A field team has a real problem to settle: How do time samples and voltage codes fail differently? They must decide what happens before they change adc bits on the device. Predict the direction first.
See the relationship first
The figure reads from left to right. The blue card is adc bits. The middle card uses this page's rule. The green card is minimum sample rate. 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 adc bits is 12.
- 2
Name the rule. fs,min=2x40=80 samples/s |90-1x100|=10 Hz |1200-1x1000|=200 Hz levels=2^12=4096 q=3.3/4096=0.806 mV; q/2=0.403 mV SNRideal=6.02x12+1.76=74.0 dB
- 3
Put in the chapter value. Set adc bits to 12. The page rule gives minimum sample rate as 80 Hz.
- 4
Read the result. Keep Hz next to the value. Use it only within the limits on this page.
Predict, then change adc bits
Try Predict what happens to minimum sample rate. Move one control, calculate, then check your idea.
Observe Bit depth partitions voltage. Sample rate partitions time. Each control can improve one axis while leaving the other failure untouched. Reset to 12 and compare minimum sample rate.
Explain Only adc bits 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. Start with the physical story
Sampling chooses moments in time. If the clock is too slow, a fast input can appear as a false slower signal. Quantisation happens after sampling and rounds each captured voltage to one of a finite number of codes.
2. Name every algebra move
Set NyquistDouble the highest useful signal frequency.
Fold a toneSubtract the nearest whole multiple of sample rate and take the magnitude.
Count levelsRaise two to the number of bits.
Find one codeDivide 3.3 V by the level count.
Translate resolutionDivide code width by 10 mV per degree C.
Estimate the ideal ceilingUse 6.02N+1.76 dB.
3. Reproduce the chapter case
|90−1×100|=10 Hz
|1200−1×1000|=200 Hz
levels=2^12=4096
q=3.3/4096=0.806 mV; q/2=0.403 mV
SNRideal=6.02×12+1.76=74.0 dB
The 1.65 V conditioned span uses 2048 codes at 12 bits, while each code represents about 0.0806 degrees C at 10 mV per degree C.
4. Try one real input
TryChange ADC bit depth and predict code width and ideal SNR. Watch the alias results stay fixed.
ObserveAt 12 bits, doubling the level count requires one more bit, but neither stored alias changes because their sample rates did not change.
ExplainBit depth partitions voltage. Sample rate partitions time. Each control can improve one axis while leaving the other failure untouched.
These are ideal converter and discrete-time bounds.
- Filter
- Nyquist equality leaves no transition band for a real anti-alias filter.
- ADC
- Noise, reference drift, non-linearity, source impedance, and jitter reduce effective bits.
- Sensor
- Code width is not the same as calibrated accuracy.
Correct, not complete: sample rate and nominal bits do not prove end-to-end measurement quality.
5. Use the result in the design
Set a passband, choose filter margin and sample rate together, then measure input-referred noise and calibrated error in physical units.
6. Record the evidence state
Keep signal band, interferers, filter response, sample clock, reference, bit depth, input span, source impedance, measured SNR, ENOB, calibration, and temperature.
7. Check yourself
Why does 90 Hz appear at 10 Hz when sampled at 100 samples/s?
What happens to code width when one bit is added?
Does a 0.0806 degree C code prove that accuracy?
The calculations reproduce the chapter's ideal sampling and ADC constants.
- Filter
- Nyquist equality leaves no transition band for a real anti-alias filter.
- ADC
- Noise, reference drift, non-linearity, source impedance, and jitter reduce effective bits.
- Sensor
- Code width is not the same as calibrated accuracy.
Correct, not complete: sample rate and nominal bits do not prove end-to-end measurement quality.
Sammy guides