A field team has a real problem to settle: How can 4 mg/LSB stay fixed while the range grows? They must decide what happens before they change full scale on the device. Predict the direction first.
See the relationship first
The figure reads from left to right. The blue card is full scale. The middle card uses this page's rule. The green card is codes per side. Follow the arrows: set the input, use the rule, then read the result and its unit.
The audit later on checks more than one number. Here, the added model uses the baseline named below and holds every other chapter value fixed. That sentence bridges the fixtures, so the numbers do not change without a reason.
Derive the baseline in four moves
- 1
Name the input. The chapter baseline for full scale is 16.
- 2
Name the rule. At ±16 g: 16,000/4=4,000 codes per side 8,000 total codes; log₂(8000)=12.97, so 13 bits At ±2/4/8 g the result is 10/11/12 bits ODR=100 Hz gives BW=50.0 Hz
- 3
Put in the chapter value. Set full scale to 16. The page rule gives codes per side as 4000 counts.
- 4
Read the result. Keep counts next to the value. Use it only within the limits on this page.
Predict, then change full scale
Try Predict what happens to codes per side. Move one control, calculate, then check your idea.
Observe Range and scale set how many code buckets are needed; sample rate sets which motion frequencies can be represented. Reset to 16 and compare codes per side.
Explain Only full scale moves here. The other chapter values stay fixed.
Check yourself
What should you do before you trust the result?
What does this small model leave out?
1. A code is one fixed-size bucket
At 4 mg/LSB, each count represents four milligravity units. A wider positive-and-negative range needs more buckets. The bucket size stays fixed; the number of addressed buckets grows.
2. Name the algebra moves
Use both signsTotal span is 2×FS.
Count bucketscodes=(2×FS×1000 mg/g)/(4 mg/LSB).
Undo powers of twoN=log₂(codes).
Round upwardA fraction of a bit still needs the next whole bit.
Bind the rateBW=ODR/2 sets the Nyquist ceiling.
3. Reproduce the widest range
8,000 total codes; log₂(8000)=12.97, so 13 bits
At ±2/4/8 g the result is 10/11/12 bits
ODR=100 Hz gives BW=50.0 Hz
Every doubled range needs one more bit because doubling a code count adds exactly one to its base-two logarithm.
4. Try the full-scale range
TryMove from ±16 g toward a narrower range while the 4 mg/LSB scale stays fixed.
ObserveNarrowing the range reduces code count. The 50 Hz bandwidth stays fixed because this control does not change ODR.
ExplainRange and scale set how many code buckets are needed; sample rate sets which motion frequencies can be represented.
This is a digital-scale ledger, not a complete MEMS model.
- Scale
- Offset, noise, nonlinearity, calibration, and temperature still change usable accuracy
- Bits
- A 16-bit register does not guarantee 16 meaningful measurement bits
- Bandwidth
- Filters and mechanical response also shape the signal before sampling
Verify register settings, orientation, clipping, noise, and real event spectra on the assembled product.
5. Test the motion claim
Use known tilts and controlled impacts at each selected range. Record clipping, code scale, noise, filter delay, and missed high-frequency content.
6. Record the configuration state
Store part revision, DATA_FORMAT, BW_RATE, range, full-resolution mode, ODR, filter, mount orientation, calibration, and raw bench traces.
7. Check yourself
Why does doubling range add one bit?
Does a 16-bit output mean all 16 bits carry range information?
Does 50 Hz bandwidth prove a drop is captured?
The 4 mg/LSB and four ranges reproduce the chapter's ADXL345-style teaching case.
- 4 mg/LSB
- Fixed nominal full-resolution scale used by the chapter
- 10–13 bits
- Code capacity, not effective-noise-free resolution
- 50 Hz
- Nyquist ceiling from 100 Hz ODR, not a complete transfer function
Correct, not complete: code arithmetic does not qualify an accelerometer product claim.
Blueprint Bina guides