Loadcells and AD620 instrumentation amplifier

I’m making a circuit with some loadcells and AD620 amplifiers.
The loadcells are similar to those sold at sparkfun: https://www.sparkfun.com/products/10245

As the loadcells are only half bridge (2x1K resistors), I set up a reference cell using 2 fixed 1K resistors connected to make a voltage divider.
Simplified schematics of the loadcell and voltage divider attached. R1/R2 would be the loadcell, R3/R4 the voltage divider.
Ports 7/4 on the AD620 is connected to +/-12V, port 5 to ground, port 3 to loadcell, port 2 to voltage divider and port 1/8 to a 300 ohm resistor to set the gain.

With only one loadcell and one amplifier - this works just fine. The voltage measured from the unloaded loadcell is pretty much the same as the voltage measured at the voltage divider, both approx half the supply voltage (6V), and the output is as expected.

However - when I try to add more loadcells to the circuit, I run into problems.
I add a new AD620, connect all ports the same way as for the previous amplifier except port 3 which is connected to a second load cell. But now - the voltage on ports 2 and 3 differ greatly (almost 1V difference) on both amplifiers. To me, this indicates that the amplifiers are drawing some current at the inputs. However the datasheet for the AD620 says Input bias current max 1nA. As far as I understand, this means that ports 2 and 3 should not draw more than 1 nA, and should therefore have not significant effect on the measured voltage from the voltage divider. Is this a correct?

Or can anyone think of any other reason for this behaviour?
I could of course make individual voltage dividers for each loadcell, but the circuit would be much simpler if I could use the same one for all of them as I am trying to do now.

Any help appreciated!

Datasheet for the amplifier: http://www.analog.com/static/imported-files/data_sheets/AD620.pdf

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Pin, not port. A port is a pair of connections such as signal+ground.

You need to post the complete circuit layout, it does sound like something simple is wrong,
but without seeing the complete circuit its impossible to say more.

Do you have adequate decoupling on the +/-12V rails? You need it, this is a really high gain
amplifier and without proper decoupling you'll get feedback via the supply rails causing
oscillation.

Thank you for looking into this :slight_smile:

Sorry for the poor schematics, attached is the full schematics.

Regarding decoupling, as you see on the schematics I only have a capacitor on the +Vs pin (0.047 microfarads, I see the standard recommendation is 0.1 microfarads but these were the biggest ceramic capacitors I had). From a google search i found a suggestion to use capacitors on both +Vs-GND and -Vs-GND (some also suggested a third capacitor +Vs to -Vs). I will try this tomorrow and see if it helps.

The components are soldered onto a prototype board. For the amplifiers, I am using DIP8 sockets. As mentioned, when I insert only one amplifier to one of the sockets (regardless of which one) it works as expected but as soon as I populate both sockets it does not function properly.
But anyways, I will do some improvements to the decoupling and see if it helps.

There is no feedback in your circuit ( nor should there be ).

In an op-amp circuit with feedback, the op-amp will drive it's output in such a way as to force its two inputs to the same voltage.

But in an op-amp circuit with no feedback, this does not happen. That is ok where you want to actually amplify the signal, as you do here. When you are building a circuit like this, which has relative small gain, you need to temporarily forget about the analysis methodology where you assume a very large gain and assume that the op-amp inputs are going to be forced to the same voltage by the feedback loop.

  • the voltage on ports 2 and 3 differ greatly (almost 1V difference) on both amplifiers. To me, this indicates that the amplifiers are drawing some current at the inputs. However the datasheet for the AD620 says Input bias current max 1nA. As far as I understand, this means that ports 2 and 3 should not draw more than 1 nA, and should therefore have not significant effect on the measured voltage from the voltage divider. Is this a correct?

This is maybe the wrong comparison. Have you compared the voltage on pin 2 of one amplifier to pin 2 of the other one ? And the voltage at pin 3 on one to pin 3 on the other one ?

And when you say "it doesn't work" when you use two load cells, what actually happens ?

Have you checked the resistance of both sides of your second load cell ? Maybe it is actually broken or defective.

What do you expect your a/d converter to do, when it gets a negative voltage, which in this arrangement, it easily can ?

michinyon:
And when you say “it doesn’t work” when you use two load cells, what actually happens ?

Have you checked the resistance of both sides of your second load cell ? Maybe it is actually broken or defective.

Now I feel stupid, your second question guided me towards the solution;
When I said ‘it doesn’t work’, I just ment that I expected the voltage on the voltage divider to be ~6V (12V / 2), but when using two amplifiers i measured ~5V. The problem was - I didn’t actually connect the second load cell. I got to insert the second amplifier and noticed the voltages was not as I expected, hence I forgot all about the second load cell - and port 3 on amplifier #2 was actually not connected to anything.
Again - connecting the second loadcell actually solved the issue. Although I understand this would cause issues - I do not understand why the voltage at the voltage divider dropped to from 6V to 5V because of this? Will the amplifier draw a current at pin 2 if pin 3 is not connected?

Also, your last question is very good;

What do you expect your a/d converter to do, when it gets a negative voltage, which in this arrangement, it easily can ?

I didn’t think of this either. My thought was that in order to get the best accuracy, I should have close to 0V on the output pin when the cell is unloaded, as I only want to measure force in one direction. But I guess you are right - I do actually run a risk of getting a negative voltage at the output pin, and I’m not sure how the Arduino would react to this.
I guess I should make another voltage divider at, say 1V, to connect to pin 5 (REF) at the amplifier and reduce the gain a bit in order to lift the equilibrium point slightly. Does that sound like a good solution?