 video lecture on opam as a triangular wave generator. At the end of this session student will be able to describe the working of triangular wave generator using operational amplifier. The last lecture we have seen the opam as a square wave generator where we got the non-synosodal nature of that wave that is square wave. Now the another non-synosodal wave that is a triangular wave can be generated using this operational amplifier. So there are two parts in the circuit where we are going to design the triangular wave generator. So the first part is the part which is producing the square wave and the second part is the part which will convert square wave into triangular wave. Now pause the video for a while and identify these two circuits using opam which will give the output as a square wave as well as output as a triangular wave okay. So first part is very simple it is nothing but the square wave generator and the second part is nothing but the integrator using opam integrator using opam which will converts the input square wave into a triangular wave. First draw the equivalent circuit for this. So first we can design a square wave generator we will connect the resistor capacitor and a voltage divider circuit to non-inverting terminal that is R2 and R1. This is nothing but the square wave generator which gives the output as a square wave. Now the same output will act as an input for the next circuit that is nothing but the integrator using operational amplifier. Connect a capacitor here here the resistor R3 capacitor C2 and one more resistor R4. So this is what the circuit which will convert the square wave into a triangular wave as a output. So here it is nothing but the output 1 we will call it as a output 2 means this circuit will give the square wave as well as triangular wave okay. Let us call it as A1 and it is A2 the output of A1 will act as an input for A2 here require 2 opams, 2 capacitors and at least 5 resistors in this circuit means the component requirement of this circuit is more. In this the output of that first opam which is a square wave and second output which is a triangular wave both waves are having the equal frequencies means the frequency of square wave and frequency of triangular wave are equal. But the frequency depends on the value of this resistor R because according to the square wave circuit the frequency depends on R inversely. There is an inverse relation between the frequency and the resistor value means overall both the frequencies depend on the value of R. This is the integral time component that is R3 C2 depends on the output square wave means R3 C2 should be equal to time t where t is nothing but the time period of the square wave. So R3 C2 value must be equal to t and here R4 is connected across C2 the R4 value should be nearly equal to 10 R3 means these extra components are required in this circuit due to which the component requirement of circuit is increased. So there is another way to design the triangular wave generator using less components. So let us see how to design it. The first A1's inverting terminal is grounded let us call it as R2 and R3. We get a square wave as a output which is nothing but Bo1. This will be extended with the resistor and one capacitor. Let us call it as C1 and R1. This is at inverting terminal and the non-inverting terminal of A2 will be grounded. This is output Vo2 which will give the triangular wave as a output and this is connected as a feedback to again A1. Now requirement of resistors and capacitors is reduced. It is a triangular wave generator using the minimum component in this circuit. Let us see the working of this triangular wave generator. So first let us call this point as point P. So A1 comparator will continuously compare this P point with respect to ground that is 0 volt and if P goes above or below the 0 volt then we can get the positive or negative saturation voltage at the output of this A1. This voltage will act as a input for the second operation amplifier A2. P will be compared by this A1 continuously with 0 and according to the P above or below that 0 value the square wave is generated at the output of the first operational amplifier. So let us draw a waveform which indicates the outputs of square wave as well as a triangular wave generator. So first this P value is above the 0 volt and we will consider that A1 will give us a positive positive saturation voltage as a output. As this output acts as a input for the next operational amplifier it will give us a negative going ramp as a output. So we will consider a V ramp signal it will give us a negative voltage up to certain value. The ramp will be a negative going ramp and due to this we can consider the conditions of this P depending on R2 and R3. R2 and R3 is voltage divider circuit in which one end is connected to the A1 and second end is connected to A2. Initially R3 will be at the positive end of the A1 and another end will be at the negative due to the A2's negative going ramp. So at a certain point the P will fall below the 0. Due to this the output of the square wave generator falls to the negative saturation voltage. And after that it will be held in the same condition due to the negative going ramp and after this condition the negative going ramp will increase towards the positive saturation voltage. At a certain level when it crosses the positive value that is plus V ramp then the output of the square wave can be switched to positive saturation voltage. And this continues it gives the output. Thus we can get the square wave and triangular wave at the output of operational amplifier A1 as well as A2. For this circuit when the output goes from the positive ramp voltage towards negative ramp then there is a positive voltage generated through R3 which is positive saturation voltage and the negative V ramp will be generated at R2. At that time P will give us the value 0 volt. So the equation can be written as V ramp upon R2 is equal to minus positive saturation through R3. The value for the positive ramp voltage can be given as minus R2 upon R3 into negative saturation voltage. So the voltage that is peak to peak output voltage is nothing but the difference between these two ramp voltages. So it is nothing but and it can be given as R2 upon R3 into P saturation. So the peak to peak output voltage can be given as twice R2 upon R3 into saturation voltage. Now the output at the integrator side or the integrator's output can be taken as equation for the integrator's output. But here V in is nothing but minus saturation voltage and VO is nothing but the peak to peak output voltage. So we will put these values in this equation from 0 up to T by 2 and minus V saturation DT into negative V saturation into T by 2. Therefore peak to peak voltage is given as V saturation upon R1 C1 into T by 2. Therefore the time period can be calculated as twice R1 C1 upon V saturation. The value of VO peak to peak from the equation of the square wave we will get T is equal to twice R1 C1 upon into twice R2 upon R3 into V saturation. So overall the time period can be calculated as 4 R1 R2 C divided by R3. So this is nothing but the time period of the triangular wave. And from this the equation for output frequency can be given as 1 by T is equal to R3 upon 4 R1 R2 into C where F0 is directly proportional to the value of R3. So the time period as the value of R3 increases the frequency will increase and if value of R3 decreases then frequency will decrease. It is the frequency of the triangular wave which depends on the value of resistor R3. This is the overall working of the operational amplifier as a square wave generator. These are the references. Thank you.