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
@@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...
Nice presentation.
No proper explaination...just saying something random things
But ram address starts at 00000H 11:01
0 1 is odd address and 1 0 is even address
Mam here we need 13 address lines but Ao is used for chip select. So we. Should use A1-A12
A1 to A13 for addressing mode and A0 for control signal
Why we use 3:8 decoder mam
Pls clarify 🤧
To accommodate more memory chips in the system. 3:8 decoder provides 8 chip selects signal, so 8 memory chips can be used.
@@kunalsalvi8382 tqsm
Why can we use 6 combinations 🥲
i think this solution is wrong
No. Of adress lines are 12 not 13.. everyone plz correct it & accordingly solve the problem
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
@@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...
It needs to be 12 address lines. That part of the solution is wrong