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Junction diodes as clipper

  1. Aug 8, 2014 #1
    In the given attachment there is a circuit given..and we have to find the output voltage waveform..

    I have attached the options also..because according to me the " when input coltage is less than 5 V , it acts as open , most of the input voltage appears in output.. while when input voltage is greater than 5 V, it conducts heavily & output voltage is +5V & it stays +5V as long as input voltage is greater than +5..so voltage is constant till the negative cycle..when diode is open & entire negative cycle appears across load.." so by this no option seems correct..ans the asn given is option (a).. but how..??
    where my explanation is wrong..?? Please help me out..
     

    Attached Files:

  2. jcsd
  3. Aug 8, 2014 #2
    this the option which was correct in this ques..if its correct.. then please explain how it is..??
     

    Attached Files:

  4. Aug 8, 2014 #3

    gneill

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    Your attached image has cropped off the value of the voltage source. It's important!

    Have you considered the effect that the load resistance has on the amount of the source voltage that appears across the clipper? Hint: There's a voltage divider involved.
     
  5. Aug 8, 2014 #4
    The voltage source has a value of 10sinωt...
     
  6. Aug 8, 2014 #5
    I don't think I am familiar with this thing..I mean what effect you are talking about??
     
  7. Aug 8, 2014 #6

    gneill

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    If your current subject of study is semiconductors such as diodes, then you should have already covered voltage dividers. Can you spot the voltage divider in the given circuit?
     
  8. Aug 9, 2014 #7
    the junction joining the 2 resistors and the diode must be the voltage divider...!!
     
  9. Aug 9, 2014 #8
    wait a second.. for voltage divider there is need to have to resistors in series..!! but here 2 resistors are'nt inseries..!! then how voltage divider is here..??
     
  10. Aug 9, 2014 #9

    gneill

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    When the diode is not conducting then it's branch effectively does not exist... so how do the resistors look if the diode branch is removed?
     
  11. Aug 9, 2014 #10
    but there's also a battery of 5 V..in series with the diode..we can't neglect it..!!!
     
  12. Aug 9, 2014 #11

    gneill

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    As long as the diode is not conducting you can neglect it. The battery will have no effect if it "sees" an open circuit due to the diode being reverse biased. It can't push current through an open circuit.

    Start by sketching the output ignoring the diode and battery entirely. Then go back and apply the clipping criterion to that waveform.
     
  13. Aug 9, 2014 #12
    thats fine..and what about for the positive half cycle..is it right that when input voltage is greater than 5 V, it conducts heavily & output voltage is +5V & it stays +5V as long as input voltage is greater than +5..so voltage is constant till the negative cycle..???
     
  14. Aug 9, 2014 #13

    gneill

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    Almost. The voltage stays at +5 V until the signal drops below +5 V, turning the diode off again. It doesn't have to 'wait' for the negative half cycle of the input.
     
  15. Aug 9, 2014 #14
    so this clearly means that the output waveform is sinusoidal with 5V as the peak voltage..??
    but in the given circuit I am attaching there's a difference..instead of the fact its almost same circuit as I asked in this discussion..
     

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  16. Aug 9, 2014 #15

    gneill

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    Yes, for the case presented in your first post that is true because the input source is a 10V sinewave that's diminished to 5 V by the action of the voltage divider. Since the clipper only "cuts in" at 5 V, it never gets to actually clip the signal.

    Yes, well this new circuit example makes the assumption that the signal presented to clipper exceeds the clipping potential V. Thus it "clips" the signal at V.
     
  17. Aug 10, 2014 #16
    But what is the difference between the 2 circuits... in first it is not waiting for the negative cycle n drops just after voltage is less than 5 V but in the 2nd case...it remains constant until the negative cycle... why is it so..??
     
  18. Aug 10, 2014 #17

    gneill

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    It looks like the same circuit layout to me! And the waveform you posted with it shows a flattened (clipped) peak during the positive half-cycle of the waveform, not a constant value until the negative half-cycle begins (when the waveform passes through 0V on its way negative).
     
  19. Aug 10, 2014 #18
    but why such clipping action is not there in the previous case..if the circuit is all same??
     
  20. Aug 10, 2014 #19

    gneill

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    The clipping action can only occur if the signal presented to the clipper exceeds the clipping level. In your second example it is assumed that the clipping voltage is less than the maximum signal magnitude (no specific values were given).
     
  21. Aug 11, 2014 #20
    is clipping level the knee voltage of a diode..???
     
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