How many bits can a memory chip with the below configuration support? For full credit, show how you got the answer.

o 8 data lines (D0-D7)

o 12 address lines (A0-A11)

o Row select (RAS) uses all 12 address lines

o Column select (CAS) uses only 10 address lines

o Three (3) lines for memory bank selection

Total Memory = 4 KB = 4096 bytes = 32768 bits

Explanation:

1. Data lines are 8 From D0 to D7

so

Total memory at single address locations is 8 bits.

2. Address lines are 12 (A0 to A11)

There are 12 address lines but 3 out 12 are for selction of chip or memory bank.

so only 9 pins are there to address the locations on one chip.

Total No. of address locations on single chip = 2^9 = 512 locations

as 1 location is 1 byte so total memory of single chip is 512 bytes.

3. Total Memory Bank

There are total 3 selection pins for memory bank.

so

Total chips = 2^3 = 8.

4. Total Memory

Total size of 1 chip = 512 bytes

Total size of 8 chip = 8x512 bytes = 4096 bytes = 4096/1024 kb = 4 kb

So total memory of system is 4 Kb = 4096 bytes = 32768 bits