Load Cell with 3 wires

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!

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…

Thank a lot for the reply!

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

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..

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: load cell readout by Rob208

You cannot connect your wires and resistors like that. You need to solder the joints.

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?

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?

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.

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:

//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);
  }
 }

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!

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…

Marmotine:
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).

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?

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.

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?

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-.

Correct.

Marmotine:
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?

I would suggest replacing both resistors with the potentiometer. The outside legs on the pot will be connected to E+ and E- and the middle leg (wiper) will be connected to an A.

What you're building here with the strain gauge and potentiometer is a Wheatstone Bridge. One of the advantages of a wheatstone bridge is that it behaves stably with changes in temperature and mixing the different resistance materials of a potentiometer and a resistor will lose that advantage. Probably not a huge issue but y'know. It will be simpler to wire up though.

Also, I don't agree with OldSteve's statement that you need to solder these joints right away. The currents you're dealing with are very low and soldering the joints won't make much of a difference. The bigger problem will be electrical noise .... but worry about that later.

Thank a lot for your answer, I really appreciate your help!

I did the resistance test on the wires and I found out that when I connect the white one and the black one, I get a higher resistance. So I assume that those are E+ and E-.

For my project, I don't need a very high precision of the load cell and the project will always be at a temperature between 0°C and 5 °C.

Are you saying that I can use only one potentiometer to replace both resistors? Something like this:

Chagrin:
Also, I don't agree with OldSteve's statement that you need to solder these joints right away. The currents you're dealing with are very low and soldering the joints won't make much of a difference.

I can't imagine just twisting the wires together. Temporary solder joints are just as fast, if not faster, and you're assured a positive connection. Wires don't need to be poked through the holes in the board for this, or twisted around each other - just 'tacked'.
Twisted wire joints are a bad habit to get into. As I mentioned, in many cases this can lead to an active signal being applied to a chip that's only intermittently powered. We all know that this is a no-no with CMOS chips.

Marmotine:
I did the resistance test on the wires and I found out that when I connect the white one and the black one, I get a higher resistance. So I assume that those are E+ and E-.

That is correct. In fact, the two lower resistances will be almost identical.

Marmotine:
For my project, I don't need a very high precision of the load cell and the project will always be at a temperature between 0°C and 5 °C.

That's not the point, You must balance the bridge to start with, so you need those resistances to be equal.

Marmotine:
Are you saying that I can use only one potentiometer to replace both resistors?

Yes, but it must be a good potentiometer, preferably a multi-turn one.

Thank for the advice!

I just want to be sure that the schematic that I made in Post #16 is good before wiring everything. So, basically, I just need to put a voltmeter between pin A+ and A- and turn the knob of the potentiometer until I get a 0 V reading.

Here's the potentiometer that I might buy, is this good?