[logo] Storage Capacity of a Compact Disc

As part of the discussion exercise on Nov. 19, you were e-mailed the following instructions and a micrograph of a CD:


Assignment:                                                       



   Estimate the digital storage capacity of a CD in megabytes.    

                                                                  

Hints:                                                            



   Assume that the smallest pit shown represents two bits of information,

     changing state at each edge.

   Remember that there are 8 bits in a byte.                      

                                                                  

E-mail me your answer, with a brief discussion of your reasoning, 

by Friday, April 26 by 5pm.                                         

[Micrograph of CD with dimension]

Micrograph of Bon Jovi CD, by permission of Michael Davidson and FSU

One solution:

Find the area occupied by one bit. In the image above I have delineated several neighboring bits to emphasize the "packing" of bits on a CD. Actually I have enclosed 2 bits in each box, a 1 bit followed by a 0 bit -- notice the longer stretches of pits or "missing" bits along each track that represent several bits set (or unset) in succession.

From this I estimate that 2 bits occupies roughly a rectangular area of 1.6 micron by 1.6 micron (micron = micrometer = 0.001 mm = 10-6 m.

So the area per 2 bits is (1.6 micron) squared or 2.5 square microns [1.3 x 10-6 mm2 per bit]. The reciprocal of this number is 800,000 bits per square mm, or since there are 8 bits per byte: 100,000 bytes/mm2.

Next we must estimate the "active" area of the CD, the area where information can be stored. By examining a disc, the following dimensions can be obtained:
-- O.D. (outside diameter) = 117 mm; I.D. (inside diameter) = 46 mm

[Sketch of CD dimensions]

The formula for the area of a disk is pi times the radius squared. Thus the area of the "O.D. disk" is 10,800 mm2 and the area of the missing "I.D. disk" is 1660 mm2. The total active area is the difference, roughly 9000 mm2 (around 14 sq. in. for those so inclined).

So the number of bytes/mm2 times the size in mm2 gives an estimate of the total capacity of a CD:
100 kb/mm2 x 9000 mm2 = 900,0007nbsp;kb = 900 Mb (megabytes).

There are some details in compact disc encoding of data that need to be incorporated to get a more accurate estimate of a compact disc's capacity. It turns out that the pits do not directly represent bits; an encoding system is used that converts 8 bits to 14 bit words. This 8 to 14 bit encoding insures that a binary 1 is separated by no fewer than two 0s. There are also 3 "merging" bits for each such block of 14 bits. Our estimate is thus too large by 18/7. In addition, there will be other contributions to "overhead" reducing the effective capacity.

One more detail is that the minimum length of a pit is 0.8 micron, not 1.6 micron, with minimum land length of 0.8 micron. Also the features "boxed" in the micrograph above represent 3 bits rather than 2... The actual capacity of a current CD is around 650 Mb.

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Last updated November 29, 2004.
Copyright George Watson, Univ. of Delaware, 1996.