3-wire Load Cell: On its lonesome and in a bridge - Page 1
Forum section: electronics / projects

author: NivagSwerdna
author_rank: Super Contributor
timestamp: October 11, 2016, 08:59:39 pm
content: I bought a set of bathroom scales which has four load cells, one in each corner.My load cells look very like these..https://www.sparkfun.com/products/10245which are specified as...https://www.sparkfun.com/datasheets/Sensors/loadsensor.pdfNow mine actually have red, blue and white wires but if white is considered common then white-blue and white-red both measure around 1k ohms so this sounds good.I sketched the bridge arrangement and it is shown in the picture. The resistors are actually 880 ohm and 68k ohms (I misread the 881 SMD marking when I made the drawing), which implies the front stage to the scale has a gain of around x77Now... I'm going to try and re-use the same bridge with my own circuit with an excitation voltage of around 3.3v (low I know but it's what I've got)I would like to model the circuit prior to building it for real and feel the specification should give me enough information but I do not know how to interpret it.If the specification says 1mv/v and I am applying 3.3v then I assume something is going to give me 3.3mv when 50kg is applied... but where would that voltage manifest itself? Is that for a single load cell and that considering it as a divider and applying say 1v you would get, measuring at white, 500mv with zero load and 449mv with 50kg applied? i.e. the blue side would decrease in resistance the red side would increase?Please can someone help me to model the bridge in Spice or similar.Thanks in advance

author: EmmanuelFaure
author_rank: Regular Contributor
timestamp: October 11, 2016, 11:10:41 pm
content: Hi,Quote from: NivagSwerdna on October 11, 2016, 08:59:39 pmBetween "A" and "B".Quote from: NivagSwerdna on October 11, 2016, 08:59:39 pm*499mV with 50kg applied, OR 501mV, depending on the polarity.That's absolutely right! Instead of simulating with spice/masturbating a numerical code without knowing anything about what you're simulating, I encourage you to take a typical high resolution ADC (LTC2400, LTC2440, MCP355x x=0,1,3, etc... If you don't have one yet, it's the best investment ever when learning precision analog) and to connect it directly to one single weight sensor. Best educational experience ever with bridge sensors!You can use a single sensor and build a fixed resistive divider with metal film resistors, and this is your wheatstone bridge. This might be your second best introduction to bridge sensors.A classical trick with this kind of half bridge sensors is to arrange them in order to use only sensors, without fixed resistors. Using two of them is easy (A third experiment here), one connected forward (red = +, blue = ground), the other connected backwards (blue = +, red = ground), and you measure the voltage between the two white wires. Following your numerical example, at full scale one sensor outputs 499mV, and the other 501mV : There's your 2mV/V at full scale.Connecting 4 ot them as a single wheatstone sensor is a little trickier. You'll find all the necessary informations here :http://arduino.stackexchange.com/questions/11946/how-to-get-weight-data-from-glass-electronic-bathroom-scale-sensors

author: NivagSwerdna
author_rank: Super Contributor
timestamp: October 11, 2016, 11:31:31 pm
content: I have plans...AD8237 and then to ADC in a PICorADS1131orPIC with OpAmps and then to Delta Sigma ADC in PICThanks for the linkBut I had hoped to do some calculations beforehand... particularly to estimate gain required.

author: EmmanuelFaure
author_rank: Regular Contributor
timestamp: October 12, 2016, 12:39:11 am
content: Quote from: NivagSwerdna on October 11, 2016, 11:31:31 pmSimplest solution, go for it. TI have also the ADS1230, same component with 20 bits ADC (18 bits for the ADS1131).Quote from: NivagSwerdna on October 11, 2016, 11:31:31 pmIn my previous comment I gave you absolutely all you need to make your calculations.

author: NivagSwerdna
author_rank: Super Contributor
timestamp: October 12, 2016, 07:34:16 am
content: Working on the basis of the 1mv/v changing the 1000 ohm / 1000 ohm balance to being 999 and 1001.Then putting them in a bridge and applying 3v3 and fully loading I get a bridge imbalance of 3.3mvMaybe that was obvious... but not to me

author: Brutte
author_rank: Frequent Contributor
timestamp: October 12, 2016, 07:48:38 am
content: Quote from: NivagSwerdna on October 12, 2016, 07:34:16 amI believe those cheap weigh sensors are made in half bridge configuration with one gauge being passive (it serves no other purpose but to compensate for temperature variations of an active gauge).So it is 1k + 1k divider with no load and 1k + 1k001 with full load.

author: EmmanuelFaure
author_rank: Regular Contributor
timestamp: October 12, 2016, 12:55:34 pm
content: A strain gauge is never passive. Glued transversally, it experiments ~30% of the longitudinal deformation. That's the "Poisson effect"https://en.wikipedia.org/wiki/Poisson%27s_ratioQuote from: NivagSwerdna on October 12, 2016, 07:34:16 amI took on my time to answer all your questions. If you didn't want answers, you should have not asked questions.I also warned you to absolutely not simulate something you don't understand. At full scale if output of a single sensor = 1mV/V, then the pair of resistances are changing from 1000/1000 to 998/1002. Not 999/1001. You throw numbers in a computer but you're not taking the time to understand how a simple resistive divider works.

author: NivagSwerdna
author_rank: Super Contributor
timestamp: October 12, 2016, 01:22:44 pm
content: I have no interest in fish; I'll deal with the non-linearity later if I get that far.Oops. That example was at 50% load. Indeed 1002/998 at 100% load.But as pointed out by Brutte it looks like one side is fixed (http://www.robotshop.com/uk/sfe-load-sensor-50kg.html) so actually more like 1004/1000 at full load.