In the circuit below :



I understand when Vin is +V , the slope of Vout is straight line .
But I don't understand when Vin is 0 V why V out is not 0 V because when Vin is 0 and there are virtual ground it mean there are voltage drop across Rin 0 V
therefore I in = IF = 0 A

And in equation IF = -C(dVout/dt)

The slope of Vout should be 0 when Vin is 0 V but in picture it's slope . What's wrong in my analysis ?

Thank you.

 

What is Vout if Vin = 0 V and the voltage across the capacitor is 1.23 V?

 

The picture is wrong.
When the input voltage is zero, there is no current to charge/discharge the capacitor.
Thus the output voltage stays at whatever value it was prior to the input going to zero.

Also for a positive input, the current directions shown are for electron flow.

 

What is Vout if Vin = 0 V and the voltage across the capacitor is 1.23 V?

I read this in my textbook they explained how this circuit work here explanation:
Iin = V in / Rin
I in = IF


When Vin is positive , slope will be as picture above untill capacitor is full and when Vin is 0 V , capacitor will discharge and slope will be as picture above . Does this circuit work and generate triangle signal only when Vin is not 0 V or clock ?

 

I read this in my textbook they explained how this circuit work here explanation:
Iin = V in / Rin
I in = IF


When Vin is positive , slope will be as picture above untill capacitor is full and when Vin is 0 V , capacitor will discharge and slope will be as picture above . Does this circuit work and generate triangle signal only when Vin is not 0 V or clock ?

When Vin is zero, the current in the resistor is zero, which means that the current in the capacitor is zero, which means that the charge on the capacitor will not change -- it will stay what ever it is.

 

when Vin is 0 V , capacitor will discharge and slope will be as picture above

No.
When Vin is 0V, the capacitor will not discharge, as there is no current into or out of the capacitor terminals.
What part of "no current" do you not understand?

Example simulation below:
Edit: Modified sim to include 3 pulses.

 

In the circuit below :

View attachment 280245

I understand when Vin is +V , the slope of Vout is straight line .
But I don't understand when Vin is 0 V why V out is not 0 V because when Vin is 0 and there are virtual ground it mean there are voltage drop across Rin 0 V
therefore I in = IF = 0 A

And in equation IF = -C(dVout/dt)

The slope of Vout should be 0 when Vin is 0 V but in picture it's slope . What's wrong in my analysis ?

Thank you.

 

Is your opamp model ideal, or are the real input parameters included. If so, maybe current leakage is discharging the cap. Otherwise I think Vout should be constant (zero slope).

Sorry about my first post.