Help with address lines

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
4,843
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.
 

Ratch

Joined Mar 20, 2007
1,070
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
 
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