RC phase shift oscillator

RC phase shift oscillator

In this case the CE amplifiers is followed by a frequency determining network ,$R_1,R_2$ combination provides dc potential divides bias and $R-E,C_E$ provides temperature stability and provids AC signal degeneration RC-RC-RC network may be used as positive feedback between input and output.The frequency determining network provides 180°  ie each RC produces 60°,amplifier produces another 180° phase shift.thus the total phase shift is 180+180=360° and satisfy the bark criterion ,.When $V_CC$ is applied ,the current through $R_L$ increases beacuase of biasing .this change capacitor C an induces voltages across $R_1$

Advantages
1) not bulky and expensive
2) pure sine wave output
Disadvantages
1) suited for low frequency
Replacing transistor by its approximate small signal model

writing loop equations
loop 1
$ßI_3R_L+I_1R_L+\frac{I_1}{jωc}+(I_1-I_2)R=0$
    $$\bbox[5px,border:2px solid black]{     +I_1(R_L+\frac{1}{jωc}+R)-I_2R+ßI_3R_L=0}$$
Loop 2
$(I_2-I_1)R+\frac{I_2}{jωc}+(I_2-I_3)R=0$

$-I_1R+I_2(R+\frac{1}{jωc}+R)-I_3R=0$
$$\bbox[5px,border:2px solid black]{ -I_1R+I_2(\frac{1}{jωc}+2R)-I_3R=0}$$
loop 3

$(I_3-I_2)R+\frac{I_3}{jωc}+I_3R=0$

$-I_1R+I_2(R+\frac{1}{jωc}+R)-I_3R=0$
$-I_2R+I_3(R+\frac{1}{jωc}+R)=0$
$$\bbox[5px,border:2px solid black]{ 0-I_2R+I_3(\frac{1}{jωc}+2R)=0 }$$

this can be put in determinent form by

$\left[\begin{array}{ccc}   (R_L+\frac{1}{jωc}+R)&-R&ßR_L\\   -R&(\frac{1}{jωc}+2R)&-R\\ 0&-R&(\frac{1}{jωc}+2R) \end{array}\right]\left[\begin{array}{ccc}   I_1\\   I_2\\ I_3 \end{array}\right]=0$
loop current  $I_1, I_2,I_3$ cannot be zero

$\left[\begin{array}{ccc}   (R_L+\frac{1}{jωc}+R)&-R&ßR_L\\   -R&(\frac{1}{jωc}+2R)&-R\\ 0&-R&(\frac{1}{jωc}+2R) \end{array}\right]=0$
=>

$(R_L+\frac{1}{jωc}+R)\left[(\frac{1}{jωc}+2R)^2-R^2 \right]+R\left[-R(\frac{1}{jωc}+2R)+0 \right]+ßR_LR^2=0$


$(R_L+\frac{1}{jωc}+R)\left[\frac{1}{j^2ω^2c^2}+4R\frac{1}{jωc} +4R^2-R^2 \right]-\frac{R^2}{jωc}-2R^3+ßR_LR^2=0$
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$(R_L+\frac{1}{jωc}+R)\left[\frac{1}{j^2ω^2c^2}+4R\frac{1}{jωc} +3R^2 \right]-\frac{R^2}{jωc}-2R^3+ßR_LR^2=0$


$-\frac{R_L}{ω^2c^2}+4R\frac{R_L}{jωc} +3R^2R_L-\frac{1}{jω^3c^3}-4R\frac{1}{ω^2c^2} +3R^2\frac{1}{jωc}-\frac{R}{ω^2c^2}+4R^2\frac{1}{jωc} +3R^3-\frac{R^2}{jωc}-2R^3+ßR_LR^2=0$


 
$ßR_LR^2+3R^2R_L +3R^3-2R^3+4R\frac{R_L}{jωc}+3R^2\frac{1}{jωc}+4R^2\frac{1}{jωc}- \frac{R^2}{jωc}-\frac{R_L}{ω^2c^2}-4R\frac{1}{ω^2c^2} -\frac{R}{ω^2c^2} -\frac{1}{jω^3c^3}=0$
$ßR_LR^2+3R^2R_L +R^3+\frac{4RR_L+3R^2+4R^2-R^2}{jωc}+\frac{-R_L-4R-R}{ω^2c^2} -\frac{1}{jω^3c^3}=0$

$ßR_LR^2+3R^2R_L +R^3+\frac{4RR_L+6R^2}{jωc}-\frac{(R_L+5R)}{ω^2c^2} -\frac{1}{jω^3c^3}=0$
$$\bbox[5px,border:2px solid red]{    ßR_LR^2+3R^2R_L +R^3+\frac{4RR_L+6R^2}{jωc}-\frac{(R_L+5R)}{ω^2c^2} -\frac{1}{jω^3c^3}=0          }$$       
to make total phase shift around loop zero 
Equating imaginary part to zero

$\frac{4RR_L+6R^2}{jωc}-\frac{1}{jω^3c^3}=0$        

$\frac{4RR_L+6R^2}{jωc}=\frac{1}{jω^3c^3}$ 
         
$4RR_L+6R^2=\frac{1}{ω^2c^2}$

$\frac{1}{4RR_L+6R^2}=ω^2c^2$
$ωc=\sqrt{\frac{1}{4RR_L+6R^2}}$
$2πfc=\sqrt{\frac{1}{4RR_L+6R^2}}$
$$\bbox[5px,border:2px solid black]{  f=\frac{1}{2πc\sqrt{4RR_L+6R^2}}}$$
is the expression for frequency of oscillation
Equating real part
$ßR_LR^2+3R^2R_L +R^3-\frac{(R_L+5R)}{ω^2c^2} =0$
here we have→
$ω^2c^2=\frac{1}{4RR_L+6R^2}$
$ßR_LR^2+3R^2R_L +R^3-(R_L+5R)(4RR_L+6R^2) =0$
$ßR_LR^2+3R^2R_L +R^3-4RR_L^2-20R^2R_L-6R_LR^2-30R^3 =0$ 
$ßR_LR^2 +R^3-30R^3-4RR_L^2-20R^2R_L-6R_LR^2+3R^2R_L =0$ 
$ßR_LR^2 -29R^3-4RR_L^2-23R^2R_L =0$   
$ßR_LR^2  =29R^3-4RR_L^2+23R^2R_L$
$ß  =\frac{29R^3}{R_LR^2}+\frac{4RR_L^2}{R_LR^2}+\frac{23R^2R_L}{R_LR^2}$
  $ß  =\frac{29R}{R_L}+\frac{4R_L}{R}+23$
$ß  =\frac{29R}{R_L}+\frac{4R_L}{R}+23$
$$\bbox[5px,border:2px solid black]{ let  k=\frac{R_L}{R} }$$
$ß  =\frac{29}{k}+4k+23$
the min value of k is obtained by differentiating with respect to k  
 $\frac{dß}{dk}=4-\frac{29}{k^2}$
Equating to zero
$4-\frac{29}{k^2}=0$
     
$4=\frac{29}{k^2}$
$k=\sqrt{\frac{29}{4}}=2.69$
The min value of ß required by the transistor can be obtained by

$ß  =\frac{29}{2.69}+4*2.69+23$
$ß  =44.5$
The phase sift oscillator using BJT will not oscillate if transmitter is chosen whose ß<44.5