BJT resistor values

Discussion in 'Homework Help' started by darylsingh, Jul 10, 2011.

  1. darylsingh

    Thread Starter New Member

    Jul 10, 2011
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    How do I find the values of R1 and R2 of this schematic? Along with VBE and Beta, I'm also given an ICQ value. All this is very new to me, is there any specific formulas for calculating this? Thanks in advance.
     
  2. darylsingh

    Thread Starter New Member

    Jul 10, 2011
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    Can someone please help me with this problem, I am really stuck and I can't find any help through my textbooks or online.
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The only think you need to do is to find the base voltage for a given ICQ (with a little help from Ohm's law) and then again using Ohms law design the voltage divider that provides that voltage to the base.
    And the current that will flow through R1 and R2 should be greater than 10*Ib.
     
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
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    If you put as much effort into solving the problem as you apparently have into finding four different ways to ask the question in separate posts on this forum, you would have solved it already!!!
     
  5. darylsingh

    Thread Starter New Member

    Jul 10, 2011
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    Sorry Adjuster, I'm new to the site and wasn't sure where exactly to post my question at first, that's why it appears in two forums. This transistor analysis stuff is very new to me and the textbook that I'm using is only confusing me more than I already am. I have been working on this problem for the past 7 hours but it hasn't made any sense to me. If possible, can you kind of outline for me in step by step bases on what I should do. Thanks for your time
     
  6. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    What textbook are you using?
     
  7. blah2222

    Well-Known Member

    May 3, 2010
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    VBE = 0.7 V
    B = 100

    If Ic(Q) is known, then you can easily figure out what VB is.

    VB = Ie*B + VBE ***Ie = \frac{(B+1)*Ic}{B} ~= Ic (B >> 1)

    Therefore, VB ~= Ic*200 + VBE.

    Now here comes the actual engineering in this design. Resistors R1 and R2 are used only for DC biasing purposes, and set the value of Vb. With this in mind, they can be large values and draw little power from the supply. Their relative ratios are the most important part.

    From here you can use KCL and try out values that you think would make sense:

    I2 = Ib + I1

    \frac{VCC - VB}{R2} = Ib + \frac{VB}{R1}

    \frac{VCC}{R2} - VB*(\frac{1}{R2} + \frac{1}{R1}) = Ib

    ***Ib = \frac{Ic}{B}

    ***VB ~= Ic*200 + VBE

    ***VCC = 20

    \frac{20}{R2} - (Ic*200 + VBE)*(\frac{1}{R2} + \frac{1}{R1}) = \frac{Ic}{B}

    If you want a simplified formula with all given values substituted:

    R1 = \frac{100*R2*(200*Ic + 0.7)}{1930 - Ic*(R2 + 200)}

    You are still going to have to chose a value for one of them to get the other.
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    But we can use ohms law directly to fond R1 and R2.
    The voltage divider current must be greater than the base current.
    In practice 5 to 30 times larger then Ib (typical 10 times larger).
    So:
    Vb = Icq*Re +Vbe

    R1=Vb/(10*Ib)


    R2=(Vcc-Vb)/(11*Ib)
     
  9. darylsingh

    Thread Starter New Member

    Jul 10, 2011
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    Thanks everyone for the help. It makes more sense now.
     
  10. stinky111

    Member

    Apr 7, 2011
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    darylsingh, I have just compleated this paper, (ANALOGUE) of which BJTs was a big part of.
    The text book I found most useful (Other than the e-books on this site) was "Electronics For Electrical Trades" 4th ed, James F Lowe.
     
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