My understanding of the instrumentation amplifier is it is used to

1. amplify the weak signal from a sensor (e.g. temp/pressure/flow rate, etc.) and

2. to remove noise from the sensor signal. Thus the instrumentation amplifier sends amplified noiseless sensor signal to the ADC (analog to digital converter) which then relays its digital copy to the microcontroller for processing.

Q1. Is the above understanding correct?

The following questions are related to the schematic diagram of the instrumentation amplifier

Q2.1 Is V1 the sensor analog signal?

Q2.2 then where is the input V2 coming from that acts as an input to the 2nd voltage follower/buffer stage of the instrumentation amplifier?

 

No. Instrumentation amplifier does not remove noise. To do that you need a noise filter.

An in-amp does two things. Firstly, it presents a high input impedance so that it does not degrade the signal from the sensor. Secondly, it uses differential inputs in order to eliminate a signal that is common to both inputs.

The main purpose of an in-amp is to reject common mode signal, and this is a characteristic called CMRR, common mode rejection ratio.

Why two inputs?
An in-amp is commonly use with sensors that can provide balanced complementary outputs. A classic example is a strain gauge in the form of a Wheatstone bridge circuit. Other examples are microphones with balanced outputs and XLR cabling. Studio and stage microphones tend to need very long cables. Each wire in the cable becomes a long wire antenna that picks up interference signals, primarily AC line frequency. This interference is likely to be common in both wires. Hence by subtraction, one can remove the unwanted signal.

 

so an instrumentation amplifier cannot be used for a temp sensor which sends an analog signal and not differential signals. Is that right?

 

so an instrumentation amplifier cannot be used for a temp sensor which sends an analog signal and not differential signals. Is that right?

NO. ANY signal is a differential signal - all voltages are the difference in potential between two points.
For a single-ended signal, the second point is ground/earth/0V or whatever you might like to call it.
The trouble is that 0V is a theoretical concept - no two 0V connections are at exactly the same voltage for all manner of reasons (current flowing in wires causing voltage drops being most common) and that leads to errors.
So the 0V that your temperature sensor is connected to is not going to be the same 0V as the ADC that is reading it, unless you are very, very careful with your wiring or pcb layout.
So the instrumentation amplifier subtracts the voltage on V1 (the temperature sensor) from the voltage on V2 (the local ground to the temperature sensor), to produce an output voltage that is referenced to the 0V at the instrumentation amplifier, which can be placed next to the ADC so that there is some hope of both their 0V connections being the same.
It is also a pretty safe bet that the noise picked up in the wire to V1 is the same as the noise picked up in the wire to V2, so by subtracting one from the other it subtracts the noise picked up in the wiring.

 

Understand that an in-amp does not remove noise from a sensor.
It eliminates interference picked up by the connecting cable from the sensor to the in-amp.

If you have a sensor with single ended output, the output of the sensor will be connected via a long cable to V1.

What is connected to V2?
V2 will be connected to GND, but not the GND of the in-amp circuit.
The GND at the sensor will be connect via the other wire in the long cable to V2 input.
In this manner, the in-amp does what it is designed to do, i.e. to eliminate the common mode signal picked by a long sensor cable.

 

An instrumentation amplifier (like any other amplifier) will ADD noise to the signal.
You use it because (you hope) it will add LESS noise than any alternative.
You still have to pay attention when calculating the noise from the source resistance.
Noise voltage = √f.√((vn^2+(R.in)^2)
where f is the bandwidth, vn is the amplifier voltage noise in V/√Hz, in is the amplifier current noise in A/√Hz and R is the source resistance.

 

Thanks for the responses. Here is what I have understood based on these responses:

1. Op-amp compares/subtracts the voltage signal of the sensor's signal pin from the sensor's ground potential.
2. Since this o/p voltage is then fed to the ADC where it compares this voltage against its own ground potential and generates the corresponding binary pattern. Ideally, great care must be taken to ensure the ADC and the sensor's return/ground lines potentials are the same i.e. 0V. Otherwise, the ADC may misinterpret the input voltage level. For example

Let's assume the sensor's ground is at 0V, but the ADC's ground is at 1V above the sensor's ground. If the o/p of the differential amplifier is 3V (which is with respect to the sensor's ground) , then the ADC will see it as 2V w.r.t its ground potential which is 1V above the sensor ground. Thus it is important to make sure both grounds are at the same potential.

3. Since the differential amplifier is using a high input impedance buffer in its first stage, this buffer won't draw a high current from the sensor/transducer bcz of its high input impedance. This makes sure that transducer/sensor doesn't get overloaded and deviate from its characteristic.

 

Think about it the other way around - it might help.
You're perfectly correct with point 3.
Point 2 is correct, and is what you do if you don't have a diff-amp.

Let's call the ADC ground 0V, because it's the only ground it knows about. It can only measure with respect to its own ground.
Let's call the sensor's ground Vrandom, because it could be all over the place.
The sensor's produces a voltage Vsensor. So the output on the sensor pin is Vrandom+Vsensor.

Connect the sensor output (Vrandom+Vsensor) to diff-amp input V1
Connect the sensor ground (Vrandom) to the diff-amp input V2
The diff-amp output is (V1-V2) with respect the diff-amp ground.
Connect the diff-amp ground to the ADC ground by the shortest track possible so that they are the same voltage. It's then safe to assume that the diff-amp ground and the ADC ground are the same
V1-V2 = Vsensor: the ADC measures the sensor voltage and ignores the random voltage at the sensor's ground.

 

Think about it the other way around - it might help.
You're perfectly correct with point 3.
Point 2 is correct, and is what you do if you don't have a diff-amp.

Let's call the ADC ground 0V, because it's the only ground it knows about. It can only measure with respect to its own ground.
Let's call the sensor's ground Vrandom, because it could be all over the place.
The sensor's produces a voltage Vsensor. So the output on the sensor pin is Vrandom+Vsensor.

Connect the sensor output (Vrandom+Vsensor) to diff-amp input V1
Connect the sensor ground (Vrandom) to the diff-amp input V2
The diff-amp output is (V1-V2) with respect the diff-amp ground.
Connect the diff-amp ground to the ADC ground by the shortest track possible so that they are the same voltage. It's then safe to assume that the diff-amp ground and the ADC ground are the same
V1-V2 = Vsensor: the ADC measures the sensor voltage and ignores the random voltage at the sensor's ground.

This explanation has helped me understand the working principle. Thanks for your help.