The function 2^2 = 4 4 is the largest no. to the power 2 ( 2 bits ) 2^8 = 256 256 is the largest number from 8 bits 2^3 = 8 8 is the largest from 3 bits etc.....
Not sure where you are getting your info.
But how is it equal to memory?
I did not see anything at that site that contradicted what I said.
I am not quite sure what you mean in the quote above. What I can tell you is that because memory devices use binary for decoding that they tend to get twice as large each time you increase there size.
Ok, let me be more clear... Lets take an example
Consider an example: Assume '&' is represented in a system as '01011011'. Now, the character '&' is represented using a 8bit number. The space required to store '&' is eight bits. So the memory has to be '8 bits', but what actually is followed is that the memory is 2^8= 256 bits. How/Why 256 is used, when it actually used space/memory is just 8 bits? This is confusing me.
That's because it's eight bits - NOT 256 - an eight bit number can represent from 0 to 255 (in decimal).
Nigel, i didn't get you properly.
So consider '&' as 1011011 - 7 Bit representation. Is this possible? I hope you understood what i asked before.
Let me take a stab at this.
Let's introduce a new term "16 bit addressing"
What does that really mean? 16 bits can represent 0-65535 decimal locations. It can also represent -32767 to + 32768 relative addressing when we store this number in 2's complement notation. We could use signed binary e.g. the highest bit represents the sign. When that happens, we get to representations for Zero. 100000000000000 and 0000000000000000. That's not fun to deal with.
I think I tried to address that:
1 add a bit 11 1>>3
11 add a bit 111 3>>7
111 add a bit 1111 7>>15
1111 add a bit 11111 15 >>31
...
11 11111111 111 11111111 1023 >> 2047 1K to 2K
111 11111111 1111 11111111 2047>> 4097 2K to 4K
Note that if you add 1 more bit which is the minimum you can add, the amount addressable doubles. You really add 1 more because 0 is a physical location.
A bit is a binary digit having the value of 0 or 1 just like a decimal digit has the value of 0 thru 9.
Well 999 is 3 decimal digits which is 10^3 or 1000 locations. We used 10 instead of 2 because 10 is the base.
Here is a datasheet for an old memory chip, a 6116: https://www.datasheets.org.uk/6116 RAM-datasheet.html Note it's totally spelled out as 2k x 8bits.
Note here: **broken link removed**
it starts getting weirder. It's number of modules x amount and then the modules have an orientation too such as 4 GB = 512 Mb x 64 bits. The computer has a 64 bit word, so that's the unit we need multiples of.
Why no. of location should be equal to power of 2.
Next... Memory increases in size because of the physical address bus on the memory.... 64k memory needs 16 bits to address every location. A0 ---- A15.
128k only needs one more bit ..... A0 ----- A17... If you want to address 4gig you need 32 bits... 8gig only needs 1 more bit, 33 bits.
you cant get half a bit.... 32.5 bits.. 6 gig ( you could address only 6 in software... but why... Just address up to 33 bits) .
Okay, is it not possible to design a memory without this power of 2, i mean 60k of memory but 16 bit of addressing bits? Why hasn't anyone done that?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?