Wednesday 6 December 2017

RC wein bridge oscillator


RC wein bridge oscillator

RC network does not produce any phase shift .therefore to obtain total shift of 0° or 360° ,a two stage CE amplifier is required
\(R_1,C_1\) and \(R_2,C_2\) acts as forward network ,voltage across the parallel combination of \(R_2,C_2\) is fed to the input of the amplifier .The frequency of oscillation is determined by \(R_1,C_2 and R_2,C_2\) .The desired frequency of oscillation can be obtained by varying 2 capacitors and resistors
The feedback network provides the positive feed back .In addition to this the resistors \(R_3 and R_4\) provide negative feedback.Hence this oscillator has better amplitude stability \(R_4\) is often a temperature sensitive resistor with positive temperature co-efficient
If the amplitude of oscillation increases the resistance \(R_2\) increases .This reduces the negative feedback which reduces the amplitude of the gain and the amplitude of oscillation is restored to stable value

Advantages
low distortion
better stability
adjustable frequency

MathJax example
Disadvantages
Costlier
used only in low frequency

\(V_f=V_o\frac{R_2||X_{c2}}{R_2||X_{c2}+R_1+X_{c1}}\)

let \(R_1=R_2=R\)

\(V_f=V_o\frac{\frac{R\frac{1}{jωc}}{R+\frac{1}{jωc}}}{\frac{R\frac{1}{jωc}}{R+\frac{1}{jωc}}+R+\frac{1}{jωc}}\)
\(=V_o\frac{\frac{R\frac{1}{jωc}}{R+\frac{1}{jωc}}}{\frac{R\frac{1}{jωc}+(R+\frac{1}{jωc})(R+\frac{1}{jωc})}{(R+\frac{1}{jωc})}}=V_o\frac{R\frac{1}{jωc}}{R\frac{1}{jωc}+(R+\frac{1}{jωc})(R+\frac{1}{jωc})}\)       
 \(=V_o\frac{\frac{R}{jωc}}{\frac{R}{jωc}+(R+\frac{1}{jωc})^2}=V_o\frac{\frac{R}{jωc}}{\frac{R}{jωc}+\frac{(Rjωc+1)}{(jωc)^2}^2}\)
\(=V_o\frac{R}{R+\frac{(Rjωc+1)}{jωc}^2}=\frac{V_oRjωc}{Rjωc+(Rjωc+1)^2}\) 
\(=\frac{V_oRjωc}{Rjωc-R^2ω^2c^2+2Rjωc+1}=\frac{V_oRjωc}{3Rjωc-R^2jω^2c^2+1}\)

 \(\frac{V_o}{V_f}=\frac{3Rjωc-R^2jω^2c^2+1}{Rjωc}\)
MathJax example



 \(\frac{V_o}{V_f}=\frac{3Rjωc-R^2ω^2c^2+1}{Rjωc}=3+\frac{1-R^2ω^2c^2}{Rjωc}\)

Equating imaginary part 

 \(\frac{1-R^2ω^2c^2}{Rjωc}=0\)
 \(=\frac{1-R^2ω^2c^2}{Rωc}=0\)
 \(1=R^2ω^2c^2\)
\(ω=\frac{1}{Rc}\)

 \(ω=2πf\)

\(f=\frac{1}{2πRc}\)
Now from real part
 \(\frac{V_o}{V_f}=3\)=> \(ß=\frac{1}{3}\)
\(|Aß|=1\)=>
\(A=3\)
therfore gain of 2 amplifier required is 3
MathJax example

No comments:

Post a Comment

Hartley oscillator

Hartley oscillator This is under the category of tuned oscillator or resonant circuit oscillator Operation When \( V_{cc}\) is appl...