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Topic: Load cell (Read 2 times) previous topic - next topic

Thestrongestduck

Sep 17, 2012, 05:24 pm Last Edit: Sep 17, 2012, 06:46 pm by Thestrongestduck Reason: 1
Hi. I'm having some trouble hooking up my load cell to my Arduino.

I'm using two wheatstone half-bridge load cells. One is used as a "dummy/reference" for the other.

I'm amplifying the voltage difference using a LM358 set up as a Differential Amplifier.



I wasn't certain about the values on Rf and Rg so I followed http://www.ti.com/lit/an/sloa034/sloa034.pdf

Here is some Data:
The load cells has a total resistance of 1965 and 1966 Ohms
I'm exciting the cells with 5V
With 150Ohm resistors on Rf and Rg the baseline Voltage on Vout = 870mV
R1 and R2 are both 1KOhm

Applying a load does not change the Value in either case.

What am I doing wrong is the Resistor values completely off or what?

Thestrongestduck

After some meditating yesterday I thought I had figured it out,

If Rf=Rg and R1=R2
   Vout=Rf/R1 x (V1-V2)

So I changed the resistor values,

Vout = 5V
V1-V2=0.100V (This i wasn't actually sure about but i figured that it'd be something like that)

Which meant that:
5=Rf/R1 x 0.100
5/0.100=Rf/R1
50=Rf/R1

Based on that I chose the new resistor values Rf=4.7K R1=100
4700/100=47 (which is close to 50 so I thought it'd be enough)

Sadly this still isn't enough gain to get any reading on the voltage change, which I find strange but I'm going to try to double it and see if that works.



RIDDICK

#2
Sep 18, 2012, 12:37 pm Last Edit: Sep 18, 2012, 09:30 pm by RIDDICK Reason: 1
1.
possibly the gain should be around 100x?

2.
maybe u should try a higher input impedance?
like in figure 6 in that application report...

3.
according to the LM358 datasheet
http://www.ti.com/lit/ds/symlink/lm358.pdf
it might be a little inaccurate (Vio=3mV)...

4.
if the input voltage difference is negative, u could try to swap the load cells (the output of the LM358 cant go lower than 5mV)...

5.
did u try to measure the difference with a multimeter?
maybe u can c some little changes, that help u to find the correct gain...
-Arne

Thestrongestduck

Yes I checked the voltage swing with my multimeter

4.1-3.4mV swing at 0-20kg

But it seems like I cant replicate that, most likely thanks to my shitty multimeter.

I also just tried a huge gain value of 470 which didn't work.

I'm going to dig out my brothers multimeter and try to get a decent reading with that.

Thestrongestduck

Using a better multimeter I get a 3mv change with 11,5Kg keep in mind that I'm still using the LM358 to subtract the values only with zero gain i.e no resistors,

I might try using 3 OPamps in a Instrumental amplifier setup and see where that gets me. 

Thestrongestduck

Uhh, my head hurts, no luck with the instrumental amplifier setup. I no longer have any idea what to do next.

MarkT

I think you need WAYmore gain and a proper instrumentation amp (very very good CMRR).  The change in resistance of a strain-gauge is measured in tiny fractions of a percent, so gains of 1000 or more aren't unreasonable.

This might be useful:
http://cerulean.dk/words/?page_id=42  there a gain-resistor of 10 ohms is used, meaning a gain of 6000.  Sounds more likely than 50!
[ I won't respond to messages, use the forum please ]

Thestrongestduck


I think you need WAYmore gain and a proper instrumentation amp (very very good CMRR).  The change in resistance of a strain-gauge is measured in tiny fractions of a percent, so gains of 1000 or more aren't unreasonable.

This might be useful:
http://cerulean.dk/words/?page_id=42  there a gain-resistor of 10 ohms is used, meaning a gain of 6000.  Sounds more likely than 50!


I agree, still I was hoping to see some small change at least with a gain of 470, but it might be the amp as well, I'm thinking of trying one of the INA amps preferably with an easy gain adjust 

MarkT

With a very small differential signal like this you cannot compromise on CMRR (common-mode-rejection-ratio), PSRR (power supply rejection ratio) and Voffs (input offset voltage).   This is exactly the situation a precision instrumentation amp like the INA125 is designed for, most opamps will have way too much input offset and offset variability to have a hope here - you'll just end up amplifying the input offset voltage to saturation or have a highly temperature-sensitive output value.
[ I won't respond to messages, use the forum please ]



Thestrongestduck

YES! I got it to work. Using two LM358Ps set up as a three Op Amp Instrumentation Amplifier. The problem I had with that circuit before must have been the low resistor values I used, This time having done the math, and using high resistor values (100K, 10K, 150K, 6.7K) I got it to work really nicely. Next thing to do is to try the two Op Amp Instrumentation Amplifier, Why you ask? Well because then i only need to use one LM358.

I'm very happy right now. And thanks to all of you guys for helping me out.

Thestrongestduck

While looking at the two op amp instrumentation amplifier circuit(2.3): http://www.ti.com/lit/an/sloa034/sloa034.pdf

I noticed that the equation: Vo=(sig+ - sig-)(1 + R1/R2 + 2R2/Rg) doesn't make sense since they state that R1=R2 and R3=R4
If R1=R2 then Vo=(sig+ - sig-)(1 + 2R2/Rg) would be true, but then there wouldn't be anything that defined the relation between (R1, R3) & (R2, R4)

Thestrongestduck

After further experimentation I get less and less certain that I managed to read a weight change, Seems more like I've built a great EMI antenna and amplifier, Putting a weight on the sensor does change the value, but it seems I can accomplish the same change by resting my finger on the sensor.

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