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Question regarding the output of a half wave rectifier

Discussion in 'General Electronics Discussion' started by saadishere, Aug 18, 2012.

  1. saadishere

    saadishere

    6
    0
    Aug 18, 2012
    hello ,

    could you please first have a look at my screenshot.
    I do not really understand how the output of the resistor (A in the oscilloscope) has been displayed, when the input voltage across the ac power is positive , D1 is forwared biased , but D2 is then reverse biased as the anode is 0 and the cathode is positive ? right ? so no current can flow accross D2 , D2 is an open circuit.
    when the input voltage is negative , D2 is forwared biased ,and D1 is reverse biased ,D1 is an open circuit.
    how can we have this display on the oscilloscope for A , many thanks.
     

    Attached Files:

    Last edited: Aug 18, 2012
  2. Electrobrains

    Electrobrains

    259
    5
    Jan 2, 2012
    You have no (or almost no) current flowing through R1. The only current flowing, is through your measuring probe. It's very high impedance, usually 1M or 10M Ohm. Compared to that, your 1k Ohm will act as a short circuit, keeping the two cathodes on the same potential.

    Thus the resulting waveform is positive when D1 is conducting and 0V when your supply changes polarity.

    In fact, D2 is hardly active at all in the circuit. Once it is reversed polarized and will block the current and in the next moment it's just (through the "short circuit" R1) connecting to the same potential as the ground of your oscilloscope.

    It looks as if you have a small negative signal in the latter case. That's probably because D1 has a (negative) leakage current, that will flow through your high impedance probe (and a little through D2).
    Because D2 is not ideal, it will allow for that negative voltage (theoretically up to approx. 0.6V) to hang around on the probe side.
     
  3. davenn

    davenn Moderator

    13,244
    1,744
    Sep 5, 2009
    Hi saadishere
    welcome to the forums :)

    for a fullwave rectifier using 2 diodes that circuit is incorrect
    the AC would/should be coming from a centretapped source --- normally a transformer. the centretap is the 0V rail and the other 2 sides have the diodes in them and they will rectify every half cycle.
    There only 2 ways to reftify a single ended AC source as you have shown, 1 is to use a single diode ( say your top one) and have a half wave rectifier, or use a bridge rectifier ( 4 diodes) for a fullwave rectifier.

    Then when used in either of those 2 ways the 1k resistor will be the load across the DC output of the diodes.
    have a look at these 3 circuits.....

    [​IMG]
    Fullwave bridge rectifier


    [​IMG]
    Fullwave 2 diode rectifier


    [​IMG]
    Halfwave rectifier

    for the first 2 rectifiers the waveform seen on the oscilloscope will be....
    [​IMG]

    for the halfwave rectifier it will be....
    [​IMG]

    Now in each example, the oscilloscope should be across the resistor load so you can view the waveform correctly


    cheers
    Dave
     

    Attached Files:

    Last edited: Aug 19, 2012
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,174
    2,689
    Jan 21, 2010
    Actually there's a third way...

    But it's generally a poor option.

    You can use 2 capacitors to divide the AC signal and create a "centre tap" You then do what you would normally do to full wave rectify. Redrawing it will make it look like the bridge rectifier but with 2 diodes replaced by capacitors.

    I would consider this of technical interest only.
     
  5. davenn

    davenn Moderator

    13,244
    1,744
    Sep 5, 2009
    hey steve

    havent seen that way done .... I assume, maybe incorrectly, non polarised caps ?

    D
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,174
    2,689
    Jan 21, 2010
    The only time I've seen it was with valve rectifiers. It can obviously be a little tricky to have all your directly heated cathodes in parallel when some are at ground potential and others at V+.

    This was a solution, and was made more viable by the fact that currents were generally small and the voltage drop across a diode (tube) rectifier was often rather huge.

    I believe you can get away with polarised caps, but you're stretching my memory. It certainly seems at first sight that you could not.
     
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