# Why use a three opamp instrumentation amplifier?

This TI application note shows this typical instrumentation amplifier (InAmp) with three opamps (p.4): Further down the page the following two opamp InAmp is shown: The former is more common (I think) and easier to understand, but is there a good reason to prefer this one over the other? After all it's more expensive since a third opamp is needed.
Also, the two opamp version doesn't have R3 or R4 in its $V_{OUT}$ equation. Is the amplification really independent of their value?

• R3 = R4 == unity gain. The value doesn't matter (within reason) Aug 24 '11 at 13:09
• @Matt - If it's only "within reason", that means that it does matter. And just looking at the schematic my common sense says it does, too. In my answer I have a corrected equation which has R3 in it. Aug 24 '11 at 15:30
• @Matt - Besides, R3 = R4 doesn't imply unity gain. It's much more complicated than that, since RG + R1 are parallel to R4, and then there's R2 as well. Aug 24 '11 at 15:37
• I guess TI is trying to make more money ;) Aug 25 '11 at 3:31

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.

• "influences the currents through the resistors" and "passes through two op-amps" are the same thing. Aug 24 '11 at 20:37
• @endolith That's a better way of putting it, I think. There will be propogation differences in the two op-amp circuit vs. the three op-amp circuit from an input signal perspective. Aug 25 '11 at 15:01

The two op-amp design has in principle some drawbacks in comparison with the three op-amp design: the common mode input range is lower and the matching of the resistors is more critical if a high CMRR is to be achieved (Graeme, 1973). However, these drawbacks become less important at high values of overall gain

http://www.biosemi.com/publications/artikel7.htm

On the other hand, this claims the CMR is better for the two-op-amp version:

The two-op-amp configuration can provide higher CMR, especially in low-voltage, single-supply applications.

I'm not sure which is correct. Also:

The V1 signal must propagate through two op amps, but the V2 signal propagates through one op amp. When input signals contain frequencies greater than the flat portion of the op-amp gain curve (Reference 2), the V1 signal attenuates more than the V2 signal. The unequal attenuation causes the signal to unbalance, and CMR reduces at high frequencies.

http://www.edn.com/article/492092-Don_t_fall_in_love_with_one_type_of_instrumentation_amp.php#ref

R3 and R4 aren't in the equation because of $R1 = R2$ and $R3 = R4$, much like the first circuit where R1 and R3 aren't in the equation (again, because $R1 = R2$ and $R3 = R4$). The missing terms simplify out because of the equalities.