I am trying to figure out the resolution of a load cell (smallest weight change I can measure) that is connected through a load cell amp and being read through a 10 bit ADC (the DATAQ DI-149)

For example: with a 1000kg load cell will I be able to display in 100g increments or 1 kg increments or 10kg increments, etc.?

This project is going to be used to measure and graph a rocket engine's thrust over time curve.

Below are the load cell and amplifier specs:

Load cell

  • Combined Error (%FS) ≤±0.020
  • Creep (30 minutes) %FS/30min ≤±0.024
  • Temperature effect on sensitivity (%FS/10C) ≤±0.012
  • Temperature effect on zero (%FS/10C) ≤±0.020
  • Output sensitivity (mv/v) 2.0±0.2
  • Input resistance (Ω) 406+/-6
  • Output resistance (Ω) 350+/-3.5
  • Insulation resistance (MΩ) ≥ 5000 (50VDC)
  • Zero balance (%FS) 2.0
  • Temperature, Compensated -10~+40
  • Temperature, Operating -20~+60
  • Excitation, Recommended (V) 5~12(DC)
  • Excitation, Max (V) 18(DC)
  • Safe overload (%FS) 150
  • Ultimate overload (%FS) 300

Amplifier specs

  • The bridge road input impedance: 2KΩ
  • Input sensitivity: 1.5~2.1 MV/V
  • Sensor excitation voltage: 5V DC
  • Load capacity : 87 Ω (four 350 ohm sensor parallel)
  • The working power supply : 18~26V DC
  • Precision: 0.3%FS
  • Temperature characteristics: Better than 100 PPM
  • Working temperature: 0~50C
  • Load Cell Amplifier Strain Sensor Transmitter 4-20mA 0-10v 0-5v
  • 5
    \$\begingroup\$ You need to show a circuit - nobody is going to read the spec, understand the spec and design you a circuit without the parting of hard-earned cash. The cash bit was a joke but hopefully you get my drift on asking too much. \$\endgroup\$
    – Andy aka
    Mar 11, 2016 at 10:22
  • \$\begingroup\$ dont want them to design a circiut. just the maths formula \$\endgroup\$
    – Tracer Ion
    Mar 11, 2016 at 10:36
  • \$\begingroup\$ id be happy to do the maths. i dont know what the calculations are though. thats what im asking for. i'll simplify the question. \$\endgroup\$
    – Tracer Ion
    Mar 11, 2016 at 10:39
  • \$\begingroup\$ with a 10 bit adc and a 5V load cell amp and a 1000kg 2m/v load cell. what is the smallest measurable weight i can weigh \$\endgroup\$
    – Tracer Ion
    Mar 11, 2016 at 10:41
  • 5
    \$\begingroup\$ You don't simplify the question by turning it into a comment, you go back and edit your question so the answer will yield specifically what you need to know. \$\endgroup\$
    – EM Fields
    Mar 11, 2016 at 11:07

2 Answers 2


Since your load cell is an analog sensor, its resolution will be infinitesimal, but noise limited.

Then, since the ADC is slicing up the sensor's output into 1024 discrete values, the system's granularity will always be one part in 1024 regardless of the range the ADC captures.


Here: Let me smart you up on load cells:

  1. The excitation is the voltage you put into a load cell in order to get an output from it.

  2. The load cell's sensitivity is the ratio of its output to its input, stated in units of "millivolts[of output] per volt"[of input], with the full-scale rated mechanical load on the load cell.

What that means is that if you have a load cell rated for 1000kg full-scale and it exhibits a sensitivity of 2 millivolts per volt, with an excitation of 10 volts it'll put out 20 millivolts when there's 1000kg sitting on it. With 500kg sitting on it and 10 volts of excitation it'll put out 10 millivolts, as it will with 5 volts of excitation and a 1000Kg load.

And the rest of it:

In order to build up the tiny signal from the load cell into something that makes sense to feed into an ADC, you'll need to amplify that signal.

The gain of the amplifier will be determined by the ADC's full-scale input requirement and the Load cell's full-scale output, the relationship being:

$$ A_V = \frac{ADC_{\ in}}{LOAD CELL_{\ OUT}}, $$

In the case of a load cell with a 20mV output and an ADC with a 5 volt full-scale input, that gain would need to be:

$$ A_V =\frac {5V}{0.02V} = 250 $$

Now, for the nitty-gritty:)

After all of that, if everything were perfect, 1000kg sitting on the load cell would result in an output of 11 1111 1111 from the ADC and 00 0000 0000 with no load on the load cell.

That's 1024 different states representing the range from zero to 1000 kg, so that range would resolve into 1024 increments of about 976.56 grams each, which would be the resolution of the system.

In the instrumentation world, "offset" means "zero" and span means "gain", so what you'll need to do to calibrate your system, once you have it all hooked up, is to make sure there's no mechanical load on the load cell, turn on the excitation supply, crank the span control up a little until you get a reading on the ADC, and wait a while until everything gets stable. Once it does, adjust the offset pot for zero, put 1000 kg on the load cell, and adjust the span control for a reading on the ADC which reflects the resolution you want.

That is, for maximum resolution set the reading to 11 1111 1111, but if you want to resolve the reading into, say, one kilogram steps for convenience downstream, set the reading to 01 1101 1000 and the count will increase by one LSB for every kilogram added to the load. Repeat the zero and span adjustments as needed to get zero and span where you want them, and you'll be done. – EM Fields 2 mins ago edit
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  • \$\begingroup\$ does that mean if its 1 in 1024 i should be able to measure to the 1kg ? \$\endgroup\$
    – Tracer Ion
    Mar 11, 2016 at 11:01
  • \$\begingroup\$ the 10 bit adc reads 10 volt imput and its resolution is 20mv so i'm guessing it will detect 500 steps between 0 and 10 volts is this correct?? \$\endgroup\$
    – Tracer Ion
    Mar 11, 2016 at 11:14
  • \$\begingroup\$ so 500 steps into 1000kg is 2kg increments thats what i came up with. just trying to confirm if its correct?? \$\endgroup\$
    – Tracer Ion
    Mar 11, 2016 at 11:15
  • \$\begingroup\$ @TracerIon: does that mean if its 1 in 1024 i should be able to measure to the 1kg ? Not necessarily. To get a definitive answer you need to edit your question so that it reflects the load cell's excitation voltage, its sensitivity, the gain of the amplifier, and the input voltage range of the ADC. \$\endgroup\$
    – EM Fields
    Mar 11, 2016 at 11:15
  • 1
    \$\begingroup\$ @TracerIon: If the ADC can resolve 10 volts into 1024 steps, why wpuld its resolution be 20 millivolts instead of 9.77 millivolts? \$\endgroup\$
    – EM Fields
    Mar 11, 2016 at 11:24

A/Ds take a analog voltage and create a binary number, the value of which is proportional to that voltage. Since the output is binary, the total different possible output numbers is 2bits. You say you have a 10 bit A/D, so it can represent the input voltage as 210 = 1024 different levels, 0 to 1023.

With the right offset and gain in the analog circuit, the ends of the range you care about are mapped to the A/D input voltage such that the resulting output numbers are 0 and 1023. This leaves you with 1023 steps that your input range is broken into. The really basic and naïve answer is therefore that the resolution is 1/1023 of the range.

However, you're not going to get that in reality. No A/D is perfect, and the input ranges for each of the 1024 possible output codes won't be exactly evenly spaced. Generally A/D's are specified to at least be monotonic with no missing codes. That means in theory two adjacent readings may be very close in voltage, then there can be a gap of up to two nominal steps. In this case, the worst case resolution is half of the theoretical ideal calculated above. Of course you need to check the datasheet of your particular A/D to find out what it's characteristics really are.

Then there is accuracy, which you haven't mentioned but is probably important. Again, take a look at the A/D datasheet. Even though the maximum spacing between codes may be 2/1023 of the input range (for a 10 bit A/D), these may be clumped a little towards one end, both ends, the middle, etc. See the non-linearity and total error specs.

Then there are issues with the load cell. These things are notorious for having some hysteresis. Look at its datasheet carefully. Most likely, you will get a little different output voltage for the same force when the cell was previously subjected to high force than when it was previously subjected to low force.

The accuracy is likely rather worse than the A/D resolution. You have to look at the whole system.

  • \$\begingroup\$ thank you so much for the responce. i do understand what your saying. my main concern is that i want to measure the engineering units to the 0.2kg so if i cant do that with a 1000kg load cell. then i will get a 100kg load cell. i understand no adc or load cell or amp are perefect. i was just after theoritcal best case and the right calculations to work. wich i think i have done. but you have been a big help. thank you \$\endgroup\$
    – Tracer Ion
    Mar 11, 2016 at 12:31
  • \$\begingroup\$ the only thing i could find in the spec sheet was "overall innaccuracy" wich is +- 64mV \$\endgroup\$
    – Tracer Ion
    Mar 11, 2016 at 12:33
  • \$\begingroup\$ @TracerIon: What's the maximum load you expect the load cell to measure? \$\endgroup\$
    – EM Fields
    Mar 11, 2016 at 15:21

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