written 3.5 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
$w=\dfrac1{\sqrt{10}CR}\\ \therefore f_{osc}=\dfrac1{\sqrt{10}\pi CR}$