Solve Clamping Circuit Homework Now - Sedra/Smith Problem

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In summary, the conversation is about a question regarding a circuit and input signal shown in an image. The original poster is trying to solve the problem but is getting a different answer than what is given in the book. They ask for help in identifying where they went wrong and another user suggests that the diode may not be ideal and may have a low reverse breakdown voltage. The original poster later realizes that they misinterpreted the question and the answer is -5 V for a 10 V peak to peak signal with a minimum voltage of -10 V.
  • #1
roeb
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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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  • #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?
 
  • #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
 
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What is a clamping circuit?

A clamping circuit is a type of electronic circuit that is used to limit the voltage of a signal to a predetermined level. It is commonly used to protect sensitive components from overvoltage or to shift the signal to a desired level.

What is the purpose of a clamping circuit?

The purpose of a clamping circuit is to maintain a specific voltage level for a signal. This is achieved by using diodes or other components to limit the signal's voltage to a predetermined range.

What are the types of clamping circuits?

The two main types of clamping circuits are positive and negative clamping circuits. Positive clamping circuits shift the signal to a higher voltage level, while negative clamping circuits shift the signal to a lower voltage level.

How does a clamping circuit work?

A clamping circuit works by using diodes or other components to limit the voltage of the input signal. When the input voltage exceeds a certain level, the diode conducts and limits the voltage to the desired level.

What is the difference between a clamping circuit and a clipping circuit?

A clamping circuit limits the voltage of a signal to a specific level, while a clipping circuit cuts off the signal at a certain voltage level. In other words, a clamping circuit shifts the entire signal, while a clipping circuit only removes the portions that exceed a certain voltage.

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