Help with address lines

gammaman

Joined Feb 14, 2009
29
How do I figure out the number of address lines?

Say I have 2K x 16. How many address lines would this be?

beenthere

Joined Apr 20, 2004
15,808
That would be 2048 X 16. 11 lines.

gammaman

Joined Feb 14, 2009
29
Thanks! Would you mind showing me how that is calculated?

RiJoRI

Joined Aug 15, 2007
536
1K needs 10 address lines (2^10 - 1024). Every time you double the "K" you add one address line.

1K = 10 lines.
2K = 11 lines.
4K = 12 lines.

1M = 1K*1K = 20 lines.

Yeah, it can be done mathematically, but I find this method easier.

--Rich

mik3

Joined Feb 4, 2008
4,846
How do I figure out the number of address lines?

Say I have 2K x 16. How many address lines would this be?
2Kx19 means that you have a data bus 16 wide and 2000 memory locations and each memory location can store a 16bit value.

Thus you need 2000 different addresses to access each memory location.

In binary 2048=2^(11)

Thus to achieve 2000 different addresses you need 11 lines.

Papabravo

Joined Feb 24, 2006
12,301
Logarithms to the base 2 is all there is to it.

mik3

Joined Feb 4, 2008
4,846
Logarithms to the base 2 is all there is to it.
Yes, but it is not obvious for all people.

Papabravo

Joined Feb 24, 2006
12,301
I never claimed it was obvious.

Ratch

Joined Mar 20, 2007
1,068
gammaman,

2048 = 2^x ===> x = log10(2048)/log10(2) = ln(2048)/ln(2) = 11

Use the log of any base you want and get the same answer.

Ratch

Papabravo

Joined Feb 24, 2006
12,301
That's the ticket -- You 'da man!