i couldn’t understand the working of a opamp integrator . i have attached the image of the developed integrator.
how much should the vcc of the opamp be given ?
if so, whether the integrator integrates to the input voltage given at the inputs or integrates to the suply voltage ??
please explain the attached image where in the input voltage given is 1 volt thinking that it would integrate to 1 volt with Vcc of 5 volts. but when i check the output voltage across the capacitor, i goes beyond 5 volts . please explain.
If you intergrate a constant you get a variable. So one volt in will produce a ramp that will go on forever getting bigger. In practice the ramp will stop when it reaches the supply of the op amp because it can't get any bigger. This is where the circuit stops being an integrator.
The integrator holds the non-inverting input at a virtual ground, and the current through the resistor and
capacitor must be equal (since the opamp inputs take no current (assuming an ideal device).
Your input circuit provides a constant current through the resistor to virtual ground, so the capacitor
also see this current and be charged up at a steady rate until the output saturates at -5V or so.
Thereafter the virtual ground is no longer maintained and the non-inverting input voltage rises to
+1V (with a time constant given by RC). Thus eventually the capacitor sees 6V across it.
And the other thing Mark T, i connected the capacitor wrongly i guess , isnt it ??
I should have connected in the negative feedback loop nah ?
Oh yeah, there is that minor detail...