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Clamping Circuit

  1. Jan 8, 2010 #1
    1. The problem statement, all variables and given/known data

    http://img96.imageshack.us/img96/6994/iamge2.png [Broken]http://g.imageshack.us/img96/iamge2.png/1/ [Broken]

    http://img42.imageshack.us/img42/82/imagevrd.png [Broken]http://g.imageshack.us/img42/imagevrd.png/1/ [Broken]

    2. Relevant equations



    3. The attempt at a solution

    I'm following along an example problem in Sedra/Smith and I seem to have gotten stuck. I can see that v_0 = v_i + v_c. For the pictured circuit and input signal I think that v_c = -4V because it seems to me that current will only flow during the 4 volt pulse and since v_c is defined as it is, v_c = -4V. Now, to get v_0 I would think that I would just say v_0 = -4 V + -6 V = -10 V. Unfortunately, the book says the answer is -5 volts. Does anyone see what I've done wrong?

    Even if I draw it I would think that all of the >0 voltage would be shifted downward so that the most negative number is -10V. Doesn't this circuit ideally shift everything to a negative voltage?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 8, 2010 #2
    Yes, that looks like a standard negative clamp to me, the diode would draw current whenever its forward conduction threshold was exceeded. The waveform should thus be shifted negatively, as you say.

    Is that diode meant to be something other than ideal (0V threshold)? Does it have a low reverse breakdown voltage? Does the question text say anything that you have not taken account of?
     
  4. Jan 8, 2010 #3
    Hi Adjuster,

    Thanks for your reply. Looks like I misinterpreted the question. They were looking for the DC component of the signal - so -5 V for a 10 volt peak to peak signal with -10 V being the minimum voltage obtained.

    Thanks,
    roeb
     
    Last edited: Jan 8, 2010
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