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Transistor circut analysis

  1. Aug 27, 2012 #1
    Hi there! I'm having trouble understanding the transistor circuit analysis. Hope you could help me :)
    Capture.PNG

    First I need to find the active region of the transistor. What I saw in the solutions was an assumption that the transistor is in saturation region and then:

    VBB=IbRb+Vbe+IeRe
    Vcc=IcRc+Vce+IeRe

    Why did the solver assume the transistor is in saturation mode, when he is supposed to find the active region?

    Next, I have to calculate the minimal RE needed for the transistor to be active.
    This time, they did refer to it as in active region. But, the equations remained the same. I thought that in Active region, it is supposed to reverse:

    VBB=IbRb+Vbe+IeRe
    Vce-IcRc-Vcc=IeRe

    Why isn't there any change in the equations ? the BC diode is supposed to be in reverse mode, isn't it ?

    Thank you!
     
  2. jcsd
  3. Aug 27, 2012 #2

    NascentOxygen

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    Staff: Mentor

    There is nothing here to indicate an assumption of saturation. Those equations hold for both regions of operation, active and saturated.
    Yes, and that condition is implicit in setting VCE = 0. (Explanation: If VCE = 0 and VBE = 0.6, then it follows that VCE = -0.6)

    [PLAIN]https://www.physicsforums.com/images/icons/icon2.gifAlso, [Broken] may I point out that you should not be using lower-case subscripts here, as they mean something different from the capitals subscripts. For DC or average values, use upper-case subscripts.
     
    Last edited by a moderator: May 6, 2017
  4. Aug 27, 2012 #3

    rude man

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    Homework Helper
    Gold Member

    The first set of 2 equations is correct irrespective of values of the R's or power supply voltages. The second set is garbage.

    A quick check shows that this configuration IS in saturation. Approximately,
    ie = (3V-0.7V)/0.5K = 4.6 mA which when multiplied by Rc = 3K gives Vcc - Vc = 13.8V, exceeding even the Vcc power supply.

    (How did I do that? I just assumed β = ∞, a reasonable approx. with the low value of Rb).

    To get your answer, you need 3 more equations:

    Obviously,

    ie = ic + ib

    and ic = βib.

    Then you need one more. Hint: what is the current ib thru Rb? And if you know Ve do you automatically also know Vb?

    That's 5 independent equations and 5 unknowns: the three currents plus emitter and collector voltages. Set Vc - Ve = 0 and solve for Re.
     
  5. Aug 27, 2012 #4
    Thank you both for your great help!
     
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