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Diode Clipper

  1. Oct 20, 2005 #1
    How do I construct a circuit that will limit the positive half of a 20V peak-to-peak sine wave to 5.6V and the negative half to -2.5V?

    So far, I have a 1 k ohm resistor and the diode is forward biased leading to VL and Vout.
  2. jcsd
  3. Oct 20, 2005 #2


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    Staff: Mentor

    What voltage rails do you have available? Is the resulting waveform supposed to be a scaled version of the input waveform, or just clipped on top and bottom to the voltages you mention? Can it be scaled and clipped, or is it not supposed to be scaled before clipping? What are the input characteristics of whatever you are feeding your output waveform into?
  4. Oct 20, 2005 #3
    For instance, I attempted to draw the diode clipper circuit which gives out a 5V instead. The input graph is suppposed to be a sine wave and the output is supposed to be a cut off at the top of the wave.

    I am not sure if it is supposed to be scaled or not, but I am thinking no.

    Attached Files:

  5. Oct 21, 2005 #4


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    Staff: Mentor

    That looks right for the positive side clipper. BTW, the usual convention is to draw the positive rails at the upper side of the schematic, and the negative rails at the bottom part of the schematic. So in your sketch, draw the positive catch diode pointing up, with its cathode connected to +5V at the top of the page.

    Now you just need the negative clipper. That's why I asked about what voltage rails you can assume that you have. Clamping at 5.6V is pretty straightforward, since that's just about a diode drop above a 5V rail. But clamping at -2.5V is a little weird, since having a -1.9V rail to clamp to would be pretty non-standard. You can make a diode clamp of any voltage drop with a transistor and two resistors, so maybe that's what they are looking for.
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