BJT transistor analysis

Discussion in 'Homework Help' started by PsySc0rpi0n, Apr 30, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hi everyone...

    I need to calculate the Ic for the attached circuit, using Thevenin method!
    β=167.1
    Vcc = 20V
    R1 = 160kΩ
    R2 = 40kΩ
    Re = 1kΩ

    I'm calculating Vth = (40/(40+160))*20 = 4V

    Then Rth = (160*40)/(160+40)=3.2kΩ

    Then using input net (base net) I'm calculating Ib as:

    Ib = (Vth - Vbe)/(Rth + Re(β+1)) = (4-0.7)/(3.5k+1k*(167.1+1))=19.26μA

    Then using Ic = β*Ib = 167.1*19.26μA = 3.22mA but the correct result is 2.75mA... Can't figure out where am I going wrong!
     
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  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Morning psy,
    What type of BJT circuit would you say that you have posted.?

    E
     
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,184
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    Voltage divider biased transistor... Is this what you're asking


    edited;
    Ok, I got it.

    I was wrong calculating Rth. It's not 3.2kΩ, but 32kΩ.
     
    Last edited: Apr 30, 2015
  4. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I have another problem... This time is with LTSpice.

    I'm trying to simulate the attached circuit but it is taking for ever to finish the simulation.. Why?
     
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  5. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    No ground reference.!
     
  6. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    OMG... But LTSpice usualy complains about it. But didn't do it this time! Bahhh... Thanks!
     
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, if I want to write the 2 nets (input and output), can I write then like this:

    Net 1 - Input net

    Vcc = R1(Ic+Ib) + Vbe + Vγ

    Net 2 - Output net

    Vcc = R1(Ic + Ib) + Vce + Vγ

    Are these 2 equations correct?
     
  8. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Is this post #4 circuit.? and what is vy.?
     
  9. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Yes it's about post #4. Vγ is the LED voltage. I think it's called forward voltage, right? Let's consider the following values:

    R1 = 2.1kΩ
    Vcc = 12V
    hFE = 119

    We are asked to find current at R1.

    So, I'm doing:

    Ib = (12-0.7-2.5)/(2.1k*(1+119)) = 34.92μA
    Vce = 12-2.5-2.1k*(34.92μa+34.92μA*119)=0.721V
    Ic = 119*34.92μA = 4.155mA
     
  10. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Looks close enough to me.
    Have you checked it using LTSpice.?
     
  11. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Not yet checked with LTSpice...
    The problem is that the answer should be 4mA.
    This is a question that our tutor posts in our school's site (quizz) and that we must answer to it within a given deadline. A friend of me helped me and we have calculated a value of 4mA sharp and the quizz "told me" the 4mA values was correct.

    But current across R1 is Ib + Ic. So it should be greater than 4.155mA, but the quizz said that 4mA was correct!
     
  12. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    LTSpice shows
    Ib=37uA
    Ic=4.39mA
    Vbe = 0.694V
    Ve=2.09V
    Vce=0.694V

    Please post your calculation showing the Ic as 4.0mA
     
  13. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I can't find them in the calculator... probably the screen "run out" of memory to store the calculations history and deleted the oldest calcs I made. And I can't remember the values of R1, Vγ and hFE.

    Every time we attempt to answer the quizz, these values changes!

    Anyway, the most important was to know that the net equations I wrote were correct! The rest is math that the calculator does!
     
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