Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 1 (Q.No. 13)
13.
The input impedance of op-amp circuit of figure is
Answer: Option
Explanation:
Due to the presence of virtual ground at input, the resistance in the series path of input of inverting amplifier is input impedance.
Discussion:
5 comments Page 1 of 1.
Chenelyn Cedron said:
9 years ago
Let's try some math and the Miller's Theorem to solve this one.
First, the closed loop gain A{cl} of a negative feedback is:
A{cl} = (V{out})/(V{in}) = (-Rf)/(Ri)
Where Rf is the feedback resistance and Ri is the input resistance. It should be clear that Rf is equal to Z.
Using Miller's theorem,
Z{equivalent_input} = (Z)/(1 - A) = (Rf)/(1 - ((-Rf)/(Ri)).
Z{equivalent_input} = ((Ri)(Rf))/(Ri + Rf).
Z{equivalent_input} = ((10k)(100k))/(100k + 10k).
Z{equivalent_input} = 9.09k.
Letter D is the nearest choice.
First, the closed loop gain A{cl} of a negative feedback is:
A{cl} = (V{out})/(V{in}) = (-Rf)/(Ri)
Where Rf is the feedback resistance and Ri is the input resistance. It should be clear that Rf is equal to Z.
Using Miller's theorem,
Z{equivalent_input} = (Z)/(1 - A) = (Rf)/(1 - ((-Rf)/(Ri)).
Z{equivalent_input} = ((Ri)(Rf))/(Ri + Rf).
Z{equivalent_input} = ((10k)(100k))/(100k + 10k).
Z{equivalent_input} = 9.09k.
Letter D is the nearest choice.
(1)
Laksh said:
7 years ago
Wonderful explanation, thank you all.
(1)
Jyothi R said:
9 years ago
By virtual ground concept.
vs - (10k * is) = 0;.
So, vs/is = input impedence = 10k ohm.
vs - (10k * is) = 0;.
So, vs/is = input impedence = 10k ohm.
Hindosh said:
8 years ago
Thank you both for the given explanation.
Pankhuri Saxena said:
7 years ago
Well said, thanks @ChenelynCedron.
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