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How to Analyze a Three-Op-Amp Instrumentation Amplifier

Jason Sachs●May 4, 2025

There is a useful circuit called an instrumentation amplifier which has several subtle advantages. Back in 2014 I wrote an article called How to Analyze a Differential Amplifier, which is a circuit that comes up so often that every electrical engineer should just know it.

This gives you a gain of \( K \) from input to output if \( R3/R1 = R4/R2 = K \):

$$V_\text{out} - V_\text{ref} = K\left(V_P - V_N\right)$$

This article is available in PDF format for easy printing

If you don’t get the resistor ratios exactly equal — which always occurs to some extent in practice — then the common-mode voltage “leaks through” the amplifier with a gain that is proportional to the error factor \( \delta = \frac{1}{2K}\left(\frac{R_3}{R_1} - \frac{R_4}{R_2}\right) \), which describes the degree of resistor mismatch so that \( \frac{R_3}{R_1} = K\left(1+\delta\right) \) and \( \frac{R_4}{R_2} = K\left(1-\delta\right) \). This mismatch is undesirable; as I mentioned in that article: (\( V_1 \) and \( V_2 \) are equivalent to \( V_P \) and \( V_N \), respectively)

Let’s say \( V_2 = 40\text{V} + 10\text{V} \sin 377t + f(t) \). That’s 40V DC plus 10V at 60Hz AC, plus some other nasty signal \( f(t) \) we don’t even know about. Heck, maybe \( f(t) \) is some Rick Astley song being played in a loop over and over again, loudly. (Never gonna give, never gonna give…) And it’s coming from some evil voltage source out of your control. Yuck. And attached to this is a millivolt signal \( V_\text{src} \) that’s a secret coded message telling you the location of Jimmy Hoffa’s body, Amelia Earhart’s airplane, and those 13 paintings stolen in 1990 from the Isabella Stewart Gardner Museum. So \( V_1 = V_2 + V_\text{src} \), and you have access to \( V_1 \) and \( V_2 \). All you have to do is separate the signal from the crap. (Inside we both know what’s been going on…) And fame and fortune are yours! A differential amplifier will let you amplify this signal and translate it up or down in voltage relative to any reference you care about, whether it’s earth ground or a 2V reference or some other waveform you prefer.

…

In a differential amplifier with an ideal op-amp and no filter, there’s no dynamic equations that have frequency-dependent terms, so there’s no “natural” separation into common-mode and differential-mode dynamics. But we find that common-mode and differential-mode are useful perspectives: common-mode represents the average of two similar circuit voltages or currents. Differential-mode represents the difference of those voltages or currents. In our case, the common-mode voltage has Rick Astley singing loudly (I just wanna tell you how I’m feeling…), whereas the differential-mode voltage has no Rick Astley and contains only the small coded message that’s going to tell you where Jimmy Hoffa and Amelia Earhart and the 13 stolen paintings are. We want to get rid of the common-mode voltage. We want to keep the differential-mode voltage.

The good news, as I mentioned in 2014, is that this differential amplifier does a pretty good job without needing anything in the way of extra effort on our part:

The only really worrisome term is the common-mode gain, which represents the signal we’re trying to keep out (Never gonna say goodbye…) and fortunately does not increase with differential gain K. Whether we have K = 10 or K = 100 or K = 1000, the common mode gain is approximately \( 2\delta \). If we use 1% resistors, the worst-case common-mode gain due to resistor mismatch is 0.04 (when R1 and R4 are too high by 1% and R2 and R3 are too low by 1%).

You want the Grungy Algebra details? Go back and read How to Analyze a Differential Amplifier.

What’s Wrong With the Basic Differential Amplifier?

The biggest disadvantage of this differential amplifier circuit is that the input impedance is dependent on R1 and R2 — which is fine if you’re amplifying the voltage across a current sense amplifier with sub-ohm output impedance. But maybe you’re measuring the voltage difference between electrodes in some chemical or biological system, and they have an output impedance in the 100 kΩ – 1 MΩ range. That’s going to mess up the gain and the resistor matching, as well as allow current to leak in or out of the input system.

One simple way to fix this is to use a voltage buffer on each input.

Can we do better, though?

Instrumentation Amplifiers: Can We Do Better?

An instrumentation amplifier improves on the single-op-amp circuit in a couple of ways. Here is the “classic” three-op-amp instrumentation amplifier:

It basically sticks a differential preamplifier onto the front end of our differential amplifier, to apply a gain to the voltage \( V_{AB} = V_A - V_B \). The analysis of this preamp is fairly simple:

  • The op amps try to make the voltages at C and D match the voltages at A and B, respectively
  • Therefore the current \( I_1 = V_{AB}/R_6 \)
  • \( V_P = V_A + \frac{R_5}{R_6}V_{AB} \)
  • \( V_N = V_B - \frac{R_7}{R_6}V_{AB} \)
  • \( V_{PN} = V_P - V_N = \frac{R_5 + R_6 + R_7}{R_6}V_{AB} \)

So the preamp gain \( K_p = \frac{R_5 + R_6 + R_7}{R_6} \). We could, for example, choose R5 = R7 = 10 KΩ and R6 = 2.21 KΩ, for a preamp gain \( K_p = 10.05 \).

Now, you might look at this circuit and think:

  • What’s the big deal; why would I do that? How is this better than just buffering the inputs?
  • The instrumentation amplifier has more resistors, so it must be more likely to have mismatch problems… and if I want a gain of 1000, what does it matter if it comes out of one stage or two?

Well, there are a couple of advantages of doing it this way, which I’ll explain, after a bit of a tangent.

Who Invented the Three-Op-Amp Instrumentation Amplifier?

I need to take a moment to acknowledge the history of this circuit. I found out about it years ago, from The Art of Electronics, by Horowitz & Hill. It appears in several notable references, including:

But these articles and books don’t cite sources for the circuit, and they’re not the first reference.

The earliest reference I can find is Application Brief D9 from George A. Philbrick Researches Inc., from September 1963:

Philbrick was the pioneering company behind the K2W electronic operational amplifier that debuted commercially in 1952. George Philbrick developed the first analog “computor” in the late 1930s, and was involved in early op-amp projects during the Second World War with Loebe Julie, before founding George A. Philbrick Researches Inc. in 1946. Several eminent analog engineers (notably Bob Dobkin, Bob Pease, Dan Sheingold, Jim Williams) worked at Philbrick before moving on to Analog Devices, Linear Technology, or National Semiconductor. Philbrick’s op amps were, of course, initially based on vacuum tubes, but by November 1960 they announced the P2 Differential Operational Amplifier, designed by Robert Malter, based on germanium transistors and a varactor-input bridge + transformer-coupled modulation/demodulation architecture.

Bob Pease describes the P2 design in amazing historical and technical detail in Jim Williams’s 1991 book, Analog Circuit Design: Art, Science, and Personalities. The P2 was expensive but successful:

So, here’s a little do-hickey, a little circuit made up of just about as much parts as a cheap \$12 transistor radio, but there was quite a lot of demand for this kind of precision. How much demand? Would you believe \$227 of demand? Yes! The P2 originally started out selling for \$185, but when the supply/demand situation heated up, it was obvious that at \$185, the P2 was underpriced, so the price was pushed up to \$227 to ensure that the people who got them were people who really wanted and needed them. So, the people who really wanted a P2 had to pay a price that was more than 1/8 the price of a Volkswagen Beetle—that was back when \$227 was a real chunk of money!

Pease edited this chapter down to a greatly abridged version in his Electronic Design column, What’s All This Profit Stuff, Anyhow?. He goes on to say:

Still, it is an amazing piece of history, that the old P2 amplifier did so many things right—it manufactured its gain out of thin air, when just throwing more transistors at it would probably have done more harm than good. And it had low noise and extremely good input current errors—traits that made it a lot of friends. The profits from that P2 were big enough to buy us a whole new building down in Dedham, Massachusetts, where Teledyne Philbrick is located to this day. The popularity of the P2 made a lot of friends, who (after they had paid the steep price) were amazed and delighted with the performance of the P2. And the men of Philbrick continued to sell those high-priced operational amplifiers and popularize the whole concept of the op amp as a versatile building block. Then, when good low-cost amplifiers like the µA741 and LM301A came along, they were accepted by most engineers. Their popularity swept right along the path that had been paved by those expensive amplifiers from Philbrick. If George Philbrick and Bob Malter and Dan Sheingold and Henry Paynter and Bruce Seddon hadn’t written all those applications notes and all those books and stories, heck, Bob Widlar might not have been able to give away his µA709s and LM301s! And the P2—the little junk-box made up virtually of parts left over from making cheap transistor radios—that was the profit-engine that enabled and drove and powered the whole operational-amplifier industry.

In 1961, Philbrick ran numerous ads for the P2, in magazines like Electronics and Electronic Design and Electronic Industries, extolling its virtues: NO TUBES, NO CHOPPERS, NO COMMON MODE ERROR, VIRTUALLY NO INPUT CURRENT (LESS THAN ONE TEN THOUSANDTH OF ONE MICROAMP) AND ALMOST NO NOISE… NOTHING BUT PERFORMANCE

Eventually, solid-state op amps from other manufacturers like Burr-Brown entered the market; in 1962 Philbrick came out with simpler silicon transistor + resistor/capacitor op amps like the P65. These were still made with discrete components, however.

Texas Instruments announced the SN521 and SN522 integrated circuit op amps in December 1962 (Operational Amplifiers Are Getting Smaller) but these look bare-bones, and nobody appears to acknowledge their existence after the late 1960s, so they can’t have been too much of a success.

Then in 1964 and 1965 Fairchild Semiconductor debuted Bob Widlar’s µA702 and µA709 integrated circuit op amps. The µA709 originally sold for \$50, but after volumes ballooned, Fairchild dropped the price to \$5 for the commercial version.

As Electronic Design later stated in a 1967 article (The ‘709’: Model T of the op amps, Electronic Design 15, July 19, 1967):

The 709 was a smash hit from the start. Among the engineering departments all over the country that spotted the 709’s potential was one at the Bendix Corp., Teterboro, N. J. There, an engineer in the flight control laboratory brought the circuit to the attention of the man who shortly became the largest single buyer of monolithic operational amplifiers. “We spotted the 709 in December of 1965, a month after it was introduced,” says Robert Reade, Bendix’s assistant purchasing director. “The spec sheet looked good. The 709 was far better than the monolithic op-amps on the market at the time—Texas Instruments’ 524A and Westinghouse’s 161Q. We immediately pumped the 709 right into our designs with every expectation that the price would fall drastically. In 1966 we used 80,000.”

A few pages later, the magazine covered Widlar’s LM101 at National Semiconductor (The op amp conjurer strikes again, Electronic Design 15, July 19, 1967).

It’s hard to believe from today’s perspective that modern IC op amps were available in 1967. Yes, we now have lower costs, rail-to-rail input and output, and low-leakage CMOS, with faster gain-bandwidth and better offset voltage, but the building block is the same essential thing.

Meanwhile, Philbrick had published an Applications Manual for Philbrick Octal Plug-In Computing Amplifiers as early as 1956, and Burr-Brown a Handbook of Operational Amplifier Applications by 1963, with all sorts of linear and nonlinear circuit goodness, but neither mentions the three-op-amp instrumentation amplifier. Philbrick’s 1964 bingo-board-style application note with the somewhat bombastic title, Typical Operational Amplifier Applications For Modelling, Measuring, Manipulating, and Much Else, by “DHS” (Daniel H. Sheingold), mentions the circuit, highlighted below in green:

One point I want to make here is that engineering to implement technology and engineering to use technology follow slightly different tracks, both advancing rapidly as time goes by. We’ve had all sorts of useful know-how when it comes to operational amplifiers, what they can be used for, how to predict the relationship of open- and closed-loop bandwidth, and how to make them stable, for at least fifteen or twenty years prior to the introduction of the sub-\$5 integrated circuit op amp. And it’s worth remembering that. Sometimes the “old” knowledge isn’t as primitive as you might think, it’s just that the available technology to work with has been limited.

Anyway — where did the circuit in Philbrick’s 1963 design brief come from? Was it from one of the geniuses at Philbrick Researches, or something previously known from the vacuum-tube era and used as a biomedical amplifier? Or a technical paper in Proceedings of the IRE? Or undocumented tribal knowledge among electrical engineers at nearby MIT or Lincoln Labs? A forthcoming answer seems unlikely, since many of the dramatis personae of the time are now deceased.

Benefits of the Three-Op-Amp Instrument Amplifier

But I’m here to talk up the benefits of this architecture, and I’ll show my version of the circuit again, just to stick with a particular notation.

We already know this circuit supports high-impedance inputs, because the input leakage into the op amps from terminals A and B is small. Bipolar op amps are typically in the 100 pA – 10 nA range, and CMOS op amps have leakage current that is sub-picoamp.

What else do we gain? (pun intended)

  • High bandwidth — by splitting the total gain into two stages, the bandwidth is higher. Suppose we have 10 MHz GBW op amps, and we want a total gain of 100.

    • If we put all the gain in the second stage, and the first stage just acts as a high-impedance buffer, then the buffers have 10 MHz bandwidth, and the second stage has a bandwidth of 100 kHz = 10 MHz / 100, for a net bandwidth of about 100 kHz.
    • If we split the gain so that each stage has a gain of 10, then first and second stages both have a bandwith of 1 MHz = 10 MHz / 10, and the net bandwidth of the entire circuit is somewhere around 500 kHz. (More on that in another article!) That’s five times higher!

  • Good sensitivity to resistor mismatch — I’ll show why this is true in a minute. This includes

    • Extremely low common-mode sensitivity to resistor mismatch among R5/R6/R7
    • Roughly equal sensitivity to resistor mismatch among R1/R2/R3/R4 for overall CMRR, compared to the single-stage differential amplifier.

  • Overall circuit gain can be adjusted with changes to one resistor (R6). Want the gain doubled? Then just reduce R6 by half. (With the four-resistor diff-amp circuit, it’s difficult to adjust the gain since we have to adjust at least two resistors with the same relative ratio.)

To understand the resistor sensitivity, we need to analyze the common-mode behavior.

Common-mode analysis of the first stage

As I did in the single-stage differential amplifier, we can define the common-mode and differential-mode voltages \( V_\text{cm} \) and \( V_\text{dm} \):

$$ \begin{aligned} V_A &= V_\text{cm} + \tfrac{1}{2}V_\text{dm} \\ V_B &= V_\text{cm} - \tfrac{1}{2}V_\text{dm} \\ V_{AB} &= V_A - V_B \\ &= V_\text{dm} \end{aligned} $$

Then the transfer function from input to first stage output becomes

$$ \begin{aligned} V_P &= V_A + \tfrac{R_5}{R_6}V_{AB} \\ &= V_\text{cm} + \tfrac{1}{2}V_\text{dm} + \tfrac{R_5}{R_6}V_\text{dm} \\ V_N &= V_B - \tfrac{R_7}{R_6}V_{AB} \\ &= V_\text{cm} - \tfrac{1}{2}V_\text{dm} - \tfrac{R_7}{R_6}V_\text{dm} \\ V_{PN\!,\text{dm}} &= V_{PN} = V_P - V_N\\ &=\tfrac{R_5 + R_6 + R_7}{R_6}V_\text{dm} \\ V_{PN\!,\text{cm}} &= \tfrac{V_P + V_N}{2} \\ &=V_\text{cm} + \tfrac{R_5 - R_7}{2 R_6}V_\text{dm} \end{aligned} $$

This can also be viewed as a matrix:

$$ \begin{bmatrix} V_{PN\!,\text{dm}} \\ V_{PN\!,\text{cm}} \end{bmatrix} = \begin{bmatrix}\frac{R_5+R_6+R_7}{R_6} & 0 \\ \frac{R_5 - R_7}{2R_6} & 1 \end{bmatrix} \begin{bmatrix} V_\text{dm} \\ V_\text{cm} \end{bmatrix} $$

And here’s where one advantage is visible:

  • the gain from the common-mode of \( (V_A,V_B) \) to the common-mode of \( (V_P,V_N) \) is exactly one
  • the gain from the common-mode of \( (V_A,V_B) \) to the differential-mode of \( (V_P,V_N) \) is exactly zero

In other words, if we slide \( V_A \) and \( V_B \) up by one volt, \( V_P \) and \( V_N \) each also increase by one volt, and the difference between them stays unchanged.

There is some small contribution from differential-mode input to common-mode output, but it is proportional to the mismatch between R5 and R7, and will be relatively small. (Furthermore, all that does is affect the output a little bit by the differential input, which means there is a small change in differential gain, which we don’t really care about much.) If we define \( \delta_{57} = \frac{R_5 - R_7}{R_5+R_7} \), then the differential-to-common mode gain becomes

$$ \begin{aligned}\frac{\Delta V_{PN\!,cm}}{\Delta V_{dm}} &= \frac{R_5 - R_7}{2R_6} \\ &= \frac{R_5 - R_7}{R_5+R_7}\cdot\frac{R_5+R_7}{2R_6} \\ &= \delta_{57}\tfrac{G_p-1}{2} \end{aligned} $$

So, for example, if we have worst-case 1% resistors with R5 = 10.1KΩ and R7 = 9.9KΩ and R6 = 2.21KΩ, then our preamp gain is 10.05, mismatch \( \delta_{57} = 0.01 \), and the differential-to-common mode gain is approximately 0.01 × 9/2 = 0.045. This is tiny; if we have input differentials in the 10 mV range then it changes the common-mode output of the first stage by 0.45 mV, which is negligible.

Common-mode analysis: three-op-amp instrumentation amplifier vs. simple differential amplifier

Now we’re ready to look at the whole thing. We’re going to compare the common-mode gain of the three-op-amp instrumentation amplifier with the simple differential amplifier that has the same overall differential gain.

Suppose we want a total gain of 100, and have worst-case 1% resistors.

Simple differential amplifier

For K = 100 in one stage, we have R1 = 0.99KΩ, R2 = 1.01KΩ, R3 = 101KΩ, R4 = 99KΩ.

The mismatch factor \( \delta = \frac{1}{2K}\left(\frac{R_3}{R_1} - \frac{R_4}{R_2}\right) = 0.02 \), and the common-mode gain is approximately \( 2\delta = 0.04 \).

A change in the input common-mode by 1V gives us a 40mV output change, which would be the same output change as an input change of only 0.4mV; the intended circuit gain is 2500 times larger than the common-mode gain.

Three-op-amp instrumentation amplifier

We’ll pick a gain of \( K_p = 10 \) for the preamp and a gain of \( K = 10 \) for the differential amp stage:

  • R5 = 10.1KΩ, R7 = 9.9KΩ, R6 = 2.21KΩ
  • R1 = 0.99KΩ, R2 = 1.01KΩ, R3 = 10.1KΩ, R4 = 9.9KΩ

For the output stage, the mismatch factor \( \delta = \frac{1}{2K}\left(\frac{R_3}{R_1} - \frac{R_4}{R_2}\right) = 0.02 \) just as in the simple differential amplifier, with a common-mode gain of approximately \( 2\delta = 0.04 \). The input stage has a common-mode-to-common-mode gain of exactly one, so the net common-mode gain of the whole circuit is also approximately 0.04.

This means the common-mode gain of both circuits is the same, if they have the same overall signal gain (100 in both cases) and the resistor mismatch of R1/R2/R3/R4 is the same. (There is an “approximately” here since the common-mode gain of the differential amplifier is approximately 2δ, not exactly 2δ.)

As far as the R5/R7 mismatch goes, there is that small differential-to-common-mode gain of the first stage (0.045), followed by the common-mode gain of 0.04 of the second circuit, for an overall differential gain error of 0.0018, but this is 55000 times smaller than the gain of 100, much smaller than any gain errors caused by the relative matching ratio of R6 to R5 and R7, so it’s not worth worrying about.

If you don’t trust my algebra, we can confirm numerically:

resistors1 = dict(R1=1e3,
                  R2=1e3,
                  R3=10e3,
                  R4=10e3,
                  R5=10e3,
                  R6=2.21e3,
                  R7=10e3,
                  description='Perfect resistors')
resistors2 = dict(R1=0.99e3,
                  R2=1.01e3,
                  R3=10.1e3,
                  R4=9.9e3,
                  R5=10e3,
                  R6=2.21e3,
                  R7=10e3,
                  description='Mismatched resistors R1-R4')
resistors3 = dict(R1=0.99e3,
                  R2=1.01e3,
                  R3=10.1e3,
                  R4=9.9e3,
                  R5=9.9e3,
                  R6=2.21e3,
                  R7=10.1e3,
                  description='Mismatched resistors R1-R7')
resistors4 = dict(R1=1.01e3,
                  R2=0.99e3,
                  R3=9.9e3,
                  R4=10.1e3,
                  R5=10e3,
                  R6=2.21e3,
                  R7=10e3,
                  description='Mismatched resistors R1-R4 opposite direction')
resistors5 = dict(R1=0.99e3,
                  R2=1.01e3,
                  R3=101e3,
                  R4=99e3,
                  R5=10e3,
                  R6=1e10, # open circuit
                  R7=10e3,
                  description='Preamp gain 1, diffamp gain 100 with mismatched R1-R4')
def inamp3(VA,VB,Vref,R):
    R1 = R['R1']
    R2 = R['R2']
    R3 = R['R3']
    R4 = R['R4']
    R5 = R['R5']
    R6 = R['R6']
    R7 = R['R7']
    I1 = (VA-VB)/R6
    VP = VA + R5*I1
    VN = VB - R7*I1
    Oplus = ((Vref/R3)+(VP/R1))/(1/R3 + 1/R1)
    Vout = Oplus + R4/R2 * (Oplus - VN)
    return Vout

for resistors in [resistors1, resistors2, resistors3, resistors4, resistors5]:
    print(resistors['description'])
    for Vcm,Vdm in [(0,0),
                    (1,0),
                    (0,0.01),
                    (1,0.01)]:
        VA = Vcm + Vdm/2
        VB = Vcm - Vdm/2
        Vout = inamp3(VA,VB,Vref=0,R=resistors)
        print("Vcm=%.4f, Vdm=%.4f, Vout=%.4f" % (Vcm,Vdm,Vout))
Perfect resistors
Vcm=0.0000, Vdm=0.0000, Vout=0.0000
Vcm=1.0000, Vdm=0.0000, Vout=-0.0000
Vcm=0.0000, Vdm=0.0100, Vout=1.0050
Vcm=1.0000, Vdm=0.0100, Vout=1.0050
Mismatched resistors R1-R4
Vcm=0.0000, Vdm=0.0000, Vout=0.0000
Vcm=1.0000, Vdm=0.0000, Vout=0.0357
Vcm=0.0000, Vdm=0.0100, Vout=0.9869
Vcm=1.0000, Vdm=0.0100, Vout=1.0226
Mismatched resistors R1-R7
Vcm=0.0000, Vdm=0.0000, Vout=0.0000
Vcm=1.0000, Vdm=0.0000, Vout=0.0357
Vcm=0.0000, Vdm=0.0100, Vout=0.9869
Vcm=1.0000, Vdm=0.0100, Vout=1.0226
Mismatched resistors R1-R4 opposite direction
Vcm=0.0000, Vdm=0.0000, Vout=0.0000
Vcm=1.0000, Vdm=0.0000, Vout=-0.0370
Vcm=0.0000, Vdm=0.0100, Vout=1.0234
Vcm=1.0000, Vdm=0.0100, Vout=0.9864
Preamp gain 1, diffamp gain 100 with mismatched R1-R4
Vcm=0.0000, Vdm=0.0000, Vout=0.0000
Vcm=1.0000, Vdm=0.0000, Vout=0.0388
Vcm=0.0000, Vdm=0.0100, Vout=0.9804
Vcm=1.0000, Vdm=0.0100, Vout=1.0192

Here the common-mode gain of all the cases with worst-case 1% mismatched resistors R1-R4 is approximately 0.04 as predicted. (Actual common-mode gain is about 0.036 for diff gain of 10, and about 0.039 for diff gain of 100; there’s probably a factor of \( K/(K+1) \) in there somewhere.)

What’s the catch?

Okay, there’s got to be some downsides here.

First, you need three op amps rather than one, but that’s the price of having high input impedance.

Actually, there is a two-op-amp instrumentation amplifier, which provides high input impedance, and has some of the same advantages as the three-op-amp circuit… here I just have to quote Kitchin & Counts in A Designer’s Guide to Instrumentation Amplifiers:

Disadvantages of this circuit include the inability to operate at unity gain, a decreased common-mode voltage range as circuit gain is lowered, and poor ac common-mode rejection. The poor CMR is due to the unequal phase shift occurring in the two inputs, VIN1 and VIN2. That is, the signal must travel through amplifier A1 before it is subtracted from VIN2 by amplifier A2. Thus, the voltage at the output of A1 is slightly delayed or phase-shifted with respect to VIN1.

For some reason I just don’t like this circuit. I think the biggest difference of the three-op-amp circuit is that the two-stage split gives you the ability to improve overall bandwidth, which you don’t get with the two-op-amp circuit.

The second disadvantage, which applies to both the two-op-amp and three-op-amp instrumentation amplifier circuits, is that the effect of offset voltage depends on the difference between offset voltage of the two input op amps, rather than just on one amplifier, so we can expect a factor of \( \sqrt{2} \) increase if the offset voltages are random and uncorrelated. (The second stage of the three-op-amp instrumentation amplifier has less of an impact, since the signals are already amplified before they see the offset voltage.)

The third issue, which can be a disadvantage, is that you have to be aware of the voltage ranges at the output of the first stage pre-amp, and make sure that neither op amp saturates. This is easier to see graphically:

The input voltage \( V_{AB} \) is expanded by the first-stage gain to \( V_{PN} \), and it’s important to know the common-mode range of the input. For example, suppose the first-stage gain is 10, and the input voltage signal \( V_{AB} \) can be as high as ±100mV with a common-mode voltage range of -1V to +2V. Then \( V_P \) and \( V_N \) can be anywhere between -1.5V and +2.5V, so we need the Vdd and Vss range to include that range with some extra design margin to cover the effects of input offset voltage and resistor mismatch, along with whatever output voltage limitations the op amps have. (Rail-to-rail op amp outputs are usually MOSFET output stages, and can get to within millivolts of the supply rails, as long as they don’t have to drive much current.) This is not a good circuit to use with single-supply op amps, unless you can guarantee the common-mode voltage range is safely away from the ends of the supply range.

For applications which use a Wheatstone bridge topology (for example, strain gages), this is not a problem. As long as the resistances R1 – R4 of the bridge are roughly equal, the bridge output nodes A and B should be close to the middle of the supply range.

Resistor bridge circuits, especially if they are high-impedance, are ideal candidates for an instrumentation amplifier.

And that’s about all I have to say for today’s article!

Wrapup

Today we looked at the three-op-amp instrumentation amplifier, along with some historical context, and the advantages this circuit has over the single-op-amp differential amplifier:

  • High input impedance
  • High bandwidth, by splitting the total gain into two stages
  • Good sensitivity to resistor mismatch, with very low sensitivity to first-stage resistor mismatch
  • Gain adjustability with one resistor

This is an easy circuit to analyze, and it’s worth understanding why these advantages exist. Hope you find it useful!

References

There are a number of references for the three-op-amp instrumentation amplifier.

First and foremost is A Designer’s Guide to Instrumentation Amplifiers, by Charles Kitchin and Lew Counts, from Analog Devices.

Here are a few others, including the ones I cited earlier:

© 2025 Jason M. Sachs, all rights reserved.

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The 2025 Embedded Online Conference
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