How much is a 24-bit converter?

How much is a 24-bit converter?

A while ago we supervised the design of a linearization module for an R/2R digital-to-analog converter (DAC). It was an interesting piece of work. In general, the resolution was very high and for example, for audio applications, if we consider a 24-bit converter – how much is that? With this post, I just wanted to inspire some thinking…

Well, if we consider a 24-bit converter, the number of levels is simply

L = 2^{24} = 16777216

so something like 17 megalevels or so. Is that much? Well, “yes”? They say that humans have difficulties visualizing grand scales and I am not sure my picture below will help, but hopefully.

Let us assume our reference range is Mount Everest. Now, I understand that not most of you have seen Mount Everest, but yet. The world’s tallest mountain. We can then add for example Burj Khalifa to the picture, etc. (Not all of us have seen that either, but it is at least approximately 10x smaller).

If we have an object of length x and assume that Mount Everest is A = 8848 meters high, we can express the impressive number of bits (INOB) as

\text{INOB} = \log_2 \left( {2^{24} \cdot \frac{x}{A} } \right) = \log_2{2^{24}} + \log_2 x - \log_2 A = 24 - 13.1 + \log_2 x \sim 10.9 + \log_2 x

Burj Khalifa thus has an INOB of approximately 20.6 bits (and Mount Everest would be 24 bits). The picture below tries to visualize this in a good manner.


But perhaps we need one of thoste tables to make life even better.

Thing INOB
Mount Everest 24
Burj Khalifa 20.6
Airbus 380 17.1
J Jacob Wikner 11.8
A bottle 9.2
An inch 5.6
1 mm 0.9

Or if we would now think in terms of the Earth’s circumference (which is approximately 40000 km), this means that you have to take a walk (or some suitable means of transport) to walk along the equator. Every 2.3 m-ish you have to stop and place a level-indicator pole of some kind.

Well, that’s in meters, what about good-old Volt?

Now, assume we define the potential drop from top to bottom of Mount Everest to be 1 Volt. Which in turn means that the least-signficant voltage level is 60 nV.
Also going back to our audio case and say that the bandwidth of interest is 22 kHz. We can now compare the levels with the Johnson-Nyquist noise (kT).

v_{nn} = \sqrt{ k T \cdot 22000 } \sim 9.54 nV_{rms}

and the level of the quantization noise becomes

v_{nq} = \sqrt{ 60^2 / 12 } \sim 17.21 nV_{rms}

If the voltage swing of our Mount-Everest-DAC is 1 Volt, we are then sniffing around levels of the fundamental noise limit (for the given bandwidths used).