[sergey23 0]

Hello, I have circuit and plots of a points of input and output shown in the photos below.
My zenner diode is the key component in the circuit .its reversed biased voltage is 3.6V.
my input pulse is 0-3.3V.
RC is circuit is sort of integrator,also i have -2.5V.
Why i need this -2.5V?
What is analog the intuition regarding how this zenner diode functions?
 How can i see the role of the reversed voltage in it functionality?
Thanks. 

https://groups.io/g/electronics101/files/Temp/zener circuit

[1113 6]

you need -2V5 to open it

but to get more butifull shape you need -25V

[sergey23 0]

Hello,the reversed bias voltage is 3.6V.

why 2.5V?

and what is the functionality of the sennet in the circuit ?

Thanks.

[1113 6]

zenner in open state is voltage shifter. to open zenner you need put current to it.