Direct coupling circuit.. my attempt... I get too many equations...

Discussion in 'Homework Help' started by Alice24, Apr 14, 2012.

  1. Alice24

    Thread Starter New Member

    Apr 22, 2011
    15
    0
    So far, 3 equations 4 unknowns.

    Given:

    VBE1 = VBE2 = 0.7 V

    Beta 2 = 80
    Beta 1 = 50

    (I wrote the transistor formulas in the scanned page in the box)

    [​IMG]


    Any equation I add seem to add more unknowns and doesn't get me close... or I not trying hard enough?

    Can anyone please give either show me the way or at least give me a clue, please?
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    I would begin by assuming that both transistors are fully saturated.
    Hence you can calculate Q2 collector and emitter current, IC2.
    This will give you the voltage across R2.
    From this you should be able to determine IB1.
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
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    Try this approach

    5V = Ib1 * R5 + Vbe1 + Vbe2 + Ie2 * R2 (1)

    Ie2 = Ib2 * (β2 + 1)

    Ib2 = Ib1 * (β1 + 1)

    So finally

    Ie2 = (Ib1 * (β1 + 1) ) * (β2 + 1)
    (2)

    And this is all we need to solve this circuit.
     
  4. Alice24

    Thread Starter New Member

    Apr 22, 2011
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    0
    Why? In this case Vce1 and Vce2 = 0 , although we do get to make the problem so much easier to ourselves, who says we're allowed to do it?

    Ahh...I was unfamiliar with this formula!

    Are there more formulas specific to directing coupling connection I might have missed?
     
  5. MrChips

    Moderator

    Oct 2, 2009
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    What I said in post #2 would not work.
    I did some calculations and assuming both transistors are turned on would not provide enough Vbe turn on voltage.

    So here is another approach that will take perhaps two iterations.

    Assuming Ib1 is small, and Vbe = 0.6V
    we can estimate VR2 = 5 - 1.2 = 3.8V
    Hence IR2 = 19mA
    If we assume β = 100 for both transistors
    we can determine Ib1

    Use this and recalculate VR1 and VR2.
     
    Last edited: Apr 14, 2012
  6. Alice24

    Thread Starter New Member

    Apr 22, 2011
    15
    0
    But MrChips, you make a lot of assumptions. Thing is, we don't have to assume, we know VBE and Betas :) It's given to us.

    As a student (me), I don't think I am allowed to take such privileges with assumptions...

    I'm more curious about how to use the formulas I was provided!

    I am trying to construct all possible formulas to this transistor...

    [​IMG]

    As well as:

    Ie1 = Ib1 + ic1
    Ie2 = ib2 + ic2
    beta1 = ic1/ib1
    beta2 = ic2/ib2
    ie1 = ib1 (1 + beta 1)


    What am I missing?
     
  7. MrChips

    Moderator

    Oct 2, 2009
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    Sorry, I did not see that Vbe and beta were given at the top of the post.
    Then use those values.

    I think you have enough information now to solve for Ib1.
     
    Last edited: Apr 14, 2012
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Your "special" equation in post 7 are mostly incorrect.
    For your circuit

    Ib1 = ( Vb - 2Vbe)/ (R5 + R2 *( (β1 + 1) * (β2 + 1) ) )


    Ie1 = IB1 * (β1 + 1)

    Ib2 = Ie1

    Ie2 =
    Ib2 * (β1 + 1)

    Ic1 =
    Ib1 * β1

    Ic2 = Ib2 * β1
     
  9. MrChips

    Moderator

    Oct 2, 2009
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    Typos on the two lines:

    Ie2 = Ib2 * (β2 + 1)

    Ic2 = Ib2 * β2
     
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