Diode Output

Discussion in 'General Electronics Chat' started by Computer Engineer, Aug 25, 2009.

  1. Computer Engineer

    Thread Starter New Member

    Aug 25, 2009
    7
    0
    Hi my friends
    I have problem with understanding the output of the following circuit

    [​IMG]

    can you please explain the output for me :)

    Regards
     
  2. Computer Engineer

    Thread Starter New Member

    Aug 25, 2009
    7
    0
    Please Guys , Just a little bit explanation will help a lot
     
  3. rspuzio

    Active Member

    Jan 19, 2009
    77
    0
    You have a 3 V battery. The diode has a potential drop
    of around 0.7 V. 3 - 0.7 = 2.3 Hence, the diode turns
    on at 2.3 V (in your case -2.3 because of the direction in
    which you hooked up the diode and the battery), which is
    just what you see in your plot.
     
  4. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    In power supplies, diodes are used to rectify an AC output. A diode lets current flow one direction but not the other. As an example, look at the picture I have on my oscilloscope. In the first picture, you'll notice just a standard sine wave. Next, I have a diode with the cathode to the negative ternimal. Lastly, I have the cathode to the positive ternimal which chopped off the positive AC. If we were to put a capacitor it would charge from the peak of the cycles and discharge on the 0V. This leaves a steady DC output, although there will still be some ripple.

    If you switched the diode to the opposite polarity the waveform would flip upside down.
     
  5. Computer Engineer

    Thread Starter New Member

    Aug 25, 2009
    7
    0
    Thank you guys for answering me
    I understood the 2.3 volt part :)
    but how the voltage increase to 10 volt , That i dont understand
     
  6. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    The -10 volts you see is due to the fact that during that period, the diode is reverse biased; therefore no current flow (except into the measuring device) and no voltage drop through the current limiting resistor. During the positive half cycle of your source, the diode is forward biased and is allowing current flow into the battery. The remainder of the positive 10 volt half cycle is dropped across the current limit resistor. Hope this helps.
     
  7. Computer Engineer

    Thread Starter New Member

    Aug 25, 2009
    7
    0
    very helpful thank you:D
     
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