Electronics - Operational Amplifiers - Discussion

Could you please elaborate your answers @D Sati and @Vikas. If we calculate with the formula you gave it again gives the answer as 4.8KHz. And the other thing I would like to put forth before you is that the advantage of this configuration is that the op amps high input impedance prevents excessive loading on the filters output while its low output impedance prevents the filters cutoff frequency point from being affected by changes in the impedance of the load.

The answer should be 4.8kHz. LM741 BW is in MHz range so it does not affect the overall cut-off. Additionally, the op-amp is configured as a buffer with ideally infinite input impedance. Thus cut off should be the same as the cut-off of passive RC LPF.

At cut off freq output of opamp should be 1/2^.5.
At frequency 0hz output is=vin

So at cutoff freq vout= vin/1.414

Because vout = vin/(r*s*C+1)
Or |v|=vin/((r*s*c)^2+1)^.5

So (r*s*c)^2=1
So r*s*C=1
So 1.8k ans.

I'm getting 4.8kHz i.e fc=(1/2*pi*C*R)

Sub values v get 4.822 khz.

Please any one tel me is it correct or not ?

vout = vin/(1+rcs).
1+(rcs)^2 = 2.

s = 1/rc.
Omega = 1/rc.

f = 1/2 pirc.
f = 4.8 KHz.

I think 4.8khz is correct answer. Because of LPF cutoff frequency formula is 1/2*pi*R*c.

Is it a zero crossing detector?

How can we use that normal formula of lowpass filter?

So confusing. Can anyone give us the right formula of LPF cut-off frequency?

Nisha could you please tell me how can we take that the value of Pi =1.34.

I am also getting 4.820khz.

Anyone, please explain the correct answer.