calculation HELP!

Discussion in 'Homework Help' started by bonjing, Jul 24, 2009.

  1. bonjing

    Thread Starter Member

    Dec 7, 2008
    12
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    anyone who can haelp me about this?
     
  2. millwood

    Guest

    the question is wrong.
     
    Last edited by a moderator: Jul 25, 2009
  3. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    Is there any other information given? For example, was the Vbe for the transistors used provided as part of the problem statement?

    hgmjr
     
  4. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Can you explain how/why it's wrong? Generally, answers can be right or wrong, but questions are usually good or bad. Although, occasionally a question is so bad that it is reasonable to call it "wrong".

    I think the question is worded oddly and requires a number of assumptions to be made, but it seems to ask a question that can be answered. But, it's hard to comment further on that without your reasoning explained.

    My opinion is that this question needs to be answered using the exponential relationship for a transistor emitter current versus base-emitter voltage. However, the question is vague about the type of transistor so you would have to make some assumptions about transistor type, reverse saturation current and effective thermal voltage nVT. In addition, you need to assume that an output current path is available.

    Ie=Is*(exp(Vbe/nVT)-1) would be a reasonable approximate formula to use.
     
  5. bonjing

    Thread Starter Member

    Dec 7, 2008
    12
    0
    the BJTs are turn ON and have a Vbe of 650mv
     
  6. bonjing

    Thread Starter Member

    Dec 7, 2008
    12
    0
    I try some calculation like this for the current across the driver
    I= E/R
    I= 150mV/2.2
    I= 68.2mA

    how about the current for Output BJT's
     
  7. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Why don't like the one I suggested? Ie=Is*(exp(Vbe/nVt)-1)

    If you invert this, you get the following.

    Vbe=nVt*ln(Ie/Is+1)

    now using Kirchoff's voltage law 0.15V=Ie*(0.47ohms)+nVt*ln(Ie/Is+1)

    This is one equation and one unknown, hence you can get an answer for emitter current Ie.

    However the question asks for voltage on the .47 ohm resistor, which is found using ohms law, or the following.

    V=0.15V-nVt*ln(Ie/Is+1)

    In order to get a numerical answer, you need to make some assumptions about the transistor. Unless otherwise stated, it's usual to assume a silicon BJT at room temperature. Hence, nVt=30mV and Is is in the 1-100 femtoAmp range (use 10 fA if no value is given).

    Note that the answer is transistor-dependent and temperature-dependent.
     
  8. bonjing

    Thread Starter Member

    Dec 7, 2008
    12
    0
    what In stands for?
    what shoulb be the value for this?


    thanks...
     
  9. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    That's LN for natural logarithm. Just take the inverse operation of the exponential function.
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    I think the 2.2 ohm resistors are supposed to be 2.2k ohms.
     
  11. bonjing

    Thread Starter Member

    Dec 7, 2008
    12
    0
    thanks steve

    what if I used the Ohm law for calculating the Voltage across 0.47 ohm resistor, what will be be the formula for this?
     
  12. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Ohms law is just V=IR, so R=.47 Ohms and I=Ie as found in solving the equation 0.15V=Ie*(0.47ohms)+nVt*ln(Ie/Is+1) using a numerical technique such as Newton's method, or search algorithm.
     
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