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A problem with diode limitters

  1. Nov 22, 2013 #1
    hello guys

    while studing the diode limitters

    i found two examples i solved the first one but i couldn't with the second
    1 -

    ohwNJGZ.png

    2-

    ih6plrd.png

    in the first i know that the diode while be forward biased in the negative half of the signal
    so the negative half will be limitted to -0.7 v
    and the V peak output will be = [(RL/(RL+R1)]*Vinput = 9.09v

    but i don't know what will be the difference between the first and the second problem !!

    so please help me guys ...Thanks :)
     
  2. jcsd
  3. Nov 22, 2013 #2

    berkeman

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    Staff: Mentor

    In the second one, for + input signal, will the diode be forward or reverse biased? What does that mean about current through the resistor?

    For - input signal, ask the same questions...
     
  4. Nov 22, 2013 #3
    okay in the + input the diode will be reverse biased so the current will not pass , but in the case of - input the diode will be forward biased and the current will pass

    am i right ??
     
  5. Nov 22, 2013 #4

    berkeman

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    Staff: Mentor

    Correct. And V=IR, so you know when you will have an output voltage. Good job! :smile:
     
  6. Nov 22, 2013 #5
    okay i know that , but i still don't know how will i calculate Vout

    i think Vout will be = -10+0.7= -9.3v .. is that right ??
     
  7. Nov 22, 2013 #6

    berkeman

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    Staff: Mentor

    That would be the negative peak voltage. The output waveform will follow the input waveform (minus the diode drop) when the diode is conducting.
     
  8. Nov 22, 2013 #7
    you mean that the output waveform will equal zero in the positive half and in the negative half it will equal -10v-0.7v ?? or what ??
     
  9. Nov 22, 2013 #8

    berkeman

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    Staff: Mentor

    In the negative half cycle, the output voltage equals the input voltage, minus the diode drop. Sketch that on top of the input waveform plot...
     
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