The Instrumentation Amplifier (IA) resembles the differential amplifier, with the main difference that the inputs are buffered by two Op Amps. Besides that, it is designed for low DC offset, low offset drift with temperature, low input bias currents and high common-mode rejection ratio. These qualities make the IA very useful in analog circuit design, in precision applications and in sensor signal processing.
Figure 1
Figure 1 shows one of the most common configurations of the instrumentation amplifier. Its clever design allows U1 and U2 operational amplifiers to share the current through the feedback resistors R5, R6 and RG. Because of that, one single resistor change, RG, changes the instrumentation amplifier gain, as we will see further. RG is called the “gain resistor”. If the amplifier is integrated on a single monolithic chip, RG is usually left outside so that the user can change the gain as he wishes. One example of such instrumentation amplifier is Texas Instruments’ INA128/INA129.
To minimize the common-mode error and increase the CMRR (Common-Mode Rejection Ratio), the differential amplifier resistor ratios R2/R1 and R4/R3 are equal. (See The Differential Amplifier Common-Mode Error Part 1 and Part 2 for more on this matter.)
Another potential error generator is the input bias current. Although, in most analysis, the input current into an Op Amp is considered zero, in reality that is not the case. A small input current flows into the Op Amp inputs and is converted into voltage by the input resistors. If the resistors are not equal, the voltage difference between the two generates an offset, which is amplified and transmitted at the circuit output. Because of that, R1 is designed to be equal with R3. Similarly, R2 equals R4.
How do we derive the instrumentation amplifier transfer function?
It is well known that the instrumentation amplifier transfer function in Figure 1 is
 |
(1) |
when R5 = R6, R2 = R4 and R1 = R3.
The proof of this transfer function starts with the Superposition Theorem. Let’s make V2 zero by connecting the U2 input to ground, and let’s calculate Vout1 (see Figure 2).
Figure 2
The calculation of Vout1 starts from the differential amplifier transfer function shown in equation (2). U3 is in a differential configuration. If we note the voltage levels at U1 and U2 outputs with V11 and V12 respectively, Vout1 can be written as
 |
(2) |
For the proof of equation (2) see The Differential Amplifier Transfer Function on this website.
To determine V11 and V12 we note that, if V2 is zero, the node between RG and R6 is a virtual ground. This is because U2 sets its output at such a level, so that its inverting input equals the non-inverting input potential. With this observation, one would realize that U1 is in a non-inverting amplifier configuration, with its feedback resistor network R5 and RG connected to a virtual ground.
Therefore, V11 can be deduced from the non-inverting amplifier transfer function:
 |
(3) |
In order to calculate V12, let’s observe that the current that flows through R5 and RG, IG, is the same as the current through R6. That is because there is no other current path. The currents that flow into U1 and U2 inputs are too small to be taken into consideration. Mathematically, we can write that the current through R5 and RG equals the current through R6 as in equation (4). Since the node between RG and R6 is at zero volts, V11 appears as a voltage drop on R5 and RG in series. Also, V12 is the voltage drop on R6, forcing the output of U2 to be driven below ground.
 |
(4) |
Therefore, V12 is
 |
(5) |
Replacing V11 and V12 in equation (2), Vout1 becomes
 |
(6) |
After calculations, and taking into consideration that R5 = R6, the result for Vout1 is as in equation (7).
 |
(7) |
For the second part of the Superposition Theorem, let’s restore V2 and let’s make V1 zero. We will note the output voltage with Vout2, and with V21 and V22 the output voltage of U1 and U2 respectively (see Figure 3).
Figure 3
Vout2 depends on V21 and V22 in a similar manner as Vout1 in equation (2).
 |
(8) |
This time, U2 is in a non-inverting configuration, so that V22 can be written as a function of V2 as in (9).
 |
(9) |
The circuit is symmetric, so we can write a similar equation for V21 and V22 as equation (4) for V11 and V12.
 |
(10) |
and therefore, V21 is
 |
(11) |
Replacing V21 and V22 in equation (8) and after calculations, we find Vout2 as in the following expression.
 |
(12) |
All we need to do now is to add Vout1 and Vout2 to find the instrumentation amplifier transfer function. The result is given in equation (13).
 |
(13) |
Q. E. D.
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If we take a closer look at the instrumentation amplifier transfer function, we note that, if RG is not connected and R2 = R1, the circuit gain becomes one. Changing one single resistor, RG, results in large gain variations, so it gives the analog designer flexibility in his application. This is the reason why the IC manufacturers choose not to integrate RG on the monolithic chip, and also choose to make R1, R2, R3 and R4 equal. As opposed to the differential amplifier, where the user has to change at least two resistors to change the gain, in instrumentation amplifiers one resistor does the job, bringing elegance and simplicity in the analog design.
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Categories: Analog Design, Differential Amplifier, Superposition Theorem
