Amplification of load cell signal with AD623 IC

Question

I am trying to amplify the signal from a load cell using an AD623 instrumentation amplifier.

_{(source: instructables.com)}

Please refer to https://www.instructables.com/answers/Amplifying-load-cell-signal-with-AD623-problem/ for the schematic. The common mode voltage is 4.5 Volts.

1K resistor is supposed to give 100x amplification. It is instead giving me around 50. My differential voltage is 0.8mV while my output is 42mV (52.5 gain). Note that I have placed the capacitors (0.1 uF and 10 uF from Vs to ground) but are not shown in figure. What might be going on? Is there a chance the AD623 is ruined? I don't think I'm doing anything wrong here, plus I've read a lot of blogs and haven't figured it out yet. Note that when I double the differential voltage, the amplification remains around 50. And when I change the resistor to 200ohms (500x multiplication according to datasheet), I get around 250x multiplication which is also half. So amplification is quite accurate, but half of what is said in datasheet.

Answer 1

It looks like you have a bipolar differential input signal, but your output can only be unipolar because the REF pin is tied to -VS. If your input is AC, your output would be clipped at zero, so you're measuring it as half what you expect. Try moving REF up to 5V maybe.

Answer 2

Your Common mode voltage is 4.5V as you mentioned. If you check the data sheet of AD623 you can find on page no.8 fig.22 that your common mode voltage is way more.

This is the image from AD623's datasheet http://users.ece.utexas.edu/~valvano/Datasheets/AD623.pdf

You could do one thing, reduce the excitation voltage in order to bring the common mode voltage to be near 1.8V and then have more gain as per the mV/V rating of your load cell to get the desired output.

I came across the same problem but the mV/V rating of my Load cell is very less. I had to change the amplifier package. AD8420. I am still testing.

If you find a better solution please let me know.