Explain working of RC phase shift oscillator. Give expression for frequency of oscillators.

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  • Transistorized RC phase shift oscillator is generally used at low frequencies
  • In this circuit common emitter amplifier provides 180º shift and remaining 180º shit is provided by the circuit
  • In RC phase shit oscillator voltage shunt feedback is used.

Fig1 RC phase shift oscillator circuit diagram

$$\text{Fig1 RC phase shift oscillator circuit diagram}$$

  • The h-parameter model for the above circuit is

Fig2 h-parameter model of oscillator

$$\text{Fig2 h-parameter model of oscillator}$$

  • In above circuit we can notice that load resistance $R_L$ and collector resistance $R_C$ are in parallel hence Let $R_L΄= R_L|| R_C$. Now current source (hfeIb) and the resistance in parallel $R_L΄$ can be replace by voltage source $(h_{fe}I_b \ R_L΄)$ and resistance $(R_L΄)$ in series as shown in Fig3.
  • The input resistance $R_i≈h_{ie}$ also resistance $R_3$ is chosen such that $R= R_i+ R_3$
  • The coupling capacitors acts like short circuit and we have neglected two h-parameter $h_{oe}$ and $h_{re}$
  • The capacitor C offers significant impedance at frequency of oscillation thus it is kept as it is.
  • The simplified h-parameter circuit is

Fig3 simplified h-parameter model

$$\text{Fig3 simplified h-parameter model}$$

  • Since this is a voltage shunt feedback, we should find the current gain of the feedback loop.
  • Applying KVL

    $\hspace{.5cm}h_{fe}I_b R_L΄+ R_L΄I_1+ X_c I_1+R(I_1- I_2)=0\hspace{3.5cm}……………(1) \\ \hspace{.5cm}R(I_2- I_1)+ X_c I_1+ R(I_2- I_3)=0 \hspace{5cm}…………….(2) \\ \hspace{.5cm}R(I_3- I_2)+ X_c I_3=0 \\ \hspace{.5cm}I_2=\bigg(\dfrac{2R+X_c}{R}\bigg)I_3\hspace{8cm}……………..(3)$

    • Put eq(3) in eq(2) we get,

      $I_1=\dfrac1R\bigg(\dfrac{(2R+X_c)^2}{R}\bigg){I_3} \hspace{7cm}..................(4)$

    • Put eq(4) and (3) in eq(1)

      $h_{fe}I_b R_L΄+ (R_L΄+X_c+R) \dfrac1R \bigg(\dfrac{(2R+X_c)^2}{R} -1\bigg)I_3-R\bigg(\dfrac{2R+Xc}{R} \bigg)I_3=0 \\ h_{fe}I_b R_L΄+ (R_L΄+X_c+R)\bigg(\frac{3R^2+(X_c)^2+4RX_c}{R^2} \bigg) I_3- (2R+X_c) I_3=0$

    • On simplifying,

      $\dfrac{I_3}{I_b} =\dfrac{h_{fe}R_LR^2}{2R^3+R^2X_c-3R^3R_L-R_L-4RR_LX_c-3R^3-R(X_c)^2-3R^2X_c-4R^2X_c-(X_c)^3-4R(X_c)^2} \\ \dfrac{I_3}{I_b}=\dfrac{h_{fe}R_LR^2}{-R^3-6R^2X_c-3R^2R_L-(X_c)^2R_L-4RR_LX_c-5R(X_c)^2-(X_c)^3}$

    • Since $X_c=\dfrac{-1}{jwC^2}$ all odd powers of Xc’s are imaginary thus separating real and imaginary terms,


  • For satisfying barkhausen’s criteria $I_3$ and $I_b$ must be in phase

    • Imaginary part of gain must be zero

      $\therefore$ $-4RR_IX_c-6R^2X_c-(X_c)^3=0 \\ -4RR_I-6R^2-(X_c)^2=0 \\ -4RR_I-6R^2=(X_c)^2 \\ -4RR_I-6R^2=\dfrac{1}{\omega ^2C^2}\\ f=\dfrac{1}{2 \pi C \sqrt{4RR_L+6R^2}}\\ \boxed{f=\dfrac{1}{2 \pi CR\sqrt{6+4\bigg(\dfrac{R_L}{R}\bigg)}}}$

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