In this tutorial you will learn about a Triangular wave generator circuit using OP-AMP IC741. With the use of few resistors and capacitors connected with the Operational Amplifier we will create a Square Wave as output. Here we are going to design two circuits, in which one will requires higher number of components and another is optimized one. And both circuits generate the non sinusoidal wave which is the Triangular wave using Op-Amp.
So, first lets learn briefly about the OP AMP and then create a two circuits with it and next learn how the triangular wave generator circuit works with formulas in detail.
What is an Operational Amplifier OP-AMP?
An operational amplifier (OA or op-amp), is a high-gain amplifier directly coupled, which is generally fed with positive and negative sources, which allows it to obtain excursions both above and below ground or reference point that be considered. It is especially characterized by its response in: frequency, phase change and high gain that is determined by externally introduced feedback. Due to its design, it has a high input impedance (Z) and a very low output impedance. This is the symbol:
To learn more about Op-Amp and its pins refer: Op-Amp IC LM741 Overview.
Basic Triangular wave generator circuit diagram
There are two parts in this 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. Here first part is very simple which is nothing but the square wave generator and the second part is nothing but the integrator using Op-Amp which converts the input square wave into a triangular wave. So lets design the circuit below. To know how to design a square wave generator in detail refer this article: Square Wave Generator
As you can see from the above circuit the circuit till Vo1 is the circuit for the square wave generator. The output of square wave generator is connected as an input to the integrator part using another Operational Amplifier.
Here the output of that first Op-Amp(A1) which is a square wave and second Op-Amp(A2) which is a triangular wave have the same frequencies. Which means the frequency of square wave and frequency of triangular wave are equal, but the frequency depends on the value of resistor R. because according to the square wave circuit the frequency is inversely proportional to R. Both the frequencies depends on the value of resistor R.
Frequency(Square wave) = Frequency(Triangular wave).
From the integrator part 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. And here our R4 is connected across C, which makes R4 equals to 10R3.
Time Period T=R3C2.
This circuit requires 2 Op-amps, 2 capacitors and at least 5 resistors which means the component requirement of this circuit is more. So lets design an optimized circuit for triangular wave generator.
Optimized Triangular wave generator using Op-Amp Circuit diagram
As per the above circuit First Op-Amp A1 inverting terminal is connected to Ground and Non inverting terminal is connected to Voltage divider with R2 and R3. From the Output of A1 we get a Square wave at Vo1. Which is connected to Resistor and capacitor which were connected to second Op-Amp A2 as shown in the above circuit diagram. The Non inverting terminal of A2 is connected to Ground.
At output Vo2 you will get a triangular wave and that is connected as a feedback to R2. Now the requirement of resistors and capacitors is reduced which makes it a triangular wave generator using the minimum components.
Working of Triangular wave generator based on Op-Amp
First lets assume a point P between voltage divider resistors and A1 located on the circuit. A1 has a comparator which will continuously compare this p point with respect to ground that it’s 0Volts. If P goes above or below the 0Volts then we can get the positive or negative saturation voltage at the output of Operational Amplifier A1 this voltage will act as an input for the second operational amplifier A2. P will be compared by this A1 continuously with 0volts and according to the P, above or below that zero value gives the square wave at the output of the first operational amplifier.
let us draw a waveform which indicates the outputs of square wave as well as a triangular wave generator
At first the P value is above the zero volt and we will consider that A1 will give us a positive saturation voltage as a output. As this output act as input for the next operational amplifier, it will give us a negative going Vramp as a output. So we will consider a Vramp signal which give us a negative voltage up to certain value.
The ramp will be negative going ramp and due to this we can consider the conditions of this P depending on R2 andR3. R2 and R3 is voltage divider circuit in which one end is connected to A1 and second end is connected to A2. Initially R3 will be at the positive end of the A1 and it’s another end will be at the negative due to the A2 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 +Vramp then the output of the square wave will switched to positive saturation voltage and this process continues again and again thus we get square wave and triangular wave at the output of operational amplifier A1 and A2 respectively
For this circuit when the output goes from the positive ramp voltage towards negative ramp then a positive voltage is generated through R3 which is positive saturation voltage and negative Vramp will be generated at R2 at that time P will gives us the value 0.
So, the equation can be written as
Peak to peak output voltage Voop is the difference between the +Vramp and -Vramp which is written as
By solving equation we get the Time period T as
The frequency of the triangular wave is given as fo = 1/T.
Frequency Fo is directly proportional to the value of R3 so as the value of R3 increases the frequency will increase and if value of R3 decreases then frequency will decrease.
Lets design a 2KHz Triangular wave generator circuit:
Lets consider our Op-Amp IC741 which has saturation voltage of 10V and supply voltage ±12v. Lets expect the output peak to peak voltage to be 7volts.
Assume R2= 10K ohm, Capacitor C= 0.05μF
So the values which we have are
- Vopp =7V
- Vsat = 10V
- C= 0.05μF
- Fo=2 KHz
Lets calculate R3 for triangular wave generator circuit:
Now the thing we need is to find the value of R1 which we can get from the equation of frequency by substituting the known values in it.
Now lets design a 2KHz triangular wave generator circuit diagram from all the component values which we got
In the above circuit we used a varistor/potentiometer of 50k ohm to get the R3 as 28.5. All all the other components are same as circuit.
Simulation and Output on Proteus software:
After connecting the output Vo2 to the oscilloscope in Proteus software to simulate the circuit we got a beautiful sharp triangular wave as shown below.
If you want the Proteus project file please comment below.
We also tried to connect the first Op-Amp output from this circuit to the oscillator pin B. the 2 waves are shown accurately as shown below.
If you built this circuit and want to test in live refer this article where you will be able to build a DIY Oscilloscope with Raspberry Pi Pico: DIY Smartphone Oscilloscope using Raspberry Pi Pico.
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