Electronics - Operational Amplifiers - Discussion
Discussion Forum : Operational Amplifiers - General Questions (Q.No. 20)
20.
What is the cutoff frequency of this low-pass filter?
Discussion:
24 comments Page 1 of 3.
Gayatri varpe said:
3 years ago
F = 1/2πRC.
F = 1/2π*-33k*0.001*10^-6.
F = 4.8kHz.
F = 1/2π*-33k*0.001*10^-6.
F = 4.8kHz.
(1)
Sri said:
10 years ago
vout = vin/(1+rcs).
1+(rcs)^2 = 2.
s = 1/rc.
Omega = 1/rc.
f = 1/2 pirc.
f = 4.8 KHz.
1+(rcs)^2 = 2.
s = 1/rc.
Omega = 1/rc.
f = 1/2 pirc.
f = 4.8 KHz.
Aditya said:
2 years ago
I am also getting 4.820khz.
Anyone, please explain the correct answer.
Anyone, please explain the correct answer.
Mtarvina said:
7 years ago
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.
Praveen reddy said:
8 years ago
According to me, the correct ans is 4.820 khz.
Sulley said:
9 years ago
I'm getting 4.8khz. So I think it should be the answer.
Ramkamal Adhokari said:
9 years ago
ANS: 4.8KHZ.F = 1/2φ.r.c.
ADDY said:
9 years ago
So confusing. Can anyone give us the right formula of LPF cut-off frequency?
Jsr said:
9 years ago
Cutoff frequency should be 4.8 kHz.
Lijo said:
9 years ago
Is it a zero crossing detector?
How can we use that normal formula of lowpass filter?
How can we use that normal formula of lowpass filter?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers