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enter image description here I made a very simple circuit to amplify a ~1 mV signal up to ~2 V.

The input is a 0~2 mV DC signal from a scientific instrument.

The problem:

I made the circuit, and it worked as intended. The 1 mV signal became 0.2 V, with the intended gain of 200. At home, I input a ~1 mV signal that I made by splitting 5V (wall adapter) into 4.999V and 0.001 V with resistors.

Next, I used the circuit with the scientific instrument. The input was ~ 1 mV, but the output was only 30 mV! The gain appeared to have changed to 30!

(I then double checked the gain with the split voltage setup--200 as intended. So nothing is broken, etc.)

The difference between the two cases is only signal source.

I am measuring the signal source with a digital multimeter. Both cases are ~1 mV. I tried measuring the current from the scientific instrument--it doesn't show up even on micro ampere scale--I didn't expect a current there, anyway. The scientific instrument signal is steady--I do not detect noise with my multimeter (0.1 mV precision).

Question:

Could there be some other difference between the two signals that would affect the gain?

Other bits of information:

The millivolt signal is the pressure reading output from a Shimadzu HPLC pump.

The circuit I built is copying the schematic of Figure 48 ("Optimal ground practice in a single supply environment") on p 20 of the AD623 data sheet.

In the figure, there is a cylinder-shaped symbol around the signal input. I don't know exactly what this is, but I have done nothing to my signal input. On page 19 of the data sheet, in figure 44, 45, there are schematics for RF interference protection and other input protections. Since my circuit works when I apply a ~1 mV signal from a split voltage, I didn't think there is a problem with RF interference in the environment, etc., so I didn't do any input protection.

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    \$\begingroup\$ In the 2nd case, does the instrument share a ground with your amplifier? What is the common mode voltage of the instrument output? \$\endgroup\$ – The Photon Jan 19 at 2:56
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    \$\begingroup\$ Vcm input must be centred i.stack.imgur.com/a3rGr.png \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 19 at 3:06
  • \$\begingroup\$ @ThePhoton As for grounding: the scientific instrument is plugged into the wall socket, as is the amplifier circuit. I did not think to measure the instrument output common mode voltage. I will do so today but cannot do so immediately. \$\endgroup\$ – user4396936 Jan 19 at 3:17
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    \$\begingroup\$ One or the other may be using an isolated power supply, and your two sockets might be on different circuits, so this is not enough to ensure they have a common ground. Try connecting ground between the two, and be sure the common mode level of the signal is appropriate at the input of the in-amp. \$\endgroup\$ – The Photon Jan 19 at 3:22
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 - the input common mode is optimised for signals approaching or at -Vs; for single supply operation the input common mode for very small output signals is between -Vs (well a little below that) to +Vs - 1.5V. See figure 23 in the datasheet. \$\endgroup\$ – Peter Smith Jan 19 at 17:04
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Most likely the common mode level of the instrument output is not in range for the in-amp input.

First you should ensure your two circuits have a common ground.

Then you should make sure the instrument common mode voltage is within the recommended range for the in-amp input. If the instrument output is biased near ground, the easiest way to do this might be to use a negative supply for the VS- terminal of the in-amp.

In the figure, there is a cylinder-shaped symbol around the signal input. I don't know exactly what this is,

That's a shield around the cable connecting the inputs of the in-amp to whatever source they're measuring.

If the connection from your instrument to the amplifier is short, you may not need it.

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    \$\begingroup\$ It might be worth pointing out that the common mode range for instrumentation amps is not simple: analog.com/media/en/technical-documentation/application-notes/… - Analog devices even has an online tool - analog.com/designtools/en/diamond \$\endgroup\$ – Peter Smith Jan 19 at 13:53
  • \$\begingroup\$ This tool is extremely helpful. I can see exactly how I need all my inputs to be. My common mode voltage was in the milivolt range, way too low. This tool should be the answer to my question. Will post my final solution in the next few days. \$\endgroup\$ – user4396936 Jan 20 at 0:55
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enter image description here

Notice here with an ideal split supply, the mean input between +1 and -2.7 = -0.85 This a bit more than a diode drop with Vref=0V

The next 2 examples are 5V single supple but different gains.

for G=1 the ideal Vcm= 1.5V about 1 voltage below 2.5V and must be used for the Vref. (which is your problem) .

The next for G>=10 Vcm ideal is 1.85V* (thanks* to Irfanview) and relative to 2.5-1.85=0.65 which is about the same as the Vbe drop in the simplified schematic.

So your solution is to change the Vcm range of your signal and changing Vref so that Vcm input range satisfies Fig 23 yet you can shift this with Vref= Vcm + 0.65V yet Vref must be between the 2 rails. This means you must shift your signals up or shift your negative rail down. enter image description here

Since the Vcm input = -0.25V and you had a gain of 200, while Max Output = 0.7V you would have to reduce your input by 700mV/200=3.5mV to achieve the same gain. Otherwise the input is not longer linear and the gain drops rapidly.

In other words you need better design specs

You must either reduce the gain or the signal level to stay within 700mV out max and add a Vcm input voltage and Vref that is 0.65V higher to choose the best dynamic range....

or make it easier and use a split supply that gives you wider Vcm input range.... that is you want to satisfy the IC specifications for a perfect design you want linear gain with some maximum swing and Vcm range.

enter image description here

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