BJT Matching?

Discussion in 'Homework Help' started by eng kryptonite, Feb 7, 2010.

  1. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    What degree of matching in emitter area is required for the base-emitter voltages of two bipolar devices with identical collector currents to have a difference of less than 4.6 mV if we use VT = 25.85 mV? How to approach this? Its quite confusing for me. Thanks! : )
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    What is the related theory behind this question?
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Here is how I would go about it.
    For a given current, the current density is inversely proportional to emitter area. Think of the diode equation and current ratio.
     
  4. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    If all I have are the VT & the voltage .. how would I go with the diode equation and current ratio? Can you please show some figures so I can follow up or explain a little further? I still don't understand the question :confused:
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Start with diode equations for the two transistor base-emitter junctions:
     I_1=I_s*(e^{\frac{V_{be1}}{V_T}}-1)
     I_2=I_s*(e^{\frac{V_{be2}}{V_T}}-1)
    (Keep in mind what I said previously about emitter area and current density.)
    For reasonable current levels, you can ignore the -1 term in each equation.
    Now, take the natural log of the ratio \small \frac{I_1}{I_2}. (I'm not going to do all the work for you).:)
    You should come up with an equation that contains \small Vbe1-Vbe2. Do you recognize what that difference represents? Plug it and  \small V_T into your equation, then take the antilog. What answer do you get?
     
  6. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    Thanks Ron .. The idea became much better now and I have came up with a solution thats attached. Nothing brilliant :D but now this thing makes sense to me. Sounds good to you?

    what I got is 1.78%.
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    My version of Word won't open your file.
    Your answer is ln(I1/I2). You need I1/I2.

    EDIT: ln(I1/I2)=0.178. Even if this were the correct answer, it is not equivalent to 1.78%.:confused:
     
    Last edited: Feb 9, 2010
  8. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    I know its not 1.78% .. What I got is 0.178e-1 * 100 which gave me that percentage. I converted the doc to word 2003 and it should be okay now. :D
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    \frac{I_{s1}}{I_{s2}}=e^{\frac{4.6}{25.85}}=e^{0.17795}

    Are you aware that e = 2.71828..., which is the base for Naperian logarithms?

     \frac {I_{s1}}{I_{s2}}=1.1948

    Since  I_s is proportional to emitter area, then the areas must match within ≈20%.
     
  10. eng kryptonite

    Thread Starter New Member

    Jan 25, 2010
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    Ooops :D
    I missed out the "e" part :)
    for some reason under all this stress I calculated that as (10^x) ..
    Thanks a lot Ron .. your help is appreciated once again.
     
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