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(NPN) Transistor

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data
    In circuit (figure), there is negligible potential drop between B and E, where β (current gain) is 100, calculate
    (i) base current (ii) collector current (iii) potential drop across Rc (iv) Vce
    (Ans: 11.25 µA, 1.125 mA, 1.125 V, 7.875 V)
    http://img147.imageshack.us/img147/6084/183h.png [Broken]

    2. Relevant equations

    In this case, [tex]\beta[/tex]=Ic/IB and Kirchhoff's Voltage Law.

    3. The attempt at a solution

    Applying KVL to this circuit,
    Vcc= IBRB + VCE + ICRC
    9= (100,000)IB + VCE + (1000)Ic
    Because IB= Ic/[tex]\beta[/tex]
    Therefore: 9= 1000.Ic + 1000.Ic + VCE
    ---------------------------------------This is where I get stuck..Am I doing this right?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 15, 2009 #2


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    Staff: Mentor

    That doesn't look right in the first equation. You should have two equations -- one for the base circuit, and one for the collector circuit.
    Last edited by a moderator: May 4, 2017
  4. Sep 15, 2009 #3
    Before going any further, can you explain how you got this?

    Vcc= IBRB + VCE + ICRC
  5. Sep 15, 2009 #4
    By applying KVL... (did I do it wrong?)

    You mean I should take two loops of currents? Can you elaborate please?
  6. Sep 15, 2009 #5
    I don't know why this problem assumes that Vbe is negligible, it's never negligible in a real transistor.

    What is IBRB is equal to? (assuming Vbe is ground)
  7. Sep 15, 2009 #6
    Hmm, so what you are trying to say is that if we neglect Vbe, IBRB=0......, that doesn't make sense.
  8. Sep 15, 2009 #7
    What do you think about this?
  9. Sep 15, 2009 #8
    I don't think KVL is the best thing to use in this situation. Since Vbe is essentially 0, you have a path from Vcc to Gnd through Rb. Can you calculate the current through Rb? That current is your Ib. Can you take it from there?
  10. Sep 15, 2009 #9
    What's the voltage difference between Vcc and Vbe? (assume Vbe is zero, as asked by the problem).

    Knowing that, can you find Ib using ohm's law?
  11. Sep 15, 2009 #10
    By applying ohm's law,
    or Ib= Vcc/Rb = 9*10^-5 A
    But the answer given at the back of my book is = 11.25 µA
  12. Sep 15, 2009 #11
    The book is correct......

    What's the value of Vcc, and Rb?
  13. Sep 15, 2009 #12
    Try using a value for Rb of 800k.
  14. Sep 15, 2009 #13
    oh yeah sorry... Ib=V/R = 9/800,000 = 11.25 µA........
    Ic= Ib*beta= 1.125 mA
    Vc=Ic*Rc= 1.125V
    These were easy...thanks.
    Vce=? (What about this one?)
  15. Sep 15, 2009 #14
    If Vcc = 9V
    and IcRc = 1.125V,

    what does Vce have to be?
  16. Sep 15, 2009 #15
    Vcc= -Vce - IcRc ( So we do get to apply KVL after all )
    Vce= Vcc-IcRc
    = 9-1.125
    =7.875 V
    Again, thanks for all the help.
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