I want to make a digitally-tunable gain (~1 to 1000x), low-noise instrumentation amplifier. There are integrated ones (e.g. AD8557) but the input noise is too high. My current plan is to use an ultra-low-noise in-amp (e.g. INA849) and make the external gain-control resistor digitally tunable (\$R_G\$ below, open to ~6 \$\Omega\$). enter image description here enter image description here

Digital potentiometers are no good because they generally don't cover the multiple decades of resistance values needed; Analog switch/multiplexer (e.g. TMUX1109) + multiple fixed resistors might work, but: my inputs (-IN/+IN) can have a common DC offset between ~+-5 V, whereas the multiplxer only works if either/both ends of \$R_G\$ (the "signal" path) is between ~-0.5 to +6 V.

So my question: Is there a multiplexer that works with the signal path at arbitrary voltage? I can imagine that a optocoupler-style component might work (LED + photoconductor) due to the complete isolation between control and signal, but the optocouplers I can find either have too large of a resistance, or are for digital only.

I am totally open to other designs. Thanks!

  • \$\begingroup\$ Be very aware that adding circuit complexity to the RG pins can result in really bad HF noise due to increases in parasitic capacitance. Working with devices that have a really low bias current is also important. \$\endgroup\$
    – Andy aka
    Nov 1, 2021 at 9:10

1 Answer 1


Welcome Eric,

what you plan to do is a logical step in realizing a low noise In-amp. For the switches, there are certainly plenty of options. I will go through them below:

Assuming you are using a +/-12 V supply (or similar), you also want your switches to be able to handle such common mode voltages.

CMOS analog MUXes

The simplest option is probably CMOS analog switches. E.g. the DG409 is an industry standard 4:2 MUX available from many manufacturers that can handle +/-15 V. Btw it is better to use switches on both sides of \$R_G\$, which is why I recommend using 4:2 MUXes. The reason is the switch capacitance slightly attenuates the gain of the corresponding input op-amp at higher frequencies and that will look like a fake differential signal after the in-amp. If you add similar switches to both nodes, both op-amps will experience a similar high frequency attenuation granting slightly better CMRR.

Better yet is to divide \$R_G\$ in two halfs and place the switch in between. If you want low-noise you are probably going to use rather low feedback resistances. So the switch on-resistance should be <10 Ohm. Many such parts are available, e.g. from Analog, Maxim, TI, Vishay etc. Low R switches also have higher C typically because they are MOSFETs internally.


One downside of CMOS switches is that they are referenced to the supply rails. Therefore, when the signal voltage changes, the switch on-resistance also changes. If you want to built the in-amp like you plan to do, the switches carry the full current flowing between the two input buffers. A changing switch resistance will therefore change the gain depending on the signal level. This is also known as harmonic distortion.

There are a plethora of optically isolated MOSFET switches as you suggested, the ones you are looking for a usually called solid-state relays with MOSFET output. Make sure they have this kind of output configuration:

enter image description here

The output side floats on the signal level, eliminating the dependence of the on-resistance on the signal level. These switches also have off-capacitance, so the same argument about using two switches instead of one applies.

As for CMOS switches, on-resistance can be very low (even well below 1 Ohm) at the cost of higher off-capacitance. Typical suppliers would be e.g. Toshiba, Panasonic, Omron, Vishay, IXYS etc.

If you need very low on-resistance and very low off-capacitance, mechanical relays are great but also quite complex to implement. You want sealed ones if your want to handle signals reliably long-term.

Better in-amp topology

The in-amp topology that you want to build is ideal for ICs because gain can be set using one resistor. But it is a bit worse if you want various gains, because your switches have to carry all the current flowing between the two input op-amps. Building a discrete in-amp allows you to avoid that issue and select switches with lower capacitance and lower current rating by doing the following:


simulate this circuit – Schematic created using CircuitLab

Another advantage is that you get to choose the input op-amps according to your needs. A disadvantage is that you have to also build the difference amp stage which requires extremely well matched resistors for good CMRR. But one can of course use an integrated difference amp here, or even an integrated in-amp with fixed gain.

  • 1
    \$\begingroup\$ Very thorough answer, nothing to add ... except ... that R1/5 ladder doesn't have to be switched symmetrically, especially if the following diff amp id good, but it does eat into voltage headroom. You can play R, 2R, 4R games with the resistors in it, switching one at a time for finer control \$\endgroup\$
    – Neil_UK
    Nov 1, 2021 at 7:50
  • \$\begingroup\$ @Neil_UK Interesting. I've never thought about switching the ladder non-symmetrically. Certainly useful for lower frequencies and small common-mode voltages when fine gain control is desirable. As the answer was getting quite long already straying from the original question I will leave it at that though. \$\endgroup\$
    – tobalt
    Nov 1, 2021 at 7:57
  • \$\begingroup\$ Of course the other thing to point out is that 'digitally controllable gain' will probably want to be logarithmic in nature. If you are going to switch resistors with CMOS switches, then 10k is a nice impedance. A decade or two lower and switch resistance uncertainty becomes a problem. A decade or three higher, and leakage can be an issue. Switching constant impedance 'pads' in and out can give a large variation in gain, with few switches, and low error, say a 20 dB and 40 dB pad gives you 1-1000 gain control in x10 steps, with only 2 switch sections. \$\endgroup\$
    – Neil_UK
    Nov 1, 2021 at 14:21
  • \$\begingroup\$ @Neil_UK Can you point me to the "pads" you mentioned? The main reason I need switchable gain is that the input often saturate the amp, for which we need to lower the gain and do some turning on DC offset etc. Therefore I'd need to put these switchable pads before the In-Amp -- will they degrade SNR etc? \$\endgroup\$
    – Eric M.
    Nov 1, 2021 at 18:25

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