Question

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Homework 6 In the slides & additional material folder on D2L is a schematic of a memory system comprised of 4 memory chips interfaced to a processor. The memory chips are numbered 1 through 4. Find the addresses assigned to each memory chip.

Answer 1

ADDRESS	ADDRESS INPUTs																				CHIP #
	A19	A18	A17	A16	A15	A14	A13	A12	A11	A10	A9	A8	A7	A6	A5	A4	A3	A2	A1	A0
0xE0000	1	1	1	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	1
0xE7FFF						1	1	1	1	1	1	1	1	1	1	1	1	1	1	1
0xE8000				0	1	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	2
0xEFFFF						1	1	1	1	1	1	1	1	1	1	1	1	1	1	1
0xF0000				1	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	3
0xF7FFF						1	1	1	1	1	1	1	1	1	1	1	1	1	1	1
0xF8000				1	1	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	4
0xFFFFF						1	1	1	1	1	1	1	1	1	1	1	1	1	1	1

A0 – A14 address lines are connected to all the four memory chips.

Hence lowest value on A14-A0 lines will be 000000000000000 and highest value will be 111111111111111

A15 – A19 address lines from processor is connected to IC74139 demultiplexer.

A17, A18 and A19 are input to NAND gate and active low signal can be generated at the NAND gate output if A17=A18=A19= 1. Hence throughout these three lines are logic 1 to enable the IC74139.

A15 and A16 are demultiplexer inputs. Output lines of demultiplexer (pins 4, 5, 6, 7 of IC 74139) is connected to Chip Select inputs of each memory.

Hence A16A15 = 00 will enable memory #1, 01 will enable memory #2, 10 will enable memory #3 and 11 will enable memory #4

Final bit pattern on address lines for each memory is shown in the table.

Therefore, Starting address of memory 1 is 0xE0000 and End address is 0xE7FFF

Starting address of memory 2 is 0xE8000 and End address is 0xEFFFF

Starting address of memory 3 is 0xF0000 and End address is 0xF7FFF

Starting address of memory 4 is 0xF8000 and End address is 0xFFFFF