Clipper circuits

The most general form of the diode clipper is shown in Figure below. For an ideal diode, the clipping occurs at the level of the clipping voltage, V1 and V2. However, the voltage sources have been adjusted to account for the 0.7 V forward drop of the real silicon diodes. D1 clips at 1.3V +0.7V=2.0V when the diode begins to conduct. D2 clips at -2.3V -0.7V=-3.0V when D2 conducts.



I don't understand the presence of the voltage sources, someone could explain? Thank you

 

With the voltage sources present, the diodes only conduct, therefore limiting the output, when the input voltage exceeds the sum of the voltage source and the forward voltage drop of the diode. Otherwise the diode does not conduct due to being reverse biased and the full input voltage is seen at the output. Of course the load connected at the output will have an influence on the actual voltage seen.

 

Thank you for explanation.

So with 5vp the wave is cut when it reaches +2V and -3V, right?

With only diodes and no Vsources i get limiting at 0.7V

What if in place of a voltage source i would be using an array of diode in series? With 2 i would get a 1.4V limiting

 

By Gorge, I think you've got it!

 

Be aware that a diode has a logarithmic relationship between forward voltage and current so you will not get a sharp clipping voltage at 0.7V but rather a gradual one over a few tenths of a volt. Thus the clipped signal will have a rounded appearance on an oscilloscope, not the square signal that an ideal diode would give.

 

I am aware of diode non-linearity but i don't know how to explain me this. Can you tell me more about this point?

 

Here's a plot of diode current versus voltage on a semi-logarithmic scale for a typical small signal silicon diode. You can see the voltage doesn't reach 0.7V until about 5mA of diode current.

Edit: The deviation from the ideal curve at higher currents is due to the intrinsic resistance of the diode.