Question

Explain your reasoning for the following:

a) Assume that the 4 x 3 memory given in the text book is available in a single chip. How many of these chips are needed to implement a 16x12 memory system? What is the total number of D FFs used for this memory system?

b) Find the number of cells in a memory chip that has capacity of 32 Kilobits and is organized as 16-bit cells. How many address and data pins does this memory chip have?

c) Find the size of the address space(in locations) of a CPU system with 24-bit address and 32-bit data bus.

d) Find the size of maximum addressable memory (maximum capacity) of a CPU system with 20-bit address and 16-bit data bus (in bytes).

e) How many 32Kx8 chips are required to completely fill the address range 30000H –BFFFFH with physical memory? CPU system has 20-bit address and 16-bitdata bus.

Answer 1

Greetings!!

a. In a 4x3 memory 4 address lines are there and each address can store 3 bits. In order to make 4x12 bits memory we need to arrange FOUR 4x3 chips in a row so that each address will get 12 bit stored. Such another 3 more rows are required to have 16x12 bit memory. Hence we need 16 chips of 4x3 size is required for this. 4 each in 4 rows.

Number of 4x3 chips=16

16x12 memory can store 192 bits and therefor we need 192 D flip flops.

b.

b.1 Number of cells=Number of bits=32K cells

b.2 Address lines required=32K/16

=2^15/2^4

=2^11

11 bit address

b.3 Data pins=16

c. 24x32 can have 2^24 addresses and each can store 32 bits.

ie 2^24x32 bits=16Mx32bits=16Mx4Bytes=64MB address space

d. 20x16 can address 2^20 locations and each location with 16 bits

ie Two 512KB memory makes 1Mx16 bit memory

So maximum size is 512KB.

Hope this helps

Thank You