0-16384 (14 bit ADC) is infinity?

16,384 = 2^14.

The lowest value the sensor records, however, isn't 0.

It's 2^0. Were it zero, any increase would be infinite on a percentage basis.

But it isn't.

1-2-4-8-...2^bitdepth

The range is from 0 to (2^N)-1

it is 0 to 16383

I think what 3kramd5 was getting at was that no system has zero noise. When we convert the voltage of a pixel into an ADU with the ADC, we cannot convert a fraction of an ADU. If RN is 3e- and FWC is 60ke-, then 3e- RN, although in floating point precision is 0.8192, ADUs are integer (at least, they are in todays sensors...maybe at some point we'll have cameras that can convert directly into 32-bit float RAW. ) Since ADUs are integer, you cannot convert any non-zero charge to zero...the minimum ADU is 1, or 2^0.

DSP (Digital Signal Processing) 101......

A N bit D/A converter is capable of resolving 2^N states. A 4 bit D/A can resolve 16 states, an 8 bit D/A can resolve 256 states, and a 14 bit D/A can resolve 16,384 states. I think we can all agree on this.

So with a 14 bit D/A there are 16,384 states. These states are represented as binary 00 0000 0000 0000 to 11 1111 1111 1111, or 0 to 16,383 in decimal.

The signal that we wish to measure is typically fed through an amplifier (or attenuator for large signals) so that it's maximum value will be scaled to the input range of the A/D converter. For example, lets say we have an 8 bit D/A converter that works from 0-15VDC.... if we are using it to measure a signal from 0 to 1VDC then we only get the last 4 bits of resolution toggling and we have thrown away the accuracy of the system. Scale the input signal up by 15X and now you get all bits toggling. In this system the state 0000 0000 does not represent 0 volts, it represents from 0 volts to less than 15Volts/256 (0.0586 volts). Likewise, the state 1111 1111 does not represent 15 volts, it represents from 14.9414 volts to 15 volts. Each state represents a range, not an absolute value. The state 1111 1111 is special, it also represents the overload condition where an input signal is high enough to saturate the converter.

So back to our Canon 14 bit A/D....

It's lowest possible reading is 0, it's highest possible reading is 16,383. In any circuit there is the noise floor... the lowest level of signal found in the input signal. In a well designed circuit, the resolution of the A/D converter will be less that the noise floor. What this results in is the last few bits of the A/D converter toggling almost at random. The noise comes from our amplifier, from our converter, from fluctuations in our reference voltage on the A/D converter, and from outside. When we get rid of our least significant digits that are toggling randomly, we are left with the "significant digits".

My suspicion is that Canon does not use 14 bit A/D converters, but uses 16 bit or even 20 bit A/D and throws away all but the most significant 14 bits.