Answer:

• Wein bridge oscillator is essentially a two-stage amplifier with an R-C bridge circuit.Wien bridge is a lead-lag network. The phase’-shift across the network lags with increasing frequency and leads with decreasing frequency. By adding Wien-bridge feedback network, the oscillator becomes sensitive to a signal of only one particular frequency.
• The circuit is set in oscillation by any random change in base current of transistor Q1, that may be due to noise inherent in the transistor or variation in voltage of dc supply. This variation in base current is amplified in collector circuit of transistor Q1 but with a phase-shift of 180°. the output of transistor Q1 is fed to the base of second transistor Q2 through capacitor C4.
• Now a still further amplified and twice phase-reversed signal appears at the collector of the transistor Q2. Having been inverted twice, the output signal will be in phase with the signal input to the base of transistor Q1and some part of the output signal at transistor Q2 is fedback to the input points of the bridge circuit. A part of this feedback signal is applied to emitter resistor R4 where it produces negative feedback.
• Similarly, a part of the feedback signal is applied across the base-bias resistor R2 where it produces positive feedback. At the rated frequency, effect of regeneration is made slightly more than that of degeneration so as to obtain sustained os­cillations.

Expression of frequency for sustained oscillation- In order to obtain the expression of frequency of oscillation  , let us consider the feedback network as shown below.

The feedback factor or gain is defined as :

$\beta= \dfrac {V_F}{V_{in}} $

Now , the feedback voltage is the voltage across the parallel combination of the resistor and capacitor and let the impedance across this combination is $Z_2$ and let the impedance of the series combination of resistor and capacitor is $Z_1$.

$\therefore V_F=\dfrac{Z_2}{Z_1+ Z_2}\times V_{in}$

and $\beta=\dfrac{Z_2}{Z_1+Z_2}$

Now , $Z_1=R_1+\dfrac{1}{j\omega C_1}=\dfrac{1+j\omega R_1C_1}{j\omega C_1}$

$\therefore Z_2=\dfrac{R_2}{1+j\omega R_2C_2}$

Now subsituting the value of $Z_1 \ and \ Z_2$ in $\beta$

$\beta=\dfrac{[R_2/1+j\omega R_2C_2]}{\dfrac{1+j\omega R_1C_1}{j\omega C_1}+[R_2/1+j\omega R_2C_2]}$

Putting  $s=j\omega$ in the above expression ,

$\beta=\dfrac{[R_2/1+s R_2C_2]}{\dfrac{1+s R_1C_1}{s C_1}+[R_2/1+s R_2C_2]}$

$\therefore \beta=\dfrac{sC_1R_2}{1+s(R_1C_1+R_2C_2+R_2C_1)+s^2R_1R_2C_1C_2}$

Resubstitute $s=j\omega , \ s^2=-\omega^2$

$\therefore \beta=\dfrac{j\omega C_1R_2}{(1-\omega^2R_1R_2C_1C_2)+j\omega (R_1C_1+R_2C_2+R_2C_1)}$

$\therefore \beta=\dfrac{j\omega C_1R_2[(1-\omega^2R_1R_2C_1C_2)-j\omega(R_1C_1+R_2C_2+R_2C_1)]}{(1-\omega^2R_1R_2C_1C_2)^2+\omega^2 (R_1C_1+R_2C_2+R_2C_1)^2}$

Since the phase shift introduced by the wein bridge circuit at desired output frequency should be 0 and for that the imaginary part should be zero.

$\omega C_1R_2(1-\omega^2R_1R_2C_1C_2)=0$

$\therefore \omega^2R_1R_2C_1C_2=1$

$\therefore \omega=\dfrac{1}{\sqrt{R_1R_2C_1C_2}}$

$\therefore f=\dfrac{1}{2\pi\sqrt{R_1R_2C_1C_2}}$

This is the expression for the oscillator frequency.

Now , if $R_1=R_2=R \ and \ C_1=C_2=C$

$\therefore f=\dfrac{1}{2\pi\sqrt{RC}}$

For this frequency , the feedback factor turns out to be $\therefore \beta=\dfrac{1}{3}$

Now , According to Barkhausen criteria , the loop gain should be greater than one for sustained oscillation.

$\mid A \beta \mid \geq1$

Substituting $\beta=\dfrac{1}{3}$ we get , 

$A\geq 3$

Thus , the gain should be atleast greater than or equal to 3 for sustained oscillation.