Since R1 = R2, for the 2-opamp version the equation for \$V_{OUT}\$ simplifies to

\$V_{OUT} = \left( Sig_+ - Sig_- \right) \times \left( 2 + \dfrac{2 R2}{RG} \right) \$

and indeed there's no sign of R3 or R4. So I made the calculation again, and I found the following, different equation (I don't include the derivation because too much TeX involved):

\$V_{OUT} = \left( Sig_+ - Sig_- \right) \times \left( 2 + \dfrac{R1 + R3}{RG} \right) \$

which I like better because at least we have a term R3 here. Of course if \$R1 = R2 = R3 = R4\$ both equations are equivalent, but this condition isn't mentioned with the schematic. (I'd appreciate it if somebody can confirm that my equation is indeed correct.)

Madmanguruman noted that the gain is minimum 2 for this configuration, which also shows in the above equations. I'm not sure this is a serious restriction, since instrumentation amplifiers are usually used for much higher gains than 2, especially for strain gauge and other Wheatstone bridge measurements. Gains of 100 to 500 are common.

~~IMO Madmanguruman's other observation that \$Sig_-\$ passes through two opamps is not correct: the inverting input of the top opamp is kept at \$Sig_+\$, and \$Sig_-\$ only influences the currents through the resistors.~~

It looks like the 2-opamp version is a good alternative for the classic version in most applications, since, like you said, you save an opamp.

**edit**

In integrated form you don't gain (no pun intended) much from choosing a two-opamp version. The INA122 costs USD 6.86 while the three-opamp INA129 costs USD 7.35, both Digikey prices.

It is possible, but the CMRR (common mode rejection ratio) will be worse than a monolithic InAmp. In a monolithic InAmp, the resistors R1-R3 are matched. Also, they are on the same die at the same temperature.

By the way, the above holds independent of the OpAmp model.

A Designer's Guide to Instrumentation Amplifiers has more on inner workings and rationale behind the design of InAmps.

Building an InAmp with 3 separate OpAmps could be a neat learning exercise**. You can plug in different OpAmps and compare results. You can compare CMRR with a factory-made monolithic InAmp.

** As Matt wrote in his comment. He had beat me to it.

## Best Answer

First, many InAmps are integrated into one IC, also the three-opamp versions. Then the cost of the extra opamps isn't that high. Consider the extra cost of less than a mm^2 real estate against the total cost to create an IC.

Next, I've never seen a single-opamp inamp. There are two-opamp and three-opamp versions, and there's a reason why you can't make one with a single opamp: you need high impedance for both inputs, and that means neither can have feedback.

editYou probably mean a differential amplifier, like The Resistance comments below. Notice that the 2 input opamps are configured as non-inverting amplifiers, and that there's no feedback to the input pins, like you would have for an inverting amplifier. This way you get high-impedance amplifying buffer-amps before the actual differential amplifier.