BJT analysis - trouble with voltage division

Discussion in 'Homework Help' started by CircuitZord, Oct 12, 2012.

  1. CircuitZord

    Thread Starter Member

    Oct 8, 2012
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    I'm essentially trying to obtain Vo/Vs for a NPN-PNP amplifier. I have obtained Vo. I'm trying to get an equation for Vs in terms of Vpi1 essentially. I have tried various techniques from applying a test voltage after rpi1 and obtaining the resistance seen from there to nodal analysis, and I'm always left with at least on new introduced variable.

    What should I try?[​IMG]
     
  2. CircuitZord

    Thread Starter Member

    Oct 8, 2012
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    Also, is the voltage across R4//R5 Vpi1 as well?
     
  3. Jony130

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    Feb 17, 2009
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    You want to find a voltage gain for this circuit ?
    http://forum.allaboutcircuits.com/c...stimage.org/x6bppa9mj/IMG_20121008_223325.jpg
    If so the voltage gain of this amplifier is equal Ao = Vout/Vin ≈ 1 + R2/(R1||R6) ≈ 39V/V.
    The full gain expression is not so easy to determine.
    As for the question. Voltage across R4||R5 resistor (at base) is equal.

    Vb = Vs * ((R4||R5)||Rt)/(R3 + Rt||( R4||R5)) where

    Rt - is transistor input resistance

    Rt =R + Rpi + gm*R* Rpi = R + (1 + gm*R)*Rpi1

    As for Vpi1. Simply use nodal analysis and find Vb and Ve voltage.
    Next step is to find Vpi1 = Vb - Ve
     
    Last edited: Oct 12, 2012
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  4. CircuitZord

    Thread Starter Member

    Oct 8, 2012
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    Thanks for your response, I will be looking into that now. Yes, that is the same circuit that I am inquiring about.

    Although I just realised I forgot to include ro1 and ro2, since the Early Effect Voltage was given to us, so that means my model and voltages will change.

    I am curious as to how you determined the open-loop gain so quickly. I don't understand what you did there, also, why is it Vin/Vout, shouldn't it be Vout/Vin?
     
  5. Jony130

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    So you have a lot of work to do

    It is a close loop gain. And we can do it so quickly because I know feedback theory.

    Yes my mistake. Sorry for that.
     
  6. CircuitZord

    Thread Starter Member

    Oct 8, 2012
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    Yea =\\\ - fun times!
     
  7. CircuitZord

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    Oct 8, 2012
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    Ooooh, I just realised what that Rt was! I did it before but I actually had to take a test voltage, call a current call Iin, solve it in terms of Vpi1 and then solve for Vin/Iin, so when I did that AGAIN out of confusion I realised it's the same thing you had. How on Earth do you do that in your head?!
     
  8. CircuitZord

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    Oct 8, 2012
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    Edit: Nvm dumb post.
     
    Last edited: Oct 12, 2012
  9. Jony130

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    We can find open loop gain by inspection if we ignore ro resistance.
    The circuit look like this:

    [​IMG]

    Ic1 ≈ 0.6V/R3 = 2.3μA and Ic2 Ve/R1 - Ic2 = 2.75μA
    And
    re1 = 26mV/Ic1 = 11.3KΩ ; re2 = 26mV/Ic2 = 9.45KΩ

    Everyone know that voltage gain of a CE amplifier is equal to

    Au = Rc/(re + RE)

    If so the voltage gain for Q1 is equal:

    Ku1 = (R3||RtQ2)/( re1 + R1||R2||R6 ) = 183KΩ/22.7KΩ
    8V/V

    RtQ2 = (β + 1) * re2 576KΩ

    And the voltage gain for Q2

    Ku2 = (R2+R1||R6)/re2 ≈ 480KΩ/9.45KΩ ≈ 50.8V/V

    So the overall open loop gain is equal

    Aol = Ku1 * Ku2
    400V/V


    And closed loop gain

    Acl = A/(1 + AB) ≈ 36.5V/V

    B = (R1||R6)/(R2 + R1||R6)
     
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    Last edited: Oct 12, 2012
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  10. Jony130

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  11. CircuitZord

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    Oct 8, 2012
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    That looks like a very useful program, however we have to use a small-signal model I think to solve this because they gave us like the Early Effect voltage, and etc. So it's an ugly small-signal model with ro1 and ro2 which complicates things a bit. I actually put the original circuit on PSPICE and got a voltage gain of exactly what you said, so this is good that it confirms the PSPICE model!!


    I remember in PSPICE I put the Early Effect Voltage in the BJT parameter, so if the voltage gain is coming to something similar does this mean the Early Effect Voltage really does not have much effect?
     
  12. The Electrician

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    What Jony130 showed you IS a small signal analysis.

    The results in the pdf file he attached show that the voltage gain when taking into account the Early effect is 37.1535, and without taking the Early effect into account, the gain is 37.2105
     
  13. CircuitZord

    Thread Starter Member

    Oct 8, 2012
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    Hmm, reading it again I see it now. Sorry, I must admit I didn't read it FULLY earlier as I saw a lot of matrices and just saved the links for later reading.

    I'm a bit confused by the circuit in the pdf file, the circuit looks the same as what I had originally, except the resistances are totally different and the Vcc seems to be 9V instead of 5V.

    I also noticed that the ro1 and ro2 were significantly different to what I obtained through my calculations because the bias currents Ic1 and Ic2 are quite different to what I obtained as well. I did it on PSPICE, and its calculations showed something slightly different yet again.

    I got Ic1 as 2.626...uA and Ic2 as 2.063...uA but in the pdf it's 2.222uA and 2.656uA respectively, hence altering the ro's noticeably. Probably doesn't make a huge difference to the final gain, but still curious to know.
     
  14. The Electrician

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    I think Jony130 was just giving you an example of how to carry out the calculations, not necessarily using your exact values.

    He'll probably be able to answer your questions.
     
  15. Jony130

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    Don't bother about Vcc and different resistor values in the diagram.
    This diagram helps me create the correct matrix. And later I use resistor values from your diagram in calculations.

    Can you show me how do you obtain those current values?
     
  16. CircuitZord

    Thread Starter Member

    Oct 8, 2012
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    OK, this will involve a few pictures as I break down what I did exactly:

    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
     
  17. Jony130

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    OK I see the "error". And again we have another example of a purely theoretical circuit not related to reality. Design by theorist.
    The first big mistake is to assume that Vbe will be equal Vbe = 0.7V for such a small current.
    In reality Vbe will be around 0.5V or smaller. And then it turns out that Ic1=1.85uA and Ic2 = 3.3uA and Vo = 4VDC LOL. Of course, I don't blame you but your teacher.
    And of course your solution for Vbe = 0.7V is correct.
     
  18. CircuitZord

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    Oct 8, 2012
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    If Vbe was 0.5V, wouldn't that mean these transistors are in cut-off mode?

    When I first saw this circuit I was a bit confused by the really large attenuation resistances compared to the rest of the resistors, is this normal for a basic amplifier?
     
  19. Jony130

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    For such a low collector current, for sure the BJT will work in active region.
    In reality there's no such thing as cut-off voltage for Vbe voltage.
    Normally when BJT circuit work when base current is larger then Ib > 5μA and Ic > 1mA. And then you we can assume that if Vbe<0.5V the will be in cut-off. But not in this case when we have a very low collector current.

    No, it's not normal

    I recalculated the voltage gain with ro include in calculation and this is what I get
    Voltage gain = 36.792
     
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