Find the voltage gain

Discussion in 'Homework Help' started by ADCapacitor, Mar 5, 2016.

  1. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    β=210, Vt=26mV, Vbe(on)=0.7V
    How to find Vs,
    then find the Voltage gain Vo/Vs ?
    I try to draw the small signal model to solve the problem


    Vth=15*R2/[R1+R2]
    Vth=1.62V

    Rth=R1//R2
    =5.5kΩ

    -Vth+Rth*Ib+Ie*Re+Vbe=0
    Vth=Rth*Ib*(1+β)*Re
    Ib=(Vth-Vbe)/Rth+Re(1+β)
    =8.2μA

    Since
    Re is short in Ac

    Rπ=Vt/Ib
    =3.18k ohm

    Vo=-β*ib*Rc
    How to find Vs,
    then I can get Vo/Vs ?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Why Vo is at emitter ??
     
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Current amplifier?

    The better question is how resistor has a value of 10 uF. On the other hand that same resistor is behind dc blocking capacitor so it likely does not matter.

    Ok. Question to OP. What text are you using for learning these stuff?
    My textbook has some good examples on these things and explanations for the approximations that can be used to shortcut thought the complicated math.
     
  4. anhnha

    Active Member

    Apr 19, 2012
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    With current amplifier or buffer I think it is better not to include the collector resistor.
     
  5. shteii01

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    Feb 19, 2010
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    You are most likely right. I assume that we are dealing with textbook exercise, therefore the actual exact function of the circuit does not matter.
     
  6. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    Thank you
    I make some mistakes when redrawing the circuit diagram
    The resistor with wrong label is the internal resistor of the signal generator.And the value is not given.
    I think this one is better
    [​IMG] [​IMG]
     
  7. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    I am using the lecture notes for learning these stuff.
    My teacher give us this question for practice
    Which textbook are you using?It sounds good
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    I'm not at all sure that this is supposed to be a current amplifier. Even if it is, the collector resistor could be there to either put a limit on the current output or to reduce the collector-emitter voltage to something that compensates for the Early effect.
     
  9. shteii01

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    Feb 19, 2010
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    My textbook was Electronic Circuit Analysis and Design, Second Edition, Donald A. Neamen.
     
  10. Jony130

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    Feb 17, 2009
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    But are you sure about this diagram and that the output is at emitter ?
     
  11. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    The output should be at collector
     
  12. Jony130

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    Feb 17, 2009
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    So you have a traditional Common Emitter amplifier. And your problem is Vs ?? So first try to draw the "correct" small-signal model.
    An then you will see that if internal resistor of the signal generator Rs = 0Ω Vin = Vs = voltage across rpi.
     
  13. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    [​IMG]
    Is that correct?
    Vs=Vin
    Vin=Ib*R1//R2 ?
    Vo=-βIb*Rc
     
  14. Jony130

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    Are you sure that Ib current is flowing through R1||R2 ?? If so, which current is flowing through rpi then?
     
  15. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    May be this one
    Vs = ib*Rpi ?
     
  16. Jony130

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    Yep, this time you are on the right track.
     
  17. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    thank you
    I have some further questions
    I found that some examples use this "RTH=0.1(1+β)Re" to find the Rth
    And it sometimes use Rth=R1//R2
    But why?
     
  18. Jony130

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    Feb 17, 2009
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    When we want to find the DC operation point (quiescent current) in order to simplify the calculation we use thevenin's theorem and replace Vcc and voltage divider at the base by his thevenin equivalent. And in this case Rth = R1/R2 and Vth = R2/(R1+R2)
    http://forum.allaboutcircuits.com/threads/transistor-base.81886/#post-585781
    Can you show us the example? Because (1+β)Re is a AC resistance seen from the base terminal when we looking into emitter and if we ignore r'e.
     
  19. ADCapacitor

    Thread Starter Member

    Feb 3, 2016
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    Design a bias-stable circuit . Let β = 100 and VBE(on) = 0.7 Vm , Ic=0.7mA, Vce=4V
    [​IMG]
    this example use Rth=0.1(1+β)Re to find Rth
    Is "Bias-stable" imply that we can use Rth=0.1(1+β)Re to find the Rth?
     
  20. Jony130

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    Feb 17, 2009
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    OK I see. This formula "Rth=0.1(1+β)Re" is a rule of thumb used sometimes by a designer. And this rule means that we must choose R1||R2 < 0.1(1+β)Re for "bias-stable". This rule ensure that our voltage divider voltage is "stiff". Because base current is a load for our voltage divider. And to ensure that our divider voltage do not change much after we loading it we use this "rule of thumb" to select R1 and R2. The slightly different rule of thumb is to choose voltage divider current Id = Vcc/(R1+R2) at least 10 times the base current. So we have:
    R1 = (Vcc - Vb)/(11*Ib)
    R2 = Vb/(10*Ib)
     
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