Offline
Newbie
Karma: 0
Posts: 19
|
 |
« on: September 17, 2012, 10:24:51 am » |
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.pdfHere 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?
|
|
|
|
« Last Edit: September 17, 2012, 11:46:04 am by Thestrongestduck »
|
Logged
|
|
|
|
|
Offline
Newbie
Karma: 0
Posts: 19
|
 |
« Reply #1 on: September 18, 2012, 04:57:19 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.
|
|
|
|
|
Logged
|
|
|
|
|
Brunsbüttel, SH, F.Rep.GERM
Offline
God Member
Karma: 4
Posts: 596
|
 |
« Reply #2 on: September 18, 2012, 05:37:13 am » |
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.pdfit 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...
|
|
|
|
« Last Edit: September 18, 2012, 02:30:20 pm by RIDDICK »
|
Logged
|
-Arne
|
|
|
|
Offline
Newbie
Karma: 0
Posts: 19
|
 |
« Reply #3 on: September 18, 2012, 05:50:47 am » |
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.
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Newbie
Karma: 0
Posts: 19
|
 |
« Reply #4 on: September 18, 2012, 06:04:11 am » |
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.
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Newbie
Karma: 0
Posts: 19
|
 |
« Reply #5 on: September 18, 2012, 07:49:25 am » |
Uhh, my head hurts, no luck with the instrumental amplifier setup. I no longer have any idea what to do next.
|
|
|
|
|
Logged
|
|
|
|
|
0
Offline
Tesla Member
Karma: 73
Posts: 6638
Arduino rocks
|
 |
« Reply #6 on: September 18, 2012, 10:29:27 am » |
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!
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Newbie
Karma: 0
Posts: 19
|
 |
« Reply #7 on: September 18, 2012, 11:25:34 am » |
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
|
|
|
|
|
Logged
|
|
|
|
|
0
Offline
Tesla Member
Karma: 73
Posts: 6638
Arduino rocks
|
 |
« Reply #8 on: September 18, 2012, 05:53:03 pm » |
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.
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Newbie
Karma: 1
Posts: 15
|
 |
« Reply #9 on: September 21, 2012, 03:09:35 am » |
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Newbie
Karma: 1
Posts: 15
|
 |
« Reply #10 on: September 21, 2012, 03:12:45 am » |
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Newbie
Karma: 0
Posts: 19
|
 |
« Reply #11 on: September 30, 2012, 06:52:53 am » |
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.
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Newbie
Karma: 0
Posts: 19
|
 |
« Reply #12 on: September 30, 2012, 08:52:07 am » |
While looking at the two op amp instrumentation amplifier circuit(2.3): http://www.ti.com/lit/an/sloa034/sloa034.pdfI 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)
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Newbie
Karma: 0
Posts: 19
|
 |
« Reply #13 on: September 30, 2012, 11:07:29 am » |
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.
|
|
|
|
|
Logged
|
|
|
|
|
|