Download Complete Design And Construction Of An Instrumentation Amplifier Circuit Using Op-Amp Research Materials (PDF/DOC)
ABSTRACT
Almost all types of sensors and transducers convert real world parameters like light, temperature, weight etc into voltage values for our electronic systems to understand it. The variation in this voltage level will help us in analyzing/measuring the real world parameters, but in some applications like biomedical sensors this variation is very small (low-level signals) and it is very important to keep track of even the minute variation to get reliable data. In these applications, an Instrumentation Amplifier is used. An Instrumentation amplifier amplifies the variation in voltage and provides a differential output like any other op-amps. But unlike a normal amplifier the Instrumentation amplifiers will have high input impedance with good gain while providing common mode noise rejection with fully differential inputs.
TABLE OF CONTENTS
TITLE PAGE
APPROVAL PAGE
DEDICATION
ACKNOWLEDGEMENT
ABSTRACT
TABLE OF CONTENT
CHAPTER ONE
1.0 INTRODUCTION
1.1 BACKGROUND OF THE PROJECT
1.2 AIM OF THE PROJECT
1.3 OBJECTIVE OF THE PROJECT
1.4 SIGNIFICANCE OF THE PROJECT
1.5 PURPOSE OF THE PROJECT
1.6 APPLICATION OF THE PROJECT
1.7 ADVANTAGES OF THE PROJECT
1.8 PROBLEM/LIMITATION OF THE PROJECT
1.9 PROJECT ORGANISATION
CHAPTER TWO
2.0 LITERATURE REVIEW
2.1 REVIEW OF RELATED STUDIES
2.2 REVIEW OF RELATED TERMS
CHAPTER THREE
3.0 CONSTRUCTION METHODOLOGY
3.1 SYSTEM CIRCUIT DIAGRAM
3.2 SYSTEM OPERATION
3.3 CIRCUIT DESCRIPTION
3.4 SYSTEM CIRCUIT DIAGRAM
3.5 CIRCUIT OPERATION
3.6 IMPORTANCE AND FUNCTION OF THE MAJOR COMPONENTS USED IN THIS CIRCUIT
3.7 POWER SUPPLY UNIT
CHAPTER FOUR
RESULT ANALYSIS
4.0 CONSTRUCTION PROCEDURE AND TESTING
4.1 CASING AND PACKAGING
4.2 ASSEMBLING OF SECTIONS
4.3 TESTING
4.4.1 PRE-IMPLEMENTATION TESTING
4.4.2 POST-IMPLEMENTATION TESTING
4.5 RESULT
4.6 COST ANALYSIS
4.7 PROBLEM ENCOUNTERED
CHAPTER FIVE
5.1 CONCLUSION
5.2 RECOMMENDATION
5.3 REFERENCES
It’s okay if you don’t get it now, in this article we will learn about these Instrumentation amplifiers and since these IC’s are relatively expensive than Op-amps we will also learn how to use normal Op-amp like LM385 or LM324 to build an Instrumentation amplifier and use it for our applications. Op-amps can also be used to build Voltage adder and voltage Subtractor circuit.
What is an Instrumentation Amplifier IC?
Apart from normal op-amps IC we have some special type of amplifiers for Instrumentation amplifier like INA114 IC. It is nothing more than few normal op-amps combined together for certain specific applications. To understand more about this lets look into the datasheet of the INA114 for its internal circuit diagram.
As you can see the IC takes in two signal voltages VIN– and VIN+, let’s consider them as V1 and V2 from now for ease of understanding. The output voltage (VO) can be calculated using the formulae
VO = G (V2 – V1)
Where, G is the gain of the op-amp and can be set using the external resistor RG and calculated using the below formulae
Note: The value 50k ohm is applicable only for the INA114 IC since it uses resistors of 25k (25+25 =50). You can calculate the value for other circuits respectively.
So basically now if you look at it, an In-amp just provides the difference between two voltage sources with a gain that can be set by an external resistor. Does this sound familiar? If not take a look at the Differential amplifier design and come back.
Yes!, this is exactly what a Differential amplifier does and if you take a closer look you can even find that the op-amp A3 in the above image is nothing but a Differential amplifier circuit. So in layman terms, an Instrumentation-amp is yet another kind of differential amplifier but with more advantages like high input impedance and easy gain control etc. These advantages are because of the other two op-amp (A2 and A1) in the design, we will learn more about it in the next heading.
Understanding the Instrumentation Amplifier
To completely understand the Instrumentation amplifier, let’s break it down the above image into meaningful blocks as shown below.
As you can see the In-Amp is just a combination of two Buffer op-amp circuit and one differential op-amp circuit. We have learnt about both these op-amp design individually, now we will see how they are combined to form a differential Op-amp.
Difference between Differential Amplifier and Instrumentation Amplifier
We have already learnt how to design and use a differential amplifier in our previous article. Few considerable disadvantage of differential amplifier is that it has very low input impedance because of the input resistors and has very low CMRR because of the high common mode gain. These will be overcome in a Instrumentation amplifier because of the buffer circuit.
Also in a differential amplifier we need to change a lot of resistors to change the gain value of the amplifier but in a differential amplifier we can control the gain by simply adjusting one resistor value.
Instrumentation Amplifier using Op-amp (LM358)
Now let’s build a practical Instrumentation amplifier using op-amp and check how it is working. The op-amp instrumentation amplifier circuit that I am using is given below.
The circuit requires three op-amps all together; I have used two LM358 ICs. The LM358 is a dual package op-amp that is it has two op-amps in one package so we need two of them for our circuit. Similarly you can also use three single-package LM741 op-amp or one quad package LM324 op-amp.
In the above circuit, the op-amp U1:A and U1:B acts as a voltage buffer this helps in achieving high input impedance. The op-amp U2:A acts as a differential op-amp. Since all the resistors of differential op-amp is 10k it acts as a unity gain differential amplifier meaning the output voltage will be the difference of voltage between pin 3 and pin 2 of U2:A.
The output voltage of the Instrumentation amplifier circuit can be calculated using the below formulae.
Vout = (V2-V1)(1+(2R/Rg))
Where, R = Resistor value the circuit. Here R = R2=R3=R4=R5=R6=R7 which is 10k
Rg = Gain Resistor. Here Rg = R1which is 22k.
So the value of R and Rg decides the gain of the amplifier. The value of gain can be calculated by
Gain = (1+(2R/Rg))
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