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Chapter: Linear Integrated Circuits : Applications of Operational Amplifier

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Instrumentation Amplifier using Operational Amplifier

In a number of industrial and consumer applications, one is required to measure and control physical quantities.

Instrumentation Amplifier:

 


Current flowing in resistor R is I= ( V1-V2)/R and it flow through R’ in the direction shown, Voltage  at non-inverting terminal op-amp A3 is R2 V1’/(R1+R2). By superposition theorem,

Vo= (R2/R1) V1+(1+R2/R1)(R2V2/(R1+R2)=R2/R1( V1’-V2’);

 V1’= R’I+ V1=R’/R( V1-V2)+ V1

V2’= R’I+ V1=R’/R( V1-V2)+V2;

V0= (R2/R1)[(2R’/R(V2- V1)+ (V2- V1)] = (R2/R1)[(1+2R’/R)(V2- V1)

In a number of industrial and consumer applications, one is required to measure and control physical quantities.

Some typical examples are measurement and control of temperature, humidity, light intensity, water flow etc. these physical quantities are usually measured with help of transducers.

The output of transducer has to be amplified so that it can drive the indicator or display system. This function is performed by an instrumentation amplifier. The important features of an instrumentation amplifier are

1.        High gain accuracy

2.        High CMRR

3.        High gain stability with low temperature coefficient

4.        Low output impedance

There are specially designed op-amps such as µA725 to meet the above stated requirements of a good instrumentation amplifier. Monolithic (single chip) instrumentation amplifier are also available commercially such as AD521, AD524, AD620, AD624 by Analog Devices, LM363.XX (XX -->10,100,500) by National Semiconductor and INA101, 104, 3626, 3629 by Burr Brown.

In the circuit of figure 6(a), source  V1 sees an input impedance = R3+R4 (=101K) and the impedance seen by source V2 is only R1 (1K). This low impedance may load the signal source heavily.

Therefore, high resistance buffer is used preceding each input to avoid this loading effect as shown in figure

The op-amp A1 and A2 have differential input voltage as zero. For  V1=V2, that is, under common mode condition, the voltage across R will be zero. As no current flows through R and R‘ the non-inverting amplifier.

A1 acts as voltage follower, so its output V2‘=V2. Similarly op-amp A2 acts as voltage follower having output  V1‘= V1. However, if  V1≠V2, current flows in R and R‘, and (V2‘- V1‘)> (V2-  V1). Therefore, this circuit has differential gain and CMRR more compared to the single op- amp circuit of figure 2.10.

The difference gain of this instrumentation amplifier R, however should never be made zero, as this will make the gain infinity. To avoid such a situation, in a practical circuit, a fixed resistance in series with a potentiometer is used in place of R.

Figure (c) shows a differential instrumentation amplifier using Transducer Bridge. The circuit uses a resistive transducer whose resistance changes as a function of the physical quantity to be measured.

The bridge is initially balanced by a dc supply voltage Vdc so that  V1=V2. As the physical quantity changes, the resistance RT of the transducer also changes, causing an unbalance in the bridge ( V1≠V2). This differential voltage now gets amplified by the three op-amp differential instrumentation amplifier.


RB(Vdc)/(RB+RA)= RCVdc/(RC+RT)

 

Applications of instrumentation amplifier with the transducer bridge,

o          temperature indicator,

o          temperature controller and

o light intensity meter .

 

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