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Transistor setup

  1. May 29, 2015 #1
    im so overwhelmed with this topic , im supposed to understand the common emitter setup of a transistor , and i really can't ! , my textbook is full with terminology and concepts that i haven't studied before ( which doesn't make sense ) and that's why i got stuck with this lesson . i feel like there are gaps that need to be filled before im able to grasp the main concepts within this topic .
    i have some questions here and if someone can suggest a source that explains how transistors work and how they are set up , i ll be more than grateful .
    my questions :
    what are the output and input circuits ? and what is the difference between them in terms of their function ?
    what do we exactly mean with forward and reversed bias and how do each of them affect the pn jucntion ?
    ps : if someone has any tips on how am i supposed to handle this i ll really appreciate it .
    thank you
  2. jcsd
  3. May 29, 2015 #2


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    Welcome to the PF.

    What textbook are you using in this class? What other resources are you using to learn the material (like wikipedia, etc.)?
  4. May 29, 2015 #3


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    Staff: Mentor

    BTW, explicit schoolwork questions go in the Homework Help area of the PF. But we can discuss general questions here in the technical forums. We just can't do your schoolwork for you. :smile:
  5. May 29, 2015 #4
    im surely not asking for this :)
    the textbook im using is the school's one . and its very poor , its just very superficial/shallow
    about other sources , i surely referred to some ( wikipedia and some videos on youtube ) and they helped a little bit , but i still have some questions .
    what are the output and input circuits ? and what is the difference between them in terms of their function ?
    what do we exactly mean with forward and reversed bias and how do each of them affect the pn jucntion ?
    i ll try to be more specific with the second question .
    i think i have a grasp of what happens when a pn junction is formed ( the diffusion of the charge carriers , the formation of the depletion area , and thus the formation of an electric field within the pn junction )
    my problem starts when we apply a voltage across the pn junction , i know that in case we forward bias the pn junction , a current is able to flow . but when we do the opposite the current won't .

    now i know this but what is the exact explanation of it ?
    at one point in my textbook , it says the current's ability to flow is due to the shrinking of the depletion area when the pn is forward biased .
    and at another point it says that the reason the current flows is that when the pn junction is forward biased , an external electric field , that opposes the internal one , forms .

    so the thing is , the affect of how the pn junction is biased , seems to be explained from 2 perspectives , or in 2 ways .
    and this really confuses me , is it the shrinking of the depletion area , or is it the formation of an external electric field that is opposite to the internal , or are these 2 things associated .
    im just trying to connect things , you know .
    Last edited by a moderator: May 29, 2015
  6. May 30, 2015 #5
    A PN junction is a quantum mechanical feature and figuring how they work analytically is beyond the scope of most electrical engineering. So what textbooks do is simplify. Use whatever analogy works for you. But be aware it's an analogy. For figuring what is actually happening, use the equations. For a BJT the Ebers Moll model is commonly used. But that is an overly complex model for most people to conceptualize (as opposed to doing the math).

    Commonly (but incorrectly) I consider a common emitter configuration to be a current driven current source. The transistor multiplies the emitter-base current by its beta (ß) to find the emitter-collector current. (There are lots of caveats to that approach, but it can be useful for planning purposes.) Once the circuit is laid out, use equations to find component values.

    I view a PN junction as shrinking when forward biased and growing when reversed biased.
  7. May 31, 2015 #6


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    Surely, it is not possible to fully explain the BJT`s principle function in this forum. Nevertheless, I`ll try to give some short answers to your questions:

    1.) The "output circuit" is a resistor that converts the transistor output current (it acts as a current source!) into a voltage. The "input circuitry" is - in most cases - a resistive voltage divider to bias the TWO pn junctions (BE-junction in forward direction, and the BC-junction in reverse direction).
    2.) see above
    3.) Try to understand the principle of a simple pn-diode in forward and reverse operation.
    4.) The explanation is based on carrier physics (movement of p and n carriers caused by an electric field (externally applied voltage)
    5.) At first, try to understand WHY there is a depletion area (carrier diffusion across the junction WITHOUT any external voltage). As a result, there is a carrier separation left and right from the junction which leads to a "diffusion voltage" and to an "internal" electrical field. Then, the shrinking of this area is caused by the externally applied collector voltage and the corresponding field.
    As a result, the depletion area is so small (mikro-meters) that - nearly - all emitted electrons move through the depletion zone, attracted by the collector voltage.

    Perhaps this helps a bit.
    Last edited: May 31, 2015
  8. May 31, 2015 #7


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    I agree to the above - however, perhaps it helps to give an additional explanation.
    Physically - the bipolar transistor (BJT) is a voltage-driven device (collector current IC is determined/controlled by the forward biasing VOLTAGE VBE between base and emitter).
    However, because it turns out that the current into the base (IB) is a - more or less - fixed (small) percentage of IC, in practice we often treat the BJT as a device that amplifies the current IB with a fixed factor B (or beta) according to the equation IC=B*IB.
    This is important to know - otherwise, one could be confused because some sources (books, articles) describe the BJT as a current-controlled device, in contrast to some others which speak about a voltage controlled device (physically correct).
    Last edited: May 31, 2015
  9. May 31, 2015 #8
    A DIODE is P-N junction....some decent explanation here:
    and the P-N Junction diode section provides bit more detail.
    These are the same P-N junctions in transistors.

    Also, the first few sections here:

    explain how the P-N junctions are used in three terminal [emitter,base, collector is the terminology]
    'bi polar junction transistors.

    " BJTs can be used as amplifiers, switches, or in oscillators.." and the intended use prescribes the type of external circuits, inputs and outputs that work.

    One such simplified circuit is Figure 1 here:

    In this configuration an input signal is provided between "Vin" and emitter [say ground] and the output is measured between "Vout" and the emitter. So the emitter is "common" to both; in this configuration, and as already posted, the output is 'amplified' by the Beta {gain} of the transistor; that is the output variation is say, for example 100 times larger than in the input variation, for example. The common emitter configuration is thus often used in a amplifier [gain device] role. An example might be picking up a weak over the air radio station broadcast signal in your home, the input, and the output would be a replication and amplification of that weak radio signal which is strong enough for you to hear. But the transistor gain is not linear over all frequencies and powers, so it must be 'biased', that is operated in range where the amplification is approximately linear. When it operates in a non linear portion of its gain, 'distortion' ensues; that means the output is not a faithful duplication of the originating signal....so, for example, maybe a high frequency horn sound sounds "bad".
  10. Jun 1, 2015 #9


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    For my opinion, this sounds a bit "strange" and - perhaps - can lead to misunderstandings (beginners).
    I think, it is better to say:
    * The output current equals the input current (into the base) multiplied with beta (current gain), and
    * The output voltage equals the input voltage multiplied with the voltage gain A. The value of A depends on the transconductance gm and the resistor Rc (collector path) and - if existent - the emitter resistance Re (effective for ac): Vout/Vin= - Rc/(Re+1/gm)
  11. Apr 8, 2016 #10
    For anyone else who comes across this problem of feeling like they need to have a complete or foundational understanding for wrapping their heads around transistors and the PN junction (I did).

    Try this link: http://www.allaboutcircuits.com/textbook/semiconductors/#chpt-2
    Goto chapter two. They build you up from the very bottom, starting with quantum physics (although just what you need to know to understand the rest) and electrons and holes. As well as explaining the workings of different types of semiconductor components.
  12. Apr 9, 2016 #11


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    Perhaps have a look on youtube. There are vids there which explain how transistors work at various levels from how they work on the semiconductor level up to basic biasing for use as a common emitter amplifier.
  13. Apr 9, 2016 #12

    jim hardy

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    @elia gomez Are you familiar with the concepts of voltage and current ? Ohm's and Kirchoff's laws ?
    Do not be offended by the low academic level of this post. I sometimes toss life-buoys to championship swimmers and wind up embarrassed.... i sense you just need confidence in what you already know to kick-start your ascent.

    All too common in today's "Publish or Perish" milieu.
    A picture is worth a thousand words.
    Decent beginner's explanations here
    following sketches are from there.

    In its simplest form
    common emitter means emitter is tied to circuit common.
    Base is input, collector is output.
    Input accepts a signal, output reproduces it but larger . A small change of voltage is turned into a larger change of voltage.
    The PN junction will only allow current to flow one way.

    So think of the PN junction as a valve that allows current to flow in the direction of the arrow so long as that's the direction voltage is trying to push it. That's forward biased. (speaking in terms of conventional current, not electron flow)

    When voltage is trying to push current against the arrow the junction refuses to conduct (so long as we don't break it by applying too much voltage). That's reverse biased.

    There exist terms like depletion region , majority and minority carriers, doping which are necessary to understand the inner workings of a semiconductor. But for working transistor amplifiers in your head, all you need to know is :
    silicon junctions conduct meaningfully when voltage across them reaches 0.5 to 0.6 volts in forward direction. Germanium ones conduct at 0,25 to 0,3 volts, but germanium transistors have fallen largely out of favor.

    Depletion region is a no-man's land surrounding an actual reverse biased junction where only insignificant current can flow unless there's some special circumstance in play. Remember that phrase "special circumstance".....
    Reverse bias widens the depletion region. Forward bias narrows it. When it reaches zero width conduction commences. See https://en.wikipedia.org/wiki/Depletion_region

    Now while the little circuit above will amplify a DC voltage we are usually more interested in amplifying an AC signal - from music or radio waves.
    So we add capacitors to block the DC parts of the signals.
    I trust you've had enough courses to appreciate capacitors pass only AC....

    Then we add some more resistors to make the transistor have a predictable current at zero signal ,
    that predictable zero current being called , logically enough, quiescent current.

    so that as input signal swings above and below zero the transistor current swings above and below its quiescent level
    producing substantial voltage swing at junction of collector-R3.
    Whatever is voltage at junction of R1-R2-Base, call it Vb
    voltage at emitter will be ~0.6 V less, (0.3 for germanium)
    and Ohm's law says that voltage (Vb-0.6) divided by R4 is your zero signal(quiescent) current through the transistor !
    Choose R3 to drop around half Vsupply at quiescent current

    Whew now we have biased the circuit. At quiescence, collector volts is steady around half Vsupply.

    A word about collector is in order now.
    The collector-base is another junction and it's reverse biased. So it shouldn't conduct. But there's a special circumstance at play -
    whatever current flows into the base-emitter PN junction causes more current to flow into the emitter through the Collector's depletion region. That was Shockley's great invention, the Transistor...
    That extra current is substantial, 10X to 100X the base current is typical. That's how it amplifies, by magnifying current.
    The ratio of current amplification is called by various authors Beta, hfe , and forward current gain. Again ,10 to 100 is typical.

    C3 allows AC current to flow around R4 so that small changes in input cause maximum change of forward bias of Base-Emitter junction , causing maximum achievable current swing in the transistor,
    That current swing causes varying voltage drop across R3 and that varying voltage drop is the amplifier's output(because Vsupply is assumed constant).
    C2 removes DC from amplifier's output signal

    and that's very briefly what every part of a common emitter amplifier is for.

    I hope this gets you started.
    If you're into hybrid pi modelling i apologize - you're way too advanced for this simple poor man's explanation

    no offense meant.

    old jim
    Last edited: Apr 9, 2016
  14. Apr 9, 2016 #13

    jim hardy

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    oh my goodness a necrothread !
    Oh well - Maybe it'll help somebody....
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