Is this argue true?Pleased help

Discussion in 'Homework Help' started by anhnhamoi, May 6, 2012.

  1. anhnhamoi

    Thread Starter New Member

    Apr 24, 2012
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    Hello,
    I am trying to learn Fielter but there is a problem with me.Could you explain me the a sentence in my text book.
    This is the sentence:
    "As illustrated in view B, the diode cannot conduct on the
    negative half cycle because the anode of D1 is negative with respect to the cathode."
    view B is attached in bellow.
    In this diagram I think that during negative half cycle, as far as I see, [pleased correct me if it is a silly thinking], I ONLY know the voltage between diode and capacitor is negative.But in above argue, it is said that the voltage of anode is negative with respect to the cathode.Could you explain me this?
    Thanks.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I don't see any diagram.
     
  3. anhnhamoi

    Thread Starter New Member

    Apr 24, 2012
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    Oh, I am sorry, i forgot.I have just attached.Thanks
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    When secondary voltage are positive respect to gnd the diode start to conduct as soon as the secondary voltage reach Vc + 0.7V.
    But for the negative half the the secondary voltage is now negative with respect to GND. So diode cannot conduct. Because we have a positive voltage across the capacitors. So diode in OFF because anode has negative voltage. And the voltage between diode is equal to
    Vab = Va - Vb = -5.7V - 5V = -10.7V

    [​IMG]



    [​IMG]
     
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  5. anhnhamoi

    Thread Starter New Member

    Apr 24, 2012
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    Thanks you very much for your help.It really very precious to me.
    Can I ask you some questions:
    Is my argue wrong?

    At first, in the first diagram in the positive half , call the resistance of diode is Rd
    and the capacitive reactance of capacitor is Xc=1/(jωC) then
    Vcap: the voltage drops in capacitor
    Vd: the voltage drops in diode
    Vcap = Xc/( Rd+Xc) x Vcc ≈ 0 because Xc<< Rd when starting power
    Vd= Rd/( Rd+Xc) x Vcc ≈ Vcc because Rd very large ,Rd-->∞
    the Vd is increased gradually according to Vcc and when it go to 0.7V the diode starting to conduct and Rd= 0Ω then the voltage is dropped in capacitor will be Vcc- 0.7V and the capacitor is starting to be charged.

    If the above argue is right then in the second diagram in the negative half the diode is never conduct and the circuit is open.Therefore the capacitor is also never charged.But in your expaination you give it is 5V.

    And Can I ask you a personal question?
    If possible Could you give me the name of software you used to repair picture above?It seem very good and I very like it.Thanks.
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well forget about Xc.
    For the positive half the secondary voltage must by greater then capacitor voltage + forward voltage drop.
    At the beginning when we have a empty capacitor ( 0V across the cap).
    Capacitor starts to charge when secondary voltage reach value > 0.7V.
    And the empty capacitor act just like a short circuit.
    So at the beginning the large amount of current flow and quickly charge the capacitor. When the secondary voltage reach his peak value capacitor is full charged and diode current stop to flow. Now secondary voltage start to drop so the capacitor start to discharge through the load. But diode is still turned off.
    And for the negative haft nothing had changed. The capacitor is still discharge through the load and the diode is OFF.
    But not at the beginning of a positive half the secondary voltage must reach at least V_capacitor + V_diode to be able to recharge the capacitor.
    See the picture

    I use Paint
     
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  7. anhnhamoi

    Thread Starter New Member

    Apr 24, 2012
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    This is really what I need.I thought that the empty capacitor is a open circuit because it has 2 plates seperately.Thanks.
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In resistor the current is proportional to the voltage present across the resistor. I = V/R
    But in the capacitor the current in proportional the rate of voltage change across it (proportional to how quickly the voltage across capacitor is changing). The faster the voltage change (frequency of a AC signal is high) the large the current flow through capacitor.
    I = C * dV/dt
    This means that to sustain current through a capacitor the applied voltage must change. The more rapidly voltage changes the larger the current. On the other hand if voltage is kept constant no current will flow no matter how large the voltage.
    Likewise if the current through a capacitor is found to be zero, this means that the voltage across it mus be constant, not necessarily zero.
     
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  9. crutschow

    Expert

    Mar 14, 2008
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    A capacitor is an open circuit to DC but has a low impedance to a changing voltage or AC.
     
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