See the relationship before changing it
The figure reads from left to right. The blue card is load-cell adc resolution. The middle card applies this page's rule. The green card is ideal mass step. Walk the arrows once: set the input, apply the rule, then read the result with its unit.
The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only load-cell adc resolution, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 16 bits.
- 2
Name the relationship. mass step = 50,000 g / 2^bits
- 3
Substitute with units. 50,000 / 2^16 = 0.763 g
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change load-cell adc resolution
Try Predict the direction of mass step = 50,000 g / 2^bits. Test another load-cell adc resolution, then compare ideal mass step.
Observe More bits divide the same load-cell span into smaller ideal mass steps. Reset load-cell adc resolution to 16 and compare ideal mass step.
Explain More bits divide the same load-cell span into smaller ideal mass steps.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
1. Turn a physical target into required steps
Full-scale range is the largest span the converter must represent. The smallest meaningful change is the gap that must receive a distinct code.
Put both values in gramsFSR = 5 kg = 5000 g; Δmin = 10.0 g.
Divide range by changeRequired steps = 5000/10.0 = 500.
Phoebe: Bits are only a compact way to count how many binary steps the ruler contains.
2. Convert 500 steps into bits
Substitute the ratioNmin = ceil(log2(500)).
Take the base-2 logarithmlog2(500) = 8.97.
Round upward, not to nearestceil(8.97) = 9 bits, because a fraction of a bit is not an available code.
3. Verify the boundary on both sides
Test 8 bitsq = 5000/28 = 5000/256 = 19.5 g, so it fails 10 g.
Test 9 bitsq = 5000/29 = 5000/512 = 9.77 g, so it just passes.
Test a common 12-bit partq = 5000/4096 = 1.22 g.
Compare step widths9.77/1.22 ≈ 8× finer than the minimum-passing 9-bit step.
4. Find the ideal quantization floor
Use the 9-bit stepσq = 9.77/√12.
Divideσq = 2.82 g RMS.
Use ideal SNR6.02(9)+1.76 = 55.9 dB.
These are theoretical converter limits, not the measured noise of the load cell and amplifier.
5. Keep timing as a separate gate
Name the fastest wanted eventFind fmax from the load or motor-current event that must remain visible.
Double it for the ideal lower boundChoose fs at least 2fmax.
Add an engineering marginReal anti-alias filters roll off gradually, so practical sampling usually exceeds the bare 2× bound.
6. Check yourself
1. Why is 8.97 rounded up to 9?
Only whole code bits exist, and rounding down gives 256 steps instead of the required 500.
2. Does 9-bit conversion guarantee 10 g accuracy?
No. It gives a 9.77 g code step; sensor, amplifier, reference, calibration, and environmental errors remain.
3. Can 12 bits remove bench noise?
No. More codes only lower quantization error; they do not remove analog noise already on the signal.
7. Honesty boundary
These are the chapter inputs, worked results, and named teaching assumptions.
- The calculation assumes the entire ADC range maps cleanly to 0–5000 g and treats uniform quantization independently
- Named teaching assumption
- Load-cell accuracy
- Chapter input or worked result
- bridge excitation
- Chapter input or worked result
- amplifier gain and offset
- Percentage, ratio, or gain
- effective number of bits
- Digital resolution or converter setting
- drift
- Chapter input or worked result
- vibration
- Percentage, ratio, or gain
- calibration can dominate
- Percentage, ratio, or gain
- The Nyquist rule is only the ideal lower bound
- Chapter input or worked result
Go deeper in Boundaries, Resolution Fit, and Filter Choice, then measure the production analog chain.
Phoebe guides this bridge