[S.T. / ] - BJT's bias

Discussion in 'Homework Help' started by PsySc0rpi0n, Apr 15, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    How can I simulate the Quiescent Point for a BJT, let's say 2N2222 in LTSpice?

    I have this attached file but I can't remember how to do it so that I can verify my calcs.

    Any help?
     
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  2. crutschow

    Expert

    Mar 14, 2008
    12,988
    3,226
    In the simulation command window, instead of doing a Transient simulation do a DC op pnt simulation.

    Note that the bias point for that circuit is strongly dependent on the hFE gain of the transistor which typically can vary over a range of 3 to 1 or more between different transistors of the same type.
    You should use a bias configuration that uses some negative feedback so the bias point only slightly varies with hFE.
    One way to add some negative feedback is to connect Rb to the collector instead of to Vce.
     
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,184
    3
    Oh sorry, I thought I uploaded the correct circuit.

    And for now, It's not an option to change the circuit. The purpose is actually to verify the stability of this setup.
    The correct circuit is the one attached now!
     
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  4. crutschow

    Expert

    Mar 14, 2008
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    So, do you still have a question?
     
  5. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,184
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    Hi...

    Not about that one...
    However I have other questions.

    I'm now testing a Colector-Base Bias and our teacher is asking us to calculate Rb so that Quiescent Point lands in half the Load Line!

    Do you understand what I mean?

    Rc=2k2Ω
    Vcc=15V
    Vbe=0.7V
    β=200

    I have to calculate the Load Line using the equation Vcc = Rc(Ib+Ic) + Vce <=> Vcc = Rc(Ic/hFE + Ic) + Vce.
    Then we find Vce for Ic = 0 and after this we find Ic for Vce = 0.

    I did this math and got the followig results for Load Line:
    Ic = 0 => Vce = 15V
    Vce = 0 => Ic = 6.784mA

    Then, half of the Load Line is PFR(7.5V, 3.792mA)

    With this values I can find Rb using the Entry net equation which I consider it to be Vcc = Rc(Ic + Ib) + RbIb + Vbe
    Replacing Ib by Ic/β I find Rb = 400.96kΩ

    If my calcs are correct, where can I find this point (PFR (7.5V;3.792mA)) in the attached plot???
     
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  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    But if Ic_max = 6.784mA than Icq = Ic_max/2 = 3.392mA
    Simply plot a load line for Vcc = 15V and Rc= 2.2kΩ
    (15V-V(col))/2200ohm
     
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  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I couldn't get that result... :( Never mind. I did it by hand!
     
  8. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,184
    3
    Can you show me your asc file?
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    1,097
    Here you have. And do not forget to set both, right and left vertical axis at the same scale.
     
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