Voltage divider bias

Discussion in 'General Electronics Chat' started by sudar_dhoni, Jan 13, 2010.

  1. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
    In voltage divider bias ,
    we are not drawing any electrons from the base.
    Then how can the electron current pass through the base and go to the collector and reach the positive terminal.
    Only if You draw electrons from the base by maintaining the base at positive potential,
    then only the electrons from Vcc can flow through the base and to the collector and reach the positive terminal of Vcc.

    The need for removal of electrons from the base is that.
    In the first circuit shown,
    when the base emitter is forward biased, the holes in the P region diffuse into the N region and the holes in the base are now filled by electrons as a result the base becomes the negatively charged. Now the electrons from Vcc negative terminal cannot pass through the transistor as they have to cross the base which is now negatively charged . But since the base is made positive the electrons that filled up the holes are attracted as a result the base now consists of holes and the electrons from - Vcc can now flow through the transistor .
    Since in Voltage divider bias this provision( Vbb) is not there , then how can electrons be withdrawn from the base and how can current flow through the transistor.
  2. thyristor

    Active Member

    Dec 27, 2009
    It's not clear to me what you are actually asking but, I think (maybe incorrectly) that you are asking how does voltage divider bias work if no base current is drawn.

    Well, of course, base current IS drawn but the values of R1 and R2 are chosen such that the standing current through them is much greater than any base current drawn (say at least 10 times greater).

    Ergo, in the calculations, the base current can be ignored to a first approximation since the effect of including it would have no real significance on any of the results.

    Thus, in reality, IR1 = (IR2 + Ib) but since IR1 >> Ib then IR1 = IR2 to all intents and purposes. (where IR1 and IR2 are the currents in R1 and R2 respectively, and Ib is the base current).

    The voltage on the base will be Vcc.R2/(R1 + R2)

    Therefore the voltage on the emitter will be about 0.6v less than this owing to Vbe

    Thus Ve = Vcc.R2/(R1 + R2) - Vbe

    and therefore Ie will be Ve/Re = [Vcc.R2/(R1 + R2) - Vbe]/Re

    Since the magnitude of Ib can be ignored for all practical purposes, then Ie = Ic to a first approximation and therefore the voltage drop across Rc will be Ie x Rc = {[Vcc.R2/(R1 + R2) - Vbe]/Re}.Rc

    Therefore, since Vc = Vcc - Ie.Rc, then Vc = Vcc - {[Vcc.R2/(R1 + R2) - Vbe]/Re}.Rc

    Bet you'll come back and tell me that wasn't what you were asking :)
    Last edited: Jan 13, 2010
  3. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
    You went too mathematical.
    What is asked is.....
    I'll start from transistor basics,
    Firstly there are holes in base and electrons in emitter.When the base emitter junction is forward biased, the holes from base diffuse into the emitter (i.e electrons in the valence band of the emitter fill the holes in the base.)
    As a result the base which had earlier holes and was electrically neutral now due the entry of new electrons has now become negatively charged.
    The electrons from the -terminal of Vcc cant pass the transistor as the negatively charged base is blocking it.
    Since the base is connected to positive terminal.It will attract the electrons from the base and as a result holes are once again generated in base.
    Now the electrons form -terminal of Vcc can pass through the transistor and go back to the +positive terminal of Vcc. Thus only if you draw the electrons from base it allows the collector electron current to flow. i.e the base current controls collector current .This is the principle of amplification.

    But since the base in the voltage divider bias is not connected to any positive terminal . Without drawing electrons from base how can collector current flow.
    I have another doubt which i will ask after i get a proper answer to my question.
  4. hobbyist

    Distinguished Member

    Aug 10, 2008
    In the picture in the top post

    You have the divider (R! and R2)

    Don't let R2 fool you thinking no positive voltage at the base of the transistor.

    R1 and R2 form a positive voltage value, that is fed to the base of transistor.

    Without R2, then R1 would have to be made much larger so as to lower the current needed to properly bias the base.

    Instead of having large values for R1, then a better method is to use a combination of 2 resistors to divide the supply voltage down to a value needed to properly bias the base.

    So there is still the positive voltage applied to the base, or else the transistor couldn't conduct, but the voltage is supplied through a divider.

    And thus the base current still flows through R1 but at the proper lower value needed to give proper base current flow.
  5. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
    You explained in terms of conventional current. I want an explanation in terms of electron flow.
    Moreover how can a resistor supply positive voltage.Only the positive terminal of the battery can supply positive voltage.
    I think you were referring to conventional flow. i.e current flows from +Vcc through R1 , the voltage gets dropped and that is responsible for the base bias. Even in my book the same explanation is given.But i cant understand it and it is not convincing.
  6. hobbyist

    Distinguished Member

    Aug 10, 2008
    I'll try to give an electron flow analogy.

    This is NOT the behavior of electron flow, But just an analogy, ONLY.

    Electron current flowing from the negative (ground side) of battery flows through R2 and through R1 to the positive side of the battery.

    If you remove R1 then you will have the battery positive term. touching R2 so the voltage drop across this resistor would equal the battery, voltage.

    By placing R1 back into the circuit, then electrons bounce up against this resistor, and begin to accumalate a negative charge, so the higher this R1 is the more electrons that pile up before they can go through, so now the battery positive term, is isolated from R2, because of R1 being there between, as you lower R1 value, then the avenue for electrons becomes wider, so now electrons can flow more readily, more can go through at one time, so the negative charge is reduced because less electrons has built up against the resistor.

    Its like a bottle neck in traffic, the higher the resistance is the tighter the bottle neck, so less electrons go through, but these electrons are still moving until the pile up at this resistor junction, so the more they keep on flowing and piling up the more negative it becomes, which is all based on the ratio of the resistance, the same amount of electrons per value of resistance.

    Conventional is a lot easier way to understand the theory of electronics, at an Engineering level.

    Electron current flow needs to be understood along with conventional, to fully get a grasp of circuit analysis.

    What I showed about electron flow, is JUST an ANALOGY, the electrons do not behave in the way I explained, but the analogy helps to explain the voltages produced using electron theory.
  7. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
    thanks for explaining in terms of electron flow.
    But what i asked is that the electrons must start to flow only from the base as they are blocking the electrons from the -terminal of Vcc to flow back to +terminal of Vcc.
    i.e the base current must control the collector current.
    So the electrons must start to flow only from the base.
    MY doubt will get clarified if you confirm my assumption.
    Does the electric field of the battery i.e the electric field from the positive terminal of the battery acts on the base? Also does the -terminal of the battery provide forward bias to emitter??
    If you confirm this then my doubt is cleared.
  8. hobbyist

    Distinguished Member

    Aug 10, 2008
    Using your second schematic with the NPN transistor and the voltage divider.

    Yes the positive term. of the battery influences the base, and the neg. term of battery influences the emitter, thereby forward biasing the base emitter junction, into conduction.

    Electrons are being pulled from the base through R1 to the pos. bat. term.

    Let me know if this helps make it more clear.
  9. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009

    Yes thanks for your confirmation.It helped me a lot

    I have another doubt about amplification i hope you will clear it.

    in the illustration given in the book where initially during the negative cycle electrons from signal first occupy the holes in base and thereby making the base negatively charged(additional electrons than the neutral level) and preventing electrons from emitter to reach the collector, and during the positive half cycle electrons are withdrawn from the base by the signal and creating holes and thereby causing electrons to move from emitter to collector.

    in case when an additional supply is connected what will happen is that the resultant voltage still exists on EB and it remains forward biased . As a result the holes in base diffuse into the emitter and electrons from the valence band of the emitter fill up the holes in the base and these electrons are drawn by the net voltage applied on the EB and when these electrons from base reach the positive terminal of the resultant voltage and equal number of electrons start from the negative terminal and reach the emitter and fill up the holes in the emitter which actually diffused in the emitter from the base. And simultaineously electrons from Vcc would just pass by the emitter and reach the collector since the electrons which had earlier filled the holes had made the base negatively charged and blocking the way for electrons from collector.Now in the base since the blocking electrons have been removed , new holes are created tempts the electrons from Vcc to cross the base and reach collector and go back to the Vcc.
    What My doubt is that
    if we increase the voltage on EB will the electrons which are present in the valence band of the base also be attracted along with those who filled the holes due to diffusion,
    and thereby making the base acquire a net positive charge( note that eventhough base has positive holes when the transistor is unbiased, it is electrically neutral).WHat i am asking is that will additional electrons also be removed from the valence band of the base other than those electrons which filled the holes and thereby making the base deficient of electrons i.e lesser than the neutral level.
    If this happens will the depletion region of EB be further reduced and therby decreasing Vbe.

    In the first figure : Electrons occupy the holes in the base thereby making base negatively charged.
    In the second figure:Electrons leave the base therby creating holes and making it neutral.
    In the third figure: ELectrons from the valence band of the base in addition to the electrons which had occupied the holes leave the base as a result making the base positively charged.
  10. hobbyist

    Distinguished Member

    Aug 10, 2008
    When they design a semiconductor, they use Non conducting materials, (good insulators) so that the only conduction that takes place, would be only the impurities, such as hole flow, in p type material, and electron flow in N-type material.

    So under normal conduction, there should not be any valence electrons flowing, otherwise the semiconducting properties would be compromised.

    Your learning transistors at the molecular level, which is good to know, but when you begin to analyse transistor circuitry, you will need to have a better understanding of transistor operation, from the electrical level.

    Such as voltages and currents, to bias a transistor stage, and signal currents and voltages.

    The physics properties of transistor behavior, will have less bearance on circuit analysis and design, compared to electrical characteristics.

    That's why I'm probably not a good enough help, with your questions, because I have been away from the learning transistor physics, for a long time, because now I deal with transistor analysis on electrical terms alone.
  11. sudar_dhoni

    Thread Starter Member

    Nov 9, 2009
    Could you please give a detailed explanation on this .
    If not could you please suggest a referrence book
    or any website.

    I had this doubt long long ago and still am not clear with it.
    In the transistor circuit analysis they use only conventional current.
    How can conventional current bias a transistor. Also how can it account for the leakage current caused due to temperature and also the concept of amplification. Till now i had convinced my self thinking the transistor to be valve as mentioned in this site.Please dont critisize me for being childish. But it was quite convincing.
    http://www.satcure-focus.com/tutor/page4.htm. BUt it could not explain the leakage current and current due to minority carries and several phenomenon . Could you please help me with this.

  12. hobbyist

    Distinguished Member

    Aug 10, 2008

    You are NOT being childish asking these questions.

    I commend you for having a great interest in wanting to learn transistor analysis.

    The only help I could really give is in the electrical analysis of transistor theory.

    The link you gave, is a good website for explaining this theory, do a google search as "transistor physics" that will bring up some links in that field.

    Also there are many knowledgeable people on this forum, that could explain this better, so I think that more people would respond to your question, if you started another thread, and call it

    "Leakage current in transistors?"

    I think more people will pick up on it and try to answer your questions.