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Clipper circuits

  1. Oct 19, 2014 #1
    In the given circuit. It was given that when Vi>Vr diode will be reverse biased or off..and when Vi<Vr diode will be forward biased... but I am not getting how we infer this..??It may sound a very basic question but its not clear...I am not able to decide it
     

    Attached Files:

  2. jcsd
  3. Oct 19, 2014 #2
    For Vin = 0V the diode is ON or OFF?
     
  4. Oct 19, 2014 #3
    I have no idea about this.. actually I am not getting the use of the battery Vr..!!!
     
  5. Oct 19, 2014 #4
    Battery (Vr) is nothing more than a voltage source. You are not familiar with a voltage source?

    As for the diode. Diode allows current to flow only in one direction (called the diode's forward direction), while blocking current in the opposite direction (the reverse direction).

    111.PNG

    Forward-biased diode (diode is ON)
    Vf.PNG


    Reverse-biased diode (diode is OFF)

    VR.PNG
     
  6. Oct 19, 2014 #5
    IR isn't 0. It is just small.
     
  7. Oct 19, 2014 #6
    Well yes, but I don't want to confuse the OP. And you can read from the first picture the reverse current is typically in nanoamps range
     
  8. Oct 19, 2014 #7
    yes I know Vr is voltage source ,but here it is affecting the bias of diode whether forward or reverse..but how its being decided.
     
  9. Oct 19, 2014 #8
    If we set Vin = 0V, then you can replace Vin with a short wire.
    So the circuit will looks like this:

    5.PNG
    And now can you tell me in which state the diode is, ON or OFF? And Vo is ?
     
    Last edited: Oct 19, 2014
  10. Oct 19, 2014 #9
    It seems to be in reverse bias..!!
     
  11. Oct 19, 2014 #10
    Yes, diode is reverse biased. But what about Vo voltage? Any clues?
     
  12. Oct 19, 2014 #11
    its in rever biase ..so it'll be open , so output V0 will be Vr..>!!
     
  13. Oct 19, 2014 #12
    no..actually considering the polarity of battery , V0 will be -Vr.
     
  14. Oct 20, 2014 #13
    Yes, very good Vo = -vr. So if we assume that Vr = 2V then Vo = -2V

    OK, so now let as try to find Vo if Vi is larger than 0V. For example Vi = 5V

    10.PNG

    The voltage at cathode is 5V and the voltage at anode is -2V. So voltage across the diode is equal to Vd = -2V - 5V = -7V.
    And this means that diode is OFF(Reverse-biased) and Vo = -2V.

    Now let as change the polarity of the Vi voltage source from +5V into -5V.

    5.PNG

    This time diode is forward biased because Vd is positive. Vd = Vo - Vi = -2V - (-5V) = 3V. And this is why the diode will conduct current.
    And Vo = -4.3V.
    So we can conclude that the diode will be ON only if input voltage (Vi) is 0.7V voltage lower than Vr voltage.
    Diode is ON if Vin < -2.7V
     
  15. Oct 20, 2014 #14
    but if we take Vi=1V then diode will be reverse biase even if Vi=-1V.. since Vd= -2-(-1)=-1V..so it will not conduct..!!
     
  16. Oct 20, 2014 #15
    I did'nt get this point..how can you generalise it.>??
     
  17. Oct 20, 2014 #16
    Yes exactly. Diode will conduct only if Vin is lower than -2.7V

    What you don't understand here?
    When diode conduct, the voltage drop across the diode is around 0.5V....0.7V.
    This means that Vin voltage must be 0.7V lower than Vr voltage to force diode to conduct.
    For Vr = 2V input voltage need to be lower than -2.7V
     
    Last edited: Oct 20, 2014
  18. Oct 20, 2014 #17
    ohkk ..so for solving the clipper circuits.. we should proceed in this way only..?? I mean assuming values of unknowns ( Vr & Vin)..then find the bias of diode..??
     
  19. Oct 21, 2014 #18
    No it is not necessary, but this "method" help us visualize the situation much better.
     
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