BJT amplifier

Discussion in 'Homework Help' started by full, Oct 18, 2014.

  1. full

    Thread Starter Member

    May 3, 2014
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    hello everyone,

    I do this problem in basic common emitter Bipolar junction transistor (BJT) amplifier ........

    me solution true? or there are some wrong?

    thanks :)
     
    Last edited: Oct 18, 2014
  2. full

    Thread Starter Member

    May 3, 2014
    225
    2
    why some time in Bipolar junction transistor the Ie=(1+B)Ib & Ie=BIb ?

    thanks,
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well you made a mistake R1 is 250k not 225K. Because Vth = 1.1538V not 1.25V.
    Ib = 3.48329μA; Ic = 417.995μA; Ie = 421.478μA.

    As for the gain

    Av = Rc/(re + RE) * (Rib||Rth)/(Rs + Rib||Rth) = 8.46 * 0.985 =8.3V/V
    re = 26mV/Ic
    Rib = ( β+1 )*(re + RE) = 80kΩ
    Rth = R1||R2 = 57.7kΩ

    And for BJT we have Ie = Ib + Ic but as you know Ic = Ib*β. So we have Ie = Ib + β*Ib = (β +1)*Ib.
     
  4. full

    Thread Starter Member

    May 3, 2014
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    thanks sir,
    when current collector equal current emitter (Ic=Ie)?
     
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    I guess when transistor is in Cutoff region.
     
  6. full

    Thread Starter Member

    May 3, 2014
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    thanks sir,
    is mean there isn't resistor and source in emitter region?
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    In Cutoff Ic=0, Ib=0, therefore Ie=Ic+Ib=0. Therefore Ic=Ie=0.
     
  8. full

    Thread Starter Member

    May 3, 2014
    225
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    obs:eek:

    see this ie=ic why?
    thanks sir,
     
    Last edited: Oct 18, 2014
  9. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Ib can be VERY small. So. It becomes a question of precision. How precise do you want to be?
     
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  10. Jony130

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    Feb 17, 2009
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    The emitter current is always equal to Ib + Ic. But because the beta value is large we can ignore the base current and say that Ie = Ic.
    For example if β = 100 and Ib = 10μA we have Ic = 100 * 10μA = 1mA and Ie = Ib + Ic = 1.01mA
    So we don't do the big mistake/error if we ignore the Ib current and say that Ie = Ic = 1mA
     
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  11. full

    Thread Starter Member

    May 3, 2014
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    I do another problem ,but this is difficult :confused:
     
  12. Jony130

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    Feb 17, 2009
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    Your DC current calculation are totally wrong.
    Vth is ok and Rth also. But your KVL loop (Vth - VRB - Vbe - VRe = 0) is wrong.
    You forget about Vee in your loop.
     
  13. full

    Thread Starter Member

    May 3, 2014
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    if I do KVL loop (Vth - VRB - Vbe - VRe +VEE= 0) true?
    thanks sir
     
  14. Jony130

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    No, wrong again.
     
  15. full

    Thread Starter Member

    May 3, 2014
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    ok , If I do KVL loop (-Vth - VRB - Vbe - VRe +VEE= 0)
    but the loop flowing from -Vth+ is mean positive in voltage true?

    about Av= (B Rc)/(r-bi +(1+B)Re) *Ri true?

    thanks sir
     
    Last edited: Oct 19, 2014
  16. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In PNP transistor current flow from +Vee--->Re--->emitter-base junction--->Rth--->Vth--->-Vee
    Because for PNP Ie current is also equal to Ie = Ib + Ic

    3.PNG

    So we start at the bottom of a Vee and we go from - to + so we give + sign.
    Vee - VRe - Veb - VRth - Vth = 0V

    Vee - Ie * Re - Veb - Ib*Rth - Vth = 0

    Ib = Ie/(Hfe +1) therefore

    Vee - Ie * Re - Veb - Ie/(Hfe +1) *Rth - Vth = 0

    Ie = (Vee - Vht - Veb)/( Rth/(Hfe +1) + Re ) = (12 - 10.2 - 0.7)/(12.75/100 + 0.5) = 1.75mA

    Ri ?? And where is RL ?
     
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  17. full

    Thread Starter Member

    May 3, 2014
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    thanks sir:):)

    Ri =(Rth//r-bi)

    I forget RL sorry ,

    Av= (B Rc//Rl)/(r-bi +(1+B)Re) *Ri
     
  18. Jony130

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    Feb 17, 2009
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    But in this circuit we don't have source resistance? So how can Rht has any effect on voltage gain ?
    For me the voltage gain is equal to
    Av = Rc||RL/(re + RE) = (β * Rc||RL)/(Rpi + (β+1)*Re) ≈ 2.6V/V
     
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  19. full

    Thread Starter Member

    May 3, 2014
    225
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    thanks sir ,you are great man...... :)
     
  20. full

    Thread Starter Member

    May 3, 2014
    225
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    I do this problem but in common collector :
    I think there is wrong in Ac analysis ...

    thanks,
     
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