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With the help of memory map interface the following to an 8086 based system operating in minimum mode:

a) 32K bytes of EPROM memory using 8K bytes device.

b) 32K bytes of RAM memory using 8K bytes device.

c) One 16-bit input and output port.

Mumbai University > Electronics and Telecommunication > Sem 4 > Microprocessor and peripherals

mumbai university mpa • 502  views
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Assume : Working frequency = 5 MHz

$\therefore$ Crystal frequency = 5 x 3 = 15 MHz

Step 1 - case (i) - 32 KB RAM -

Size of chip = 8 KB

= 8 x 1 KB

= $2^3$ x 1024 x 8

= $2^3$ x $2^{10}$ x 8

= $2^{13}$ x 8

we need B addressing line Ao to A12 for internal decoding.

Total memory = 32 KB

No. of chips = $\frac{32 \ KB}{8 \ KB}$ = 4 = 2(MB1) + 2(MB2)

$\therefore$ we need two memory bank.

size of memory bank = 8 KB (even) + 8 KB (odd)

= 16 KB

= $2^{14}$ x8 ---------------------{0000 0011 1111 1111 1111}= 03FFF

Memory Bank 1

starting address = 00000H

Ending address = 00000

00000 + 03FFF = 03FFF H


Memory Bank 2

Starting address = 04000H

Ending address = 04000H

04000H +03FFFH = 07 FFFH


Case (ii) 32 KB g EPROM Using 8 KB devices :

Size of chip = 8 KB

= 8 x 1 KB

= $2^3$ x 1024 x 8

= $2^{13}$ x 8

We need B addressing lines AO - A12 for internal decoding.

Total memory = 32 KB

No. of chips = $\frac{32 KB}{8 KB}$ = 4 = 2 (MB1) + 2(MB2)

$\therefore$ we need two memory bank.

size of memory bank = 8 KB (even) + 8 KB (odd)

= 16 KB

= $2^{14}$ x 8

Memory Bank 1

Ending address = FFFFF H

Starting address = - 03FFF H

FFFFF H- 03FFF H = FC 000 H

Memory Bank 2

Starting address = EAOOO H

Ending address = +03FFF H

EAOOO H +03FFF H = EAFFF H

Step 2: Memory addressing -

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