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How many address lines and data lines are required from 256K 16 memory system?
In a 256K x 16 memory system, the memory has 256K (256 * 1024 = 262,144) addressable locations and each location holds 16 bits of data. To calculate the number of address lines needed, we find the base-2 logarithm of 256K, which is 18 (since 2^18 = 262,144). For the data lines, since each location holds 16 bits, 16 data lines are required. Thus, the system requires 18 address lines and 16 data lines.
Why it is necessary to demultiplex AD0 to AD7 address lines?
Demultiplexing the AD0 to AD7 address lines is necessary to separate the address and data signals in microprocessor systems, particularly in those where a single set of lines is used for both functions. This allows for more efficient use of the bus, as the same lines can carry address information during one phase of operation and data during another. By demultiplexing, the system can clearly distinguish between when it needs to read or write data versus when it is accessing specific memory addresses, ensuring accurate communication and reducing the risk of errors. Additionally, it helps in simplifying the design of the bus architecture and enhances overall system performance.
A microprocessor has a data bus with 64 lines 32 lines address bus.what will be the Maximum number of bits stored in memory?
2^32 is amount of blocks that address bus could locate. and each blocks is 64bit because data bus has 64 lines. then maximum number of bits stored in memory is (2^32)*64 bit. By: Mohammad Saghafi Email: mohammads1364@yahoo.com
Why ad0-ad7 lines are multiplexed?
The AD0-AD7 lines are multiplexed to optimize the use of limited I/O pins in microcontrollers and microprocessors. By combining address and data functions on the same lines, the system can reduce the number of physical connections needed, allowing for a more compact design. This multiplexing requires additional control signals to differentiate between address and data phases, but it ultimately enhances efficiency in data transfer and system performance.
When did Family Lines System end?
Family Lines System ended in 1983.
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 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 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 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 and data lines are required from 256K 16 memory system?
In a 256K x 16 memory system, the memory has 256K (256 * 1024 = 262,144) addressable locations and each location holds 16 bits of data. To calculate the number of address lines needed, we find the base-2 logarithm of 256K, which is 18 (since 2^18 = 262,144). For the data lines, since each location holds 16 bits, 16 data lines are required. Thus, the system requires 18 address lines and 16 data lines.
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 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 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 address lines can access a 16MB memory?
for 16 MB memory has 24 address lines
How many address lines are necessary to address two megabytes of memory?
Let N be the number of addresses line 2 megabyte = 2*1024 =2048 N = log (size in bytes) /log 2 N= log 2048/log 2 N=11
How many nuMBer of address lines required for 1 MB memory?
Firstly we need to convert Mb's into bits i.e 1Mb=1024x1024 = 210x210 =220 That means there are 220 memory locations and we will need 20 address lines.
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.
