Engineering Solve Clamping Circuit Homework Now - Sedra/Smith Problem

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The discussion revolves around a homework problem involving a clamping circuit from Sedra/Smith. The user initially calculates the output voltage, v_0, as -10V based on their interpretation of the circuit, but the textbook states the answer is -5V. It is clarified that the problem requires finding the DC component of the signal rather than the minimum voltage. The user acknowledges the misunderstanding and confirms that the correct answer reflects the DC component of a 10V peak-to-peak signal. The conversation highlights the importance of accurately interpreting circuit problems in electronics.
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Homework Statement



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

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

Homework Equations





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?
 
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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?
 
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
 
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