 Hello and welcome to the video lecture on Oscillators Using Operational Amplifier Part 2. At the end of this session, students will be able to describe the working of Wendbridge oscillator and they will be able to describe the working of quadrature oscillator. So before moving towards the concept, you have to recall the following concepts. Basics of operational amplifier that is IC741. You should know the working of oscillator and you should know the working of operational amplifier as a integrator. So before moving towards actual concept, we will revise the oscillator concept first. So according to the previous lecture, the oscillator consists of the two main blocks. First is amplifier and second is feedback circuit and to give the sustained oscillation for any oscillator, the criteria should be fulfilled which is Barkhausen's criteria and it is having the two conditions. The first condition is the gain must be unity and the second condition is the total phase shift around the circuit should be 0 degree or 360 degree. So based on this criteria, we have seen RC phase shift oscillator in the last lecture. Now in this lecture, we will see some another sinusoidal oscillator circuits. So first we will see the circuit of Wendbridge oscillator. Wendbridge oscillator is most commonly used because of its simplicity and the frequency stability. So let us see the circuit of Wendbridge oscillator. For that we will require the operational amplifier. We will provide the VCC and minus V double here, one resistor and a capacitor in series, one feedback resistor which is RF, then one more RC network we will take which will be parallel R1 resistor. So let us call it as R2 C2 and it will be R3 C3. So this is the circuit of Wendbridge oscillator. As you can see here, the Wendbridge is connected between the input and output terminals of the operational amplifier which is 741. So for the sake of convenience, we will modify this circuit. So let us draw the equivalent circuit for this. So according to this diagram, we will draw the equivalent circuit. So here it is 741, we are going to supply VCC as well as minus V double E. As you can see here at the negative terminal, there is a feedback resistor RF and R1 is connected here which is grounded. Now at the positive terminal that is at the non-inverting terminal, the parallel RC network can be shown which is nothing but R2 C2 and the series RC network which is nothing but R3 C3 is connected. So this is the equivalent circuit for this Wendbridge oscillator. Now you can see here there are two parts of this circuit. So the first part is nothing but the part of amplifier. The second part is nothing but the feedback circuit which is taken from the output and given to the input that is non-inverting terminal of the operational amplifier. So it is nothing but the feedback circuit. So if you see carefully this feedback circuit in this, the parallel RC network and series RC networks are connected. So the parallel RC network act as a low pass filter and series RC network act as a high pass filter. So the combination of this both circuit will overall give us the result that it should not allow the too high or too low frequency to be passed. But at the particular frequency the output VO will be maximum and that frequency is nothing but the frequency of oscillation which is indicated by FO. The second condition that is a phase shift around the loop should be 0 degree or 360 degree totally and this is achieved when the bridge is balanced and the bridge is balanced at the resonant frequency. So in the case of Wainbridge oscillator the frequency of oscillation is nothing but exactly the resonant frequency. The equation for FO can be given as 1 upon 2 pi root of R2, R3, C2, C3. So it is nothing but the equation for frequency of oscillation for this circuit. We will consider the values of R1, R2, R3 is equal to R and the values of capacitor C2 and C3 is equal to C. By putting these conditions in this equation we will get the frequency of oscillation FO is equal to 1 by 2 pi RC. At this particular frequency the gain achieved can be given as AV is equal to 1 by beta is equal to 3. So for the design of the amplifier we will require the values of R1 and Rf and with the help of this equation we can find out the values of R1 and Rf. So let us see how AV is nothing but the gain of the amplifier. Now you will see here carefully then it is nothing but the non-inverting configuration of the OPAN. So in the non-inverting configuration the gain is given as 1 plus Rf upon R1 and this value is nothing but 3. So Rf upon R1 is nothing but 2 that is the Rf value must be twice of R1. So with the help of this equation we can design the resistors required for feedback as well as the non-inverting sorry inverting terminal connection. This is pole about Wainbridge oscillator. Now let us see about the quadrature oscillator. Second circuit is quadrature oscillator. It will produce the sine as well as cos wave. So it has the two outputs first is sine wave, second is cosine wave and the phase difference between these two is nothing but 90 degrees. So it is called as a quadrature oscillator. The quadrature oscillators are widely used in the telecommunication circuits for measurement purposes such as special purpose voltmeters. So let us see the circuit of quadrature oscillator first. For that circuit we will require the three operational amplifiers as well as sorry the two operational amplifiers as well as for this quadrature oscillator design we will require two operational amplifier as well as three RC networks. So let us see the circuit. First we will draw the operational amplifier with name A1 let us call it R1 this is first output the output is extended with resistor here is a second operational amplifier indicated with name A2. It is also having one capacitor it is offset minimizing resistor here is second output. So as you can observe this is the circuit of quadrature oscillator. In the circuit there are three RC networks used first is R1 C1 second is R2 C2 and third is R3 C3 pause the video for while and identify the circuits found by A1 and A2. So the circuit found by A1 is nothing but the non-inverting integrator and the circuit found by A2 is nothing but purely inverting integrator. So these are the two circuits which are found by A1 and A2. These two circuits form the amplifier factor of this oscillator and this is nothing but the feedback circuit. So R3 C3 form the feedback circuit and the output of A2 is given to A1 through this voltage divider circuit that is R3 C3. Now the first condition of the gain at the particular frequency the maximum gain can be achieved here A v is equal to 1 by beta is equal to 1.414 and with the help of this equation and with the help of the integrator circuit we can design the desired values of R1 C1 as well as R2 C2. Now in this circuit we cannot predict that where we can get the exact sine wave suppose we will consider we will get the sine wave at Vo1. So the another cost that is the cost wave can be obtained at the another output. Now the second condition of achieving the phase shift around the loop that is 0 degree or 360 degree can be achieved in this circuit like that. First A2 amplifier will give us minus 90 degree or 270 degree phase shift at the output and the remaining 90 degree and minus 270 degree can be obtained with the help of the feedback circuit as well as the operational amplifier A1. So A1 and the feedback circuit will give us the remaining minus 270 degree or 90 degree phase shift. Thus we can obtain the maximum gain 1.414 as well as the phase shift condition that is the phase shift around the loop is 0 degree or 360 degree. The frequency of oscillation in this circuit can be calculated as 1 by 2 pi Rc where the resistor R1, R2 and R3 are equal to R means all the three resistors are having the equal value and all the capacitors are having the equal value. So thus for this condition the frequency of oscillation of this oscillator is 1 by 2 pi Rc. The particular frequency we can design the quadrature oscillator circuit by calculating the appropriate values of R and C. These are the references. Thank you.