Interfacing Memory With 8086 Microprocessor Problem 1

[@padmajamisra2700]

Nice presentation.

[@ryder2674]

No proper explaination...just saying something random things

[@adityagurav5646]

But ram address starts at 00000H 11:01

[@muskanchhabra4929]

Mam here we need 13 address lines but Ao is used for chip select. So we. Should use A1-A12

[@Ankit_2002]

A1 to A13 for addressing mode and A0 for control signal

[@naveenbolla1003]

0 1 is odd address and 1 0 is even address

[@yuva2916]

Why we use 3:8 decoder mam Pls clarify 🤧

[@kunalsalvi8382]

To accommodate more memory chips in the system. 3:8 decoder provides 8 chip selects signal, so 8 memory chips can be used.

[@yuva2916]

@@kunalsalvi8382 tqsm

[@manavjain7940]

i think this solution is wrong

[@najmashaik143]

No. Of adress lines are 12 not 13.. everyone plz correct it & accordingly solve the problem

[@kanupriyaagarwal4523]

There are 2 eproms and 2 rams each of size 4k*8, so if u add the size of two eproms OR two RAMs they add up to 8K*8 which is 2^3 * 2^10 = 2^13 thus 13 address lines

[@najmashaik143]

@@kanupriyaagarwal4523 but We divided the memory into two 4k*8 rams or EP roms.. hence the address lines will be seperately calculated & interfaced to specific memory chips...

[@yuva2916]

Why can we use 6 combinations 🥲

[@unstoppableguy7896]

It needs to be 12 address lines. That part of the solution is wrong