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What is the output of this UJT relaxation oscillator circuit?

  1. Jun 30, 2012 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data


    This is a relxation oscillator that's realized by UJT. The graph is included.

    What is the output of the circuit?


    http://img821.imageshack.us/img821/2274/ujtok.jpg [Broken]


    2. Relevant equations


    http://www.allaboutcircuits.com/vol_3/chpt_7/8.html


    3. The attempt at a solution


    Well, to find the output I need VEB or VBB which are not given to me...therefor unless I make the assumption that VEB = 0.7 volts this cannot be solved. And I can just say that

    Vout = VR2
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 30, 2012 #2

    I like Serena

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    Hey Fp!

    Seems to me that without the specification of the UJT you are indeed limited in what you can do.

    What you can do, is give the shape of the output signal.
    You can also give the frequency of the output signal.
    And you can calculate the value of the capacitor.
     
  4. Jun 30, 2012 #3

    Femme_physics

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    Yes I know the value for the capacitor :) But thanks for confirming it's a dead end-- I thought as much!
     
  5. Jul 1, 2012 #4

    NascentOxygen

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    Everything good again with your teacher, FP? Now that the dust has settled. :wink:
    When designing with active devices, you'll never be "given" precise specifications, for the simple reason that the specs vary from one device to the next, from one production batch to the next, and from manufacturer to manufacturer. Generally, the best you can hope for is a range, or, next best, a typical value. With this, the designer does the best she can. As for VBB, you are as good as given it; if you know it to an accuracy of 10% then consider it is known.
    You feel that's an unreasonable assumption to make? :uhh:

    It's not clear whether the question sheet referred to the allaboutcircuits site, and its figures for the 2N2647, or whether you added that on your own initiative, but I expect those figures should be fine for the calculations here. Note that the amplitude and width of pulses for triggering may not be of major importance, as the signal is either certain to be more than adequate for the need, or else it will be fed to a buffer or attenuator, so your not being able to specify an output level to an accuracy of 3 decimal places is not the calamity you may think. :cool:
    10/10 :rofl: :rofl:
     
  6. Jul 2, 2012 #5

    Femme_physics

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    Thanks Nascent, I can write an explanation about the assumption I made at worst case scenario.

    Pretty much, I just think he needs more experience teaching. He's still a young guy.


    While we're on the issue of UJT and seeing it's such a short thread, I thought to throw in another question...


    http://img403.imageshack.us/img403/5214/question99.jpg [Broken]

    Given:
    Ic = Ie
    Vp = η x VBB + 0.63
    Vv = 1.2 volts
    VEB = 0.65
    η = 0.4

    Calculate the frequency of the circuit pulses. Neglect the discharge time of the capacitor.



    ANSWER:

    I wasn't sure how to answer it, so I picked at a notebook I acquired of someone who already took my electronics course and this is what he wrote:

    http://img441.imageshack.us/img441/5263/answer99.jpg [Broken]

    Where is the formula for t0 and t1 taken from? And what does it mean?
     
    Last edited by a moderator: May 6, 2017
  7. Jul 2, 2012 #6

    NascentOxygen

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    The formulae will fall into place once you have figured out how it works. :smile:

    First, you'd better identify the <oh, no, not this again> EMITTER of that BJT. :tongue:

    Next question to ponder, what is that ZENER doing?

    You'll soon fix that; you're aging him fast. :smile:
     
    Last edited: Jul 2, 2012
  8. Jul 2, 2012 #7

    Femme_physics

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    *sets hair on fire and jumps from Everest!*


    Well... The zener defines the voltage at the cross-section above the zener to the ground. *cleans snow from hair*... The emitters are the arrows. There is one emitter for the NPN transistor and one emitter for the UJT
     
  9. Jul 2, 2012 #8

    NascentOxygen

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    You have a good understanding of the UJT oscillator, so move your focus to the BJT in this modified oscillator.

    Providing the zener is adequately fed, it will maintain the base of the NPN at a constant voltage. What would follow on, as a consequence of this, to that NPN's voltages and currents?
     
  10. Jul 2, 2012 #9

    Femme_physics

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    Before I answer this, let's clear the details....

    This is not an NPN, this is a PNP!

    And this is not a BJT, it's a UJT!

    Or...am I blind?
     
  11. Jul 2, 2012 #10

    NascentOxygen

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    You're right. :smile:

    PNP and NPN transistors are examples of Bipolar Junction Transistors.
    This circuit is a UJT relaxation oscillator, but it is augmented by a PNP BJT.
     
  12. Jul 2, 2012 #11

    Femme_physics

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    I'm not sure what the last part means...if there's no BJT defined in the problem, then there is no BJT. How can this logic be flawed?
    Bet let me try answering your original question

    Well, it will affect the amount of current passing through the circuit but this has to be calculated to figure out
     
  13. Jul 2, 2012 #12

    NascentOxygen

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    There is sufficient information on the schematic to enable you to calculate the PNP's emitter current and voltage.

    You can see that a resistor of the original UJT oscillator (that you examined in an earlier thread), has been replaced by a PNP transistor in the circuit you are examining here.

    A PNP transistor is also known as a BJT. (The same goes for an NPN, it also is a BJT.)
     
  14. Jul 2, 2012 #13

    Femme_physics

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    Can't they just call things one name? :( Fine.

    I did find the current IE all by myself :)

    http://img600.imageshack.us/img600/4098/ieic.jpg [Broken]

    And Ie = Ic according to the problem definition


    The problem is that they're asking for the " frequency of the circuit pulses"... I'm unsure how to relate it all, and that new formula is really confusing
     
    Last edited by a moderator: May 6, 2017
  15. Jul 2, 2012 #14

    NascentOxygen

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    I'll accept your word that IE is 3.2mA. Your approach looks right.

    As you know, IE is approx equal to IC.

    So, where does the path for IC lead to, after it emerges from the PNP transistor?
     
  16. Jul 2, 2012 #15

    NascentOxygen

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    P.S. I shall have to depart shortly (in 10 mins). Back in 9 or 10 hrs.
     
  17. Jul 2, 2012 #16

    Femme_physics

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    well, to the UJT or the Capacitor, depends on the time.

    ahh...drat ;) Well, we'll rehash this tomorrow then if you will...thanks!
     
  18. Jul 2, 2012 #17

    NascentOxygen

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    Just as with the basic UJT oscillator, for most of the time that source of current is doing nothing but charging up the capacitor. So, the only change we have essentially made here is to replace the resistor in the basic RC timing network with a constant current source. How is that going to affect the shape of the waveform at the EMITTER of the UJT?

    The importance of encountering alternative technical terms was brought home to me as a student. We had an exam, and I was pleased with it as we had thoroughly covered the work just a few days before. But some of the other students were not happy with it. One of the questions asked us to "explain the need to bias a Bipolar Junction Transistor", and my classmates claimed we hadn't been taught that topic. I realized they were right, we had thoroughly explored biasing NPN transistors, but the lecturer may not have even once referred to them as BJTs. Though of course our textbooks did. :wink:
     
  19. Jul 3, 2012 #18

    I like Serena

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    Hi Fp.

    Do you have formulas for a capacitor being charged?

    You should have something like:
    Q = C V
    where Q is the charge on the capacitor, C is the capacity, and V is the voltage across the capacitor.

    And you should also have something like:
    ΔQ = I Δt
    where ΔQ is the change of charge on the capacitor, I is the constant current, and Δt is the time interval that the capacitor is charged.

    In your case the capacitor is charged from the Vv voltage to the Vp voltage.
     
  20. Jul 3, 2012 #19

    I like Serena

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    Btw, I liked how you found Ie. Good! :smile:
     
  21. Jul 3, 2012 #20

    Femme_physics

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    Last edited by a moderator: May 6, 2017
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