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B Input signal in a transistor

  1. Jun 26, 2017 #1
    Hello ,

    This doubt arose while studying chapter on semiconductor devices in grade 12 of high school physics .

    Please clear a doubt in as simple
    language as possible . In order to be
    specific with my question I need a
    diagram which is exactly as mentioned
    in the video below . Kindly see

    Note : I do not wish to understand the working of a transistor .

    I would just like to understand what type of input current (AC/DC) IB enters the base. ?

    You may look directly at two times , at 7:32 and at 13: 40 . Here the instructor applies an input signal Vi across the battery Vbb and says that the input signal varies from Vbb+Vi to Vbb-Vi i.e she superimposes the two voltages . Is it correct to superimpose the voltages ?

    Is the input current IB entering the base an alternating current ?

    I would like to understand , what happens when a AC source is applied parallel to DC battery ?

    What is the role of capacitor in this diagram?

    If two DC batteries are in parallel , then the bigger battery dominates and current flows to the smaller battery from positive terminal .

    But what happens when AC signal is put across DC ? Is it a simple addition just like shown in the video ?

    Please give a simple explanation as I have a very basic understanding of AC .

    Many Thanks
     
  2. jcsd
  3. Jun 26, 2017 #2

    sophiecentaur

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    An ideal battery is a Voltage Source so it maintains that voltage whatever it is in parallel with. two batteries in parallel will be a smoke making machine as both will 'insist' on their own PD. It is not a good way to design practical circuits with batteries that are fighting each other and delivering large currents to resolve contentious voltages.
    The diagram in the video that shows an AC signal applied across the battery is nonsense. She has tried to make the circuit too simple and not realised her error. It is more normal not to use a battery but to provide base bias with a resistor network which is fairly high impedance and then the AC coupled input just adds to the bias volts.
    The currents through the bias network and the input signal will add in a way that follows the various resistor values in the circuit. I recommend that you look at some simple transistor circuit images that a Google search will give you. A common emitter configuration is fairly easy to identify and you will see a wide variety of ways in which an AC signal is added to the bias volts or current.
    PS an isolated transistor is best treated as a current operated device but, once a modern, high gain transistor is inserted into a circuit, it is often easier think in terms of voltage amplification. This can be a problem until you get more familiar with actual circuit designs and the many short cuts that can be taken in everyday circuit designs.
     
  4. Jun 26, 2017 #3

    davenn

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    the DC voltage will produce an offset for the AC. This is called a DC Bias

    That is, instead of the AC cycling above and below 0V , it will now cycle above and below whatever the DC value is


    main-qimg-5e73fe9bc855a1f5a7abd6a703f7f27b?convert_to_webp=true.png


    Dave
     
  5. Jun 26, 2017 #4

    Tom.G

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    As others have pointed out, the presenter of the tutorial has taken a shortcut. It is much more realistic if the capacitor from the input, Vi, is connected to the Base, not to the Battery. Again, in reality, there would be a resistor in series with Vi.
    The input capacitor is briefly explaned at 13:45 in the video.
    In that case, the Base voltage is indeed the sum of the AC and DC voltages.

    For another way of looking at it, RB and the added resistor in series with Vi could be considered the internal resistance of a "non-ideal" battery and signal source.
     
  6. Jun 26, 2017 #5
    I thought , if the two sources were in series then we could superimpose the two voltages .

    What would happen if Vi were in series with Vbb ? Would the DC voltage still produce an offset for AC ?
     
  7. Jun 26, 2017 #6

    davenn

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    an AC source isn't going to work in series with a DC supply
    putting an AC source in series with say, a battery, would not be a good thing for the battery
     
  8. Jun 26, 2017 #7
    Ok .

    Could you please explain in a very simple language how DC biasing occurs when an AC source is applied across a DC battery ?
     
  9. Jun 27, 2017 #8
    One more thing , what is the role of capacitor ?
     
  10. Jun 27, 2017 #9

    davenn

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    here's some reading for you

    https://en.wikipedia.org/wiki/DC_bias

    https://en.wikipedia.org/wiki/Biasing

    https://forum.allaboutcircuits.com/threads/need-explanation-for-dc-bias-coupling-and-offset.64321/

    it stops DC going back into the AC source .... that is a form of DC decoupling


    Dave
     
  11. Jun 27, 2017 #10
    Sorry . I tried reading the links but almost everything has gone over my head .

    Could someone else help me understand what is going on at the leftmost part of the circuit ?

    How are the two voltages superimposed ?
     
  12. Jun 27, 2017 #11

    davenn

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    what specifically do you want to know about it ?

    one way is shown in the left part of that circuit
     
  13. Jun 27, 2017 #12
    This does not seem an effective way for you to try to learn about something so basic to transistor circuits. You most likely don't realize this, but you are asking folks here to guess at how much you might possibly know, so that they can phrase their answer and have even a prayer you'll understand. Either you need to provide a lot more information on your level of knowledge, or else pick a more effective learning strategy.

    If you think you might really find yourself interested in electronics (vs. just this one confusing thing you have been shown), you could find a basic curriculum (via book, online tutorial, etc.) that lets you start with what you do understand, and move on step by step with each lesson. You could speak to your teacher & see what he/she suggests for books or tutorials; or you could ask here. There are many excellent primers; if they deal with electronics rather than purely physics, they don't have to be textbooks necessarily.
     
    Last edited: Jun 27, 2017
  14. Jun 27, 2017 #13

    sophiecentaur

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    It's not impossible to use a transformer in series with the bias. That would produce a base voltage of VDC +VAC. Transformer coupling is very suitable in some circumstances.
    Nothing. The (ideal) battery prevents the AC source having any effect. Even a real battery will effectively be shorting out the AC input to ground.

    You are being exposed to too many things at once, I think and it is all "going over your head".
    Adding signals: Basically you can add Voltages in series or Currents in Parallel. Its KVL at work in real life.
    The left part of the circuit is not correct in the video, at the point where she tries to add the DC and AC so do not try to 'explain' what is said. I always suggest that people should challenge their teacher when stuff like this happens. (I know it's not always possible to access teachers but it is quite possible to challenge in a polite way.)
     
  15. Jun 27, 2017 #14
    It is not a compulsion to add to a thread without actually adding something useful to it .

    Even though most of your thoughts are right , I do not find them relevant in this thread .

    I request you to not spoil this thread further by trying to justify your post .
     
  16. Jun 27, 2017 #15
    I agree :smile: .But somehow or the other I need to understand things .
     
  17. Jun 27, 2017 #16

    sophiecentaur

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    But you need an effective way. There are dozens of sources about transistor circuits. You need to do a search and find a tutorial that suits your level. Using the Internet effectively is a skll that everyone needs.
     
  18. Jun 27, 2017 #17
    I'm sorry if what I wrote seemed offensive; it was not meant personally; your experience is a common one on forums. @sophiecentaur is giving you essentially the same advice I gave.
     
    Last edited: Jun 27, 2017
  19. Jun 27, 2017 #18
    You still haven't got my point . The question is not about transistors :rolleyes: .

    The questions I have asked are not be found in tutorials/books.This is why I put it here at PF . I have been a member for sometime and I understand what is to be put here and what not .

    Anyways no point in dragging it further .You are a terrific member with a lot of credibility :smile: .May be I couldn't put across my doubt properly or maybe this question was not suitable for this sub-forum .
     
  20. Jun 27, 2017 #19

    sophiecentaur

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    If your basic question is not about transistors then perhaps you could restate it in another context? You are quoting from a flawed video, which is not best for understanding.
     
  21. Jun 27, 2017 #20

    Buckethead

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    The instructor in the video is misleading you into thinking that you can just force an AC signal onto the battery voltage and this is not correct. Here is the correct way to understand what she is trying to do:

    Ignore the AC signal for the moment and realize that the battery (a DC voltage) is being used in order to "turn on" the transistor. The transistor will not work until it is turned on. To turn on a transistor you must apply about 700 mv between the base and emitter and this is where the battery comes in. This is called "biasing" the transistor. Also it would make things clearer if you imagine the battery to be a 700mv voltage source. This will be just the right amount to turn on the transistor and make the base to emitter junction conduct current.

    What she is trying to explain in the video is that if you change the voltage of the battery by adjusting it somehow so that it varies from just slightly above 700mv to slightly below 700mv, then this will cause the transistor to turn on a little more and a little less. When this happens the output will also conduct and it will be inverted as well.

    That is all she is trying to explain. Where she made the mistake is in mis-leading you into thinking that you can just force an AC signal across the battery and in doing so force the battery to change its voltage as I explained above. In the real world you cannot do this without damage to the battery or to the signal generator. So like I said, just make believe the batter is somehow changing its voltage from about 650mv to 750mv in a sign wave sort of way, and you will see this "signal" on the output.
     
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