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Zener diode assumption

  1. Jun 20, 2012 #1

    Femme_physics

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    General basic electronics questions

    I have 3 questions

    QUESTION 1
    --------------

    http://img35.imageshack.us/img35/4011/zener.jpg [Broken]

    No matter what value of R I have, as long as the source voltage (16V in our case) bigger than the zener's diode voltage specification (6V in our case), the zener diode will conduct in this case (despite being backwards biased). Is that a correct assumption to make?



    QUESTION 2
    --------------

    I'm asked if this connection is CE (Common Emitter)

    http://img543.imageshack.us/img543/4290/50887855.jpg [Broken]

    I would say of course not, because here the arrow, which represents a diode, is pointing from the collector to the base. In common emitter, the emitter is common to both the base and the collector.

    Such as:

    http://img23.imageshack.us/img23/816/ce2f.jpg [Broken]


    QUESTION 3
    --------------

    Given a sinosoidal signal V(t) = 2.2 sin 314t. This signal rides on a direct current whose value is 12V. You may say that the average voltage of the signal is 12√2 V

    ANSWER
    ----------

    This one is a bit confusing to me. I wanna say "NO, for a full rectified wave, the average value would be 2U(max)/pi"
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 20, 2012 #2

    NascentOxygen

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    Q1. You are correct.

    Q2. Your justification seems confused. The emitter is a fixed terminal of the transistor device itself. The location where a circuit requires that terminal of the transistor is shown on schematics as the terminal sporting the arrow in the symbol. The emitter of a transistor is always its emitter, even when you have it wrongly soldered into a circuit you are constructing. :smile:

    Q3). Try sketching the composite waveform, and graphically demonstrate its average. (Rectified? You didn't say that the question uses the word "rectified". Better be clear about that minor detail. :smile: )
     
    Last edited: Jun 20, 2012
  4. Jun 20, 2012 #3

    Femme_physics

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    :)

    Yes, that's what I thought. :) Hmm, perhaps I didn't explain myself right. Basically, in terms of schematics, Common Emitter depends on the arrow mark on the drawing. If the arrow is pointing outside from the emitter, it's CE, any different location of the arrow would mean it's not CE.


    By the way-- does the arrow signify a diode? I remember my teacher saying that, although lately I lost some of the confidence in him.


    To be honest this question is a bit confusing. It's supposedly a "sinosoidal wave", but it "rides" on a direct current!? What does that mean? Which signal am I supposed to draw? My best guess is that they mean a half-rectified waveform

    http://img23.imageshack.us/img23/6868/waveforma.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  5. Jun 20, 2012 #4
    Is common emitter configuration dependent on the direction of the arrow or is it common emitter only when emitter junction is common to both the input and output terminals ? can't the direction of arrow in emitter terminal depend on p-n-p or n-p-n type ?
     
  6. Jun 20, 2012 #5
    By 3 rd question diagram do you mean the waveform shown in the attachment below ?
     

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  7. Jun 20, 2012 #6

    NascentOxygen

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    No, nothing of the sort. :frown: You had better review your notes on common emitter configuration.

    If the arrow points in towards the base, that means its the emitter of a PNP transistor. If the emitter arrow points out, it's the emitter of an NPN transistor. The arrow always denotes the emitter.

    Were you to use just 2 of the transistor's 3 terminals, then, yes, you will find the behaviour between the base and emitter terminals to be as a diode. (So it should be, since it's a PN junction.)

    If he was reading this, he might reassess that 100% mark you recently scored. :uhh:

    You draw a sinewave symmetrical about an axis, as usual. But instead of that axis being 0 volts, you label it 12 volts and redraw your 0 volt line further down the page. :smile:
     
  8. Jun 20, 2012 #7

    Femme_physics

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    I am aware of it, but we're not talking about PNP or NPN, we're talking about what is a common emitter, and in this case the emitter must point outside. So common emitter only transpires in NPN...no?


    But we always use 3 when it's conducting, one's the base, the other "collects" the current and the other "emits" the current...

    Hehe ;)

    Hmm...how do I know how further down to draw that zero?
     
  9. Jun 20, 2012 #8

    I like Serena

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    Hey Fp! :smile:


    From wikipedia (bolding mine):

    "In electronics, a common-emitter amplifier is one of three basic single-stage bipolar-junction-transistor (BJT) amplifier topologies, typically used as a voltage amplifier. In this circuit the base terminal of the transistor serves as the input, the collector is the output, and the emitter is common to both (for example, it may be tied to ground reference or a power supply rail), hence its name. "



    Is the emitter (the one with the arrow thingy in it) connected to ground or power?



    As you saw in your last thread about VBE=±0.7 V, the arrow thingy behaves like a diode.

    If you have a voltage drop in the direction of the arrow, the transistor is conducting.
    Furthermore that voltage drop is 0.7 V.
    The extra side effect is the the connection between collector and emitter is also conducting.

    If you have a voltage drop opposite to the direction of the arrow, the transistor is blocking.
    Just like a diode. ;)



    Can you draw a graph of a "direct current whose value is 12V"?
    (Note that this is not a sinusoid.)

    And can you also draw a graph of a "sinosoidal signal V(t) = 2.2 sin 314t".
    What is its amplitude?
    (You're allowed to add little drawings of flowers and such. :wink:)

    Can you finally draw a graph that sums both graphs?
    What is its average?
     
    Last edited: Jun 20, 2012
  10. Jun 20, 2012 #9
    Femme_physics,

    No, it is not. The resistor can be so high that it starves out the diode. That means it reduces the current to its reverse saturation value so it does not enter the reverse breakdown region, and operate as a zener anymore. The zener must have a minimum amount of current to operate as a zener. How much depends on the zener.

    The inputs and output have to be designated before the question can be answered.

    What is the average of a constant value current/voltage? What is the average of a sinusoidal without any offset? A rectified sine is not sinusoidal anymore. Averages are additive.

    Ratch
     
    Last edited: Jun 20, 2012
  11. Jun 20, 2012 #10
    True :)
     
  12. Jun 20, 2012 #11

    NascentOxygen

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    We shouldn't be, but you started it! :tongue2: Your reference to the direction of the arrow somehow being relevant here, when it has no relevance, is what led to this diversion.
    Maybe your teacher used a phrase something like "to the outside" and you thought he was talking about the arrow's direction? But he wasn't talking about the arrow's direction; that has no bearing on Common Emitter, or Common Collector, or any configuration. Better refer to your class notes or textbook for the distinguishing feature of a Common Emitter configuration.
    No. Common Emitter is as valid for PNP as NPN.
    True. But suppose you needed a small diode, and none were available, you could employ a transistor but leave one of its leads (say, the collector) unconnected, and you'd have yourself a PN junction, AKA a diode. :smile:

    When relating the BE junction to it's 0.6V drop, it might be better to refer to it as a PN junction (which it is and always will be) rather than a diode (a very strange one when it's involved with transistor action).
    If your scale is 1 cm = 1 volt, about 12 cm down. :wink:
     
    Last edited: Jun 20, 2012
  13. Jun 20, 2012 #12
    2) is a common collector amp because the collector is common to input and output.
     
  14. Jun 21, 2012 #13

    Femme_physics

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    In question 2 in the original question drawing? Apparently not.

    I imagine the emitter is regardless the lowest one (without the arrow in this case), and the collector is always the upper one (with the arrow in this case).

    But it's not really a diode, is it? It's an electrical construct that acts like a diode, I think.

    One step at a time :) Is this alright? ->

    http://img543.imageshack.us/img543/272/avgm.jpg [Broken]

    (Ahh...forgot flower!)

    About how high, and how do I see it in calculations?

    That's a question directly from a test, they weren't designated and I can only answer "true or false" on whether it is CE or not CE.



    Because how can an emitter on a PNP be common to both? In PNP, it emits current towards the collector.


    I don't have class notes about it and we technically finished the material, which I think it complete bollocks since we didn't really go over some important stuff like this, that appeared for instance in last year's test (we're doing an external test not composed by our teacher).


    Well, if you say so... I thought that PNP looks more like common collector since ->

    http://img152.imageshack.us/img152/9350/pnpnpn.jpg [Broken]

    I hadn't thought of that. Do people do it a lot, or is it purely for theory?
    Seems like using a diode would be cheaper.

    Alright, I wanna see if my idea for a graph is correct so far so I'll wait with it :)
     
    Last edited by a moderator: May 6, 2017
  15. Jun 21, 2012 #14
    Last edited by a moderator: May 6, 2017
  16. Jun 21, 2012 #15
    And the emitter is always the one with the arrow regardless of its position!
     
  17. Jun 21, 2012 #16
    Transistor is like two "diodes" working in opposite direction and ya when emitter to base junction is reverse biased considering collector to base junction reverse biased it acts in "offset mode" !

    But if collector- base juntion is forward biased and emitter base junction is reverse biased it won't work in offset mode and there will be some of the current flowing !

    so both the biasings have to be considered !
     
  18. Jun 21, 2012 #17

    Femme_physics

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    Last edited by a moderator: May 6, 2017
  19. Jun 21, 2012 #18
    arrow is always located in the emitter and points towards the direction of conventional current !
     
  20. Jun 21, 2012 #19
    In your question there was never a "rectifier" word used i guess ? it was only written that ac signal rides over dc signal ?
     

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  21. Jun 21, 2012 #20

    I like Serena

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    Yep. In the original question drawing, the emitter is not connected as a common emitter.


    Well, the arrow thingy always identifies the emitter, but I think you already got that by now.


    It's really a diode.
    Actually, physically it's 2 diodes lumped together into 1 component.


    Not it's not alright - there's no flower in it!

    But yes, this is actually the last drawing I asked for.
    The only thing wrong with it is what you marked as average.
    Did you mix up average with RMS?

    Oh, and I see 2 sinusoids in there.
    What's the purpose of the 2nd one?


    Yeah, a diode is cheaper, but not by much. ;)
    They are both very cheap.


    Well, now you know it needs flowers!
     
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