BJT Transistor Problem 2!!

Discussion in 'Homework Help' started by de1337ed, Mar 16, 2011.

  1. de1337ed

    Thread Starter New Member

    Mar 6, 2011
    23
    0
    Hey guys, Ive been struggling with this transistor problem and I can't seem to figure out the problem. I am suppose to find the state that the transistor is in, and no matter what I do, it seems like I'm getting the wrong answer. I start off by assuming saturation and making the left side into a thevenin equivalent. This causes the Vs = Vth = 2.92V, and the Rb = Rth = 12.078kOhm.
    Also β=100
    KVL(1):
    2.92V - Vb - 0.6V - Ve = 0
    ->2.32 = (12.078kOhm)Ib + (1kOhm)Ie
    ->2.32 = (12.078kOhm)Ib + (1kOhm)(Ib + Ic)
    ->2.32 = (13.078kOhm)Ib + (1kOhm)Ic

    KVL(2):
    15V - Vc - 0.2V - Ve = 0
    ->14.8V = (3kOhm)Ic + (1kOhm)Ie
    ->14.8V = (3kOhm)Ic + (1kOhm)(Ic + Ib)
    ->14.8V = (4kOhm)Ic + (1kOhm)Ib

    Now you do a system of equations with (1) and (2) and you get a negative number for Ib, which isn't possible. And I'm very confused with what the issue is.

    I also tried assuming that it was foreward active, but when I do this, and find the relation between Vce and Vs, I notice that the Vth of 2.92 is large enough for the transistor to be in saturation. So I believe that it should be saturated, but the math isn't working out correctly. Please help me!:mad:

    Thank you.
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    A simple analysis of this circuit, ignoring base current, shows that the transistor is not saturated.

    The base bias potential divider unloaded output voltage of 15V*15kΩ/(15kΩ+62kΩ) = 2.92V. This is NOT sufficient to produce saturation in this circuit.

    The emitter potential would be 2.92V - 0.60 = 2.32V, so that a 1K emitter resistor would give 2.32mA emitter current. The collector current would be β/(1+β) times less, or about 2.30mA

    The true collector current would be a bit lower still, certainly not enough to cause saturation with a 15V supply and 3kΩ collector load.

    Knowing this, go back and find a more exact solution.
     
  3. de1337ed

    Thread Starter New Member

    Mar 6, 2011
    23
    0
    What clues would let you know that a transistor is/isn't saturated? Also, if you look at the diagram, and if you try and calculate Vce by subtracted Vo2 from Vo1, with the Vth given, you get a Vce that is less than 0. I thought that if this occurs, the transistor must be in the saturation region.

    If I'm going about this wrong, then can you tell me how to correctly find the equation for the foreward active region for this transistor? Because another part of the question is to find the Q-point and I wouldn't know how to do that unless i do Vo1-Vo2 to get Vce, then finding the necessary Vs to make the Vce = (15-.2)/2 = 7.4V.
    Thank you.

    edit;;;
    okay, so what I did so far is:
    2.92 - Vb - 0.6V - Ve = 0
    ->2.32 = IbRb + (β+1)IbRe
    ->Ib = 2.05 x 10^-5 A
    ->Ic = βIb = 2.05mA.

    Am I doing this correctly? I feel like this doesn't really prove that it is in foreward active, and not saturation though. Because Ic/Ib will of course be 100... How would I actually prove that it is foreward active? Thank you.
     
    Last edited: Mar 16, 2011
  4. Vahe

    Member

    Mar 3, 2011
    75
    9
    If you assume forward active mode, you will assume that the base-emitter junction is forward bias and take Vbe=0.7V and use the fact that IC=beta*IB. After you have calculated all currents and voltages, you have to show that VBC (the voltage from the base to the collector) is less that 0.4V. This is to make sure that the base-collector junction is reverse biased. In forward active mode BE junction is forward biased and BC junction is reverse biased.

    If the transistor is in saturation, both the BE junction and the BC junction have to be forward bias. It turns out that the forward voltages in full saturation for these junctions are not the same due to construction. So one usually sees that Vbe,sat=0.75V and Vbc,sat=0.55V and therefore Vce,sat=0.2V. If you assume saturation, you can no longer assume that IC=beta*IB. In fact, you should be able to calculate IC, IB, etc without using the beta (since beta is only for forward active mode anyway). After you calculate IC and IB you will see that their ratio IC/IB is less that beta.

    Cheers,
    Vahe
     
  5. Vahe

    Member

    Mar 3, 2011
    75
    9
    I just finished some calculations and I am getting
    Vb=2.68V, Vc=9.10V and Ve=1.98V. So as I said in my earlier post Vbc is definitely less than 0.4 so it is reverse biased and Vbe=0.7V and forward biased; therefore the transistor is in the forward active mode. I am also getting Ib=19.65uA, Ic=1.965mA and Ie=1.98mA. All calculations were done assuming beta=100.

    Cheers,
    Vahe
     
  6. de1337ed

    Thread Starter New Member

    Mar 6, 2011
    23
    0
    I'm not quite understanding how you got those numbers, can you show some work (like KVL's etc)? And I don't understand where you are getting the 0.7 thing, our class assumes that the Vbe cutoff is 0.6V. Thank you. :cool:
     
    Last edited: Mar 16, 2011
  7. de1337ed

    Thread Starter New Member

    Mar 6, 2011
    23
    0
    Oh, and also, can you explain to me how to find the quiescent point in this situation? thank you.
     
  8. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    Here is a quick analysis before going deeper into the the beta calculations.

    To check for linear operation, assume Vbe = 0.65v.

    (15v. x 15K) / (15K + 62K) = 2.92v. = VB
    (2.92v - 0.65v.) = 2.27v. = VE
    (2.27v. / 1K) = 2.27mA. asume to be IC.
    (2.27mA x 3K) = 6.81v. = Vd across 3K collector resistor.
    (2.27v across RE and 6.81v across RC = 9.08v. total)
    So (15v. - 9.08v.) = 5.92v. = VCE.

    Therefor (VE + VCE) = VC = 8.19v positive at the colector terminal.

    And 2.27v. positive at the emitter terminal.

    and 2.92v. positive at the base terminal.

    The collector is more positive than the base, reversed bias, the base is more positive than the emitter. forward biased.

    and VCE is greater than fractional voltage.
     
  9. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The base-emitter voltage of every transistor is a little different. It is affected by base current and by temperature. But in this circuit the emitter resistor value is high enough to make the Vbe differences negligible.
     
  10. Vahe

    Member

    Mar 3, 2011
    75
    9
    If I show the work then there won't be anything for you to do :)
    When the base-emitter junction of a BJT is forward biased, you will see a voltage of 0.6-0.8V across the junction. Not all textbooks use the same numbers, so use whatever you are comfortable with. In the end it does not really matter.

    --Vahe
     
  11. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The datasheet for a transistor shows the base-emitter voltage for a typical one at vatious collector currents and at various temperatures. But even if transistors have the same part number some will have lower and some will have higher voltages than the typical one.
     
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