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How many address lines are required for connecting CPU to 00 to 0F?
Wiki User
∙ 16y ago
00 to 0F represents 16 addresses Binary representation is 1111 therefore you need 4 address lines to connect to all addresses
Wiki User
∙ 16y ago
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How many address lines would be required for a 2K?
In the 2k*16 , the 11 address lines are required and the 16 input-output lines are required..
How many no of address lines required in 1MB memory 111622 or 24?
How many no of address lines required in 1MB memory 11,16,22 or 24 u haven't specified correct options! 20 address lines will be required because 1 MB is 1024 KB that is 1024*1024 Byte which is equivalent to (2^10)^2 bytes if ur memory is Byte addressable then address lines required will be 20.
How many address lines and data lines are required for a2Kx8 memory?
A 2K X 8 memory requires 11 address lines and 8 data lines
How many address lines and data lines are required for a 128k x 8 memory system?
17 address lines and 8 data lines. 2^17=128k
How many address lines are required if micro processor has to access 2kb memory?
2kb=2*1024=2048 2^11=2048 therefore 11 address lines are required
How many address lines are needed to access 256KB of main memory?
The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
How many address lines are required for 1 kilobyte of ram?
10. 210 = 1024.
How many nuMBer of address lines required for 8 MB of memory?
It takes 23 address lines to address 8 mb of memory.
How many address lines are requried for 12mb?
There are 24 address lines required for 16 mb. That covers 12 mb. The next step down is 23 address lines, which is 8 mb. The 8085 and 8086/8088 cannot address 12 mb. Only the 80286 and higher can.
How many Address lines and data lines of microprocessor and microcontroller?
Microprocessor has 16 address lines and microcontroller has 20 address lines
How many address lines are required to address 64 kbytes of memory?
You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.
How many address lines are require to access 128MB of memory?
ANSWER There are 2128 combinations of addresses. This is about 3.4 x 1038 locations. Assuming each address holds a 32-bit word, that's 1.2 x 1039 bytes. That's a LOT of memory.
