be honest, today there are few people that are willing to even try to explain something to you.

    People like You, Jim and countless others, make the students life more easy, where new concepts are not that naive.

    I will definitely check out that book, thank you for recommendation.
     
  21. Oct 18, 2011 #20

    es1

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    Slight change in 'forward gain' due to early voltage. This changes both the collector and the base current technically. If we assume the feedback mechanism hold Ie constant and the biasing network was correctly picked to hold Vb constant.

    Ie is actually the controlled variable here. As the emitter and collector current are very close in a forward active high gain NPN usually we just consider them to be the same.

    This link does a good job of running through the math.
    http://www.stanford.edu/class/ee122/Handouts/EE113_Course_Notes_Rev0.pdf [Broken]
    Start at section 6.

    Google for "common emitter feedback" and you'll get a ton more links which explain it in various ways.

    The above calculations assume an ideal NPN and therefore B is constant but it is pretty easy to see from their math what would happen if B was reduced due to early effect.

    So I think the answer to the question you were really trying to ask, what breaks as I increase the pot's resistance, is B is reduced. Or, the NPN becomes a poorer amplifier even though it is still forward active.
     
    Last edited by a moderator: May 5, 2017
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