Your circuit working with adjusted values!

All the best,
Robin

 

http://en.wikipedia.org/wiki/RC_time_constant

RC represents 63% of the total charge time!

Hmm so it means of RC constant of 10ms represent 63% of the total charge time. That means for for 100% of the charge time is 15.8ms (10ms/63)*(100) ? so anything below 63Hz is alright ?

Not sure if I'm understanding it correctly..

 

Hmm so it means of RC constant of 10ms represent 63% of the total charge time. That means for for 100% of the charge time is 15.8ms (10ms/63)*(100) ? so anything below 63Hz is alright ?

Not sure if I'm understanding it correctly..

Nope, you do understand
I too have fallen into this trap before!

All the best,
Robin

 

opps correction. so the cutoff frequency is (1/2*pi*RC) .. approximately 15.9Hz....

So anything below 15.9hz will work nicely.... Just tested it and it is correct as theoretical value !!!

Wow really appreciate for your help !!!!

Anyway what can I do to explain the DC offset occurring when it hit over that cutoff frequency due to the RC value ?

 

I think it goes like this:
Initially your input source is starting at a low negative voltage. This results in the output (as it is inverting) to start all the way saturated at 9V. Then your input source steps up so the output of the integrator begins to fall. But because the RC circuit has a small rate of decay the input source changes before the integrator can go all the way fully down. This is due to the capacitor and resistor network taking along time to charge.

 

The RC Constant is for 0.7 of the charge time. So that sounds about right. Try replacing the capacitor with one that is 0.5uF (or even 100n)

0.7 is for a passive RC integrator. An op amp (ideal) integrator charges to 1.0 times (equal to) the -DC input voltage in one time-constant.