Instrumentation Amplifiers - a quick overview

Introduction

Instrumentation amplifiers (often shortly called INAs) are a workhorse in precision measurement electronics. Whenever you need to extract a small voltage difference in the presence of large common-mode signals they are the tool of choice. In my own projects, I use them regularly in measurement bridges (such as Wheatstone bridges for strain gauges or resistive sensors), in vacuum pressure gauges (for example the Wheastone bridges used for Pirani gauges) and in various other sensor front-ends where both accuracy and high input impedance are crucial.

The idea behind instrumentation amplifiers builds on the classical difference amplifier, but with added refinements that address the practical pitfalls of resistor imbalance, gain adjustment, and input loading. The first monolithic instrumentation amplifier chips appeared in the late 1960s and early 1970s, driven by advances in integrated circuit fabrication that allowed manufacturers to match and laser-trim resistor networks on a single die. Today, devices like Analog Devices AD620 or Texas Instruments INA series make it possible to achieve gains in the hundreds or thousands with excellent common-mode rejection ratio (CMRR), something nearly impossible with discrete resistor networks. This makes them especially easy to use in conjunction with analog to digital converts after the INA output - with a very small number of external discrete components.

Recap: The difference amplifier

So first let’s take a look at the difference amplifier again.

As one can see the resistors form two voltage dividers - once for the non inverting input for input voltage $V_a$ and once on the inverting input for $V_b$. First lets take a look at the voltage on the non inverting input. Here we apply the voltage drop from $V_a$ to $V_o$:

[ \begin{aligned} R_{24} &= R_2 + R_4 \\ I_{24} &= \frac{U_B}{R_2 + R_4} \\ U_{ni} &= R_4 I_{24} \\ &= U_B \frac{R_4}{R_2 + R_4} \end{aligned} ]

Note that this path also carries a leakage current - it places a load on your inputs. This means that this circuit is not a high impedance circuit!

On the inverting input the resistors $R_2$ and $R_4$ form a voltage divider again - this time from voltage $V_A$ to $V_O$:

Setting up the equations again:

[ \begin{aligned} R_{13} &= R_1 + R_3 \\ I_{13} &= \frac{U_A - U_O}{R_1 + R_3} \\ U_{inv} &= R_3 I_{13} \\ &= (U_A - U_O) \frac{R_3}{R_1 + R_3} \end{aligned} ]

Now as usual we know that the operational amplifier drives the output till it reaches equilibrium between the two inputs, i.e. that $U_{inv} = U_{ni}$. This yields

[ \begin{aligned} (U_A - U_O) \frac{R_3}{R_1 + R_3} &= U_B \frac{R_4}{R_2 + R_4} \\ U_A - U_0 &= U_B \frac{R_4(R_1 + R_3)}{R_3(R_2 + R_4)} \\ U_0 &= -U_B \frac{R_4(R_1 + R_3)}{R_3(R_2 + R_4)} + U_A \end{aligned} ]

Now one usually sets - for simplicity - a balanced configuration:

[ \begin{aligned} R_3 = R_4 &= R_{34} \\ R_1 = R_2 &= R_{12} \\ \to U_O &= -U_B \underbrace{\frac{R_{34}(R_{12} + R_{34})}{R_{34}(R_{12} + R_{34})}}_{1} + U_A \\ &= U_A - U_B \end{aligned} ]

On the first glance this already looks like the circuit we want to utilize to amplify voltage differences for example from a measurement bridge. Unfortunately it has a few pitfalls:

• As we have seen the voltage divider on the non inverting side places a load on our sensor or bridge output. This may be undesireable since usually for bridges or sensors like piezos you would like to have very high impedance inputs (i.e. the measurement system should not disturb the measured system).
• The gain of the circuit is pretty hard to adjust - it is of course possible but you have to keep the balance between both resistor chains.
• Balance also introduces another problem: The common mode gain that adds a term $G_{cm} V_{cm}$ to our output. Since we are working with real and not ideal resistors even more expensive ones have deviations of the real resistance - usually at least in the one digit percentage range. It’s very hard to balance such a circuit with discrete components though it’s simple when manufacturing the resistors on a chip die and being able to perform laser trimming during the manufacturing process.

Instrumentation amplifier

This leads us to the instrumentation amplifier that is based on the concept of the difference amplifier - but with an additional input stage. You could build the following circuit of course using discrete components - but then you again get problems with balance mismatches at the different resistors as well having to balance 6 resistors and adding a gain defining 7th resistor. It’s much easier to just use readily available instrumentation amplifiers.

As we can see in the blue dotted rectangle we just have again the difference amplifier as discussed before. We denote the input voltages as $V_a’$ and $V_b’$ for easier setup of our equations. We again assume this operated in balanced operation and thus

[ U_O = U_a' - U_b' ]

So let’s focus on the input circuit on the left. Again we can see that the resistors form a divider ladder

As we can see the outputs of both operational amplifiers in the top and bottom section drive the inputs of the difference amplifier $U_a’$ and $U_b’$. They form the upper and lower end of the resistive divider. The constraint that the operational amplifiers try to reach that their inputs are balanced ($U_{inv} = U_{ni}$) sets the condition that $U_{inv,A} = U_a$ and $U_{inv,B} = U_b$ (except for saturation). This condition and the resistor $R_1$ define the total current in the divider:

[ I_G = \frac{U_A - U_B}{R_1} ]

The same current has to flow over the whole divider (since there is no source or sink attached between $R_2$ and $R_3$. For balance reasons we will assume that $R_{23} = R_2 = R_3:

[ \begin{aligned} R_G &= R_2 + T_1 + R_3 \\ I_G &= \frac{U_{o,A} - U_{o,B}}{R_2 + R_1 + R_3} \\ I_G &= \frac{U_{o,A} - U_{o,B}}{R_1 + 2 R_{23}} \\ \frac{U_{o,A} - U_{o,B}}{R_1 + 2 R_{23}} &= \frac{U_A - U_B}{R_1} \\ U_{o,A} - U_{o,B} &= (U_A - U_B) \underbrace{\frac{R_1 + 2 R_{23}}{R_1}}_{G} \\ U_A' - U_B' &= G (U_A - U_B) \end{aligned} ]

Now applying the known relation for the difference amplifier in the second stage ($U_O = U_A’ - U_B’$) we can conclude:

[ \begin{aligned} U_O &= U_A' - U_B' \\ &= G (U_A - U_B) \end{aligned} ]

As we have seen above we only have to tune the single resistor $R_1$ to tune the gain factor of the whole circuit. Usually readily available instrumentation amplifiers use only one external resistor $R_1$, all other resistors are built on-die into the chip - which allows the manufacturer to manufacture them with way lower tolerances than one can do for discrete components - and for very high precission instrumentation amplifiers the manufacturer can perform laser-trimming of the resistors during manufacturing to balance them.

Conclusion

We have seen how and why instrumentation amplifiers work - the principle is pretty straight forward. You can build a similar circuit with your operational amplifiers and mathematically they are equivalent (except for the gain) to a simple difference amplifier built with a single operational amplifier. The gain you get is:

• High input impedance. This is crucial for example for measurement bridges or piezo sensors (for the latter - do not forget transient protection).
• You can select the gain factor in an extremly simple fashion by just an external resistor. And you don’t have to tune other resistors - a change of the single external resistor does not change common mode voltage or any other balancing of the circuit
• The manufacturer can provide a way more stable and balanced configuration of the feedback resistors than is possible with discrete components (again important for common mode voltage)

When choosing an instrumentation amplifier the most important parameters that you have to look at are:

• Common mode rejection ratio (CMRR)
• Input offset voltage
• Input bias current
• Noise
• Gain bandwidth product and bandwidth

This article is tagged: Tutorial, DIY, Electronics, Physics, Basics, Measurements, OpAmp