Question

Choose the value of RF and C for a 5kHz input signal to obtain good differentiation.

(a) RF = 1.6×10^3Ω,  C1 = 33×10^-6F

(b) RF = 1.6×10^3Ω,  C1 = 0.47×10^-6F

(c) RF = 1.6×10^3Ω,  C1 = 47×10^-6F

(d) RF = 1.6×10^3 Ω, C1 = 10×10^-6F

The question was posed to me in a national level competition.

Enquiry is from Differentiator topic in division Operational Amplifier Applications of Linear Integrated Circuits

Answer 1

Correct answer is (b) RF = 1.6×10^3Ω,  C1 = 0.47×10^-6F

Explore the finest explanation: For a good differentiation, the time period of the input signal must be larger than or equal to RF C1  i.e. T ≥ RF×C1

Given f=5kHz , T=1/f = 1/5kHz = 2×10^-4 –> Equ(1)

RF×C1 = 0.4µF×1.6kΩ =7.52×10^-4 –> Equ(2)

Hence Equ(1) ≥ Equ(2).