# How to Design LM358 Op-Amp Practical Integrator

Here LM358 op-amp based practical integrator is designed. Practical integrator are circuits which integrates or adds the input voltage over the period(inverse of input signal frequency) of the input signal to produce added up output voltage. Integrator circuit can be passive(using resistor and capacitor) or active integrator(using op-amp). Here active integrator is designed with LM358 op-amp which is a common op-amp(operational amplifier).

Recommended Prerequisites

The theory of basic/ideal and practical integrator was previously explained in details. See the following tutorials.

The theoretical practical integrator circuit diagram is shown below.

But in practice many things has to be considered to design an integrator. The nature of input signal and desired output signal, the input signal amplitude, the output signal swing, biasing of the op-amp, type of supply(single or dual supply) effects the design of an integrator. As such the above theoretical practical integrator circuit will only apply for limited conditions.

For the LM358 op-amp based integrator design, let us consider that the input signal amplitude is 100mV and frequency is 1KHz. Let us use single supply voltage 5V for LM358 op-amp. Also let us consider that the output signal should swing around 2.5V. Let the DC gain of an integrator be 10.

The LM358 based integrator circuit diagram is shown below.

Below is the calculation of the above circuit values.

Since the DC gain of the integrator is 10 which means,

$$A = \frac{R_{F}}{R_{1}} =10$$

Let $$R_{1} = 1k\Omega$$ then $$R_{F} = 10\times1K\Omega$$ = $$10K\Omega$$

For proper integration if  $$f=1KHz$$ is signal frequency and $$f_{a}$$ is the breakover frequency of the integrator then,

$$f \geq 10 f_{a}$$

and therefore,  $$f_{a} = \frac{f}{10} = \frac{1000}{10} = 100Hz$$

Now for practical integrator we have,

$$f_{a} = \frac{1}{2\pi R_{F}C_{F}}$$

or,  $$C_{F} = \frac{1}{2\pi f_{a}R_{F}}$$

or,  $$C_{F} = \frac{1}{2\times3.14\times100\times10\times10^3} = 0.159\mu F$$

therefore, $$C_{F} \simeq 0.1\mu F$$

Next we calculate the input coupling capacitor. Let the lower cutoff frequency be 10Hz then,

$$f_{cc} = \frac{1}{2 \pi R_{1} C_{1}} = 10Hz$$

or, $$C_{1} = \frac{1}{2\times3.14\times1\times10^3\times10} = 9.95 \mu F$$

therefore, $$C_{1} \simeq 10\mu F$$

The biasing resistor R3 and R4 sets the bias voltage at the midpoint of 0V to 5V which is 2.5V. The value of R3=R4=4.7Kohm.

Next we calculate the capacitor C2 which sets the cutoff frequency for the noise from the supply. The cutoff frequency should be set as low as possible. Let the cutoff frequency be 7Hz. Then we have,

$$f_{c} = \frac{1}{2 \pi (R_{3}||R_{4}) C_{2}} = 7Hz$$

where, $$(R_{3}||R_{4}) = \frac{R_{3}\times R_{4}}{R_{3}+R_{4}} = 2.35K \Omega$$

therefore,

$$C_{2} = \frac{1}{2 \times 3.14 \times 2.35 \times 10^3 \times 7} = 9.68 \mu F \simeq 10 \mu F$$

For a square wave of 100mV amplitude and frequency of 1KHz input the output is a triangle wave. The following picture shows the input and output signal on oscilloscope(computer simulation) for the above designed LM358 op-amp integrator circuit.

Next see how to build this integrator on breadboard and test it with Matlab/Simulink in real time LM358 Op-Amp Integrator Test with Matlab Simulink Oscilloscope.