I'm curious as to exactly how in-amps are designed at a transistor level. They're often shown as being made of three op amps, but due to how specifically the amplifiers are being used I expect that there may be some differences between them, to optimize performance for certain things, and some things (the reference side of current mirrors, perhaps) might be shared between them.

I strongly doubt that they just copy the same op amp design three times and add resistors.

So, to sum it up into a question: How are instrumentation amplifiers designed, at a transistor level? How are the three component amplifiers specialized for their limited tasks, and what, if any, circuitry is shared between them?


2 Answers 2


From the top of my head:

There's basically a few things you'd want to have from a "perfect" amplifier, but which are hard to realize within a single one:

  • High Common-Mode rejection ratio (CMRR)
  • High Input Impedance
  • Low Output Impedance
  • High Gain
  • Low Noise figure
  • Low Output bias

In a three-Opamp differential amplifier (and I'd assume that things like the INA128 actually are made of three opamps!), the input impedance of the output opamp doesn't really matter – so you can use something with a lower input impedance, but with a high output drive strength. In fact, I'd speculate that it might even make sense to use BJTs for the input stage differential amplifier of that third opamp – you'd be sinking exactly what need, and:

That third opamp would ideally have a high CMRR – and it's, I've been told[citation needed] a bit easier to use laser-trimmed on-die resistors to make this thing a little more symmetrical if these resistors are lower value due to more current flowing through them.

So, wild guess: Third opamp input differential stage: BJTs, rest FET, with a relatively fat FET pair at the output.

The two input Opamps wouldn't need as much CMRR (in fact, none, as long as they react identically), but a high input impedance – an ideal use case for FET inputs.

Friis' noise formula tells us that these two mostly define the noise figure of the overall circuit, so it's at least likely the stages after the input stage are also BJTs. A significant amount of the overall voltage gain might happen here (for exactly Friis' noise formula reasons).

I mentioned laser-trimmed resistors: Since you need to really get the resistors in an instrumentation amplifier right, the streaks of weakly doped silicon that make up resistors on ICs are in this case designed to be "zappable" with a laser during production – meaning they can be adjusted after / while being measured by calibrated equipment.

Because I can:

The three opamps for which I could find die shots (which I'm not competent to interpret:

  • \$\begingroup\$ I'd assume it may be possible to skip the output current gain stage on the input amplifiers, no? Even a BJT-input op amp doesn't need that much bias current. \$\endgroup\$
    – Hearth
    Commented Mar 11, 2019 at 19:02
  • 1
    \$\begingroup\$ Also, some in-amps have a sort of "isolation-amp-lite" feature in that they allow you to set a DC bias on the output (particularly useful for high-side current sensing, for example); is this specialization only something that goes into the output op amp, or do the input ones need to be modified to make this work, or... Honestly, this should maybe be its own question entirely. \$\endgroup\$
    – Hearth
    Commented Mar 11, 2019 at 19:05
  • \$\begingroup\$ I'm not competent enough to interpret those die shots either, but they sure do look fascinating. \$\endgroup\$
    – Hearth
    Commented Mar 11, 2019 at 19:42

Regardless of the IC, be it digital or analog, there are a quite a few simplifications that can be made because:

  • The internal nodes don’t have to drive pads.

Driving a pad requires a minimum load of ~2pF and a few nH. while internal nodes can be a few tens of fF. While internally you can get away with a few pA (in some cases even fA), externally you rarely go below a few mA. Driving any pad always requires comparatively massive transistors.

  • Internal circuitry is fixed, and thus well-defined.

A pad can connect to nearly anything. The designer has to consider a wide spectrum of possibilities. That is not the case for internal circuitry.

While “true” op amps require 3 stages to achieve very high gain and very low output impedance, depending on the application it’s possible to design amplifiers with only one stage. The output impedance only has to be low enough for the internal loads, and the gain high enough to satisfy a design objective. It’s rather common to have internal integrated amplifiers with gains of 1000 or less, and output impedances of 10kohm or more.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.