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Electronics with Transistor (another problem)

  1. Sep 6, 2011 #1

    Femme_physics

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    Sorry I'm jumping from one problem to another, I just have plenty of time today and I got my scanner :smile:


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

    Find Ic, Ie, Ib

    (circuit drawing and data in the solution)

    3. The attempt at a solution


    http://img263.imageshack.us/img263/3285/findcurrents.jpg [Broken]

    Is that correct for Ie and Ib?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 6, 2011 #2

    ehild

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    It is correct, but use scientific notations or mA units instead of the lot of zeros.
    Note that kΩ multiplied by mA is also Volt. If you plug-in resistances in kΩ units, you get the currents in mA-s.

    ehild
     
  4. Sep 6, 2011 #3

    Femme_physics

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    Yea but does it really matter? I mean, either way it's correct. I'll still get full points for the right score.


    True. I just sometimes worried I'll get confused with the milis and the kilos and whatever, so I stick to the originals :smile:


    Anyway, I solved it and will post the full scan later^^ thanks.
     
  5. Sep 6, 2011 #4

    ehild

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    It is all right. But engineers do not like to write out so many zeros.

    ehild
     
  6. Sep 13, 2011 #5

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  7. Sep 13, 2011 #6

    I like Serena

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    Good mooorning Fp! :smile:

    All your calculations are correct.
    Except the last 2 clauses.

    You seem to have left out part of the problem when you stated that Ic=0...?

    Anyway, if Ic is 0, you obviously have a different circuit, and then you cannot assume that Ib is unchanged.
     
    Last edited: Sep 13, 2011
  8. Sep 13, 2011 #7

    Femme_physics

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    Goo morning! :smile:

    How come? when I typed beta = Ic/Ib

    I defined Ic as 0.
     
  9. Sep 13, 2011 #8

    I like Serena

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    You just calculated Ic to be 1.599 mA.
    And beta was given as 30.
    So what did you do?
     
  10. Sep 13, 2011 #9

    Femme_physics

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    Yea that was in the old calculations, but then they have redefined IC = 0 so I had to adapt?
     
  11. Sep 13, 2011 #10

    I like Serena

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    Btw, since you mentioned I should have a stamp, I've got one now! :wink:

    However, I'm still waiting for a solution to put it on...
     
  12. Sep 13, 2011 #11

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    They can't *just* redefine that!
    But what I suspect, is that the battery on the right should be gone.
    Did you add that yourself?
     
  13. Sep 13, 2011 #12

    Femme_physics

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    I still don't see what's wrong? Oh and there are words that I can't translate in the question. that's why there's a little gap in the question. I'll try though.

    They said "Find Vce at the section point (when Ic = 0)"

    And the last question is to "find Ic at the saturation point when Vce = 0"
     
  14. Sep 13, 2011 #13

    Femme_physics

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    No, that's what the question says...really!
     
  15. Sep 13, 2011 #14

    I like Serena

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    What does the question say exactly?


    As I see it there are only 2 ways for Ic to be 0.

    One if you remove the voltage on the collector, so no current can flow.

    And the other possibility is if you remove the voltage on the base, so the transistor behaves like an open switch.
     
  16. Sep 13, 2011 #15

    Femme_physics

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    I wrote that in 2 posts above u

    I suppose it could just be one of those questions where you can give a literal answer without calculations. You know, there is such thing as easy questions with a simple one-liner answer?
     
  17. Sep 14, 2011 #16

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    Aha!
    A transistor has several modes of operation.

    Apparently the question asks something about the "Saturation" and "Cut-off" points of the transistor.

    Saturated means the base current is so high that the "transistor switch" is fully closed, meaning Vce is 0.

    Cut-off means the base voltage is so low, that the "switch" is fully open, and Ic is 0.

    The
     
  18. Sep 14, 2011 #17

    Femme_physics

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    Aha :smile: so my answers are correct?
     
  19. Sep 14, 2011 #18

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    I think they want you to redo the calculations, for both cases (which will be much easier).

    Saturation is equivalent to removing resistor Rb, and then calculate Ic.

    Cut-off is equivalent to removing the battery on the left, and calculate Vce.
     
  20. Sep 14, 2011 #19

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    Answers? :confused:

    Afaik you only did Ic=0, which is the cutoff.
    You can't calculate beta like you did, because Ib would be zero too.
    But yes, in Cutoff, Vce would be 12 V. :smile:
     
  21. Sep 14, 2011 #20

    Femme_physics

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    so I'm not allowed to use the transistor formula in this case?

    Then what can I?

    [/quote]

    Where'd u come up with that?
     
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