I am designing a simple resistance measurement circuit. Resistance range is from \$10m\Omega\$ to \$10k\Omega\$ and desired accuracy is 0.1%.

The circuit provides constant current to the resistance using a transconductance amplifier. An instrumentation amplifier is used to measure the voltage drop across the resistor. The current through the unknown resistance has to be sinusoidal, as per the requirements of my assignment.

When I connect both the instrumentation amplifier inputs to ground, I get a voltage at the output of \$2.22mV\$:

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I've found that this is due to the offset input voltage. To overcome this, I have included a voltage source on the inverting input (pins 1 and 5 - the offset null pins - are not enabled in the spice model). This minimizes the output voltage offset:

enter image description here

I tested the circuit using a sinusoidal signal of \$1\mu V\$ RMS (the lowest I'll be working with), and the output is perfect:

enter image description here

Now the problem:

When I connect the inputs across the unknown resistance, with the constant current source (transconductance amplifier) supplying current to that resistance, even though the voltage drop across it is still \$1\mu V\$ RMS, there is a mysterious increase in the output of the instrumentation amplifier:

enter image description here

As the simulation shows, this increase happens in final op-amp of the instrumentation amplifier. I'm completely confused by this. Why has the output increased all of a sudden, even though the voltage drop being measured is the same? How can I get the accurate output?

  • \$\begingroup\$ You are skipping over the input amps. Your simple offset correction doesn't work for varying voltages. Look at the voltages on the input amps under your varying conditions, and calculate what result they should produce. Try adding an offset to the two input amps as well as the output amp. \$\endgroup\$ Oct 12, 2015 at 17:24
  • \$\begingroup\$ The voltage probes on the two input amps are showing the correct voltage, without any offset. I guess since they are just buffers, the offset doesn't matter. Am I correct in assuming that? \$\endgroup\$
    – Hassaan
    Oct 12, 2015 at 17:47

1 Answer 1


This is caused by the common mode voltage at your input.

A real differential amplifier has some sensitivity to the common mode voltage at the input that results in a signal at the output even though there is no differential input voltage.

It can be caused by mismatch of the resistors or the result of the offset voltage of the opamp changing with the common mode voltage.

Since you are doing a simulation with "perfect" resistors it is presumably caused by the opamp.

You need to research CMRR (Common Mode Rejection Ratio - Wikipedia) - it will be specified on all opamp data sheets.

In the real world you wouldn't be able to use such a small signal as the 1uv that you are using because of such imperfections. The resistors would not be perfect either and would also contribute to the CMRR.

If you rearranged your circuit such that the resistor under test is floating (i.e. not connected to ground) and the input to the amplifier was connected to ground there would be no common mode variation and you would not have the problem.

  • \$\begingroup\$ So I just disconnect the ground from the constant current source, and connect a ground to the inputs of the instrumentation amplifier? \$\endgroup\$
    – Hassaan
    Oct 12, 2015 at 17:50

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