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Homework Help: Transistor : Output voltage of the amplifier

  1. Dec 27, 2017 #1
    1. The problem statement, all variables and given/known data
    Example2.jpg

    2. Relevant equations


    3. The attempt at a solution

    This is a solved example given in my Physics textbook . I am not understanding a couple of things .

    1) How is Iin calculated as Vin/Rin ?

    Why is the book not taking potential difference between Base Emitter into consideration ?

    Instead shouldn't Kirchoff's Voltage rule be applied in the Base Emitter loop ?

    I think it should be Iin =( 1mV -VBE )/2KΩ

    2) Shouldn't the diagram be showing Vout across 100kΩ ?

    I don't understand why book considers Vout to be VCE ?

    Please help me understand these issues ?
     
  2. jcsd
  3. Dec 27, 2017 #2

    Delta²

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    ##V_{BE}## is equal to input voltage which probably is small, much smaller than 1mV, so ##1mV-V_{BE}\approx1mV##.
     
  4. Dec 27, 2017 #3
    I don't think this is the case .

    I think in a silicon PN junction potential difference between Base and Emitter is about 0.7 V .

    This is quite large as compared to the bias battery .
     
  5. Dec 27, 2017 #4

    Delta²

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    You might be right , I am not an EE and don't know much about transistors, just said to share my thought on this.
     
  6. Dec 27, 2017 #5
  7. Dec 27, 2017 #6

    Charles Link

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    To me, it looks like a problem that is poorly presented. e.g., to forward bias ## V_{BE} ## , you normally need ## V_{BE}=.7 ## volts. They also don't specify the collector bias voltage. Perhaps someone else has a solution, but my recommendation is to not spend much time on this one as is. This "common emitter" electronic configuration is a good one to know, but this problem looks to me to be a poor one to use to try to learn it.
     
    Last edited: Dec 27, 2017
  8. Dec 27, 2017 #7
    Thanks .

    Suppose instead of 1mV battery there is a 1 V battery , then do you think what I did in the original post Iin=( 1V -VBE)/2KΩ is correct ?
     
  9. Dec 27, 2017 #8

    Charles Link

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    I looked at the problem in the "link" you had posted. For that one, with the book's example, a ## V_{BE} ## of 18.5 Volts I think is totally inaccurate. I expect if you hooked it up in the lab, you might find that ## I_B ## would be larger than what is computed using the value ## \beta=200 ##, and you would find ## V_{BE}=.7 ## volts. And then I think you might have removed the "link": https://www.physicsforums.com/threads/find-the-state-of-transistor.935500/#post-5909999 ## \\ ## Edit: And on this one. a base voltage of 1 volt might work, but they still need to specify the collector voltage. Also ## V_{in} ## (ac) usually gets coupled in with a capacitor, and ## V_{out} ## (ac) is usually also coupled with a capacitor.
     
  10. Dec 27, 2017 #9
    Thank you .

    Do you think post 7 is correct ?
     
  11. Dec 27, 2017 #10

    Charles Link

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    That part works, but please see my edited additions to post 8.
     
  12. Dec 27, 2017 #11

    cnh1995

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    I don't think this problem is practical and as Charles says, not much helpful in learning CE configuration.
    The battery needs to be in series with the input ac source (theoretically). In this circuit, as per circuit theory, the battery can't dc-bias the transistor as the ac source is directly connected across the BE junction. Even if the battery were a 10V one, it couldn't dc-bias the transistor in this diagram.
     
  13. Dec 27, 2017 #12

    Charles Link

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    See also my post 8. IMO, this textbook looks like it may contain numerous errors. I encountered that type of thing from time to time in my college days. The best solution would be for the OP @Jahnavi to learn the material from a book that presents it better and with fewer mistakes, if that is at all possible.
     
  14. Dec 27, 2017 #13
    @Charles Link , do you also think this is correct ?
     
  15. Dec 27, 2017 #14

    cnh1995

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    I have seen many junior college physics books in India showing wrong biasing circuit for CE amplifier. @Jahnavi, I believe that includes the NCERT textbook as well.
    I agree that you should study amplifiers from some standard electronis books (or COP by H C Verma).
     
  16. Dec 27, 2017 #15

    Charles Link

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    The way they show it, if you use ## V=1 ## volt, it would work, provided the ## V_{in} ## was ac coupled (with a capacitor).
     
  17. Dec 27, 2017 #16

    cnh1995

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    Yes, but I wasn't including a coulping capacitor in my reasoning. If there isn't one, the battery is ineffective.
     
  18. Dec 27, 2017 #17
    But then it would be something like an input AC signal placed parallel to a DC battery (1 V)

    What would then be the net effective voltage applied across Base Emitter ?

    This is so confusing :cry: .
     
  19. Dec 27, 2017 #18

    cnh1995

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    That Iin wouldn't be the true Iin as it won't flow through the base. It only flows through the battery. The base current will be determined by the input ac voltage, which will be the same even if you remove the battery.
     
  20. Dec 27, 2017 #19
    I was assuming AC input signal was not present .
     
  21. Dec 27, 2017 #20

    cnh1995

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    The net BE voltage will be the input ac voltage. The battery is redundant as far as dc biasing is concerned.
    Edit: assuming no coupling capacitor.

    Ok, but when you connect the ac source, Iin≠IB.
     
  22. Dec 27, 2017 #21

    Charles Link

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    Usually coupling capacitors are used on the ac signals, so that the bias voltage would not be short-circuited by the ac input connection. Meanwhile, oftentimes the ac signal can be a very small signal, sometimes from sensitive sensors where the signal needs to be amplified. A ## V_{DC}=1 ## volt could destroy some sensors. The ac coupling with a capacitor is quite necessary in these amplifiers.
     
  23. Dec 27, 2017 #22

    gneill

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    I suspect that the given circuit diagram is flawed; Perhaps the original intention was to represent the bias network's essential small-signal equivalent circuit, but something went wrong after the illustrator got hold of it. The 1mV apparently should be associated with the input source, not a base bias battery. Or, that battery originally was meant to be the small signal source, and the "Input voltage" source is a mistaken evolution of what should have been just a text label to a labelled source.

    Note that the way that input source is connected the base bias circuit as shown would have no effect: the base-emitter voltage would be entirely determined by that source. It would need capacitive coupling to isolate it from the bias network and superimpose the signal on the bias.

    That said, the given solution clearly assumes that the 2 kΩ resistor is in series with a 1 mV input signal and the base of the transistor, since they use it to calculate the base current, ##I_{in}## . So I'd consider take the circuit to be a small signal simplification that should probably have looked more like this:
    upload_2017-12-27_10-48-14.png
     
  24. Dec 27, 2017 #23
    Thank you .

    Even if that is the case , wouldn't applying KVL give IB=( 1mV -VBE )/2KΩ ?
     
  25. Dec 27, 2017 #24

    QuantumQuest

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    Let me give my two cents recalling what I learned in Electronics some 30 years ago. For an NPN transistor of Si material the threshold in order to become conductive is ##0.7 V##. Given the diagram, there is only ##1 mV## source and the input signal cannot pass through the junction before the sum of the voltages on the resistor of ##2 K\Omega## plus the ##1 mV## from the source gets over the threshold of the NP junction##^*##. Then the transistor will get conductive and the output will be a bigger signal regarding amplitude - as this is the basic purpose / application of a common emitter circuit i.e voltage amplifier. So, for 1) you have to take the total conductive voltage and divide it by ##2 K\Omega## in order to find ##I_{in}##. For 2) ##V_{out}## is taken across the ##100 k\Omega## resistor - it is the amplified alternating signal. The DC voltage source gives the bias voltage between collector and ground.

    EDIT: ##^*## As pointed out by qneill, a coupling capacitor in series is needed in order to get the sum of voltages of input signal plus ##1 mV## DC from the source, which is absent in the diagram provided. Apparently, there are various things missing / wrong with this diagram and the provided solution in OP.
     
    Last edited: Dec 27, 2017
  26. Dec 27, 2017 #25
    If no coupling capacitor is used just as in the original example , how does AC input signal short circuit the DC battery ?

    Sorry , I am fairly new with AC and transistors .

    I only understand "shorts" in a DC circuit .
     
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