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Diode Clipper

  1. Sep 14, 2008 #1
    Hey everyone, I need some help with this problem. A triangular wave generator produces a waveform (triangular wave with peaks of 10 V). I need to develop a circuit that clips the voltages at 5 V (ie, they become flat at 5V). We're allowed to use DC voltages of +10 V and −10 V as well as resistors, capacitors (although I don't see why we would need to) and standard signal diodes (so in other words, no zener diodes).

    I was trying to do something similar to what was instructed here (3rd circuit shown): http://www.allaboutcircuits.com/vol_3/chpt_3/6.html

    I might just need a kick in the head or something to get my head around this. My general problem is that I set it up like the one that clips at 2V and -3V - however since I'm limited to 10V DC batteries, I get clipping at 10.7V... I was wondering if it would be possible to perhaps use resistors to dissipate 5.7 V?

    Also, it's been a while since I did this so in the link I gave is the 1 Kohm resistor there just to be a load resistance and make sure the diodes don't burn out? Or does it play some other role?

  2. jcsd
  3. Sep 14, 2008 #2


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    Can't you use more than one diode to get the clamp voltage you are looking for?
  4. Sep 16, 2008 #3
    I suppose, but that wouldn't give me exactly 5 V... would it? Are you just suggesting I stack the diodes?
  5. Sep 16, 2008 #4
    I was thinking, since I'm constrained to use 10 V batteries what if I made some voltage dividers and stuck the output voltages where I want the specialized batteries to be? Essentially fashion a 5.(whatever) battery out of a 10 V battery and put 110 Ohm resistor as the load resistor and 140 Ohm resistor with the 10 V battery.
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