measurementresistors
I have a load cell whose resistance is varying between 318.0 and 318.8 Ohm. I have used a voltage divider but still the output voltage is rarely changing when the resistance changes.
Is there any electric circuit that I can use in order to model the resistance variation in terms of Voltage variation?
Like gbarry says. The next higher value you find in the Digikey offering is 2 kΩ, so you would need a 6386 Ω resistor in parallel. That is between one end and the wiper, not across the full resistance. 6386 Ω is also an odd value, so you may want to compose that using different standard values as well.
Now the 1523 Ω, how precise does that have to be? You represent it in 4 significant digits, that's 0.1 %. Potmeters are often 20 %, more precise ones can be pretty expensive. So our 2 kΩ can be anywhere between 1600 Ω and 2400 Ω. In the first case your parallel resistor should be 31650 Ω, in the second 4170 Ω. That's a wide range. You could use a 33 kΩ trimming potmeter. Use a multiturn is you want to get the 1523 Ω really precise.
Note that placing another resistor over the potmeter's wiper distorts its linear characteristic.
The greenish curve shows the resistance as a function of the rotation in % for a 2.2 kΩ resistor parallel to a 5 kΩ linear potmeter. The purple curve shows that resistance for the 2 kΩ potmeter plus the 6386 Ω, so that's not so bad after all. The rule for the most linear characteristic is to have the ratio resistor/potmeter as large as possible.
OK, problem solved. The bridge is connected like this. Only one resistance in the load cells is variable, the other is fixed.
Why the confusion above? I was measuring resistance of a load cell which came from a different scale. The cells looked pretty similar, therefore I thought they were the same. But they were not! Eureka!
simulate this circuit – Schematic created using CircuitLab
Best Answer
If the resistance change is indeed so small (which I doubt; it's only 0.25 %), then you need a Wheatstone bridge:
The amplifier is an instrumentation amplifier, so the feedback is internal. Rx is your variable resistor, and R3 is a fixed 318 Ω resistor. R1 and R2 are equal, their actual values are not that important, let's pick 10 kΩ for them. Then the voltage at the inverting input of the instrumentation amplifier will be 2.5 V, while the voltage on the non-inverting input will vary between 2.5 V (Rx = 318 Ω) and 2.497 V (Rx = 318.8 Ω). That's only a 3.14 mV difference, and you'll need a 1000 \$\times\$ gain to get 0 V to 3.14 V out of this.
The 0.25 % variation is smaller than the tolerance of a 1 % resistor, and even 0.1 % resistors will give you a noticeable offset. If you pick 0.1 % resistors for R1 and R2 then the theoretical 2.5 V may deviate by as much as 2.5 mV, that's 80 % of your full range. So you'll have to trim the resistors so that the output is 0 V at a 318 Ω resistance.
If you use 100 kΩ resistors for R5 and R6 and a 10 kΩ multiturn for the potmeter you'll be able to adjust the reference voltage between 2.48 V and 2.52 V.