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### Topic: Load cell (Read 4324 times)previous topic - next topic

#### Thestrongestduck

##### Sep 17, 2012, 05:24 pmLast 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

#1
##### Sep 18, 2012, 11:57 am
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 pmLast 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

#3
##### Sep 18, 2012, 12:50 pm
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

#4
##### Sep 18, 2012, 01:04 pm
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

#5
##### Sep 18, 2012, 02:49 pm
Uhh, my head hurts, no luck with the instrumental amplifier setup. I no longer have any idea what to do next.

#### MarkT

#6
##### Sep 18, 2012, 05:29 pm
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 will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### Thestrongestduck

#7
##### Sep 18, 2012, 06:25 pm

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

#8
##### Sep 19, 2012, 12:53 am
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 will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### Saielectrosystem

#9
##### Sep 21, 2012, 10:09 am
Please follow this link
http://cerulean.dk/words/?page_id=42&cpage=1#comment-14815

#### Saielectrosystem

#10
##### Sep 21, 2012, 10:12 am
This is link for the brake out board for arduino load cell

http://tenettech.com/search.php?orderby=position&orderway=desc&search_query=loadcell&submit_search=Search/

#### Thestrongestduck

#11
##### Sep 30, 2012, 01:52 pm
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

#12
##### Sep 30, 2012, 03:52 pm
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

#13
##### Sep 30, 2012, 06:07 pm
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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