Information capacity of Ternary-based system over Binary-based

Question

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.

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).