Clippers in Diodes: Current Flow Explained

[M. next]

Observe the attached figure, and the associated graph, shouldn't be like the opposite of this graph?

If it is positive the first half the current should pass the diode and it is on, no?

 

[NascentOxygen]

The output of the clipper is v_o. The clipper removes the positive pulses from v_o. The diode conducts on the positive peaks, limiting their swing at 0.6v

 

[M. next]

Why does it remove the positive pulses?

 

[M. next]

Why not the negative one?

 

[NascentOxygen]

Why not the negative one?

The diode doesn't conduct when it is reverse biased, so it is effectively not present for negative excursions of v_o thus negative pulses are not clipped and they remain unaffected.

The input is on the left, the output is the voltage on the right. Maybe it would be clearer if the line marked "—" were to be labelled "ground". https://www.physicsforums.com/images/icons/icon6.gif

 

[M. next]

Please elaborate more, I know that the diode doesn't conduct when it is reverse biased.
Here we have Vi as input voltage, and as you can see in the figure it should be forward biased for positive input voltage, and thus we should remove the negative pulse - not the opposite.

 

[NascentOxygen]

The diode conducts on the positive peaks so they don't make it to the output.

 

[M. next]

Hmm, probably my question is not very clear, right?
The diode allows the passage of voltage if it is in a way that goes along with the pointer head "arrow like" and this " | " or in other words if "Vp-Vn>0".
Let me post my question in a different way: here if voltage from input passes through the diode when it gives a positive pulse, will it be forward biased or not? And why? With details please.

 

[M. next]

And thanks for your time in advance