sagarpm,
Yes. Its cutoff frequency is 3.42 KHz.
Let Rp = 1k in parallel with 100K.
Then Rp = 0.990099K
fc = 1/[(2PI)(47X10^-9)(.990099X10^3)] = 3,420 Hz.
Regards,
Kral
Thanks for reply..
Gain in pass band will be 1+ 100/1= 101 right ?.
Is this a more stable filter or the one in inverting configuration more stable. I mean the one in which the input is given to one end of the 1 K and Non-Inverting pin being grounded. Any comparisons...
Regards
Sagar
sagarpm,
Your calculated va;ue of 101 for the DC gain is correct. Stability of the circuit depends on the circuit depends on the open loop gain vs frequency characteristics of the op-amp. For most of the op amps commercially available, your circuit will be stable, since the dominant pole will be close to zero frequency.
regards,
Kral