I want to design a circuit which outputs a signal ranging from 0-70 mV using a variable resistor ranging from 80.51 Ohm - 183.05 Ohm and a supply voltage of 12 V. My idea so far has been to make a wheatstone bridge differential amplifier using an instrumentation amplifier with unity gain to subtract a constant offset voltage from the varying input signal. From what I have read, an in-amp is much more preferred than using a single op-amp because of its improved CMMR. I am not sure if it is superior with only unity gain, though.

As seen in the image, I tried drawing this circuit in LTSpice using LT1168 as In-amp. R42 would be the variable resistance, and it will be located on a separate PCA from the rest of the components. Hence, I fear more noise might be picked up on the positive input. I calculated that R43 would either need to be 17313.87 Ohm or 0.85 Ohm i.o.t. have 70 mV difference between the two edge cases for R42. I thought it would be better to go for the first option so as to limit the power dissipation over the resistors.

Hence the offset to be subtracted is formed by voltage divider R21 + R22. With R42 being 183.05 Ohm the output from the In-amp should in theory be 70 mV, and 0 mV when R42 = 80.51 Ohm.

When simulating, however, I only get 500-600 mV as output for any of the cases. It must be said that I am a beginner to LTspice, so I am not sure how accurate the simulation is, but it still gets me concerned. Will it be impossible to achieve such low voltages using this method, assuming that 0.1% resistors and suitable filtering are used? I would gladly accept suggestions for alternative ways of solving the problem as well as any advice for attenuating as much noise as possible.

Wheatstone bridge differential amplifier


You have to keep the inputs within the input common mode range.

Typically with an instrumentation amplifier that depends on the gain as well since there are internal nodes that can saturate for some designs, however in this case the manufacturer lists the range for G=1 as +1.9V to 12V-1.4V = 10.6V. Obviously your inputs are both within a couple hundred mV of 12V so this won't work.

It will be easier if you can ground one side of the low resistances and use bipolar supplies such as +12/-5V.

As to an instrumentation amplifier having an advantage, that depends on whether you expect significant common mode voltages to be present and if said voltages are varying. In a typical Pt100 RTD signal conditioner (which this sounds like it is, maybe I'm guessing wrong) that's usually not true since the reference half of the bridge is local to the amplifier. A simple non-inverting op-amp amplifier would be fine in that case.

Noise from the circuit can be minimized by using low resistances (reducing the Johnson-Nyquist noise) and using a low noise op-amp or in-amp. Noise from external sources is typically attenuated by using a filter in front of the amplifier input, which might be a simple RC low-pass or something more sophisticated. In most of the situations I run into (but not all) you can almost ignore the internal circuit noise since external noise is much more significant.

  • \$\begingroup\$ Thank you! I was not aware of the input common mode range being a problem. \$\endgroup\$ – Lord CH Oct 18 at 14:06
  • \$\begingroup\$ Could you elaborate on what you mean by grounding one side of the low resistances and why this would help? Regarding low-pass filtering, how am I to decide which capacitor value to use? I guess that by having as high time constant as possible would decrease the cutoff frequency. But are there other disadvantages than increased response time? And what type of capacitor is best suited for high precision (film type, tantalum...)? \$\endgroup\$ – Lord CH Oct 18 at 14:20
  • \$\begingroup\$ The idea is to keep both the input voltages near ground, with supplies that are bipolar, so the input voltages are at least several volts from each supply. Long time constant means large C and/or R. Leakage current can be a problem with the C, and op-amp input bias current can cause errors with large R. Also if you actually did have a situation with high AC common mode noise, mismatch between the RC on each input can cause degradation of CMRR because one input is attenuated more than the other. \$\endgroup\$ – Spehro Pefhany Oct 20 at 2:06

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