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Transistors circuit - Basic questions

  1. Aug 30, 2011 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    Given the following transistoric circuit. The transistor acts in the active region. In addition: VBB = 0.6V

    http://img855.imageshack.us/img855/9901/newcircuit.jpg [Broken]


    3. The attempt at a solution

    Before I get there I have two questions:

    1) What does it mean that the transistor acts in the active region?
    2) VBB is not the drawing. Why does it matter to me what it equals to if I can't see it?
     
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  3. Aug 30, 2011 #2

    gneill

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    When a transistor is "acting in the active region", it means that it is operating in the region of its characteristic curves where it is behaving essentially linearly. That is, it is not biased to cutoff (no collector current flowing) or to saturation (where small changes in base current are ineffective at altering the collector-emitter current).

    VBB is the designation for the base voltage with respect to circuit ground.
     
  4. Aug 30, 2011 #3

    Femme_physics

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    Hmm...I'm not sure what you mean by "not biased to cutoff"...in fact that explanation kinda confused me. Can you try to write it differently maybe it would help, please?


    Hmmm.

    I've circled in red where I think VBB is, VEE is and VCC is. Is that correct?

    http://img269.imageshack.us/img269/8585/cccircledinredok.jpg [Broken]
     
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  5. Aug 30, 2011 #4

    gneill

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    Biased to cutoff means that the base voltage drops below the threshold required for base current to flow; the base-emitter junction behaves like a diode, and you need to forward bias a diode in order for current to flow. If the base-emitter voltage (Vbe) drops below this forward bias threshold (about 0.7V for a silicon transistor), then base current will stop and and the collector-emitter current will be shut off, too. The transistor then "looks like" an open circuit for collector-emitter current, and the base can no longer actively control that current.

    Similar things happen at what is called "saturation". So much base current flows that the transistor is "fully on" -- the tap's open as much as it can go! No further increase in base current can cause more collector-emitter current to flow, and that current becomes essentially insensitive to small changes in base current.

    Regarding the VCC, VBB, and VEE designations, have a look here:
    attachment.php?attachmentid=38412&stc=1&d=1314719225.gif
    They designate the DC voltage with respect to ground at the collector, base, and emitter terminals of the transistor. It is perhaps an unfortunate fact that traditionally the supply voltage that provides the supply voltage for the circuit has also come to be called "Vcc"... so one needs to be a bit careful to distinguish between that supply voltage and the VCC that exits at the transistor terminal when a collector load resistor stands in the path between them. Usually the distinction is made clear by statement or context in a given problem.
     

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  6. Aug 31, 2011 #5

    Femme_physics

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    Maybe I have problem with my English here, but "forward bias a diode"? don't understand what it could mean.

    Bias means "prejudice" or "influence opinions"....are we trying to pretend electrical components have feelings?


    We're actually treating it as an ideal transistor, so I think it means that the 0.7V doesn't exist?

    In fact, just to clear things up, is my view of where the 0.7V is correct?

    http://img825.imageshack.us/img825/7752/17888088.jpg [Broken]




    Ah this clears it :smile: thank you. I'll try to work on the exercise soon^^
     
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  7. Aug 31, 2011 #6

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  8. Aug 31, 2011 #7

    gneill

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    That's more or less it :smile: In electronics a bias is the potential difference that exists across some component, usually one designed on purpose by some arrangement of resistors (and sometimes other components) in concert with the circuit's power supply. This is referred to as the "bias network".

    You recall that a diode only conducts when its anode terminal is at a greater potential than its cathode, right? When the anode is positive with respect to the cathode the diode is said to be forward biased and will allow current to pass through. On the other hand, when the anode is negative with respect to the cathode no current will flow and the diode is said to be reverse biased.

    Most ideal transistor models that I've come across keep the (approximately) 0.7V junction drop in them. This is because the base current can be very sensitive to very small changes in the base voltage, and the collector current variations are β times that. You might want to refer to your course materials to see what they are suggesting in this section.
    Yes, that's it for the 0.7V. But keep in mind that it's a voltage drop that's internal to the transistor (it is in fact the voltage drop of a forward biased junction inside the transistor). In practice, when the transistor is operating in the active region this voltage varies a bit from 0.7V depending upon the magnitude of the base current. More sophisticated transistor models will take this into account. An external bias can make this voltage less until the base-emitter is no longer forward biased, shutting off the base current and hence the collector-emitter current.

    You should make a point of indicating the emitter lead in your diagrams by placing the arrow on it per the conventional symbol. Note that the direction of the arrow also indicates whether the transistor "sandwich" is comprised of layers arranged NPN or PNP, which makes a difference in the required biasing arrangements for the part. The arrow indicates the direction that current will flow when the base-emitter is forward biased. Your original diagram indicated an NPN transistor, so current will pass from the base to the emitter when it's forward biased, and collector-emitter current direction is from the collector to the emitter.

    Your original image showed the voltage source Vcc connected above the collector load resistor R3 (at the little circular connection point at the top of the circuit). So you can think of having a 16V battery (Vcc is given as 16V) connected between that point and the ground node. Again, keep in mind there's the possibility of confusing VCC the collector potential (w.r.t. ground) and the power supply Vcc.

    Yes, that's more or less it.
     
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  9. Sep 1, 2011 #8

    Femme_physics

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    Oh, OK. First time I hear that!

    Thanks for the reminder, and the language tutorial :smile:

    I've sent him an email yesterday, and we have a class tomorrow, so I'll find out :approve: but I believe we should ignore it.


    Right. So my drawing is accurate? For instance, if a current flows through the emitter to the collector or vise versa...is there a voltage drop?

    You're right, I forgot to do it this time, although we were told in our course we'd only see NPN.

    Yes I copied it exactly.

    Are you sure? Like this?

    http://img52.imageshack.us/img52/4740/likethist.jpg [Broken]

    But why did they mark a plus and a minus like in the original image if that's not where the voltage source is?





    I'll keep that in mind, though I fail to understand why they'd call it in similar names.
     
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  10. Sep 1, 2011 #9

    gneill

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    The power supply connection (Vcc) looks like this:
    attachment.php?attachmentid=38474&stc=1&d=1314880458.gif
    Note that Vcc is not the same as VCC.

    Ah! That + and - are designating where the output of the circuit is considered to be. I suppose that at some point you'll be asked to find the voltage at the collector (with respect to ground), and perhaps, with some small signal being applied at the base -- you will see that that signal has been both amplified and inverted.

    I doubt that it came about through a committee vote :smile: It's one of those things that are put down to "historical reasons".
     

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  11. Sep 1, 2011 #10

    Femme_physics

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    Do the two ground locations touch each other, too? Like this?

    http://img42.imageshack.us/img42/8331/ummgs.jpg [Broken]

    I find it rather weird how you translated the original circuit to that. Let me apply your logic (or what I think is your logic) to another problem.


    Does this,

    http://img40.imageshack.us/img40/2902/original1n.jpg [Broken]

    Really means that?

    http://img35.imageshack.us/img35/8210/meanthatmeanthat.jpg [Broken]



    Like, if you wanna connect this circuit to some other circuit or device?

    I have the strangest feeling my mind has been both amplified and inverted just now.

    :wink: :wink: :smile:

    Thanks.

    Oh alright. Trust old people and convention to overcomplicate things.

    Anyway, thanks for now, I'm gonna go to class and BBL.^^ You're great help gneill!
     
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  12. Sep 1, 2011 #11

    gneill

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    Yes, exactly. The "ground" node is a common connection point throughout a circuit.
    Yes. Note that my "circuit translation" consisted of making explicit the implied power supply, Vcc.
    Yes, precisely.
    No problem. Always glad to help :smile:
     
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  13. Sep 2, 2011 #12

    Ouabache

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    Hi FP!! :smile:
    Are you familiar with http://en.wikipedia.org/wiki/Voltage_divider" [Broken]?
    (just follow the link)
    With this knowledge you can calculate
    precisely what Vbb is. (Hint: it is not 0.6V)

    I suspect your instructor meant to say Vbe = 0.6V (the turn on
    voltage between the base & emitter junction of your transistor)

    Armed with Vbb, using KVL, you could make an equation
    from that point through base-emiitter junction, through Re to ground potential.
    There would be only one unknown variable in that equation.
    By solving, you would know the value of that variable. :cool:
     
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  14. Sep 2, 2011 #13

    Femme_physics

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    But there's a difference between this:

    http://img42.imageshack.us/img42/8331/ummgs.jpg [Broken]

    And that?

    http://img683.imageshack.us/img683/4518/diff1.jpg [Broken]

    Right?

    Which one is it?

    My instructor said it's best not to use 2 Vcc's. But rather to call the arrows to the ground point you drew here

    http://img840.imageshack.us/img840/900/tocall.jpg [Broken]

    As Vc, Vb and Ve.

    Vcc, Vbb, and Vee are generally the names of voltage sources for the colllector/base/emitter respectively. That's what he told me, anyway.

    I "think" I do. I may have used it in the past not knowing the name.

    You're right. I talked with him about it today and there's a misprint - it's VBE = 0.6V

    Alright then!
    http://img851.imageshack.us/img851/8580/900o.jpg [Broken]
     
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  15. Sep 2, 2011 #14

    gneill

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    They are electrically identical. Any points connected by a conductor (wire) must be at the same potential, and are the same node electrically. The ground connections represent a common conduction path for anything connect to it via the "ground" symbol.
    That works fine. There are various conventions used by different books, authors, schools, etc.. As long as you understand what particular potential or current is being referred to there should be no problem. Choose a naming scheme and stick with it throughout a given problem.
     
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  16. Sep 2, 2011 #15

    gneill

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    The assumed value of Vo = 8V seems a bit odd to me. Usually one wants to determine the operating point of the transistor, and hence the output voltage, from the biasing network conditions rather than the other way around. Certainly if I do so using typical assumptions about the transistor, I don't arrive at 8V for Vo.

    Is there more to this problem than has been shown so far? I haven't seen an actual statement about what specifically is to be determined.
     
  17. Sep 2, 2011 #16

    I like Serena

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    In your second loop you write:
    Sum V = 0; Vcc - R3Ic - REIE = 0

    But you're forgetting the voltage VCE here, which is not zero.

    However, if you work it out properly, there's still something wrong with the numbers in this circuit. They are wrong to make physical sense, as gneill already noted.

    Note that you found a beta = 0.1111.
    Actually you made a mistake (as I already noted), but you should know yourself that beta is an amplification factor, typically greater than 10.

    "Trust your physical cues!" :smile:
     
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  18. Sep 4, 2011 #17

    Femme_physics

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    Thank you. I have a test in hydraulics tuesday. I'll reply to these messages IN EXTENSIVE DETAILS on wednesday. Real thanks. You guys rock. :smile:
     
  19. Sep 5, 2011 #18

    Femme_physics

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    You're right, I can't believe I forgot this simple fact.

    Will do :smile:

    Find Beta
    Find voltage on Re (VRE)
    Find voltage on R2 (VR2)

    Ah, I thought that's the link I might have been missing.


    Oh. 0.1111 can never be a logical result for beta?
     
  20. Sep 5, 2011 #19

    I like Serena

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    Hi Fp! :smile:

    For a regular transistor it's not a logical result.
    I can't speak for all transistors though.

    If you correct your calculations you'll find a reasonable result for beta.
    It's such a nice round value, it has to be right! :smile:
    And you'll also find reasonable results for your other values.
     
  21. Sep 5, 2011 #20

    Femme_physics

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    Duly noted! Though I'm once again finding myself building too many equations and too many unknowns. Any hints?
     
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