Would you check my transistir circuit solution

Discussion in 'Homework Help' started by CrktMan, Jan 3, 2006.

  1. CrktMan

    Thread Starter Active Member

    Nov 29, 2005
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    Please see the circuit diagram as attached.
    Given values are: β = 100, VBE= 0.7V, VCE= 0.3V and VA= infinite

    Find: Ic & Vb

    My solution: By a quick observation of the circuit we can find that transistors is operating in satuartion region not in active region. Because Ic = αIe ~ 1 mA; so, Vc is close to zero volt wich is less than zero.

    One finds -5 + 5*Ic - Vce(satuation) = 0

    or, Ic = (5-0.3)/5 = 0.94 mA
    So, Ib = Ie - Ic = (1 - 0.94) = 0.06 mA

    β(forced) = Ic/Ib = 0.94/0.06 = 15.67, which is smaller than given β = 100

    Vb = 10 - 10*0.06 = 4.4 V

    I am not feeling good with my solution!!!
     
  2. JoeJester

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    Apr 26, 2005
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    Cktman,

    Don't look so deep into the problem.

    Write, on the diagram the GIVENS. If you have a Voltage at a point on the diagram, write it at the point.

    The extra givens are distractors to what your trying to do.

    The good news is you are 100% correct with your instincts.

    I'm not sure how you arrived at:
    The rest of the calculations look pretty good.
     
  3. n9352527

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    Oct 14, 2005
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    I think because Ve is not at ground then the equation that you used to get the Ic is not valid (there's no point in asking for Vb if Ve is at ground because Vbe is given as 0.7V, is there?). If you redo the Ic considering the Ve, then you would get

    5k*Ic + Vce(sat) + Ve = 5

    If you follow this with the equations describing the relationship between Ib and Ic and Ib with Ve, you would end up with three simultaneous linear equations which you could solve for Vb and Ic.

    * By the way, your original Ic equation should have been -5 + 5*Ic + Vce(sat) = 0, notice the sign. However, this is a moot point since the equation is not valid anyway.

    You also made an error in deriving Ib, like Joe pointed out. Remember to include Ve at your Ib equation too.
     
  4. CrktMan

    Thread Starter Active Member

    Nov 29, 2005
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    n9352527,

    Your points look to be very instrumental here. And with my new calculations, it appears that Ve is not indeed grounded. Please correct me further here:

    Vbe = Vb - Ve
    0.7 = 5 - 10*Ib - Ve
    Ve = 4.3 - 10*Ib....... (A)

    Vce(sat) = Vc - Ve
    0.3 = 5 - 5*Ic - Ve
    Ve = 4.7 - 5*Ic ...........(K)

    Solving (A) & (K) we get:
    4.3 - 10*Ib = 4.7 - 5*Ic

    or, 5Ic - 10Ib = 0.4
    or, Ic - 2Ib = 0.08 .......(L)

    We have, Ic + Ib = Ie = 1 mA .... (D)

    Solving (L) & (D) we get: Ib = 0.31 mA and Ic = 0.69 mA
    Thus Vb = 5 - 10Ib = 1.9 V and Ve = Vb - Vbe = 1.9 - 0.3 = 1.2 V

    Looks to be totally different results!!!
     
  5. hgmjr

    Moderator

    Jan 28, 2005
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    Shouldn't the final equation be:

    Solving (L) & (D) we get: Ib = 0.31 mA and Ic = 0.69 mA
    Thus Vb = 5 - 10Ib = 1.9 V and Ve = Vb - Vbe = 1.9 - 0.7 = 1.2 V

    You had the right answer, just a hiccup in substituting the actual value into your equation.

    hgmjr
     
  6. JoeJester

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    Apr 26, 2005
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    Man oh man.

    I took a much simplier view of the problem. I wrote the attachment the first day I read the original post.
     
  7. hgmjr

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    Jan 28, 2005
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    CrktMan,

    After looking at this problem more closely, it would seem that there is a basic flaw in the problem statement.

    Assuming that the 1 milliamp current source that appears in the emitter of the transistor is only intended to indicate that 1 milliamp is flowing out of the emitter and into ground then there would seem to be a problem with the given information.

    Since β = 100, then the transistor's alpha is 100/101 which is 0.99. That would yield an Ic of 0.99 milliamps since Ie is 1 milliamp. The voltage drop across the collector load resistor would be 0.99 times 5K which equals 4.95 volts. That would seem to set up a basic conflict since 5V - 4.95V is 0.05 millivolts. That would leave no voltage for Vce(sat).

    In short, I am not sure if the problem as stated could ever be physically realized.

    At least that is my current thinking. I am open to the interpretations of others. I will follow with great interest the discussion as it continues to develop.

    hgmjr

    P.S. I think I see where I may have gotten off track. Ic is 0.94 milliamps as cktman and joejester have already determined. That means that the additional emitter current must be coming from the fact that the base is being overdriven by a factor of around 47. That still makes the 1 milliamp suspicious. I note that joejester's simulation shows an emitter current of 1.43 which makes sense to me.
     
  8. JoeJester

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    Apr 26, 2005
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    hgmjr,

    That small error [13 uA] could be accounted for in alot of ways. It could be blown off as meter calibration or whatever.

    This wouldn't be the first question that was magical ... I've seen questions that were obviously cut and pasted, with a parameter changed, yet the related parameters didn't. Instructors can have some magical components in their repertoire.
     
  9. CrktMan

    Thread Starter Active Member

    Nov 29, 2005
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    Joe thank you brought that in again. The question is whether it's acceptable to consider Ve as the ground potential when an independent current source is biasing the emitter? Although when we write KVL, it's considered to be so!!!

    Thank you jgmjr for correcting the error!
     
  10. CrktMan

    Thread Starter Active Member

    Nov 29, 2005
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    Further to the question: This question was asked in a very respectable and professional exam. I did not change anything at all.

    Other point, as an statement to the examinee, is given that examinee can make any statement regarding the circuit from the practical stand point.

    It's obvious that we have to work with a forced β, not with β = 100, because transistor is not operating in the active region rather in the saturation.

    CrktMan
     
  11. JoeJester

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    Apr 26, 2005
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    Crtkman,

    I'm sure you reported the question as you received it. I have no doubts.

    I've seen tests in licensing in this country [U.S.] where ... due to a copy and paste, the component becomes a magical component ... defying our knowledge base.

    KCL is one of the cross checks I do ...

    I personally thought you looked too deep into the problem, as I stated in my first post.

    Had the question used a named transistor, I would have used that in the simulation, but it wasn't available. The simulation became interpetive ... where you could apply the same formulae used on the test to confirm the thinking applied.
     
  12. JoeJester

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    Apr 26, 2005
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    To add to my looking too deep into the problem statement.

    When we look too deep, we start applying other formulae that can contradict the question. This can shake the confidence of someone new in electronics. Shaking one's confidence is alot easier than one might suspect.
     
  13. CrktMan

    Thread Starter Active Member

    Nov 29, 2005
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    I agree with you Joe. It does very much so. Even in this case I shoke myself saying that 'damn, I even do not know how to write a simple KVL equation'!

    CrktMan
     
  14. hgmjr

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    Jan 28, 2005
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    Cktman,

    I hope you will let us know what the correct answer to the problem is once the answers to the exam are announced.

    hgmjr
     
  15. CrktMan

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    Nov 29, 2005
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    Unfortunately I would never probably know the answer, because it's one of the previously taken exam question and the authority never reveals any answers.
     
  16. JoeJester

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    hgmjr,

    You said it all with your tagline ....
     
  17. CoulombMagician

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    Jan 10, 2006
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    Interesting that the drawing does not have the assumptions mentioned in the original post.

    Let's look at the constraints and degrees of freedom. Looking at the transistor as a "blackbox" three terminal device there are three terminal currents and three terminal voltages( relative to some prescribed reference point ).

    1. KCL gives us one current constraint, Ie = Ic + Ib
    2. KVL gives us one voltage constraint Vbe + Vec + Vcb = 0.

    3. Transistor model gives us another current constraint Ic = beta*Ib.
    4. Transistor model voltage constraint Vbe = 0.7v
    5. Problem constraint Vce = 0.3v, Vce is usually a dependent variable and this will come back later but for now accept this as a constraint from the problem statement in the original post.
    6.The ideal DC current source adds a current constraint Ie = 1 mA but the voltage across the source is arbitrary.
    7. The Vcc = 5V tied across the current source and the two resistors adds another voltage constraint, namely that V(10k) + Vbe + Vec + V(5k) = 0
    8. The 10k resistor adds a constraint between Ib and VCC - Vbase.
    9. The 5k adds a voltage constraint bewteen Ic and VCC - Vcollector.

    So we have 6(transistor) and 1(current source voltage) = 7 degrees of freedom and 9 constraints, problematic.

    You can see from constraints 4,8,9 that KVL can't be satisfied around the loop of constraint 7 because of constraints 3 and 6. Ib and Ic would have to be independent (violates #3) to satisfy 7 if 5 and 6 were true. Another way to look at it is that VCC - 5k*1mA*beta/(beta+1) ~ 50mV < VCEsat = 300mV.

    I think the problem as stated in the OP is overconstrained. Even if the transistor were operating in the reverse biased quadrant( positive voltage across the current source > VCC) the beta would not be this big. Any ideas?
     
  18. n9352527

    AAC Fanatic!

    Oct 14, 2005
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    Your results seems correct. Except as hgmjr pointed that the last calculation should be:

    Ve = Vb - Vbe = 1.9 - 0.7 = 1.2 V

    I assume that you just wrote a wrong number, because your result of 1.2V is correct.

    Ve is definitely NOT at ground potential. I've never seen an active (non-zero current) current source with zero potential difference across it, and I would be very surprise if I've ever seen one in my lifetime. I also pretty sure that that is a current source and not just an indication of the Ie value as hgmjr assumption (the symbols and notations are different between those two).

    There are two pitfalls in solving this problem, the first and most obvious one is assuming that Ve is at ground potential. The second one is assuming that the relationship of Ic = hfe * Ib is valid.

    1. Let us prove that if we assume Ve is at ground the problem is in fact not solvable:

    Assume Ve = 0

    5 = 5*Ic + Vce(sat)
    5 = 5*Ic + 0.3
    Ic = 0.94mA

    Ib + Ic = 1mA
    Ib = 1 - 0.94
    Ib = 0.06mA

    These results concurs with the attachment that Joe posted.

    Substituting the Ib:
    5 = 10*Ib + Vb
    5 = 10*0.06 + Vb
    Vb = 4.4V

    Now, if Vb is 4.4V and Ve is at ground that means the equation:

    Vb = Ve + Vbe , must be valid.

    4.4 = 0 + 0.7 , which is not!!!!

    So the assumption that Ve is at ground potential is therefore NOT VALID!

    2. The second assumption that Ic = hfe * Ib which would lead to Alpha of 100/101, is also not valid because we know that the transistor is in saturated region and NOT in linear region (as CrktMan already pointed out).

    3. If we consider that Ve is not at ground potential, then:

    5 = 5*Ic + Vce + Ve (1)
    5 = 10*Ib + Vbe + Ve (2)
    Ic + Ib = Ie (3)

    Substituting the values

    5 = 5*Ic + 0.3 + Ve
    5 = 10*Ib + 0.7 + Ve
    Ic + Ib = 1mA

    Solving the simulataneous linear equations, we would get:

    Ic = 0.69mA
    Ib = 0.31mA
    Vb = 1.9V
    Ve = 1.2V

    Which completely satisfy the problem. I don't think there is anything magical/wrong/unsolvable/incomplete in the problem. It is just a matter of avoiding the pitfalls (acknowledging that Ve is not at ground potential!) and working out the solution. Apart from those pitfalls this is a straightforward problem for basic analogue design course.

    Of course, the biggest pitfall of all is looking at the problem objectively. I think this is true for all engineering problems out there, over simplified/complicated view or approaches of a problem would only lead to erroneous solution and I do try, even though personally it is always tempting and difficult, to avoid these two differing views or approaches. I guess what I am trying to say is, just look at the problem objectively, work out an appropriate approach and don't try to over simplify or over complicate the problem. If there is a solution that satisfies the problem and all the available constraints then most probably that solution is the right one. Remember that we are not infallible, all of us.
     
  19. hgmjr

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    Jan 28, 2005
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    It appears that the solution put foward by N9352527 makes the fewest interpretive assumptions about the transistor circuit and the given parameters. That is the symbol for an "ideal" DC current source shown between the emitter and ground making the impedance between the emitter and ground infinite. All of his calculations appear accurate. It is hard for me to argue with his results.

    Unless I have missed something, the solution that assumes that the emitter is tied directly to ground seems to lead to contradicting results with the figure showing 1 milliamp of emitter current and analysis leading to a value somewhere in the vacinity of 1.43 milliamps per joejester's simulation.

    What have I missed, joejester, coulombmagician, cktman?

    hgmjr
     
  20. JoeJester

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    Apr 26, 2005
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    To be honest with you, I missed a couple of really big items.

    I saw that V C-E was 0.3 and V B-E as 0.7 I took those to be meter readings and not design specifications.

    The next biggest mistake I made was the 1 mA current generator between the emitter to ground, with the current flowing into the ground terminal. I didn't see the arrow clearly in the circle and thought it was a current meter.

    Those two mistakes made my conclusions out of this world, a flat out misconception.

    Now it has me wondering if the two resistors are tied to a +5V supply.

    So as I was revising the simulation, it came to me that the transistor would have to overcome the 1 mA current generator before the transistor could begin to conduct.

    The current source internal resistance is set to infinite. Changing it to 1M didn't make much of a difference.

    Anyways ... what am I missing in the simulation?
     
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