Zener Diode Clipping Graph Help

[Ronaldo95163]

So we started looking at Zener Diodes in our electronics course.
What I am having trouble with is the interpretation of the graph.

For the example our diode has a Zener Voltage is 6.2V. I understand what happens in the positive half but when the direction changes is where I get confused.

Why is the pd across the diode 0.7V after clipping? If the polarities switch on the negative cycle of the input signal why would it be -0.7V and not 0.7V?

 

[Hesch]

If the polarities switch on the negative cycle of the input signal why would it be -0.7V and not 0.7V?

Well, you are the one to decide the sign of the measured voltage.

So if you choose the forward voltage to be +0.7V, the reverse voltage will be -6.2V.