Digital Electronics - Memory and Storage - Discussion

what is meant by 2218?

@All.

it is just the model number of the given RAM, you need to check the chip size they have given it is 16k x 1.

Hence, we know that chip size is equal to a multiple of the depth size of the chip (16k) and the width of the chip (1bit).
Hence it is equal to 16k only, we know that 2^n, where n is the number of address lines used, 16k (16*1024) can be written as (2^4) + (2^10) after solving we get 2^ (4+10) equals to 2^14 after refer this to 2^n we get the n=14 as the answer.

The size of the memory is N*M

where N is the address lines and M is word length
no of registers/memory location required is 2^N

Given memory capacity is 16k
thus 2^N=16K
1K=1024 memory locations
thus16k=16*1024=16384
now 2^N=16384
After factorising 16384 by 2 we ll get N AS 14
SO ADDRESS LINE REGUIRED IS 14.

2^4= 2 x 2 x 2 x 2 = 16.
2^10= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024.
2 to the power of 4 = 16.
2 to the power of 10 = 1024.
So: add the two power 4 + 10 = 14.

2^10 is always referred as a kilo, right so i.e. K.

16 = 2^4 so we add up 10+4 = 14 address bits.

Are address bits and address lines are the same thing ?

Then what is the address location?

16k = 16*1024 = 2^4*2^10.

So total 14 address lines require.

How is it?

Anybody, Please explain in detail.

Here K means Kilobytes.

So, 1 K = 1024 bytes.

Ram is only 2gb.

How to increase the RAM?

According to me, 10 is the right option.