I am trying to apply gain for a 0.0 - 16.7 mV signal with an INA122P, with a single 10 V supply for all the circuits, with Vref driven at 0.0 mV.

Here is the circuit that I am using, no Rg applied, for which the datasheet said that the default gain of the INA122P is 5.

I have simulated the sensor with resistors to create a Wheatstone bridge, with a differential voltage output of around 0.0 - 16 mV, which I have confirmed by measuring with a multimeter.

INA122P with wheatstone bridge

Here is the measurement of Vdiff and the output voltage:

Vdiff Voutput (Gain = 5)
0.0mV 9.8mV
1.4mV 15.5mV
5mV 26.2mV
10mV 50.8mV
11.9mV 59.9mV

My question is, why is the output of 0.0 mV and 1.4 mV Vdiff not resulting in a 5x gain?

In the example, 1.4 mV * 5, so it is supposed to be 7 mV. It seems to have an offset of around 1-2 mV Vdiff. Looking at the INA122P's datasheet, it might have to do with the following graph, because I believe my Vcm is sitting at around 5.005 V and it might not provide linear output, but I'm not quite sure how to refer to this graph; that's one speculation.

enter image description here

Things I have tried:

  • I have tried to connect both the INA122P's positive and negative inputs from the same voltage divider of the Wheatstone bridge to provide 0.0 Vdiff, and it still provides 9.8 mV.
  • I was thinking to lower the Vcm by plugging both positive and negative input supplied from the ground of my circuit (maybe bad practice), which provides 2.7 mV, which is lower.
  • The TLC271 single-supply buffer creates a 5.9 mV increase to Vref, achieving 0.0 mV when I removed the buffer.

Some similar questions I have observed:
INA122 output provides saturation around 0,3V It seems that my circuit provides better performance, but I still need to confirm: why is there an offset of around 9.8 mV when I apply 0.0 Vdiff?

Is there something I overlooked? Should I apply some decoupling capacitors?

Another similar question I have checked, made me think to apply 1 MΩ resistors: Common mode rejection range for INA122

Also, here is my breadboard, after I removed the op-amp buffer:

INA122P Breadboard Circuit

  • 2
    \$\begingroup\$ Are your output voltage measurements with respect to 0 volts? What is the offset on the reference pin? \$\endgroup\$
    – Andy aka
    Commented Aug 13, 2022 at 13:02
  • 1
    \$\begingroup\$ The picture has a pot on the Vref pin. Remove it and connect pin 5 (Vref) to pin 4 (GND) directly, not to the ground bar. Then when probing the output, probe directly at the chip pins. Measuring millivolts can be tricky, there are little voltage drops everywhere. \$\endgroup\$
    – Mattman944
    Commented Aug 13, 2022 at 13:20
  • \$\begingroup\$ @Andy aka, the offset on the reference pin is 0.0mV, I put the potentiometer to the smallest ratio, which results in 0.0mV from the readings of the multimeter, not sure if there are other considerations when measuring from a multimeter. I will try to probe it with an oscilloscope. \$\endgroup\$ Commented Aug 13, 2022 at 13:29
  • 2
    \$\begingroup\$ Just like in your other question, try giving the INA122 a negative voltage rail and see if the problem goes away. \$\endgroup\$
    – Hearth
    Commented Aug 13, 2022 at 15:57
  • 3
    \$\begingroup\$ This is usual feature of rail-to-rail outputs – it usually can not go strictly equally to any rail under load, and Ref input of InAmp creates load for OpAmp. INA122 output can go only as close as 0,1 V to any rail (datasheet). There is a rule: don't believe to 1st page of datasheet :) Simplest solution, i think, is to set zero potential slightly more than zero with that potentiometer, and, if signal then goes to ADC, sample it relative to Vref of INA, this is common technique. \$\endgroup\$
    – Vladimir
    Commented Aug 13, 2022 at 23:13

1 Answer 1


Where are the two rightmost resistors wired?

Is there something I overlooked? Should I apply some decoupling capacitors?

Yes. One should always use a "decoupling" capacitor, the very nearest supply pins of the op-amp, on the breadboard.

... apply 1 MΩ resistors?

1 Meg resistors are not needed. You have already 100 kOhm.

Made with FREE simulator microcap v12

Simulation in two configurations does not show anything wrong.

Bipolar supplies. enter image description here

Unipolar supply + Vref offset. enter image description here

  • \$\begingroup\$ What circuit simulation are you using if i may ask? And yes, the image doesn't show decoupling capacitor, but i used the decoupling capacitor. \$\endgroup\$ Commented Aug 14, 2022 at 11:17
  • 1
    \$\begingroup\$ Decoupling capacitor in the schematic ... we always forget in simulation. But I don't see it on the breadboard very nearest power pins of op-amp supplies? The simulator used is FREE, see homepage microcap v12 spectrum-soft.com/download/mc12cd.zip \$\endgroup\$
    – Antonio51
    Commented Aug 14, 2022 at 14:20
  • \$\begingroup\$ Note also that when you are measuring some "mV", this must be done with more care than usual "V", because of the "junction" phenomena (Seebeck effect thermocouple) instrumentationtools.com/topic/thermocouples download.tek.com/document/LowLevelHandbook_7Ed.pdf \$\endgroup\$
    – Antonio51
    Commented Aug 14, 2022 at 14:30

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