 Welcome to the session on RC phase shift oscillator. Learning outcomes are at the end of session students will be able to analyze the RC phase shift oscillator circuit. Contents are like this, what is an oscillator that we have seen and what are the requirements of an oscillator that also we have seen in the previous video lecture. For RC oscillator you must satisfy the requirement that is requirement of oscillations nothing but Markovson's criteria, which means that loop gain that is A into beta must be greater than unity and second condition that is phase shift must be 360 degree. For that we have used an amplifier stage which provides a phase shift of signal. Therefore we must use common emitter configuration of amplifier of a transistor. So we can see output is phase shifted at the output of amplifier and that is given to the feedback circuit which provides again 180 degree phase shift which can be obtained through passive elements like inductor and capacitor. So again we will get 180 degree phase shift and that is added with the input signal and therefore there will be a positive feedback of a signal. So we can say here total 360 degree phase shift is obtained and this is the basic principle of RC oscillator circuit. RC oscillators are nothing but audio oscillators which provides the output frequency of signal in the range of 20 hertz to 20 kilohertz. There are two types of RC oscillators that is RC phase shift oscillator and Vain bridge oscillator. In this lecture we will see RC phase shift oscillator circuit. Basic principle of RC phase shift oscillator as you know we are using inverting amplifier, output voltage is given to the input of CR network, we are using three CR networks so that 180 degree phase shift is obtained. Here each CR combination provides 60 degree phase shift that output voltage is V2 and that V2 is again provided to the input of amplifier that is V2 equal to VIN. This we can say here basic principle of RC phase shift oscillator gives the phase shift with feedback network nothing but CR combination and we are using amplifier nothing but a transistors amplifier. Figure 3 shows the circuit diagram of RC phase shift oscillator where transistor is wired as an amplifier R1 and R2 provides the biasing, RC controls the corrector voltage RE and CE will provide the temperature stability for the amplifier stage. This output of amplifier is given to the feedback network consisting of three CR stages. This output of feedback network is provided to the base of transistor through resistor R3. Therefore, we can say here feedback is employed here with the voltage shunt fashion. So variations in the base current that is at the base of transistor is amplified in the collector circuit and thus you will get changes in the output and that is used as a again input to the amplifier stage therefore there will be a regeneration of signal hence oscillations are obtained at the output. As we are providing feedback and we are using the transistor to avoid the loading effect of on transistor amplifier we are selecting the value of R3 minimum nothing but it is R-HRE nothing but input impedance of transistor amplifier. So hence total phase shift is obtained and now we will see the derivation for gain and the frequency of oscillation. We have seen the circuit and equivalent circuit of that is in figure 4 which is nothing but H model of transistor phase shift oscillator. Transistor amplifier is now converted to HRE HRE into V0 HFE into IB 1 by HOE this is basic H model and these are CR stages of feedback network as value of HRE and 1 by HOE is very very small that is negligible therefore HRE is neglected and as you know HOE is very very small but therefore 1 by HOE value will be very large and this 1 by HOE is in parallel with RC so parallel combination of both will give you the value of RC therefore only RC will remain over here. Now this is a current source we are converting now this current source into a voltage source and that is why we have the value of RC in series with this voltage source. So now value of voltage source will become HFE into IB into RC and in series with that RC these are the remaining stages. So value of this voltage source is HFE IB into RC through this 3 loops current is flowing and that current is dominant current or maximum current. So current flowing through the first loop will be I1, second will be I2 and third will be IB. So this is your figure 5 which is showing the simplified equivalent circuit. Now you can apply KVL to these 3 loops and obtain the equation for these 3 loops. For loop 1 you will get the equation like this R plus RC plus 1 by j omega c into I1 minus R into I2 plus HFE into RC into IB equal to 0 for the second loop minus I1R plus 2R plus 1 by j omega c minus R into IB equal to 0 for the third loop I1 will be 0 minus R into I2 plus 2R plus j omega 1 by j omega c into IB equal to 0. Here to solve such equation we will take help of the matrix and we will find its determinant to find the values of required quantity. We are interested to find the condition for gain nothing but value of HFE and second thing we must find the value of frequency of oscillations. That is why arrange the coefficients of I1, I2 and IB in the matrix form. So this is for loop 1, for loop 2 and for loop 3 respectively. If you solve this equation or if you solve this matrix its determinant can be calculated and equated to 0. So here you get this equation where you have some real part and this imaginary part. So green color will show you the imaginary part of the equation and red color will show you the real part of the equation. If you equate this imaginary part to 0 you will obtain the frequency of oscillation value for the RC phase shift oscillator and for the real part you will obtain the value of HFE nothing but gain of amplifier. First I have equated imaginary part to 0 and therefore I cut the value of Xc like this that is Xc equal to under root 6R square plus 4R into RC putting the value of Xc as 1 by 2 pi RC and so you will get the value of frequency of oscillation as 1 by 2 pi RC under root 6 plus 4 RC upon R. If you assume R equal to collector resistance then that value is nothing but 1 by 2 pi RC into under root 10. Now I equate the real part to 0 and obtain the condition for HFE. Here in this equation we can replace Xc square nothing but we can replace the value of Xc by the value in equation 1 that is under root 6R square plus 4R into RC and then you will obtain the further equation. If you solve this equation for the value of HFE you will obtain HFE equal to 23 plus 29R by RC plus 4RC by R. Again here for R equal to RC HFE should be greater than 56 and hence we can say that for oscillations we must use or we must obtain the gain of amplifier greater than 56. Now recall in RC phase shift oscillator how much phase shift can be provided by the feedback network? Yes and the answer is 180 degree. Now the applications of RC phase shift oscillator can be used in the musical instruments voice synthesis also used in GPS units where all this instrument will work on the audio frequency range. Also it can be used as a test equipment which operates in the audio frequency range. These are the references used. Thank you.