**1 Answer**

written 2.6 years ago by |

The circuit diagram RC phase shift oscillator is shown below:

The three section of RC network produce extra phase difference of 180°(60° by each section) and the transistor in CE configuration add mother 180° phase shift. So total phase shift around the loop is 0° or 360°. In the last section hie of Transistor add to R' thus giving the net Resistance R.

The AC equivalent circuit is shown below:

Now,

KVL in Fig-3 in each loop we have

$(R_L+R-j/wc)I_1-RI_2+h_{fe}R_LI_3=0\cdots(1)\\ RI_1+(2R-j/wc)I_2-RI_3=0\cdots\cdots(2)\\ 0.I_1-RI_2+(2R-j/wc)I_3=0\cdots\cdots(3)$

Simplifying these three equation by determined method we have,

$\begin{vmatrix} (R_L+R-j/wc) &-R &h_{fe}R_L \\ -R& 2R-j/wc &-R \\ 0 &-R &2R-j/wc \end{vmatrix} =0$

By solving determinant & equating imaginary port is equal to zero we have

$\dfrac{-4R(R+R_1)}{wc}-\dfrac1{wc}\left(3R^2\dfrac1{w^2c^2}\right)+\dfrac{R^2}{wc}=0 \rightarrow w^2=\dfrac1{C^2(4RR_L+6R^2)}$

Ordinarily RL is taken equal to R1 we are

$w=\dfrac1{\sqrt{10}CR}\\ \therefore f_{osc}=\dfrac1{\sqrt{10}\pi CR}$