Consider a 16G-byte byte-addressable memory board built from 1G-byte byte-addres

Question

Consider a 16G-byte byte-addressable memory board built from 1G-byte byte-addressable memory chips (1G=230 ).

a. [0.5pt] How many address lines go into the board?

b. [0.5pt] How many data lines go into the board?

c. [1pt] What is the size of the decoder used on the chip to decode an address?

d. [0.5pt] How many chips are needed to build the board (assuming no error check is implemented)?

e. [0.5pt] What is the size of the decoder on the board?

f. [1pt] Suppose the most significant address lines are used to refer to the chip number. On which chip is the following memory location found? 10 00100000 00000000 00000000 00000000

Explanation / Answer

a)

Given that total memory 16G-byte= 16 X 230 bytes= 24 X 230 bytes =234

Therefore the memory contains 234 addressable memory locations.

We require, 34 bits are required to refer 234 addresses.

Hence, 34 address lines go into the board.

b) Each address refers on byte location (data in memory location) . 1 byte=8bits.

Thus, the output requires 8 data lines.

c)

Since the decoder must match one of the 230 locations.

Decoder of n inputs can match one of the 2n locations.

Therefore the size of the decoder is 30X230. That is, the decoder has 30 input lines.

d) Since the 16G-byte board is built from 1G-byte chips, 16 chips are required.

e)

To refer 16( 24 )chips on the board, the decoder should have 4 input lines.

Therefore, the size of the decoder is 4 X 16

f)

Given address is

10 00100000 00000000 00000000 00000000.

0000= chip 1

0001=chip 2

…

…

1111=chip 16

The first 4 significant bits are 1000 it refers to chip 9.

Therefore the given memory location is found in chip 9.

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