Simplifing my Circuit

Discussion in 'General Electronics Chat' started by ahmedzica, Jan 9, 2012.

  1. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    [​IMG]

    Let Rb= 1k , Rc=100 and Vcc= 5v

    Can anybody simplfy this circuit and explain why is Vcc-V(Rb)-V(Be)=0

    Thanks
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    The above equation states that if you begin with the voltage applied to the circuit, 5V in this case and then you subtract the voltage drop across the resistor Rb and then you subtract the voltage across the base-emitter junction, you should end up with 0 volts.

    hgmjr
     
  3. wmodavis

    Well-Known Member

    Oct 23, 2010
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    It is also called Kirchoff's Voltage Law. The sum of voltage drops around a loop is equal to zero.

    [SIZE=+1]Kirchhoff's Voltage Law - Introduction[/SIZE]


    Kirchhoff's Voltage Law - KVL - is one of two fundamental laws in electrical engineering, the other being Kirchhoff's Current Law (KCL).
    • KVL is a fundamental law, as fundamental as Conservation of Energy in mechanics, for example, because KVL is really conservation of electrical energy.
    • KVL and KCL are the starting point for analysis of any circuit.
    • KCL and KVL always hold and are usually the most useful piece of information you will have about a circuit after the circuit itself.
     
  4. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    I already know that. But I'm wondering if we used KVL why ignored V(Re) as in the left loop

    Thanks for your help
     
  5. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Because Vbe is assumed constant and non-dependent on Ic, which it practically is.
     
  6. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    can you tell my which loop are you loooking at
     
  7. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    I really appreciate your efforts. and I know how to use KVL. But I can't imagine how this circuit would be if we replaced this Vcc (1 Terminal ) with 2 Terminal and Removed that ground over there
     
  8. Adjuster

    Well-Known Member

    Dec 26, 2010
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    If Vcc were replaced by a connection to another network, you would have another more complex circuit. If there is a particular circuit you are thinking about, perhaps you might want to explain what you have in mind

    What are you actually trying to analyse - the simple transistor circuit considered as a load on Vcc?
     
  9. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    Yes I want to calculate Ic and Vce
     
  10. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    The loop is the one you posted first, Vcc-V(Rb)-V(BE)=0.
    Another one would be Vcc-V(Rc)-V(CE)=0. If you know that Ic=β*Ib then you should be able to calculate the rest you need.
     
  11. Adjuster

    Well-Known Member

    Dec 26, 2010
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    At low input voltages the current may be estimated approximately by supply current I of (Vcc) = (β+1)(Vcc-Vbe)/Rb.

    Since here Rb is only 10 times Rc it is likely that, with a reasonably high gain transistor, above a certain voltage the transistor will saturate. This is because (β+1)(Vcc-Vbe)/Rb will exceed Vcc/Rc. At higher input voltages therefore, I(Vcc) ≈ (Vcc-Vbe)/Rb + Vcc/Rc
     
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  12. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    Oh i got it. But why saying (β+1) is it only β? and after saturation will we cut of the exceeded wave?
     
  13. ahmedzica

    Thread Starter New Member

    Jan 9, 2012
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    I got it, huge thanks
     
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