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Clipper circuit: Sign of output voltage

  1. Jul 2, 2012 #1
    I have attached the image of the clipper circuit below.
    When Vin is positive the diode is reverse biased and the output and input are at the same loop. When Vin is negative the diode is forward biased and Vo=0.
    My problem is finding the sign of the output, although I know that Vo and Vin are numerically equal. The output waveform of this clipper circuit shows that Vo= Vin. My question is why. If we consider the loop containing Vo and Vin, they must be of opposite polarity so that the sum of the voltages around the loop is zero. If they are equal then the sum of all voltages around the loop is not zero. i have failed an exam just because of this.The waveform i get is the exact opposite of the real one beacause i think that Vo=-Vin.
    We are being taught basic electronics and dc circuits at the same time, and i think dc and ac circuits must be a prerequisite of this electronics course. Is this ok.

    Attached Files:

  2. jcsd
  3. Jul 2, 2012 #2


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    Science Advisor

    A wire cannot have more than one voltage on it at the same time.

    So, the input voltage and polarity in the diagram have to be the same.

    Viewed in parallel like this, they do have the same voltage and polarity (although this may spell trouble for the diode) but if you move around the circle formed by the input voltage and the diode, then you have a source voltage and opposing it you have a voltage across a diode. This gives no net voltage around the circle.

    Kirchoff's Law is safe. It would still be safe if you added a resistor in series with the diode.
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