Hey friends, welcome to the Kohiki.com ALL ABOUT ELECTRONICS. So, in this article, we will understand the instrumentation amplifier.

**Instrumentation Amplifier Explained**

## Why Instrumentation amplifier is prefered over the differential amplifier in certain applications

Now, this instrumentation amplifier is one kind of differential amplifier but it has very high gain as well as very high common-mode rejection ratio and very high input impedance. And this instrumentation amplifier is used in industrial applications as well as in test and measurement circuits.

So, now in this article, let’s understand how this instrumentation amplifier is different from the differential amplifier which is designed using the op-amp, and in certain industrial applications why this instrumentation amplifier is preferred over the differential amplifier.

So, first of all, let’s see some problems that we can face if we use this normal differential amplifier (Instrumentation Amplifier) in certain industrial applications. So, now in many industrial applications, it is required to measure certain parameters like temperature, flow, humidity, etc. And these parameters are measured using the transducers like RTD, Strain Guage, and Thermocouples.

**Bridge Circuit**

So, these transducers convert the measured parameter into electrical signals. And many times to convert this measured parameter into electrical signals, these bridge circuits are used. For example, if you take the case of RTD, in RTD as the temperature changes the resistance of RTD will also change. So, if we connect this RTD at one of the arms of this bridge circuit then the resistance of this particular arm will change according to the temperature. And accordingly, the voltage that is generated at the particular node will also change.

So, if we measure the voltage across these two terminals A and B, then the voltage will change according to the temperature. And this generated differential voltage can be amplified using the differential amplifier. Now, many times in industrial applications it is quite possible that the location where these measurements are carried out and the locations where these sensors are deployed are quite far away from each other.

So, in this case, it is quite possible that the noise or the interference signal will get superimposed over the measured signal. And to avoid the problem of this noise or interference, many times the signal is measured using the differential measurement technique. For example, if we take the case of this bridge circuit, the output that we are getting through the bridge circuit is also a differential signal. And this differential signal further can be amplified using the amplifier.

**Op-amp As Differential Amplifier**

So, as you can see over here, the output of this bridge circuit is given to this differential amplifier. So, the advantage of this differential amplifier is that whatever common mode noise or the common-mode interference signal which appears at the two terminal of this differential amplifier (Instrumentation Amplifier) should ideally get eliminated. And only the differential signal across the two terminals should get amplified. So, if Va and Vb are the voltages that appear across the two terminal of this differential amplifier, then the output of this differential amplifier can be given by this expression.

So, at the non-inverting terminal, the voltage will be equal to this R4 divided by R3 plus R4 times the voltage Va. And that voltage will get amplified by the factor of 1 plus R2 divided by R1. Similarly, the output voltage because of the input at this inverting terminal will be equal to, minus R2 divided by R1 times this voltage Vb.

So, if we rearrange these terms then we can write this same expression like this. Now, here, if we assume that the ratio of this R4/R3 is equal to this ratio of R2/R1. in that case the output voltage can be given by the expression, R2 divided by R1 times the voltage, Va minus Vb.

So, in this case, as you can see, the output voltage V out will be the difference between the two input signals multiplied by the gain of this circuit. Now, so far we have assumed that there is no common-mode voltage present at the input terminals. But if the common-mode voltage is also present along with the differential input voltage in that case, in general, the output voltage of this differential amplifier (Instrumentation Amplifier) can be given by this expression.

So, the differential input voltage will beget amplified by the differential gain, while the common-mode voltage will get amplified by the common-mode gain. Now, in this case, suppose if the common-mode voltage is also present then we can say that the voltage Va will be equal to this common-mode voltage Vcm plus Vd/2. Likewise, this voltage Vb will be equal to cm minus Vd/2.

So, if we assume that the ratio of R4/R3 is equal to R2/R1, in that case, the output voltage V out will be equal to R2 divided byR1 times this voltage, Va minus Vb. And if we put these two values of Va and Vb, then the output voltage Vout will be equal to R2 divided by R1 times this differential voltage Vd.

So, if we assume that this op-amp is ideal, in that case, this circuit should amplify only the differential input signal. And it should reject the common-mode signal. But even if we assume this op-amp as ideal, then also because of the mismatch between these resistor pairs, the common-mode signal will also present at the output terminal.

So, even if there is a one percent error in any of the resistors, then also the common-mode rejection ratio will get affected. And the common-mode signal will present at the output terminal. And this particular scenario will become critical when you differential input signal is very small and the common-mode signal which is present at the two input terminal is larger than the differential signal. So, let’s take some values of these resistors as well as the input voltages and let’s see how the mismatch between these resistor pairs can affect the output voltage.

So, here let’s assume that the resistor R1and R3 are 1 kilo-Ohm. While resistor R2 is equal to 10 Kilo -Ohm. And here we are assuming that there is a 1percent error in the resistor R4. So, instead of 10 Kilo-Ohm, the value of thisR4 is equal to 10.1 Kilo-Ohm. And the differential input to this differential amplifier is equal to 10 mV. while the common-mode input to this differential amplifier is equal to 1V.

So, in this case, let’s find out the output voltage of this differential amplifier. Now, we know that the output of this differential amplifier (Instrumentation Amplifier) can be given by this expression. And if we put the value of these four resistors in this expression then we will get the output voltage Vout as 10.009 times this voltage Va plus 10 times this voltage Vb. So, now let’s put the value of this Va and in terms of this common-mode voltage and the differential input voltage.

So, if we put the value of this Va and Vbthen we will get the output voltage Vout as 0.009 times common-mode voltage plus 10.0045times this differential input voltage. And if we compare this expression with the expression then we can say that the common-mode gain of this circuit will be equal to0.009. While the differential gain will be equal to 10.0045. And if we find the common-mode rejection ratio that is the ratio of this differential gain and the common-mode gain then that will be equal to roughly around 61 dB.

So, ideally, this common-mode rejection ratio should be infinite, but because of the slight mismatch in any of these resistors, the value of this common-mode rejection ratio has been reduced to 61 dB. Now, let’s see how it will affect the output voltage. So, here the value of this common-mode voltage is equal to 1V. So, this common-mode voltage will get amplified by this factor. And because of that, the output voltage Voutwill be equal to 9mV. While differential input to this circuit is10 mV.

So, it will be amplified by the factor of10. So, at the output, we will have 9mV because of this common-mode signal and almost 100 mV because of this differential input signal. So, as you can see 9 percent error will get introduced in the output voltage. And it will further increase if there is a mismatch between more than 1 resistor or if the common mode signal that is present at the input terminal is more than 1V. So, in that case, this error will further increase.

So, as you can see over here, the mismatch between the resistor can affect the output drastically and it can reduce the common-mode rejection ratio of this differential amplifier (Instrumentation Amplifier). So, the one way to solve this problem is to use the monolithic ICs for this differential amplifier. So, one can use readily available differential amplifier ICs like AD 629 and INA 106.

So, in these ICs the resistor is fabricated in the IC itself. And in these ICs the ratios of these resistors are matched quite accurately. So, using these ICs you can increase the common-mode rejection ratio and you can minimize the error in the output voltage. So, even if we use such monolithic ICs for this differential amplifier, then also we will face the problem of the input impedance.

Because if you see over here, the input impedance at this inverting terminal is equal to R1. While the input impedance of this differential amplifier (Instrumentation Amplifier) at this non-inverting terminal is equal to R3 + R4. Instrumentation Amplifier So, if these input impedances are comparable to the resistance of this bridge circuit, in that case, the differential input that is available at the two input terminals will also reduce.

**input impedance**

So, it is highly required that the input impedance of this differential amplifier should be very high. So, one way we can solve this problem is to use the buffers before the inputs. So, if we use the buffer circuits then we can increase the input impedance of this differential amplifier.

So, in this way, by using these buffer circuits along with these monolithic ICs, we can solve the problem of this input impedance as well as the problem of this mismatch between the resistor pairs Instrumentation Amplifier. But the problem with these monolithic ICsis that gain that is provided by these ICs are fixed. So, you cannot change the gain of this differential amplifier. And in your application.

**Instrumentation amplifier circuit explained with the derivation**

if you want to change the gain then using these ICs you cannot change the gain. So, one way we can avoid this problem is by using the non-inverting op-amp instead of these buffer circuits. So, the same circuit can be represented like this. So, in these circuits, the initial gain can be provided by these non-inverting buffers. But the problem with this circuit is that, along with the differential input signal, the common-mode input signal will also get amplified.

So, you will not get any improvement in the signal to noise ratio. Apart from that in this circuit, you also need to match these four resistors, because if there is a slight mismatch in these resistors the once again the common-mode rejection ratio of the circuit will get compromised. Instrumentation Amplifier So, these are the problems that you face with these differential amplifier circuits (Instrumentation Amplifier). And the one way we can avoid this problem is by using the instrumentation amplifier. So, in terms of the circuit, if you see, there is only one difference between this circuit and this circuit.

So, over here this resistor Rg is common between the two op-amps. Now, the advantage of this configuration is that only the differential input signal will get amplified by these op-amps. While the common-mode signal will pass as it is. So, this instrumentation amplifies available in IC format and except for this resistor Rg, all the resistors are fabricated internally. And because they are fabricated internally, they are highly matched. So, just by connecting this resistor Rg externally, we can set the gain of this instrumentation amplifier (Instrumentation Amplifier).

So, for this configuration, if we assume thatR5 is equal to R6 and the ratio of this R4/R3 and R2/R1 are matched in that case, the output voltage Vout can be given by this expression. So, now let’s derive this expression and during the derivation, Instrumentation Amplifier let’s also understand how this circuit will amplify the differential input signal and it will pass the common-mode signal as it is. And using this how we can improve the common-mode rejection ratio of the circuit.

Now, in this circuit, let’s assume that the voltage at this node is equal to VB’ and the voltage at this node is equal to VA’. So, the output voltage Vout can be given by the expression R2/R1 times this voltage (VA’ – VB’), provided the resistance R4/R3 that is equal to R2/R1. So, first of all, we need to find the difference between these VA’ and VB’ voltages.

Now, if you observe this circuit, the voltage at this node will be equal to VB because here we are assuming that all the op-amps are ideal. So, whatever voltage appears at this terminal, the same voltage will also appear at this terminal. Likewise, the voltage at this node will be equal to voltage VA.

So, from this we can say that the current that is flowing through this resistor Rg will be equal to VB- VA divided by this resistor Rg. And let’s say this current is equal to Rg. So, we can say that the current Ig will be equal to VB- Va. divided by this resistor Rg. Now, the same current Ig will also flow through this resistor R5 and R6, because here we are assuming op-amps are ideal. So, no current is flowing into the op-amp terminals.

So, we can say that the same current Ig is flowing through this resistor R5, Rg, and R6. And using this we can find the difference between this voltage VB’ and VA’. So, we can say that VB’ -VA’ will be equal to this current Ig times resistor (R5 + Rg+ R6). And if we assume that this resistor R5 is equal to R6, in that case, we can say that VB’ – VA’ will be equal to Ig times (2*R5+Rg).

So, now if we put the value of this current from this expression then we can write VB’ -VA’ will be equal to (VB -VA), multiplied by 1 plus two times R5, divided by this resistor Rg. Or we can say that VA’ – VB’ will be equal to (VA – VB) times, one plus 2 times R5 divided by this resistor Rg.

So, this is the voltage that will be present between the two input terminals of this differential amplifier. And the output voltage Vout will be equal to R2 divided by R1 times this voltage (VA’ -VB’). That is equal to R2 divided by R1 times, one plus 2 times this resistor R5, divided by Rg times this voltage (VA- VB)So, in this way, this differential input will get amplified by this differential amplifier.

So, now let’s also understand how this instrumentation amplifier will pass the common voltage as it is and it will only amplify the differential input signal. So, now let’s say, the common-mode voltage VC is present at these two terminals. So, the same voltage will also present at these two terminals. Now, because the voltage at these two terminals is the same, so no current will flow through this resistor Rg.

And because of that, we can say that no current is also flowing through this resistor R5 and R6. And in a way, we can say that the same voltage is also present at this terminal and this terminal as well. so, we can say that for the common-mode voltage these two amplifiers (Instrumentation Amplifier) will act as a buffer or unity follower. While it will amplify only the differential input signals.

So, in this way by using this instrumentation amplifier the common-mode rejection ratio of the circuit can be further increased. And these instrumentation amplifiers (Instrumentation Amplifier) is particularly useful in an environment where the differential input signal is very small and the common-mode input signal which is present at the two terminal is larger than the differential input signals. And that is why this instrumentation amplifier is preferred over the differential amplifier in industrial as well as the test and measurement circuits.

**FAQ**

## YouTube Video

So Here Are A YouTube Video Based On Instrumentation Amplifier, Which Was Uploaded By Ekeeda

So, I hope in this article you understood the importance of this instrumentation amplifier and why this instrumentation amplifier is preferred over the differential amplifier in certain applications.

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