Part 4 · Data Conversion

Ideal SNR — where \(6.02N + 1.76\) comes from

Key point: a full-scale sine against uniform quantization error \(\sigma_e^2=\mathrm{LSB}^2/12\) algebraically yields \(\mathrm{SNR}_{\mathrm{dB}}=6.02N+1.76\). That is a ceiling, not a promise.

Along the way

1
Noise power first. From Part 3: \(\sigma_e^2=\Delta^2/12\) with \(\Delta=\mathrm{FSR}/2^N\).
2
Sine power. Full-scale sine has \(A=\mathrm{FSR}/2\) and average power \(A^2/2\). Ratio of those two powers is SNR.
3
FSR cancels. Bigger range → bigger LSB → more noise, but also more signal. The dB line depends only on \(N\).
4
Each bit ≈ +6 dB. \(20\log_{10}2\approx 6.02\). Step \(N\) and watch the amber dot jump. That is the whole “6 dB per bit” slogan.
5
Ceiling, not a datasheet. Assumes no clip, no DNL/INL, busy-signal white \(e\). Real parts quote SINAD, always worse. Coherent sampling is how you even measure this on an FFT.

Dither, overload, and slow DC (the uniform-(e) model lies). Still: every datasheet SNR discussion starts here.

Next: Part 5 — analog noise density also folds.

SNR vs bits (ideal quantizer)

Blue curve = formula · amber dot = your \(N\) · each bit ≈ +6 dB

Bits \(N\) FSR (V) — scales LSB only

LSB—

σ_e = LSB/√12—

SNR—

≈ dB / bit6.02

\[ \sigma_e^2 = \mathrm{LSB}^2/12 \]

→ —

\[ P_{\mathrm{sine}} = A^2/2,\quad A=\mathrm{FSR}/2 \]

→ —

\[ \mathrm{SNR}_{\mathrm{dB}} = 10\log_{10}(P_{\mathrm{sine}}/\sigma_e^2) = 6.02N + 1.76 \]

= —

FSR cancels: larger range → larger LSB → more noise power, but also more signal power.

Try it: step \(N\) by 1 — SNR jumps ≈6 dB. Then open the Accuracy Translator and match ENOB ↔ SNR on the nomograph.

Deep link: Accuracy Translator.