 to the video lecture on OPAM as triangular and tooth wave generator. At the end of this session, students will be able to design the triangular wave generator using operational amplifier and they will be able to describe the working of south wave generator using operational amplifier. So, before moving towards the design problem of triangular wave generation, you should recall the working of triangular wave form generator using OPAM and you should go through the calculation of design parameters in the triangular wave generation using operational amplifier. Problem statement is design a triangular wave generator for the given frequency 2 kHz and the output peak to peak voltage is expected to be 7 volt. We have to consider the OPAM with saturation voltage Vsat is equal to 14 volt and the supply voltage is plus or minus 15 volt. Assume R2 is equal to 10 kilo ohm and capacitor value is 0.05 micro farad. So, let us see how to design the triangular wave generator using this given operational amplifier. The given quantities are output peak to peak voltage at 7 volt, nothing but 14 volt, supply voltage is given plus or minus 15 volt, then the capacitor value we will assume 0.05 micro farad, value of R2 is 10 kilo ohm and we have to design that triangular wave generator frequency 2 kilo hertz. For this we will first calculate value of R3 because the output peak to peak voltage can be given as 2 R2 upon R3 into V saturation. So, already we know the value of V saturation and output voltage peak to peak. So, we can write where the V saturation is given with 14 volt point. By solving this we get the equation like R3 by R2 is equal to 4. Now the value R2 where we have assumed it as 10 k, so I can write R3 directly as 14 kilo ohm. So it is the parameter R3, then we have to consider the output frequency which is nothing but 2 kilo hertz. The equation for output frequency is given as R3 upon 4 R1 R2 C. Now we have the values of R1 sorry we have the values of R2 and R3. So you can directly take the ratio R2 R3 like 4, otherwise you can write the values of R2 and R3. So by calculating this we get R1 C1 is equal to 2 kilo hertz sorry R1 we will get R1 C1 is equal to 0.5 millisecond where the frequency is 2 kilo hertz. So this is nothing but the time integral time constant T. Now we can write the value of C1 which is assumed as 0.05 micro farad R1 can be calculated as 0.5 millisecond divided by C1. So we get R1 is equal to 10 kilo ohm. Now we have the values of R1, R3, R2 and capacitor C or we can call it as C1. So let us draw the circuit which will generate the triangular wave with the help of these parameters. According to the previous lecture we will require two operational amplifiers where the supply voltage is given as plus 15 and minus 15 volt. It is grounded and we will connect the voltage divider circuit which is having R3 and R2 resistor in that R3 is of 40 kilo ohm. So for that better way to connect 50 kilo ohm pot in the circuit R2 is nothing but the 10 kilo ohm resistor and it is giving the output square wave at this point. Now this output will act as an input for the next circuit which is nothing but the integrator where R1 is 10 kilo ohm and C1 or C is nothing but 0.05 micro farad. It will be at inverting terminal of open A2. Here also we are giving 15 volt and minus 15 volt supply. Now this will be grounded and this is taken as the feedback and it is connected to R2 and here we will get the triangular wave as a output. So this is nothing but the design of a triangular wave generator for the particular frequency of triangular wave which is 2 kilo hertz and the output peak to peak voltage is expected to be 7 volt. First we will see the difference between the triangular wave and south tooth wave. The triangular wave means a wave form which is having the equal rise and fall time means the time consumed to switch from negative ramp voltage towards positive ramp voltage and positive ramp voltage towards negative ramp voltage is equal. That is nothing but the triangular wave form. But in the south tooth wave form you will find the unequal rise and fall time means the rise time can be faster than the fall time like this or vice versa means a fall time is faster than the rise time. And we can generate the south tooth waves using triangular waves just by modifying its rise time and fall time. We will take the reference of triangular wave generator. We will require again 2 operational amplifiers. So consider we are giving plus 12 volt and minus 12 volt. Then I will consider the voltage divider circuit having R 1 and R 2. Then this will act as a input for the next circuit that is integrator which is having R 1 and C 1 as a components. And here we are getting the output voltage V 0 which is nothing but the square wave. Now this C and R 1 are the components for integrator formed by operational amplifier A 2 and this is the non-inverting terminal. Its feedback is given to again R 2. Now there is a slight modification in the circuit. So the non-inverting terminal of A 2 is connected with the part let us say R 4 which is further connected to minus V E and plus V C C supplies. Means we can generate a south tooth wave from the triangular wave just by injecting or inserting some DC level at the non-inverting terminal of operational amplifier A 2. Now depending on the position of R 4 the DC level will be inserted at the output of A 2 that is a triangular wave and it will be affected to the duty cycle of the square wave and further we can achieve the south tooth wave having the unequal rise and fall times. For example if the R 4 is at center then we will get exactly 50 percent duty cycle of the square wave at the output of operational amplifier A 1. So the fall time and rise times are equal and we will get exactly the triangular wave. If we consider that R 4 is shifted towards minus V E then due to this the certain DC level will be added to the output such that the duty cycle of the square wave at the output of A 1 will be less than 50 percent and it can be indicated like this. So as it is less than 50 percent the on time will be less. Now stop the video for while and think on how we will get the output at A 2. So with the reference of this square wave which is having 50 percent less than 50 percent duty cycle the fall time is less and the rise time is more and in the second case where the R 4 is shifted towards the VCC can indicate it with the square wave having the duty cycle more than 50 percent means here the on time will be more and in this case the fall time is more than the rise time. The output of A 2 depends on the position of R 4 which will add the DC level in the output and accordingly the square wave duty cycle will be changed. The frequency of the output of A 2 will be reduced when it will be shifted towards V E and plus VCC but the amplitude does not depend on the position of R 4. This is the working of sawtooth wave generator using operational amplifier. These are the references. Thank you.