anonym_nep
- Jan 18, 2017
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- Joined
- Jan 18, 2017
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Consider a 386 system and it has 40 address lines, calculate the total memory it can address. Design a logic circuit to address three 1024 MB RAMs and two 512 MB RAMs .
As the 80386 is a 32 bit system an the address lines addres words (32 bit = 4 byte) this seems correct.=> 2^40 x 4 bytes
1024 MB requires 30 address lines for bytewise addressing. If the memory were addressed wordwise, not bytewise, then you'd use less address lines. In this contect, 30 seems correct.one 1024 MB can be addressed by 30 address lines, so three 1024 can be addressed with 32 address lines
No need to get stuck. 512 MB = 1/2 * 1024MB. What does this mean in terms of address lines?for a 512 MB ........stuck from here
The 80386 doesn't know about memory chip sizes. You'll have to devise a way to create chip select signals for the single memory chips from the address lines. Hint: look for address decoder...Design a logic circuit to address three 1024 MB RAMs and two 512 MB RAMs .
Consider a 386 system and it has 40 address lines, calculate the total memory it can address. Design a logic circuit to address three 1024 MB RAMs and two 512 MB RAMs .