Calculate Rth, Eth

Discussion in 'Homework Help' started by anhnha, Jul 29, 2012.

  1. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    Hello,
    I am getting stuck on the determining Rth and Eth in the red section of the circuit.
    I tried to calculate Rth, Eth by Thevenin's Theorem but I cann't get result.I am confused if I need to replace transistor by transistor equivalent model.
    I have tried the following way:
    In the base-emiter loop of figure 1:
    Vcc = Ib.Rb + Vbe+ (β + 1) Re
    and then I have the circuit like firgure 2.
    Can I conclude Rth= (β + 1) Re, Eth= Vbe ?
    Could anyone help me?
     
    Last edited: Jul 29, 2012
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Something is not quite right with your equation.

    Vcc = Ib*RB + Vbe + Ie*RE


    And additional we all know that in BJT

    Ie = Ib + Ic = Ib + β*Ic = (β +1) * Ib

    So we end up with this

    Vcc = Ib*RB + Vbe + (β +1) * Ib *RE

    So if we solve for Ib we get this

    Ib = (Vcc - Vbe)/( RB + (β+1)*RE )

    But this has nothing to do with Thevenin's Theorem. Also I do not think that we can call this Rth and Vth.
    Because only the dynamic resistance seen from base into the BJT is equal Rd = (β+1)*(RE + re)(β+1)*RE
    DC current resistance seen from the base is equal to

    Rin_dc = Vin * ( (β+1)*Re )/ (Vin - Vbe) )
     
    Last edited: Jul 29, 2012
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  3. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    Thank you,
    But how can I get these value, Rd, Rin_dc?
    I am learning VOLTAGE-DIVIDER BIAS and I want to apply Approximate Analysis.
    The condition is βRe > 10R2, therefore I need to calculate resistance seen from base into the BJT.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well we can solve for Rin_dc using this diagram

    [​IMG]

    And Rin_dc = Vin/I1 So

    Ie = ( Vin - Vbe)/Re

    I1 = Ie/(β+1)

    I1 = (Vin - Vbe) / ( (β+1)*Re )

    Rwe_dc = Vin / I1 = Vin / (Vin - Vbe) / ( (β+1)*Re ) = Vin * ( (β+1)*Re )/ (Vin - Ube)

    Now let as assume the Vin = 1.6V ; Vbe = 0.6V ; β = 99; Re = 1KΩ
    So DC current resistance is equal to
    Rin_dc = Vin /I1 = 1.6V/10uA = 160KΩ
    And if we increase Vin to 2.6V the DC resistance will change to
    Rwe_dc = Vin /I1 = 2.6V/20uA = 130KΩ

    So the change in input voltage ΔVin = 2.6V - 1.6V = 1V causes a change in input current ΔI1 = 20μA - 10μA = 10μA
    So the AC current resistance (dynamic resistance) is equal to
    rd =ΔVin/ΔI1 = 1V/10uA = 100KΩ
    And as you can see rd is equal to (β+1)*Re
    rd = (99 + 1)*1K = 100KΩ
     
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  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    All so I want do add that we can replace our BTJ circuit with equivalent circuit, which looks like this

    [​IMG]

    But this circuit is valid only when we want to find I1 current.
     
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  6. WBahn

    Moderator

    Mar 31, 2012
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    You KNOW this equation is wrong. You don't need to know anything about the circuit it is describing, either. Why? Because the units don't work out! You have a voltage equal to the sum of a voltage, a voltage, and a resistance.

    No point going forward until the units work out in what you've got to that point.
     
  7. anhnha

    Thread Starter Active Member

    Apr 19, 2012
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    Sorry for my mistake.
    It must be Vcc = Ib.Rb + Vbe+ Ib.(β + 1) Re
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    No need to apologize for the mistake. I'm merely emphasizing the value of always, always, always checking units.

    Even if you are a committed Units Nazi such as myself, you will occasionally make mistakes that were catchable but that you then fail to catch because you forget about the 'always' part. In fact, in this case your mistake might have been only related to an error in transcribing the equation from a worksheet to the forum post. We tend to not be as careful doing something like that, because we aren't actually trying to work the problem. I've found that I make enough such transcription errors that I try to check units when writing reports and posts as much as I do when working the problem. It has kept me from letting a number of embarrassing mistakes be seen by other eyes (but not all of them, of course).
     
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