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Topic: Load Cell with 3 wires (Read 21769 times) previous topic - next topic

Marmotine

Hi everyone,

I bought a load sensor (SN-10245) with 3 wires (black, white, red) and I also bought the Load Cell Amp HX711 which need a 4 wires input.

I want to use only one load cell, but I saw on Internet that these load cell need another load cell to work.

I think that it is possible to change my 3 wires load cell to a 4 wires by using a half bridge.

I wonder if someone could explain me how to use a half bridge, and if someone have any electrical circuit plan for me.

Thank a lot!

   

Wawa

Connect the load cell to +E, -E, and the + input.

Connect a 1k resistor from +E to the -input.
Connect a 1k resistor from -input to -E.

The -input has now a fixed voltage divider with two 1k resistors.
The +input has the load cell (= variable voltage divider).
Leo..

Marmotine

#2
Mar 16, 2016, 04:54 pm Last Edit: Mar 16, 2016, 05:04 pm by Marmotine
Thank a lot for the reply!

Here's what I understood from it. Tell me if it is good!


Wawa

Yes, correct.
It's indeed a half-bridge.
Make sure you have the load cell wiring right.
Some have non-standard coloured wires.
Find the two wires with the largest resistance with a multimeter.
Those are +E and -E.

If weight goes negative, swap +A and -A.
Or swap +E and -E of the load cell.
Leo..

Marmotine



I have done exactly like my schematic, but the program keep saying me a '' 0,0 g '' reading.

I took a some voltage measurement:

VCC / GND: 4.2 V

E+ / E- : 4.2 V

Resistor 1: 2.1 V

Resistor 2: 2.1 V

A+ / A- : 2 V


My multimeter is very basic (GMT-12A) and I when I am trying to measure a resistance, I can't get a reading between the minimum and the maximum of the needle course.

What is the problem?

I am using this program: https://codebender.cc/sketch:260485#load%20cell%20readout.ino


OldSteve

You cannot connect your wires and resistors like that. You need to solder the joints.
Please do not PM me for help. I am not a personal consultant.
And others will benefit as well if you post your question publicly on the forums.

Marmotine

I'm just doing tests to make sure everything works before welding permanently. I do not have a breadboard. Does it really change anything I do not solder yet?

OldSteve

#7
Mar 16, 2016, 11:10 pm Last Edit: Mar 16, 2016, 11:11 pm by OldSteve
I'm just doing tests to make sure everything works before welding permanently. I do not have a breadboard. Does it really change anything I do not solder yet?
I wouldn't do it. That's a good way to kill electronic components. It's possible for power connections to be intermittent while inputs are active.
You would do better to buy a breadboard, or at least to solder pin headers onto the PCB, use jumper leads with sockets, then solder any other connections with temporary joints. It's soldering, not welding, by the way. Welding is a whole different thing.
While you have things connected the way you do, you cannot be sure of anything.

Edit: And you should post your code, (between code tags of course), not send us to another site to read it.
Please do not PM me for help. I am not a personal consultant.
And others will benefit as well if you post your question publicly on the forums.

Marmotine

#8
Mar 16, 2016, 11:22 pm Last Edit: Mar 17, 2016, 03:59 am by Marmotine
Quote
It's soldering, not welding, by the way. Welding is a whole different thing.
English is not my first language, Google translate....

About the project itself, does the half brige schematic is good? Because I just want to be sure that I'm doing it the right way before I solder anything. 

EDIT: I saw on Internet another half bridge and I wonder what is the ''A Output'' and the ''B Output''? Does this half bridge have better chances of success?



Here's the code:

Code: [Select]
//Load cell input onto HX711 board using a Blend Micro board
//Robert Cundall March 2016
//wire DAT to pin 8, CLK to pin 5, GND to GND and 2.7-5V to VIN

#include "HX711.h"
#define calibration_factor -2220.0
//change this calibration_factor to match the output reading to the applied load
//if you know the mV/V output from a load cell for the full range output (FRO) use the following formulae to calculate the calibration_factor
//calibration_factor = 4440000 / ((2/mV/V)x FRO)

#define DOUT 8
#define CLK 5
float value;
HX711 scale(DOUT, CLK);

void setup() {
 
  Serial.begin(9600);
  pinMode(13,OUTPUT);
  scale.set_scale(calibration_factor);
  scale.tare(); //Assuming there is no weight on the load cell at start up, this resets the output to 0
  }

void loop() {
  Serial.print("Reading: ");
  value=scale.get_units(), 0;
 
  Serial.print(value);
  //scale.get_units() returns a float
  Serial.println();
 
  delay(250);// 1/4 second delay
  //LED 13 switches on if the load >2000 and off if it goes below that
  if(value>=2000){
  digitalWrite(13,HIGH);
  }
  else{
  digitalWrite(13,LOW);
  }
 }

Marmotine

#9
Mar 17, 2016, 03:12 am Last Edit: Mar 17, 2016, 03:44 am by Marmotine
I found out on Internet that I should use 4 of these Load Cell.

Then, should I try to build a quarter bridge? Anyone have a schematic for this?

I thought of something like this, but I don't understand why, on the schematic, the strain gauge has only 2 wires.




Thanks a lot!

Wawa

#10
Mar 17, 2016, 05:43 am Last Edit: Mar 17, 2016, 05:44 am by Wawa
You have a load cell with three wires.
Two "strain gauges" in one.
Forget about the picture in post#9.

That load cell in your picture usually has red as "output".
So red to A+, White to E+, Black to E-
Connect resistors as I told you in post#1.

You would have picked this up if you had measured the load cell, as explained in post#3.
I would still do that.
Leo..

Paul__B

English is not my first language, Google translate....

I saw on Internet another half bridge and I wonder what is the ''A Output'' and the ''B Output''? Does this half bridge have better chances of success?

That is precisely the same circuit.  "Output A" and "Output B" are your A+ and A-.  But you need to check the resistance of the three wires as Wawa explained in reply #3.  Whatever the colours are, the wires with the highest resistance between them are E+ and E-, while the other wire which has the same resistance to either of those, is A+.

The two 1K resistors also go to E+ and E- with their common terminal to A-.  There is only one way to wire it (well, actually E+ and E- may be swapped for A+ and A-, but it has exactly the same effect), you merely need to get the three wires from the sensor correctly identified (by measuring resistances).

Marmotine

Thank for your explanation!

Since I do not have a lot of experience with electronic, I wonder how I could measure the resistance of a wire using my multimeter GMT-12a



When I found out which wire is E+, E- and A+, then on the schematic of post #8:

E+ became the white wire (on the schematic)
E- became the black wire (on te schematic)
A+ became the red wire (on the schematic)

Do I understand right?

Chagrin

Set your multimeter to the Ohm position (shows "X1K" in your picture) and measure the resistance between the three wires. Assuming the red/white/black wires are used in the typical strain gauge fashion then the resistance between white/black should be twice the resistance between white/red and black/red.

I wanted to mention that your two, 1K resistors need to be nuts-on identical resistance for this to work right and the chances of them having identical resistance is very low. Ideally you'd use a ~5K, multiturn potentiometer instead of two resistors but any potentiometer would be better than the 5% tolerance resistors you're using.

When at rest (and with your load cell amp unpowered) the the resistance between E+ and A+, E+ and A-, E- and A+, E- and A- must be identical. When powered on the voltage between A+ and A- must be as close to zero as possible (and a potentiometer would let you adjust that). With a load on the strain gauge the voltage change between A+ and A- the load cell amp is going to be measuring is a change in voltage in the 10s of millivolts.

Marmotine

Ok, if I understand right, I do not need to connect the load cell to the Amp to measure the resistance. The two wires that will indicate the higher resistance when I put them in pair should be the E+ and E-.

About the potentiometer, if I understand right, I can use a 5k potentiometer (and connect the left and center pin of it) to replace of one of the 1k resistor. Then, I could ajust the pentiometer to get about 0V between A+ and A-.

Should I replace the two resistor with a potentiometer or just one would be fine? Should I use the big button looking potentiometer or should I use a fine tunning potentiometer?

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