# Clipper circuit: Why some of the portion of the positive input cycle remains un-clipped due to the diode cut-in voltage

I understand that when the input waveform is positive, the diode does not conduct hence the positive half cycle of the input waveform gets clipped. Also, when the input waveform is negative, then the diode conducts hence the negative waveform appears as the output. However, I am unable to understand why some of the portion of the positive half cycle remains unclipped. I understand, that it is because of the cut in voltage of the diode which is 0.7 volts. However, I am unable to understand the mechanism. It would be very helpful if someone could make me understand why some of the portion of the positive input cycle remains unclipped. • Link to the source of this "information" so that others may know to avoid it. Apr 8, 2018 at 10:00
• tutorialspoint.com/electronic_circuits/… Apr 8, 2018 at 10:08
• The section replicated here appears to be from page 27 of that PDF. Take a look at page 29: the output waveform presented there is identical to this one (but makes more sense in that context). Seems like mixed-up images to me. Apr 8, 2018 at 17:24
• All diodes have a reverse leakage current , which Ileak*R load= +Vm Apr 9, 2018 at 1:28

Andy's answer is great but it does show the ideal type diode. In order to show the effect of the diode's "cut in voltage" as the OP called it I show this simulation of the real world diode and the effect it has on the output voltage. • Add voltage across the diodes also.
– G36
Apr 8, 2018 at 10:31
• Nice pictures - what sim is it? It looks too neat to be LTS but looks can be deceiving LOL. Apr 8, 2018 at 10:47
• @Andyaka - I used LT Spice from Linear Technology. A long time ago I changed the workspace background color to white from the previous black color. The picture above was made by taking screen shot snips form the LT Spice window and pasting up a composite in Paint. I did that to mimic the picture that you showed in your answer. Apr 8, 2018 at 11:24

It would be very helpful if someone could make me understand why some of the portion of the positive input cycle remains unclipped

Quite simply, the diagram is wrong but, surprisingly, the words below it are correct (assuming an ideal diode). Where did this incorrect picture come from?

This is what should happen with an ideal rectifer: - • tutorialspoint.com/electronic_circuits/… Sir, from this source. Even I have found a few errors in this pdf. However, for sake of coherent learning, I am referring to this source. Sir, I am having difficulty in understanding the clipping operation of diodes with cut-in voltage. Apr 8, 2018 at 9:36
• I'd throw that document away if it makes such basic errors. I can't recommend any reading material other than what is present on stack exchange. Apr 8, 2018 at 9:41
• Yes, sir. This makes my work a little difficult. I have an exam the next week. I thought that I would be able to brush up my concepts from these series of tutorials. But it seems that the errors in this study material are too many to ignore. Apr 8, 2018 at 9:50
• It says practical output waveform and I am sure it wants to talk about a real diode Apr 8, 2018 at 11:36
• @PlasmaHH I think you are missing the main point of the question. Apr 8, 2018 at 11:39

There is an error they show you the voltage across the diode not the voltage across the resistor. • Sir, then what should be the output? And sir, what is the relationship between the voltage across the diode and the output voltage? Apr 8, 2018 at 9:56
• But where is the output? And why do you call me sir?
– G36
Apr 8, 2018 at 9:59
• The output voltage is the voltage across the resistor. You are teaching me something. Out of respect, we refer to our teachers as sir or madam. And in formal situations, we call an unknown person by 'sir' or 'madam', which is also a sign of respect. This is the norm in India. Apr 8, 2018 at 10:05
• OK. Assume for a moment that the input signal is a sine wave swings between Vm = +2V and -Vm = -2V. Now try to analysis the circuit assuming that the diode will start to conduct if Vf = 0.6V. For a positive half-cycle, the diode doesn't conduct because it is reverse-biased. So there is no current through the diode and therefore also no current in resistor either. So the Vo is ?? Do the same for a negative half-cycle.
– G36
Apr 8, 2018 at 10:12
• The general shape of the graph is the voltage across the diode, true, but the polarity and phase are wrong. Apr 8, 2018 at 19:03