Tough BJT? Is for me.

Discussion in 'Homework Help' started by Sleepcakez, Jul 4, 2010.

  1. Sleepcakez

    Thread Starter New Member

    Jun 26, 2010
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    [​IMG]I have the following circuit.
    No values are given for R1 or R2, and we do not know Vth or Rth.
    I cannot figure out how to find Rth.

    I've seen this equation used but I cannot derive it myself so I don't really know what is going on.
    They say Rth=.1(Beta+1)Re

    Edit: I'm using Thevenin to find values for R1 and R2
     
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  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You find the Thevenin equivalent for the voltage divider combination of R1, R2 and Vcc.

    Vth=R2/(R1+R2)*Vcc

    Rth=R1||R2=R1*R2/(R1+R2)

    Looking into the base you see a series DC voltage VBE≈0.6 to 0.7 V, and an equivalent resistance of (1+β)(re+RE). Resistance re is the dynamic emitter resistance which may be neglected when RE is sufficiently large.

    Given these conditions you can then find the base current IB and so on.

    I_B=\frac{V_{th}-V_{BE}}{R_{th}+(1+\beta)(r_e+R_E)}
     
  3. Sleepcakez

    Thread Starter New Member

    Jun 26, 2010
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    The thing is I'm not given any values for R1 and R2.
    I was told to find values for R1 and R2 to obtain a bias stable circuit with the Q-point in the center of the load line.
    We calculated Icq and Ibq no problem.
    I'll include an excerpt of the solution manual.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    What are you using as the DC bias for beta? There won't be any one correct answer, there will be a range of correct answers.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Then you need to provide whatever other information the question includes. Simply giving the general schematic as shown would not enable the calculation of any values - there is no explicit design criteria.
     
  6. Sleepcakez

    Thread Starter New Member

    Jun 26, 2010
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    My bad Beta is 150.

    I'm not really looking to solve anything. I want to derive the formula they used to find R thevenin. I absolutely cannot figure it out.
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The solution method you provided is what you should follow - once you are given the specific details for the design criteria. Is the problem that you don't understand the method given?
     
  8. Sleepcakez

    Thread Starter New Member

    Jun 26, 2010
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    I don't understand how you calculate R Thevenin without resistor values for R1 and R2. The easy way out would be Rth = R1||R2, but thats not the case.

    Basically I want to know how they derived
    Rth = (.1)(B+1)Re
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The "trick" in the solution is (as you say in your first post) to make a guess at a reasonable value of Rth in relation to the resistance looking into the base. In the case of your solution the suggestion is to make Rth equal to 1/10th of (1+β)RE which is probably a reasonable approach. As Bill Marsden points out there is a range of values for R1 & R2 you could work on to set the required bias condition.

    Another (similar) starting estimate for the total bias resistor series value is to set the current flowing in the divider chain (R1 & R2 in series) to about 10 times the base current.

    Sometimes there may be other constraints such as having a minimum input resistance value for the amplifier stage - which could have some bearing on the slection of the values of R1 & R2.
     
  10. Sleepcakez

    Thread Starter New Member

    Jun 26, 2010
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    So it goes something like this

    10RbIb = IeRe >
    RbIb = .1(IeRe) >
    Rb = .1(IeRe/Ib)>
    Rb = .1((IbRb(Beta+1))/Ib) >
    Rb = .1(Beta+1)Re

    So the normal assumption is that the voltage in the base is 10 times the voltage in the emitter?
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    Have you figured out the midpoint you are after?

    Figure the voltage when the transistor is fully on, figure the voltage when the transistor is fully off, figure out the mid point from that.
     
  12. Sleepcakez

    Thread Starter New Member

    Jun 26, 2010
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    For finding Icq
    10V = 5V + Ic(Rc+Re)
    Icq = 3.57143 mA
    Ibq = 3.57143/150 = .0238mA

    From there it was finding Rth and Vth and then solving down into R1 and R2. I just really didnt understand the equation
    Rth = .1(Beta+1)Re.
    Escpecially when I saw elsewhere
    Rb<= .1(Beta)Re

    I think I'm cleared up on what I need to do so far.

    I appreciate the help.
     
  13. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well, I think that if we talk about dc current we cannot mix it with ac parapets. And certainly re is ac parapets.
    So equation for Ib should look like this:

    I_B=\frac{V_{th}-V_{BE}}{R_{th}+(1+\beta)(R_E)}

    Or if we do the math
    Vcc - I2• R2 - I1• R1=0
    I2=IB+I1
    I1•R1 - Ube - Ie•RE=0


    I_B=\frac{R_1*V_{cc}-V_{BE}*(R_1+R_2)}{R_E*(1+\beta)*(R_1+R_2)*(R_1*R_2)}

    No, current that is flow through voltage divider mus be at least 10 times large then base current.
    Or with slightly different point of view but withe the same result, we make impedance of a voltage divider (impedance looking into the voltage divider) Rth = (R1||R2) small (at least 10 times) compared with the dc impedance looking into the base (β+1)*RE
     
  14. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yeah - I really was asleep on the job there!
     
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