Hello, can someone help explain on how to solve this.
#Bits required to address 4M X16
merry fern
Not sure if these are correct. Please correct me if I'm wrong, I'm still confused.
1a. 4M x 2 bytes = 22 x 220 x 2 = 223 total bytes, so 23 bytes are needed
1b. 4M words = 22 x 220 = 222, so 22 bits are required for an address
Haven't answered #2 yet. 😅
safe elbow
byte = 8 bit
word = 16 bit usually
dword = 32 bit
qword = 64 bit
4M x 16 word = 4M addressable so 4 * 2^20 = 2^22 = 22 bits
8 bytes = 23 bits (x2)
512K x 8 bits? ram chip / 16-bit word
256 MiB / 4096 KiB
256 * 1024 K / 512 K * 8
256 * 2 /8 = 512 / 8 chips = 64 chips
1 / 256k chip needed per word or "less than 1" or "1" ?
2d. 16M word so 16 * 2^20 = 2^4 * 2^20 = 24 bits
256k word = 2^8 * 2^10 = 18 bits + "6 bit for the chip selector" multiplexer
18+6 = 24
merry fern
Thank you let me review your solutions
safe elbow
each chip might have a 18 bits + 1 bit on/off selector from the multiplexer (mux)
merry fern
22 bits is 1a right not 1b?
safe elbow
23 bits for 8-bit addressable
22 bits for 16-bit addressable
you need one more bit for byte, because you have to choose between high and low byte of the word
AX => AH or AL ?
merry fern
oh ok, but for 1B i got 22 as well.. word aaddresable?
safe elbow
1a is 23
1b is 22
safe elbow
2b is either a fraction or 1
merry fern
how to solve 2A-2C?