I do not fully understand the working of your wheatstone bridge. Normally you have 3 known resistor values in the bridge and based on the voltage of between the two measuring points you can calculate the last resistor value. In your case I see 4 strain gauges. The weight goes on all four of them ? Or just one ?
So if you're stressing the four gauges with the same load, you're actually measuring the difference in measurement between the four of them and maybe that's not linear at all given a certain weight.
Besides that the output of the instrumentation amplifier can swing between both rails, in your case I assume it's 5V and GND. If something happens to that 5V, let's say it drops to 4.8 V. The maximum output of the amplifier will be 4.8V regardless of what difference you put at the input.
As a last remark, look at the datasheet of the AD623 on page 14 fig.40. I'm not sure what that graph means, but if the output voltage drops that dramatically because you draw 1.5 mA,. You'll need to add a buffer, but I might be misinterpreting this graph.
I would split the circuit in blocks and do some tests. Replace the wheatstone bridge by two trimpots and reproduce the input voltage difference and see how the amplifier reacts. Add some load to the output and see how it reacts. Measure the voltage differential on the wheatstone bridge with a multimeter and see if it's coherent with your calculations. Finally measure what happens with the voltage rail when you see the issue occur.
Just found this link http://designtools.analog.com/dt/inamp/inamp.html?inamp=AD623%205V
What's the common voltage of your bridge? If it's around 2.5 V, the calculator will not go beyond the 3.6 V output voltage
The Strain gauge setup should be correct, it is a full bridge instead of the quarter bridge you are describing. I have a hard time myself to wrap my head around how the full bridge is working, but all gages are active and two are compressed and two expanded. That gives a good sensitivity and also other benefits like linearity and automatic compensation for thermal expansion.
Regarding the graph (fig. 40). It is not crystal clear as you said, but do the analog input on the Arduino draw 1.5 mA? Thats all the AD623s output pin is connected to... Better measure it..
That calculator was however a very good find! Common mode voltage was a new concept to me. My common mode voltage should be 2.5 Volts. That means that the maximum output is 3.6 V which is exactly what I get.
I think that was the issue. Thank you!
unless it is some kind of special one.
This is a special one.
The decoupling capacitors are absolutely essential, or the amplifier will oscillate. The inputs are limited to 3.5 volts in this particular circuit. Pay close attention to the circuit design notes and limitations in the AD623 data sheet, and keep in mind that plug-in breadboards are notorious for circuit instabilities when used with op amps.
See this thread, especially the last couple of posts. http://forum.arduino.cc/index.php?topic=248765.0
There are rather severe limitations on the output voltage range, depending on the power supply configuration and common mode input voltage. In your circuit you cannot expect to get the full range on output (see Figure 23 of the AD623 data sheet).
Yes the capacitors will be installed soon. And every wire and component is soldered on a prototype board so that should be ok. I learned about breadboards limitations the hard way
That was a helpful thread, wish i had found it sooner.
Figure 23 in the AD623 datasheet says that a CMV of 1.9 V gives the largest output (5V) which would be ideal for me.
Just a fast thought, if I connect the loadcell to the 3.5 V supply instead of the 5 V (but keep the AD623 feed at 5V, My CMV would drop to 1.75 V instead of 2.5 V. Is that correct or am i missing something?
However, if the common mode voltage can't be altered in any way, I think a simple solution to the problem is to chose the gain so that max load = 3.6 volts. It will be a lower resolution but still acceptable.