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BJT voltage divider bias beta variation

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Given, β=100, Is = 6E-16
    See attachment.
    a) Find min value of Rb such that the BJT is in active region
    b) For the Rb found, what is the base-collector voltage if β=200


    2. Relevant equations



    3. The attempt at a solution
    Assuming, VCE min for active region operation as 0.75V, I got Ic = 0.87mA
    since β=100, Ib = 0.0087mA
    Vbe = Vt ln (Ic/Is) = 0.728V

    using Ib = (Vth-Vbe)/Rth {thevenin equivalent}, I got Rbmin as 7K. The voltage divider voltage is 0.75v and Rth is 2.1k.
    But now, when I plug in Ib = Vth-Vbe/Rth, I get 0.00104mA. Which Ib is correct?

    I can't figure out how to calculate the beta change. Where do I use the beta?

    Ib is 0.0087mA, with new beta Ic will be 1.74mA. which will make IcRc = 3.48v, which doesn't make sense.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

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  3. Jan 31, 2012 #2

    rude man

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    Rb has to furnish the base current plus the current thru the 3K. So how did you arrive at that strange equation for Rb? You know Vcc and you calculated ib and Vbe correctly, so what is the right equation for Rb?
     
  4. Feb 1, 2012 #3
    I can't figure out the equation for Rb.

    I assumed the Vth as 0.75v (voltage divider output)

    So, if Rb=7k, I get Vth as 0.75v

    Equation for Rb = (Vcc-Vth)/Ix
    Ix=0.25mA (this is the current flowing thru the voltage divider).

    To supply the desired Ib, Rth should be 2.1k
    Rth = Rb(3K)/(Rb+3k)
     
  5. Feb 1, 2012 #4

    rude man

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    I don't know what Vth is. Stop worrying about Thevenin equivalents. That theorem is not appropriate here.

    Instead, write an equation summing currens to zero at the base.
     
  6. Feb 1, 2012 #5
    Whenever I see voltage divider bias, I immediately think thevenin.

    Anyway,

    Vcc-Vbe/Rb = Vbe/3k + Ib

    I know, Ib = 0.0087mA and Vbe = 0.728v

    I get, Rb = 7.05K

    How do I use the new β?
     
  7. Feb 1, 2012 #6

    rude man

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    Almost.
    (Vcc - Vbe)/Rb = Vbe/3K + Ib. Which I think is what you meant anyway. So, so far, so good.

    Rb = 7.05K is correct.

    With the new beta of 200, what do you compute Ic to be? Vc? Realizing that Ic can't exceed 2.5V/2K and Vc can't be < 0V, what do you conclude about Ic and Vc?
     
  8. Feb 2, 2012 #7
    Max Vce is Vce sat, which is about 200mV generally.

    This will result in an Icmax of Ic = (2.5-0.2)/2K = 1.15mA

    IS that it , you just assume Vce?
     
  9. Feb 2, 2012 #8

    rude man

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    I would give you an A at this point.

    Saturation voltage is a tricky business, it depends a whole lot on the particular type of transistor. At saturation, the base-collector junction is also forward-biased. Generally, Vc_sat is lower than 200 mV. More like 50-100 mV. But you see the picture vey well now.
     
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