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# Why is capacity of RAM is a power of 2?

vishalhakData Transfer Rate

The speed with which data can be read from a CD ROM drive. 150 kilobytes per second was the original standard rate; 2x = 300 kb/second; 4x = 600kb/s (etc.); 12x = 1.8 mb/s; 16x = 2.4 mb/s.

good info......can u tell me why ram capacity always come in 2s power

eg 512=(2)9

1024=(2)10

Because memory is measured in terms of bits or bytes.enigmacan u tell me why ram capacity always come in 2s power

eg 512=(2)9

1024=(2)10

A byte is eight bits, resulting in the possibility of 256 values (28). A kilobyte is 1,024 (210) bytes (the term "byte" is often defined as a collection of bits rather than the strict definition of an 8-bit quantity.)

Or this can be the reason that RAM works on binary numbers (0,1) so it's memory is always measured in power of two. Correct me if I'm wrong

hi friend's

my point of view is that

as computer understand only binary language i.e 0 & 1

so for addressing purpose in memory if the size is in power of 2 than we can address all registers using n bits where n is 2^n=size of memory.

eg..if size of memory is 1024kb (i.e there are 1024 registers with 8 bit each) than we require 10 bits ( i.e 2^10 =1024) for addressing this memory area .

pls correct me if i am wrong...

my point of view is that

as computer understand only binary language i.e 0 & 1

so for addressing purpose in memory if the size is in power of 2 than we can address all registers using n bits where n is 2^n=size of memory.

eg..if size of memory is 1024kb (i.e there are 1024 registers with 8 bit each) than we require 10 bits ( i.e 2^10 =1024) for addressing this memory area .

pls correct me if i am wrong...

big k is right or u can say everyone is right

correct...your topic answer is absolutely okfaizaanhi friend's

my point of view is that

as computer understand only binary language i.e 0 & 1

so for addressing purpose in memory if the size is in power of 2 than we can address all registers using n bits where n is 2^n=size of memory.

eg..if size of memory is 1024kb (i.e there are 1024 registers with 8 bit each) than we require 10 bits ( i.e 2^10 =1024) for addressing this memory area .

pls correct me if i am wrong...

A nonvolatile memory having a non-power of two memory capacity is disclosed. The nonvolatile memory device includes at least one plane. The plane includes a plurality of blocks with each of the blocks divided into a number of pages and each of the blocks defined along a first dimension by a first number of memory cells for storing data, and along a second dimension of by a second number of memory cells for storing data. The nonvolatile memory has a non-power of two capacity proportionally related to a total number of memory cells in said plane. The nonvolatile memory also includes a plurality of row decoders. An at least substantially one-to-one relationship exists, in the memory device, for number of row decoders to number of pages. Each of the row decoders is configured to facilitate a read operation on an associated page of the memory device