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Given the datasheet here, I can't explain my observations with my AD620AN test circuit.

The circuit:

  • Supply: +5v / 0v (Well regulated)
  • 1k potentiometer across the supply, with the wiper going to both the reference pin and the negative input
  • +ve/-ve inputs: K-type thermocouple. (41uV/K)
  • Gain resistor: 100 Ohm (Gain of 500).

It seems that the gain is strongly dependent on the voltage of the reference pin. If I set the pot at precisely 2.5v, I get roughly the expected gain. But even slighly higher or lower and the gain reduces significantly. (For example, just +/- 0.05V change in reference reduces gain by about 10% for output signals of +/- 100mV). I've been taking measurements with a standard multimeter.

I haven't done meticulous testing with this, but none the less have lost a fair bit of hair. Is there something obviously wrong with what I'm doing? I must have run through the figures in the datasheet two dozen times. Is it the circuit at fault? Or my measurements? Or both?

AKR
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Jodes
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1 Answers1

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Notice the small block diagram in Fig.2 if the AD620 datasheet.

simplified block diagram a 3-OpAmp instrumentation amplifier from the AD620 datasheet

Here's a larger diagram from a datasheet of a similar InAmp with a 3-OpAmp topology (LT1920).

block diagram of a 3-OpAmp instrumentation amplifier from the LT1920 datasheet

Notice that VREF is not a high impedance input. In order not to skew the differential output stage, VREF should be connected to a low impedance output. A voltage divider by itself is not a low impedance output. However, the voltage divider can be buffered.

enter image description here

I have built and tested a similar circuit. (In my case, the VREF came a source other than a potentiometer, though.)

Nick Alexeev
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    I got bit by this error once. I had connected the ref pin of an instrumentation amp straight to a potentiometer. It has interesting effects. – Connor Wolf May 21 '13 at 00:42