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I think I found a mistake

  1. Jul 14, 2012 #1

    Femme_physics

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

    http://img854.imageshack.us/img854/6386/vbbvbbvbbvbb.jpg [Broken]

    Ic = Ie
    Veb = 0.7 volts

    2. Relevant equations

    http://www.allaboutcircuits.com/vol_3/chpt_7/8.html

    3. The attempt at a solution

    A resource I'm relying on claims that

    http://img37.imageshack.us/img37/586/pakoya.jpg [Broken]

    I made the drawing there showing it doesn't make sense, because there are different currents flowing in R1 and R2.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 14, 2012 #2

    I like Serena

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    The current through the base of a transistor is always pretty small.

    Often it is neglected to make the calculations easier.

    I see you also have Ic=Ie.
    These 2 currents are also not exactly the same.
    It should be Ie=Ic+Ib.
    So here the base current is also neglected.

    How big do you think Ib is (approximately)?
     
  4. Jul 14, 2012 #3

    Femme_physics

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    The exercise tells me to assume Ic = Ie , that's why I wrote it :)

    I'm told that the base current is negligible
     
  5. Jul 14, 2012 #4

    I like Serena

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    So if we treat Ib as being zero, then I1 is the same as I2...


    Btw, I believe you have a formula ##I_e = β I_b##, with β the amplification factor.
    For a BJT transistor, β usually has a value of about 100.

    In other words the base current is about 100 times as small as the emitter current.
     
  6. Jul 14, 2012 #5

    CWatters

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    What he said...

    It's common for a circuit designer to choose the current flowing in the base bias circuit R1 and R2 to be at least 10 times the expected base current so that the base current can be ignored (engineers 1/10th rule). Sometimes worth assuming that they have done this, solving the rest of the problem to work out the collector current and then go back and check your assumption was correct.

    In this case if we assume the base current can be ignored then IR1 is about 0.1mA

    So the base is at about 8V and the emitter 8.7V

    So Ie is about 12-8.7/R3 = 1.1mA

    If the transistor is half decent it will have a gain of 100 so the base current about 1.1mA/100 = 0.011mA which is indeed ten times less than 0.1mA.
     
  7. Jul 14, 2012 #6

    Femme_physics

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    Ahh...yes.... since IB is irrelevant the same current flows... true...

    Thanks :) I'll solve it tomorrow and see if I get the same results.
     
  8. Jul 16, 2012 #7

    Femme_physics

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    Last edited by a moderator: May 6, 2017
  9. Jul 16, 2012 #8

    CWatters

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    Looks ok. For completeness you should probably comment on the discharge time through R4. Negligible in this example but might not allways be.
     
  10. Jul 16, 2012 #9

    I like Serena

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    Hey!
    There is a little squirrel in there!!
    That looks alright. :)
     
  11. Jul 17, 2012 #10

    Femme_physics

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    Thanks you two :)

    And yea, everybody loves the Ice Age squirrel!^^
     
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