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Zener Diode Clipping Graph Help

  1. Sep 12, 2015 #1
    So we started looking at Zener Diodes in our electronics course.
    What im having trouble with is the interpretation of the graph.

    For the example our diode has a Zener Voltage is 6.2V. I understand what happens in the positive half but when the direction changes is where I get confused.

    Why is the pd across the diode 0.7V after clipping? If the polarities switch on the negative cycle of the input signal why would it be -0.7V and not 0.7V???
     

    Attached Files:

  2. jcsd
  3. Sep 12, 2015 #2

    Hesch

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    Gold Member

    Well, you are the one to decide the sign of the measured voltage.

    So if you choose the forward voltage to be +0.7V, the reverse voltage will be -6.2V.

    2000px-Kennlinie_Z-Diode.svg.png
     
    Last edited: Sep 12, 2015
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