Some researchers are trying to get a memory cell capable of having 3 states instead of 2.

1) How many memory cells, in principle and as a rough estimate, does a typical 1 megabyte memory chip has? is it 8*1024 cells?

2) If you have N memory cells, each has x logical levels, what is the number of possible representations that we can get out of it? is it x^N representations?

Let's take Binary system as an example. Say we have 8*1024 memory cells. We can store only one of the possible 2^N representations in each cell, so at the end the number of stored bits is just N=8*1024. Now, for Ternary system, the gain is log2(3) ~ 1.58 times so we can store ~ 1.58 * 8 * 1024 "bits" or 8*1024 "trits".

Correct? EDIT: sorry I meant 8*1024*1024.

  • $\begingroup$ 1MB is 8x2^20 bits, not 8x2^10. $\endgroup$ Dec 18, 2013 at 8:33

1 Answer 1


Binary (two states)

1 Megabyte = 1,048,576 bytes = 8,388,608 bits = 8,388,608 cells. See https://en.wikipedia.org/wiki/Megabyte and https://en.wikipedia.org/wiki/Byte. 1 byte = 8 bits.

There are $2^{8,388,608}$ different possible values that can be stored in this much memory.

Ternary (three states)

Now a cell can have 3 possible values. If we have 8,388,608 cells, each of which can have 3 possible values, then the entire memory can hold $3^{8,388,608}$ possible values. This is the equivalent of $\lg 3^{8,388,608} = 8,388,608 \times \lg 3 \approx 8,388,608 \times 1.58 \approx 13,295,629$ bits. This corresponds to $13,295,629/8 = 1,661,953$ bytes.

Therefore, this many ternary cells can store the same amount of data as 1,661,953 of ordinary binary cells (as 1,661,953 MB of memory).

  • $\begingroup$ I think the $$2^{8,3888,608}$$ is the "possible" values that can be stored, but how many values actually can be stored? See, the 1.58 gain from binary to Ternary is clear, but the actual number of values that can be stored in either of them is what is not so clear. $\endgroup$
    – student1
    Dec 18, 2013 at 20:30
  • $\begingroup$ To clarify. Suppose you have 20 binary cells. Assume we limit our representations to: 0 to (+2^20)-1 which is decimal 1048757. What is stored is only one of these values, say 500 decimal. This means that the storage capacity of these cells is 1 decimal number. Using Ternary, the capacity is 1*1.58 = 1.58 decimal numbers. Right? $\endgroup$
    – student1
    Dec 18, 2013 at 20:53

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