Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 3 (Q.No. 16)
16.
The inverting op-amp shown in the figure has an open loop gain to 100. The closed loop given 
is
isAnswer: Option
Explanation:
Applying KCL inverting node
.
Discussion:
9 comments Page 1 of 1.
Hariharan said:
2 years ago
I think the correct answer is -9.
Zulu Chungtia said:
6 years ago
-9 is the answer.
(1)
Anjana said:
7 years ago
Gain=A/(1+AB),
Here B= -1/10
Reciprocal of( -Rf/R1),
A=100.
Here B= -1/10
Reciprocal of( -Rf/R1),
A=100.
Sunitha said:
8 years ago
Answer is option D. I agree.
Komal said:
8 years ago
Formula for closed loop gain for inverting amplifier is:
Acl = -(Rf/R1)/(1+((1+Rf/R1)/Aol)),
Acl = -10/(1+(11/100),
= -9.
Acl = -(Rf/R1)/(1+((1+Rf/R1)/Aol)),
Acl = -10/(1+(11/100),
= -9.
Waqar said:
8 years ago
Yes, your are Right @Mukund.
Mukund said:
8 years ago
Its answer would be option B ie -9.
This question is asked in gate 2001.
This question is asked in gate 2001.
Raju palakkunnu said:
1 decade ago
You are right if op-amp open loop gain is infinity (or too large), only then the virtual ground concept exist. But here the open loop gain is limited (100).
So we can't use that formula. The answer is D.
So we can't use that formula. The answer is D.
Shailesh said:
1 decade ago
We can't calculate it directly?
Because its inverting amplifier and its gain is -Rf/R1 = -10.
Because its inverting amplifier and its gain is -Rf/R1 = -10.
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