Multistage amplifier problem

Discussion in 'Homework Help' started by epsilonjon, Mar 5, 2012.

  1. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Question: Calculate the dc voltage at the collector of Q1 if the capacitor C3 becomes shorted.


    [​IMG]

    Attempt:


    Assuming a stiff voltage divider at the base of Q1 I get VE1=10k*10/(10K+47k) - 0.7 = 1.05v. Therefore IE1 = 1.05mA. If I assume that the dc collector and emitter currents for Q1 are roughly the same, then I think we have this situation:

    [​IMG]

    Applying KCL at node VC1 and doing the maths yields VC1=3.77v.

    Could someone tell me if this is a correct method of solving the problem? The solutions manual I found for the book does it in some weird way which doesn't make sense to me, and gets a different answer too :confused:

    Thanks!
    Jon.
     
    Last edited: Mar 5, 2012
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    My answer will look like this

    Vc1 = 10V * (R6||((Hfe+1)*R8)/ ( (R3||R5) + (R6||((Hfe+1)*R8) ) - 1mA * R3||R5||(R6||((Hfe+1)*R8)

    (R6||((Hfe+1)*R8) = 9.375K

    R3||R5 = 4.27K

    Vc1 = 10V * 9.375K/(4.27K + 9.375K) - 1mA * 4.27K||9.375K = 6.870V - 2.933V = 3.937V

    Or if we simplify the circuit


    Vc1 = 10V * R6/(R3 + R6) - 1mA * R6||R3 = 6.8V - 3.2V = 3.6V

    But now Q2 is in saturation so our calculations are incorrect.
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    So we now that for Vc1 3.9V Q2 is in saturation. So to find new Vc1 value we need use Thévenin's theorem.

    [​IMG]


    Vth = 10V * R6/(R6 + R3||R5) - 1mA * R3||R5||R6 = 7V - 2.99V = 4.01V ( I use superposition)

    Rth = R3||R5||R6 = 2.99K

    And now situation look like this

    [​IMG]

    So now we need to solve for Vb.
    But it will be easier to solve for Ve first because Vb = Ve + Vbe
    In saturation the only equations which holds for the BJT is
    Ie = Ic + Ib

    Ie = Ve / Re

    Ib = ( Vth - (Vbe + Ve) /Rth

    Ic = (Vcc - (Vce_sat + Ve) / Rc


    Solve this for Ve I get this result

    [​IMG]

    And if I assume Vbe = 0.65V and Vce_sat = 0.1V

    Ve = 2.08V
    and Vb = 2.08 + Vbe = 2.73V

    So my final answer to you question

    Calculate the dc voltage at the collector of Q1 if the capacitor C3 becomes shorted


    is Vc1 ≈ 2.73V
     
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    Last edited: Mar 5, 2012
  4. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Hi Jony130 and thanks for your answers :)

    I see now where I went wrong in my initial attempt. I took the input resistance at the base of Q2 to be βR8 which is completely wrong (I was getting confused with the AC case). I corrected for this and get the same answer as you in your first post, which takes Q2 into saturation.

    I've just finished Thevenizing the circuit and I get the same as you again :) .

    Thanks for your help!!!

    Interestingly the solutions manual gives an answer of VC1=4.03v with some very strange workings out! :s
     
  5. Jony130

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    Feb 17, 2009
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    can you post the answer (picture) from the book?
     
  6. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Sure thing:

    [​IMG]
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For sure there must be a error in the book.
    This answer is definitely not a solution for your question.
     
  8. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Yeah i'm really not sure what is going on with that one. It's a good job we have kind people like you to help out :D
     
  9. Brownout

    Well-Known Member

    Jan 10, 2012
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    The book answer is for C2 shorted. The orginal question was for C3 shorted. Sure you read the question correctly?
     
  10. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Ah yeah you're quite right. My book definitely says C3 shorted, but the solution does say C2 shorted. However, the answer they give wouldn't make sense for C2 shorted either. And all the resistor values they refer to are the same as in my diagram. I can't work out what they're doing! But nevermind, i'm going onto the next question anyway :p

    Thanks again!
     
  11. Brownout

    Well-Known Member

    Jan 10, 2012
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    You got me there. I see what they are doing. In the first part, they are calculating an equivilant resistance for the transistor r'ce as vce/ic. In the second part, they are using that equililant resistance in a voltage divider equation with (r'ce+r4)||rin(q2) at the bottom leg, and R3 at the top leg.
     
  12. Brownout

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    Jan 10, 2012
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    However, the top leg of the divider should be R3||R5.
     
  13. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
    1
    Can I just ask one more quick related question? :D

    If I have this circuit:

    [​IMG]

    The question is, if R1 opens what happens to Q1? And what will be the Q1 dc collector voltage?

    I am sure that it will go into saturation, but my book (the same book as before) says that it will go into cut-off and VC1 = VEE. (The answer for this question is at the back of the book, but with no working).

    I really think this is incorrect, but could someone just confirm for me? I have only been learning about transistors for a short time so i'm not yet confident, and i'm just reading from books so don't have teachers to help.

    Thanks again!!! :D
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    If we assume Hfe = 150 and Vbe = 0.65V

    Ib = ( 30V - 0.65V) / ( 330K + 151*34K) = 5.4uA

    And Ic = 806uA

    But

    Ic_max = 30V/(22K + 1K + 33K ) = 536uA

    So yes BJT is in saturation
     
  15. epsilonjon

    Thread Starter Member

    Feb 15, 2011
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    Thanks, that's what I got :)

    Stupid book!
     
  16. Brownout

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    Jan 10, 2012
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    It probably gave the answer for R2 open.
     
  17. epsilonjon

    Thread Starter Member

    Feb 15, 2011
    65
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    Yeah must be. This book is so filled with errors it's criminal :mad:
     
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