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Clipper circuit in series and its output

  1. Jun 6, 2012 #1
    Hello all,

    I've been having problems regarding this attached clipper circuit and its output; i don't understand how the first part remains zero in the output wave form, isn't the dc source -2V here?, if it is then in the 1st positive half cycle AC would turn the diode on but not DC since according to DC its reverse biased. So since ac voltage is more than the negative dc voltage the diode turns on and (it acts as a short circuit-yes we are considering an ideal diode here) so shouldn't the output across R show something? why would it be zero? Honestly speaking, I really don't understand the way teachers explain this concept to us...it just seems more complicated....anyways, I would really appreciate it if someone could explain this clearly to me...

    Thanks in advance :)
     

    Attached Files:

  2. jcsd
  3. Jun 6, 2012 #2
    Imagine that instead off AC voltage you connect variable DC voltage source. And we set this voltage to 0V. So the diode is reverse bias, and no current will flow in the circuit.
    If no current is flow through resister there is no voltage drop across resistor.
    So the output voltage is 0V.
    Now if we increase input voltage to 2V. Our input DC voltage cancel voltage from DC source in series with the diode.
    Voltage at diode anode is equal to
    Va = Vin - 2V = 2V - 2V = 0V
    But if we further increase the input voltage and voltage at anode reach 0.6V the diode start to conduct the current in direction.
    +Vin ----> DCsoure_ Diode --> resistor---> -Vin
     

    Attached Files:

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