 Good morning, dear friend. My name is Sachin Gengze, I am Professor and Head of Department of Electronics Engineering at Valchan Institute of Technology, Solapur. In today's lecture, we are going to discuss about a simple square wave generator. After completing this session, you student will able to explain working of a simple square wave oscillator and then you will be able to design a square wave oscillator using operational amplifiers. As we know, wave form generator is a circuit which generates repetitive wave forms of fixed amplitude and frequency. We can decide what frequency we want, what amplitude we want and accordingly we can design the wave form generator. The popular type of wave form generators are available which can generate sinusoidal signal, square wave signal, triangular wave signal or even a sawtooth signal. In today's lecture, we are going to discuss about a very simple and popular square wave generator. If we are using a PAM, then the square wave generator can very easily design using a PAM. As we know in case of a PAM, we can make the output of the PAM to go into saturation and then we can design a square wave generator where the output of the PAM is forced to swing between plus V saturation and minus V saturation and then if we take this output, that is nothing but a square wave. The circuit which we are going to discuss or the circuit which we are going to look at is of course a square wave generator. This is also called as a free running or r stable multivibrator because it has two states, one state when the output of the PAM is in plus V saturation and the other one when it is in minus V saturation and then the transaction from these two states or the transaction among these two states takes place without any external input. So, a very simple circuit here can be used as a square wave generator. As you can see, it can be designed using operational amplifier. One can use a popular PAM like 741 and along with that what I require are few of the resistors and a capacitor over here. So, using only few components and a PAM, I can design a square wave generator. Now, let us first have a look at how this circuit can generate or let us first understand how this circuit generates a square wave. Now, for that let us we have to first assume that the voltage initial voltage across this capacitor is 0. So, initially when there is no power supply connected or the power supply is not on the voltage across this capacitor is 0. So, as soon as I make the power supply on that is plus V c c and minus V e e at that initial point the voltage at across the capacitor is 0 which is nothing but the voltage at V 2 and then I can say that the voltage at inverting terminal is 0. Now, as the voltage at the inverting terminal is 0, however there is some voltage present at the output of the op amp and that is because of the output offset voltage. Now, let us make an assumption that this output offset voltage which is initially present at the output of the op amp is positive. So, the part of that positive voltage is fade back and then it appears at the V 1. So, now I can say that the V 1 which is a non-inverting input of the op amp is a positive voltage. However, the voltage V 2 which is at the inverting of the op amp is a 0 and then the non-inverting input is greater than the inverting input and that is why the output of the op amp goes into the plus V sat. So, as a step number one of analysis I say that the output of the op amp V 0 is plus V sat. Now, when the output of the op amp is plus V sat the part of the output is fade back to the input. Now, if I apply the voltage divided rule over here please understand this V 0 or the voltage at this point is plus V sat then the voltage at the point V 1 can be found out as V 1 is equal to plus V sat into R 1 divided by R 1 plus R 2. So, the voltage at this point V 1 is definitely going to be less than the V sat and it is equal to plus V sat into R 1 divided by R 1 plus R 2 and this voltage is fixed because the output of the op amp is fixed to the plus V sat. Now, meanwhile what happened to the capacitor C? Now, if this voltage V 0 is plus V sat the capacitor C start charging towards the plus V sat with the polarity shown this as a positive and this as a negative the capacitor start charging. So, the voltage across the capacitor C start increasing towards a positive and which is nothing but the voltage at the V 2. So, I can say that the voltage at the V 2. So, the situation is that the voltage at the V 1 is fixed to plus V sat into R 1 divided by R 1 plus R 2 the voltage at the V 2 is slowly rising and going towards the plus V sat because the maximum voltage over here is plus V sat. Now, at some instant of point. So, this voltage V 1 is fixed to plus V sat into R 1 divided by R 1 plus R 2 this voltage V 2 is rising towards plus V sat at some instant of point this voltage V 2 is will become just greater than V 1 the voltage at the point V 2 becomes greater than V 1 that is the voltage at the inverting terminal becomes that more than that at the non inverting terminal and at this particular instant of time the output of the op amp becomes minus V sat because the voltage at the inverting terminal is more than that of the non inverting terminal output of the op amp becomes minus V sat. Now, as soon as the output of the op amp becomes minus V sat the voltage at the point V 1 which is nothing but the non inverting input can again be found out with the same voltage divider rule V 1 is equal to now it is V 1 is equal to minus V sat into R 1 divided by R 1 plus R 2. So, this voltage is negative this is minus this is minus V sat into R 1 divided by R 1 plus R 2 meanwhile the voltage minus V sat is nothing but which is coming the voltage minus V sat is also going over towards the V 2 and now what happened to the capacitor the capacitor which was previously charging towards the positive now start charging towards the minus V sat because this voltage is minus V sat through this resistor R the capacitor now start charging towards the minus V 2. So, what is the situation V 1 is fixed because this output voltage is fixed that is minus V sat. So, the V 1 is equal to minus V sat into R 1 divided by R 1 plus R 2 the voltage at the point V 2 which is nothing but the voltage across capacitor is now going towards minus V sat. So, a point will come or the instant will come when the voltage at the point V 2 or this voltage V 2 falls below V 1 because V 1 is minus V sat into R 1 divided by R 1 plus R 2, but V 2 is going towards minus V sat and at some instant of point this voltage V 2 falls below that of V 1 or in other word the voltage at the inverting input is a less than that at the non-inverting input and at this instant of time the output again change the state from minus V sat to plus V sat because now the voltage at the non-inverting input is greater than at that at inverting input and output again become plus V sat and when output becomes plus V sat this voltage again becomes plus V sat into R 1 divided by R 1 plus R 2 and this and this continue. If we look at the waveforms you can see that the waveform at the output V 0 is nothing but it switches between the plus V sat and minus V sat and the voltage across the capacitor is when the output is plus V sat it is rising towards plus V sat when the output is minus V sat it is going towards the minus V sat right. So if I take the output at V 0 then that output is nothing but a square wave so this is a very simple way to generate the square wave. Now the period of the square waveform is given by as the capacitor C and resistor R are involved in the in the charging and discharging period the period of the period of the square waveform is given as 2 R C log of 2 R 1 plus R 2 divided by R 2 and if I say frequency then the frequency is nothing but the 1 upon time period which is equal to given by this formula 1 upon 2 R C natural logarithm of 2 R 1 plus R 2 divided by R 2. Now if I select the value of R 2 and R 1 such that R 2 is equal to 1.16 R 1 then the frequency can be calculated as F 0 is equal to 1 upon 2 R C right. Now with this let us take a pause and try to design a square wave generator for say 1 kilo hertz. I think you are ready with the answer let us find out how 1 kilo hertz square wave generator can be designed. Let us assume R 1 is equal to 10 K so I assume this R 1 is equal to 10 K I am selecting R 2 is equal to 1.16 time R 1 then R 2 I can find out is 11.6 K and if 11.6 K R 2 is not available I can use a potentiometer of 20 K then I have to assume some value of capacitor let us see is equal to 0.05 micro farad but I know that F 0 that is the frequency equal to 1 upon 2 R C then R is equal to 1 upon 2 F 0 C so F 0 I already know that I am trying to design this for 1 kilo hertz so I can substitute the value and I can find out the value of R is equal to 10 K. So this is how a very simple by a simple method a square wave generator can be designed using operational amplifier, a capacitor and few of the registers. The references used for this is Op Amp and Linear Integrated Circuit by Ramakan Ghai Pat that is a very popular book. So thank you very much for joining me joining my lecture and we will continue with this oscillators in next lecture also. Thank you once again.