Design 8086 microprocessor based system with following specifications

(a) Microprocessor 8086 working at 10 MHz in minimum mode

(b) 32 KB EPROM using 8 KB chips

(c) 16 KB SRAM using 4 KB chips

Explain the design along with memory address map.

Mumbai University > Computer Engineering > Sem 5 > Microprocessor

Marks: 10M

Year: Dec 2014

Figure below shows 8086 microprocessor in mimimum mode working at 10 MHz.

Step 1: Total EPROM required = 32 KB

Chip size available = 8 KB

∴Number of chips required=32KB/8KB=4

∴Number of sets required=Number of chips /Number of banks=4/2=2

SET 1: Ending address of SET 1 = FFFFFH

SET size = Chip size x 2 = 8 KB x 2 = 16 KB

i.e. 0000 0011 1111 1111 1111=03FFFH

Starting address = Ending address – SET size

$\hspace{2.5cm}$ =FFFFFH–03FFFH

$\hspace{2.5cm}$ =FC000H

SET 2: Ending address of SET 2 = FBFFFH (previous ending - 1)

SET size = Chip size x 2 = 8 KB x 2 = 16 KB

i.e. 0000 0011 1111 1111 1111=03FFFH

Starting address = Ending address – SET size

$\hspace{2.5cm}$ =FBFFFH–03FFFH

$\hspace{2.5cm}$ =F8000H

Step 2: Total RAM required = 16 KB

Chip size available = 4 KB

∴Number of chips required=16KB/4KB=2

∴Number of sets required=Number of chips/Number of banks=2/2=1

SET 1: Starting address = 00000H

SET size = Chip size x 2 = 4 KB x 2 = 8 KB

i.e. 0000 0001 1111 1111 1111=01FFFH

Ending address = Starting address – SET size

$\hspace{2.5cm}$ =02000H+01FFFH

$\hspace{2.5cm}$ =03FFFH

SET 2: Starting address = 04000H (previous ending - 1)

SET size = Chip size x 2 = 4 KB x 2 = 8 KB

i.e. 0000 0001 1111 1111 1111=01FFFH

Ending address = Starting address +SET size

$\hspace{2.5cm}$ =02000H+01FFFH

$\hspace{2.5cm}$ =03FFFH

Step 3: Memory Map

EA=Ending address, SA=Starting address, EB=Even Bank, OB=Odd Bank.

Step 4: Final Implementation: