View Full Version : Unsolved Load cell with ina12x + pic18f2550
11-16-2011, 01:07 PM
I've got a litle dudes about a proyect for amplification the signal of a load cell.
I'm, working in a project where 30 microcontrollers read the weight of a load cell and send it via GSM/GPRS.
So my dude is what amplifier to use (INA125, INA122 or INA126)??.
By now, I have tried INA125 with this settings:
Single Power supply. Vcc = 5v
When this amplifier gets the maximum weigth of the load cell (it gets 4,22 v on the output). This voltage is used for a microcontroller PIC18F2550 to make a A/D convertion, but the problem is when I get these 4,22 (max) the PIC doesn't understand that this is the maximum level (for the PIC, the maximum level would be 5 v of Vcc).
Is there any way to fix 4,22 v from the INA125 at the input of the Analog Reference to the PIC ?
Please, would be desirable to use another INA like INA122 / INA126. What INA would be the best for this design (Load Cell to amplifier to A/D module all with 5 v and single power supply)??
Thanks in advance
Beau Schwabe (Parallax)
11-16-2011, 01:53 PM
How do you have the VRef configured on the INA125? With a 5V supply, according to the PDF you can only use the 2.5V ref setting....
"Positive supply voltage must be 1.25V above the desired reference voltage. For example, with V+ = 2.7V, only the 1.24V reference (VREFBG) can be used."
11-16-2011, 02:12 PM
Thanks for your answer.
I don't have Vref configured in INA125. 5v from the power supply directlyto the bridge and output V- and V- to the INA125.
I assume that you input the loadcell output into pin 6 for the V+ and pin 7 for the V-. Also the ground of the loadcell and the ground pin on the INA125 [pin 3 and pin 5] have to be tied together [as does the ground on the microprocessor].
Using only 5 volts as a supply to the loadcell does not give much output. If the loadcell can handle 10 volts the output would be twice.
Why don't you use the VrefOut of the INA125 to supply the bridge? I realize that this is a more complex circuitry; especially as you probably have to use a TIP29C or equiv. if the bridge is composed of resistors less then, say, 1000 ohms. But since the bridge output is ratiometric; that is, the output is directly proportional to the bridge supply, it makes sense to make the supply well regulated.
Beau Schwabe (Parallax)
11-16-2011, 02:49 PM
Figure #6 page 13 of that same data sheet , looks to be the configuration that you should use.
11-16-2011, 04:30 PM
thanks for both answers.
I was using #5 page 12 but with 5v not 3v asthe datasheet says. I can't use #6 page 13 because my load cell must to be fed with 5 v min to 18v max.
I have tested too Vref 5v with a power supply of 12v and the output is the same 4,22v (max). 4,22 vol is not a good reference for the PIC because PIC understands a Vref of 5v of its own power supply (not those 12 v of the power supply of the INA125. The ground is the same of course)
If I use 12 v to power supply INA125 I could use then 10v Vref to fed the load cell, but in this scenario, the PIC can't understand Vref=10v because it is above 5v of the max power for PIC and I would break the PIC microcontroller.
1. If I use 5v Vref from INA125 (a power supply of 12v) to load cell, I get max 4,22 v once amplification is done.
2. If I use 10v Vref from INA125 (a power of 12v) to load cell, I can't make a correct Vreference for PIC because it must be 5v as limit.
Please, do you have a way to make a correct reference for PIC?
Thanks for your help
Beau Schwabe (Parallax)
11-16-2011, 04:56 PM
Adjust your gain resistor between pins 8 and 9 with the 4.22V configuration.
I'm sorry but I don't understand your problem as stated in reply #6.
The best I can suggest is to disconnect the bridge/amplifier from the ADC/microprocessor in order to test the analog amplified output of the loadcell.
If you can cycle the applied load on the cell you can then determine if the bridge/amplifier if working. If you can't cycle an applied load to the cell, you can simulate the load by shunting one arm of the loadcell bridge with a high value resistor. You will have to determine which of the two bridge arms to shunt which will give a positive output from the amplifier. Shunting either the supply voltage and the V+ arm or the ground and V- arm will result in a positive output. A shunt resistor in the range of megaohms should give a reasonable simulated output. If you know the value of the bridge arms a shunt resistor value can be determined which will give a specific output. I can link you to a source but if you go to micromeasurements web site [ manufactures of strain gages] you will find that they are an excellent resource for such information.
Boy, was I wrong on the approximate value of a shunt resistor. So lets first give some assumptions:
The load cell is the normal, strain gage based, wheatstone bridge arrangement cell; with an unamplified full-scale output of, say, 3 millivolts/Volt.
If so, the strain gages, making up the arms of the wheatstone bridge are, most probably, 350 ohms.
[There are other assumptions as to how the gages are arranged [strained] but the shunting is only to tell if the cell/amplifier are working.]
To get a simulated output of about 1/2 the full-scale load you can use a shunt resistor of about 60k ohms.
By-the-way; if the load cell is not new it may pay to check to see if the output at zero load is, indeed, about zero. This is hard to do as the normal output is too low to show up on a normal digital voltmeter. Still, with the load cell disconnected but powered check that the voltage from each output is 1/2 the supply voltage. If so set the voltmeter at the lowest scale [200 mV] and take a reading between the outputs. There should not be hardly any voltage.
Sorry about the shunt value given in the previous message. Memory, you know ----.
11-17-2011, 08:50 AM
Thanks Drei and Beau for your answers and suggestions. They are very usefull for me.
The unamplified output for the load cell is 1,8 mv/v. I have tested it with no load and it works perfect for zero load, giving at the output for INA125 0 v aprox. The PIC understands it well and maps the Analog Value to 0 (in a scale from 0 to 1023).
The load cell is for a max weigth of 100 Kg, and I had adjusgted to work at 80 Kg max. To make it, I have set the Gain Resistor of INA125 to 100 Ohm, in this scenario, when I put on 80 Kb or more in the load cell, I get the maximum reading at output of INA, that is 4,22. These means that with a load of 80K and Vref=5v:
1. WIth Gain Resistor = 100 Ohm I get 4,22 Ohm at output.
2. With Gain Resistor more than 100 Ohm I get an output reading less than 4,22. It is normal because I decrease the gain.
2. With Gain Resistor less than 100 Ohm I get an output reading of 4,22, it means that the maximum voltage at the output of INA125 with 5v of Vref is 4,22.
I have read the datasheet again and on page 3 #Table "SPECIFICATIONS: VS = +5V" it says that ther output for 5v (V+)–0.8 Maybe it could explain why I'm getting max 4,22 v (4,98v power supply - 0,8 typical gap at output = 4.18 aprox my 4,22v that I get). If this was correct and was a correct understanding, I believe that I should adjunt via software in the PIC microcontroller the settings for mapping 5v as max 4,22v.