Voltage divider bias electron current

Discussion in 'General Electronics Chat' started by sudar_dhoni, Jan 26, 2010.

  1. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
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    I have posted the circuit diagram and have myself marked the direction of electron flow in voltage divider bias . Please verify it.
    If my diagram is correct , then you must wonder why i have'nt marked the electron current throught R2. The reason is as follows.

    The negative terminal of Vcc pushes the electrons in emitter to the base.
    There a few of them come out of the base and the rest comes out of the collector. This is Ib and Ic respectively.
    They both then combine and reach +terminal of Vcc.
    Then equal amount of electrons Ib + Ic come out of the negative terminal.
    Since initially the emitter has supplied electrons equal to Ib + Ic , it must get back the same amount of electrons to maintain the concentration of free electrons in the emitter, so the elctrons coming from - terminal of Vcc (Ib + IC) must go back to emitter. So no electrons goes to R2.
    IS My explanation correct??
     
  2. ifixit

    Distinguished Member

    Nov 20, 2008
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    There is current in R2 also because there is a barrier voltage of approximately 0.6 volts across the base-emitter junction. IR2=Vbe/R2.
     
  3. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
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    I cant understand it . A bigger explanation please.
    Are you dealing with conventional or electron flow.
    I want electron flow.
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    Look at the voltage across R2. That will show the direction of current and its magnitude as suggested in post #2.
     
  5. ifixit

    Distinguished Member

    Nov 20, 2008
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    If you, temporarly, remove the transistor you can see that there is a path for electrons to come from the ground and pass up through R2 and R1 then back to the battery. Now with the transistor in, some electrons pass through the emitter, as you discribed, but some go to R2 as well.

    Search this site for more details on transistor function.
     
  6. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
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    IF you remove the transistor its totally different because current starts to flow from battery so it will flow through R2.

    But if you insert transistor the flow of electrons starts from the emitter due to the push given by the negative terminal of the battery.

    My doubt in the first thread is quite satisfying.
    But actually do electrons flow through R2 , But how is that possible according to my first explanation in the first thread ??
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Current will (must) be flow through R2 because there is a voltage difference between base and negative terminal of voltage source and there is a close path for R2 current.
    The Vb=Ie*Re+Vbe so current flowing through R2 is equal
    IR2 = Vb/R2
     
  8. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
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    You are explaining mathematically.Thats ok.
    But according to my theory in the first thread , how can the electrons flow through R2.
    My explanation is in the first thread is correct according to my knowledge.
    Please talk in terms of electron and not conventional.
    Conventional current is mentioned in all the books but electron current is not mentioned in any book.
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Ask yourself, what voltage is at base of BJT.
    And if there is a voltage on the base and we have closed loop for current then current mus flow through R2.
    Because force that cause the current to flow is a voltage.
     
  10. ifixit

    Distinguished Member

    Nov 20, 2008
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    Electric current flow is equivelent to electron flow just as a river current is referring to water flow.
     
  11. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Your explanation works only if R2 is not in the circuit. Electrons from the battery will flow both through the emitter of the transistor and through R2. Therefore, the electrons flowing from the battery I(battery) = Ie + I(R2).

    The electrons flowing through the emitter are divided into base electrons, Ib, and collector electrons, Ic. The base electrons join with the R2 electrons in R1. This gives I(R1) = I(R2) + Ib.

    The electrons flowing through R1 are joined by the electrons flowing through the collector before returning to the battery.

    I(battery) = I(R1) + Ic
    I(battery) = I(R2) + Ib + Ic : substitute I(R1) = I(R2) + Ib
    I(battery) = I(R2) + Ie : substitute Ie = Ic + Ib

    That is the same relationship we started with.
     
  12. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
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    Your explanation is very good.
    But in transistor action the electrons start to flow from emitter and not from battery.The battery pushes electrons in emitter (Ie) .Some of which comes out of the base as (Ib) and flows through R1 and the rest comes out of the transistor as (Ic) and then they join to form Ie and reach positive terminal of the battery. Equal amount of electrons come out of the -terminal ( Ie). Now to maintain the concentration of free electrons in the emitter all these electrons have to flow to emitter. So no electrons flows through R2. Is this correct?

    Your explanation is from the battery. I want an explanation from emitter,because if you start from the battery , you are not treating the transistor as a special device i.e special device means it must supply electrons . But instead you are treating it just like an ordinary conductor which allows current to flow through it.
     
  13. Wendy

    Moderator

    Mar 24, 2008
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    You are letting the direction of current flow confuse you. In most cases it doesn't matter.

    The thing that is important is the resistance from Base to ground through the emitter. This is figured by Re * ß.

    Treat it like a resistor in parallel with R2, and you will be much closer to the answer.
     
  14. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
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    Your explanation works if you consider the flow of electrons starting from the battery. But actually the flow starts from the emitter . Then this explanation wont work out. I have mentioned what is happening if we start the flow from the emitter in the previous thread.
    According to that theory your explanation wont work.
     
  15. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    This is not true BJT is not a "current source", emitter don't create the electrons. BJT is device that control flow of a current by help of a base current .
    And electrons always starts flow form the voltage source, electrons are injected into the emitter region by emitter bias supply .
     
    Last edited: Jan 28, 2010
  16. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
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    How does the BJT achieve this.
    Please explain in terms of electron hole movement.
     
  17. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Try this.

    Replace the battery with a resistor. If the electron current is initiated by the emitter of the transistor, the circuit should be the same. This is, however, not the case. Without the power supply, no electrons flow.

    Even at a device physics level, a voltage is required across the base emitter junction,Vbe, and across the collector base, Vbc, junction for the device to operate. Otherwise, Ib = Ic = Ie = 0.
     
  18. beenthere

    Retired Moderator

    Apr 20, 2004
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    From your PM, you seem to believe that current in the PN junction must stop when the existing free electrons have combined with the available holes. That is not going to be the case, as there is an external source of voltage and current in the circuit - http://www.allaboutcircuits.com/vol_3/chpt_2/6.html
     
  19. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
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    If there were no external source ( base voltage and base lead) then there would be no flow of electrons from the emitter to collector since the base has now become negatively charged due to diffusion and electron hole combination. Now to continue the motion of electrons from E to C we give a base lead and suck the elctrons from base and therby creating holes and now allowing electrons to flow from E to C. The more you suck the more number of electrons go from E to C. This is amplification process.
    I asked 3 more doubts in my private message.
    Could you please post it here as the message that i sent to you is not in my sent items and its showing empty.
     
  20. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
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    The electron does not automatically start from the emitter. You misunderstood my explanation. I told that the -terminal of battery pushes it and thereby the flow of electrons start from emitter.

    In transistor action the electrons start to flow from emitter and not from battery.The battery pushes electrons in emitter (Ie) .Some of which comes out of the base as (Ib) and flows through R1 and the rest comes out of the transistor as (Ic) and then they join to form Ie and reach positive terminal of the battery. Equal amount of electrons come out of the -terminal ( Ie). Now to maintain the concentration of free electrons in the emitter all these electrons have to flow to emitter. So no electrons flows through R2. Is this correct?

    If according to you power supply supplies the electrons first and it just flows through the transistor casually , then please give me a detailed explanation on what for the transistor is doped and why is it used.
     
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