Transistor Problem

Discussion in 'General Electronics Chat' started by ahmedzica, Jan 10, 2012.

  1. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    I've a circuit like that

    [​IMG]

    I'm trying to determine R1,R2,Re and Rc
    When Ic= 10mA and, Vce=8v and Bdc=75

    I've done my best and I got 3 equations

    Here is them

    14= 10Rc+10Re
    22-Vb= 10 Rc + 7.3
    Vb= 10 Re + 0.7

    and i cannot go on :(

    Can anybody help,please?
     
  2. colinb

    Active Member

    Jun 15, 2011
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    Don't forget your units in the equations. Otherwise you'll get wrong answers. If you just insert a lone “10”, it is not the same as “10 mA”.
     
  3. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    I see but I'm just wrinting my equations . I've 3 equations with 4 unknws so I'm tring to get the fourth one
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    What about Vcc ??
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Try this this is the rule of thumb.

    R1 = (Vcc - Vb) / (11 * Ib)

    R2 = Vb / (10 * Ib)


    For good thermal stability we choose Ve to be grater then Vbe.
    So for good thermal stability we usually choose Re=(0.1 ... 0.4)*Vcc/Ic or Ve large then 1V (Ve>>Vbe)

    Rc = (Vcc - Vce - Ve)/Ic


     
  6. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    Vcc = 22v volt
     
  7. russ_hensel

    Well-Known Member

    Jan 11, 2009
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    One equation ( at least ) should include some parameter of the transistor. Think about it.
     
  8. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    I already did :(
     
  9. Jony130

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    Feb 17, 2009
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    So for Vcc = 22V and beta = 72 ; Ic= 10mA; Vce=8V

    Rc + Re = ( 22V - 8V) / 10mA = 1.4KΩ

    I choose Rc = 1.2 KΩ and Re = 200Ω

    Ve = Re * Ie = 200Ω * 10.138mA ≈ 200Ω * 10mA = 2V

    Vb = Ve + Vbe = 2.65V

    Ib = Ic/beta = 10mA/72 ≈ 139μA

    R1 = (22V - 2.65V) / (11 * 139μA) ≈ 13KΩ

    R2 = 2.65V / ( 10 * 139
    μA) = 2K
     
  10. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    How come Rc=1.2 KΩ and Re= 200 KΩ!! isn't anyway but this i.e away from guessing ?
     
  11. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    200Ω*** Sorry
     
  12. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    From this

    Vcc = Ic * Rc + Vce + Ie*Re

    And if we assume Ic = Ie

    Rc+Re = ( Vcc - Vce)/ Ic
    = ( 22V - 8V)/10mA = 1.4KΩ

    So I choose Rc = 1.2K and Re = 1.4K - 1.2K = 200Ω
     
  13. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    I know I'm asking about the latest stop why choosing Rc=1.2 based on what! and should RE=>10R2 in order for this Voltage divider to be stiff
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Usually we choose Rc > Re for for several reasons.
    The first one and perhaps most important reason is that Re reduces the amplifier output voltage swing. And the voltage gain is equal to Rc/Re.

    I use some other condition to achieve the desired voltage divider stiffness.
    I choose voltage divider current much larger than the base current (10 * Ib typical). This is almost the same condition as yours.
     
  15. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    oh ok thanks :)
     
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