Electronic Devices - Oscillator Circuits - Discussion
Discussion Forum : Oscillator Circuits - General Questions (Q.No. 1)
1.
Calculate the value of C1 = C2 for the Wien bridge oscillator to operate at a frequency of 20 kHz. Assume R1 = R2 = 50 k
and R3 = 3R4 = 600?
and R3 = 3R4 = 600? 
Discussion:
10 comments Page 1 of 1.
Prachi said:
2 years ago
Why is R3 and R4 not involved in the equation? Please explain me in detail.
(2)
Tshepo Ishmael MAKARA said:
5 years ago
Wien bridge oscillator is designed to supply a frequency of 10khz.
Assume R1=R2=70k ohm and R3=2R4=500 ohm also assume that the capacitor value is equal, determine the capacitor value that will produce the frequency.
Assume R1=R2=70k ohm and R3=2R4=500 ohm also assume that the capacitor value is equal, determine the capacitor value that will produce the frequency.
(1)
Mike said:
7 years ago
Why is R3 and R4 not involved in the equation?
Monisa said:
7 years ago
Very useful, Thanks all.
Yoga said:
7 years ago
F=1/2*π*√ (r1*r2*c1*c2).
Wellwisher said:
9 years ago
It can be obtained from the frequency of oscillation for the wienbridge.
Muhammad Talib Faiz said:
1 decade ago
Since we know for Above circuit,
f = 1/{2*pi*Sqrt(R1C1R2C2)} where 'f' is resonance frequency.
When C1=C2 and R1=R2 then,
f = 1/(2*pi*c*R).
By Rearranging and solving the equation.
C = C1 = C2 = 1/(2*Pi*20k*50k).
= 159pF.
f = 1/{2*pi*Sqrt(R1C1R2C2)} where 'f' is resonance frequency.
When C1=C2 and R1=R2 then,
f = 1/(2*pi*c*R).
By Rearranging and solving the equation.
C = C1 = C2 = 1/(2*Pi*20k*50k).
= 159pF.
Srinivasan said:
1 decade ago
f=1/(2*pi*sqrt(R1*R2*C1*C2)). Since R1=R2=R=50k, C1=C2=C, therefore
20k=1/(2*pi*50k*C), C=159.15pF
20k=1/(2*pi*50k*C), C=159.15pF
Student said:
1 decade ago
C =[1/2*pi*R*f]
=159.155 nF!
=159.155 nF!
(1)
Agashya said:
1 decade ago
f=[1/2*pi*RC]
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers