I don't understand the picture well. If the input voltage is greater than VDD, the current flows upward.
Does the current flow and voltage as much as ㅡ 0.6 ㅡ VDD from the input voltage? Or does the voltage take only as much as 0.6 minus the input voltage and the current flows?
And in the ground part, if the input voltage is -2, the ground voltage is 0V.
Isn't it 0-0.7-(-2)??
I modified the drawing that I sent. 1 2 3 and Vdd ground
Briefly explain
When current flows in the forward direction from the pn junction, there is a threshold voltage value to cross p-->n.
Usually it is 0.6V, but you can consider it equal to 0.6 or 0.7
1. If the Vdd terminal is ground as shown in the picture above, I have drawn 3-- Vout.
When +Pulse i1 flows, up +0.7V----1
-When Pulse i2 flows, it becomes -0.7V down-----2
In other words, since the input voltage changes direction by +Pulse -Pulse only the threshold voltage value due to the diode, waveform 3 vout is created where the top and bottom are the same.
That picture is the new Vout drawn in 3.
This is easy to understand.
If you don't understand this, there is no easier way to explain it.
In the original text, + Vdd is added here, so when it is +pulse, 0+ Vdd+0.7V,
In case of -Pulse, 0-(0.7)V is displayed on the original drawing output V out
Thank you