12-bit or 16-bit ADC for a 0.2 mV Requirement?
An eight-channel 12-bit SAR ADC with VREF = 5.000 V, a 20 pF internal sample capacitor and a multiplexer on-resistance of 200 Ω scans eight sensor channels at 20 kS/s each. Each channel arrives through a 10 kΩ series resistor from a remote sensor; nothing is buffered, and the only capacitor on the board is 100 nF placed at the sensor end of that lead, not at the ADC pin. The sample window is configured to 1.5 µs. The sensors deliver 0–4 V with 2 mV rms of broadband noise over 0–10 kHz, and the specification calls for resolving 0.2 mV at the sensor. A vendor offers a pin-compatible 16-bit version of the same converter for three times the price.
1. Check whether the 1.5 µs sample window is long enough. Quantify the error if it is not.
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The sample capacitor charges through the 10 kΩ series resistor plus the 200 Ω mux switch: τ = (10 000 + 200) × 20 pF = 204 ns.
For a 12-bit converter, settling to ½ LSB means settling to 0.5 × (5/4096) = 0.61 mV of a 5 V step, i.e. e^(−t/τ) ≤ 1.22 × 10⁻⁴, so t ≥ 9.01τ = 1.84 µs.
Available time is 1.5 µs = 7.35τ, leaving a residual of 5 V × e^(−7.35) = 3.2 mV ≈ 2.6 LSB.
And the worst case is a full-scale step between adjacent channels on the mux - precisely what a scan does every 50 µs. So every channel transition carries a few LSB of settling error that is correlated with the previous channel's value: it looks like crosstalk and it will not average away.
Fix before anything else: buffer the channels (or drop the series resistance to ≤1 kΩ), and move the 100 nF to the ADC pin, where it supplies the sample charge locally. Note the 100 nF at the sensor end is useless for this - the 10 kΩ sits between it and the converter.
2. Determine the effective resolution the system can actually deliver and state whether the 16-bit part helps.
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Quantisation: LSB = 5.000/4096 = 1.22 mV, so the 12-bit converter alone is ~6× coarser than the 0.2 mV requirement.
Noise: 2 mV rms against a full-scale sine of 5 V peak-to-peak (1.768 V rms) gives SNR = 20 log(1.768/0.002) = 58.9 dB, i.e. ENOB = (58.9 − 1.76)/6.02 ≈ 9.5 bits. The analog noise, not the converter, sets the resolution floor.
Verdict on the 16-bit part: it does not help. With 2 mV rms of input noise the extra codes are noise, and a 16-bit converter fed through the same 10 kΩ unbuffered path needs 11.8τ = 2.4 µs of settling - the 1.5 µs window becomes more than twice as wrong. Buying it converts a solvable analog problem into an expensive one.
What actually works: average. The 2 mV rms of noise dithers the 1.22 mV LSB, so N samples reduce the noise by √N: N = 100 gives 2 mV/10 = 0.2 mV, and 100 averages at 20 kS/s is 200 readings/s per channel - fine for a pressure or temperature input. After settling is corrected, this meets 0.2 mV with the 12-bit part, provided the ADC is monotonic and the noise is uncorrelated with the signal.
3. Rank the three proposed actions - buy the 16-bit ADC, add 100× averaging, buffer and re-terminate the input - and justify the order.
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1. Fix the input network (mandatory, first). Buffer the channels or cut the source resistance, and put the charge reservoir at the ADC pin. This removes a 2.6 LSB error that occurs on every channel change. That error is deterministic and signal-correlated, so it will not average down by √N - averaging it merely produces a stable wrong answer. Nothing else is worth doing until this is fixed.
2. Add 100× averaging (meets the requirement). After step 1 the error budget is dominated by 2 mV rms noise, and 100 averages bring it to 0.2 mV, matching the specification with the 12-bit converter. Cost: throughput falls from 20 kS/s to 200 S/s per channel - acceptable only if the measurand is slow, which must be confirmed against the process time constant.
3. Reject the 16-bit ADC (until proven necessary). It buys resolution the analog front end cannot support, triples the cost, and tightens the settling requirement to 2.4 µs. It becomes justified only if, after buffering and averaging, the system still needs better than ~0.1 mV - at which point the limiting factor will be the sensor and its wiring, not the converter, and the 3× budget is better spent on shielding, twisted pairs and a lower-noise excitation.
Order of work matters because resolution added ahead of an unresolved analog error is money spent on a false reading.