# Instrumentation Amplifier Circuit using Op-Amp

**Instrumentation Amplifier Circuit using Op-Amp:**

Virtually all sensors and transducers convert real-world parameters like light, temperature, and weight into voltage signals that can be interpreted by our electronic systems. The variation in voltage levels serves as a valuable tool for analyzing and measuring real-world properties. Nevertheless, in certain scenarios, such as biomedical sensors, these variations can be extremely subtle, referred to as low-level signals. In such cases, it becomes imperative to meticulously monitor even the slightest changes to ensure the acquisition of reliable data. This is where the utilization of an Instrumentation Amplifier finds its application across various domains.

Instrumentation amplifiers, often referred to as INOs or in-amps, share the ability to amplify voltage variations and yield a differential output similar to other operational amplifiers (op-amps). However, what sets instrumentation amplifiers apart is their capacity to offer high input impedance, sufficient gain, and the capability to effectively suppress common-mode noise through fully differential inputs, which sets them apart from standard amplifiers. If this concept seems somewhat complex at the moment, don’t worry. In this discussion, we will delve into the realm of instrumentation amplifiers, and despite their higher cost compared to op-amps, we will also explore how we can construct an instrumentation amplifier for our specific needs using a conventional op-amp like the LM385 or LM324. Additionally, we will touch upon the creation of voltage adder and voltage subtractor circuits using op-amps.

**What is an Instrumentation Amplifier IC?**

Aside from standard op-amps, we have various unique amplifiers for instrumentation, such as the INA114 IC. It’s nothing more than a couple of ordinary op-amps coupled for a specific application. Let’s have a look at the INA114’s internal circuit diagram in the datasheet to learn more about this.

As you can see the IC takes in two signal voltages V_{IN}– and V_{IN}+, let’s consider them as V1 and V2 from now for ease of understanding. The output voltage (V_{O}) can be calculated using the formulae

V_{O}= G (V2 – V1)

Where G is the gain of the op-amp and can be set using the external resistor R_{G }and calculated using the below formulae

G = 1+ (50kΩ/RG)

**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, if you look at it today, an In-amp is just a device that supplies the difference between two voltage sources with a gain that can be controlled by an external resistor. Is this something you’ve heard before? If not, go back and look at the Differential amplifier design.

Yes, a Differential amplifier does just that, and if you look closely, you can see that the op-amp A3 in the preceding image is nothing more than a Differential amplifier circuit. In layman’s words, an Instrumentation-amp is a type of differential amplifier with additional benefits such as high input impedance and easy gain adjustment. These benefits are due to the other two op-amps in the design (A2 and A1), as we will see.

**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 circuits and one differential op-amp circuit**. We have learned about both these op-amp designs individually, now we will see how they are combined to form a differential Op-amp.

**Difference between Differential Amplifier and Instrumentation Amplifier**

In our previous article, we learned how to create and utilize a differential amplifier. Due to the input resistors, the differential amplifier has an extremely low input impedance and a very low CMRR due to the high common-mode gain. The buffer circuit in an Instrumentation amplifier will overcome these issues.

We also need to modify a lot of resistors in a differential amplifier to change the gain value, whereas in a differential amplifier, we can vary the gain by merely adjusting one resistor value.

**Instrumentation Amplifier Circuit Diagram 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.

I used two LM358 ICs to make the circuit, which required three op-amps in total. We’ll need two of the LM358s for our circuit because it’s a dual package op-amp, which means it contains two op-amps in one package. Similarly, three single-package LM741 op-amps or one quad-package LM324 op-amp can be used.

The op-amps U1:A and U1:B in the above circuit operate as a voltage buffer, allowing for a high input impedance. The U2:A op-amp is a differential op-amp. Because the differential op-resistors amp’s are all 10k, it operates as a unity gain differential amplifier, with the output voltage equal to the difference 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))

**Simulation of Instrumentation Amplifier Circuit**

As you can see the input voltage V1 is 2.8V and V2 is 3.3V. The value of R is 10k and the value of Rg is 22k. Putting all these values in the above formulae

Vout = (V2-V1)(1+(2R/Rg))= (3.3-2.8)(1+(2x10/22))= (0.5)*(1.9)= 0.95V

The output voltage is 0.95V, which is consistent with the simulation above. As a result, the above circuit has a gain of 1.9 and a voltage difference of 0.5V. So, in this circuit, the difference between the input voltages is measured, multiplied by the gain, and the output voltage is produced.

You’ll also see that the input voltages V1 and V2 appear across the resistor Rg, which is owing to the Op-amps U1:A and U1:negative B’s feedback. This guarantees that the voltage drop across Rg is equal to the voltage difference between V1 and V2, causing equal current to flow through resistors R5 and R6, resulting in the voltage on pin 3 and pin 2 on the op-amp U2:A being equal. If you measure the voltage before the resistors, you can observe the actual output voltage from the op-amps U1:A and U1:B, whose difference equals the output voltage in the simulation.

**Testing the Instrumentation Amplifier Circuit on Hardware**

Enough Theory lets actually build the same circuit on a breadboard and measure the voltage levels. My connection setup is shown below.

I utilized the breadboard power supply that we had previously constructed. This board has the ability to deliver both 5V and 3.3V. I’m powering both op-amps with 5V and using 3.3V as the signal input voltage V2. Using my RPS, I set the other input voltage V2 to 2.8V. The gain of the circuit will be 1.9 because I used a 10k resistor for R and a 22k resistor for R1. The difference voltage is 0.5V, and the gain is 1.9, resulting in an output voltage of 0.95V, which is measured and presented in the image with a multimeter. The video below demonstrates the entire operation of the instrumentation amplifier circuit.

Similarly, using the formulas given above, you can alter the value of R1 to set the gain as needed. Because the gain of this amplifier can be readily regulated with a single resistor, it is frequently employed in audio volume control circuits.

I hope you were able to follow the circuit and learned something beneficial. If you have any questions, please post them in the comments box below or use the forum to get a quicker response.