Transistor help

Discussion in 'Homework Help' started by Garurumon, Oct 28, 2013.

  1. Garurumon

    Thread Starter Member

    Mar 17, 2013
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    Could someone check if the relations written in attached file are correct?

    I was confused, if I have some base current, but I have resistors on collector and emitter of that transistor, will the current through the transistor still be (1+Beta)*Ib, or will it be limited by those transistors, never minding the Ie=(1+Beta)*Ib equation?
     
  2. Garurumon

    Thread Starter Member

    Mar 17, 2013
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    I see the mistake on T1 base current
     
  3. Garurumon

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    Mar 17, 2013
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    and in the voltage divider xD I just wanna know about the last question
     
  4. Jony130

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    Feb 17, 2009
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    What is the last question ?
    This equation Ie = (1+Beta)*Ib holds until Ie < Vcc/(RC+RE)
     
  5. Garurumon

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    Mar 17, 2013
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    you just answered it, thank you..

    But Ib still flows through Ic if not..?

    I have to find Vi1 and Vi2 on the image, and I do not know what currents flow through Rc and Re.
     
  6. Jony130

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    Feb 17, 2009
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    IB never flows through Ic.
    IB current flows from Vcc --->R1 --->RB--->Base-emitter T1--->Base-emitter T2--->RE ---->GND
    And first use Kirchhoff law to find Vi2 voltage
     
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  7. anhnha

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    Apr 19, 2012
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    Jony, did you mean that the equation only holds if transistor is in active region?
    And, isn't Ie always smaller than Vcc/(RC+RE)?
     
  8. Garurumon

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    Mar 17, 2013
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    So, through Rc flows Vcc/(Rc+Re), and through Re flows Vcc/(Rc+Re)+Ie1 ?
     
  9. Jony130

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    Yes.

    In real circuit yes. But not always in calculation when you use "active region" equation.
     
  10. Jony130

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    No, this image explains everything.
    [​IMG]
     
    Last edited: Oct 28, 2013
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  11. anhnha

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    Apr 19, 2012
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    I think you meant that is incorrect as transistor operates in saturation region.
    In saturation region both junctions are foward biased but this does not make sure that VCE always larger than zero.
    Is that right?
     
  12. Jony130

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    I'm not quite sure what you mean by writing "but this does not make sure that VCE always larger than zero"
    In saturation Vce is always smaller than < 0.6 but larger then 0V
     
  13. Garurumon

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    Mar 17, 2013
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    yes, but i have limiting resistors, that's what confuses me! if I take that Ic=beta*Ib, then I get Vi2 voltage about 1kV!
     
  14. anhnha

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    Apr 19, 2012
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    If so, I am confused with this.

    Is Ie always smaller than Vcc/(Rc +Re)?

    Ie = (Vcc - Vce)/(Rc + Re) and Vce always larger than zero => Ie < Vcc/(Rc +Re) always, right?
     
  15. Jony130

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    For what values??
     
  16. Jony130

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    Try to solve this circuit

    [​IMG]


    For Vcc = 10V ; Rb =10K; Rc = 1K; Re = 100Ω ; and Vbe = 0.6V and Hfe = 100
     
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  17. Garurumon

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    Mar 17, 2013
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    so, what does Ic equals to here? And what is the voltage above Re, and bellow Rc?
     
  18. anhnha

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    Apr 19, 2012
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    Vcc = Ib*Rb + Vbe + Ie*Re

    => Ib = (Vcc - Vbe)/( Rb + (Hfe +1)Re) = (10 - 0.6)/( 10K + (100 + 1)0.1K) = 0.468mA

    Ie = (Hfe + 1)Ib = 101*0.468mA = 47.268mA
    Ic = Hfe*Ib = 100*0.468mA = 46.8mA
    Ve = Ie*Re = 47.268mA*0.1K = 4.7268V
    Vce = Vcc - Ic*Rc - Ie*Re = 10V - 46.8mA*1K - 47.268mA*0.1K = -41.5268V :D

    If Vce < 0 how current flow from C to E?
     
  19. Jony130

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    This result (Vce = -41.5V) give as one important hint.
    Our BJT is saturated and Ic = Hfe*Ib don't hold anymore.
    Because if Ic = Hfe*Ib = 46.8mA >> Ic_max = Vcc/(Rc+RE) =10V/1.1K = 9mA transistor is in saturation.
    So to solve tis circuit we need to assume Vce_sat voltage and use this equation
    Ie = Ib + Ic.
    Let as assume Vce_sat = 0.1V and Vbe = 0.7V.

    Ie = Ve/Re
    (1)

    Ib = (Vcc - Vbe - Ve)/RB (2)

    Ic = (Vcc - Vce_sat - Ve)/Rc
    (3)

    And now if we solve this for VE we have this

    \Large Ve = (\frac{Vcc - Vbe}{RB} + \frac{Vcc - Vce_{sat}}{RC}) * RB||RC||RE = 0.975676V

    So

    Ve = 0.975676V

    Vb = Ve + Vbe = 0.975676V + 0.7V = 1.675676V

    Vc = Ve + Vce_sat = 1.075676V

    And the currents

    Ie = Ve/Re = 9.75676mA
    Ic = (Vcc - Vc)/Rc = 8.924324mA
    Ib = (Vcc - Vb)/RB = 0.8324324mA
     
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  20. anhnha

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    Apr 19, 2012
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    Wow, that makes lot of sense now.
    Just one more question.
    In saturation Vce is vey small but is it constant in the region?
     
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