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An 8085 Microprocessor based system uses a 4 k x 8 bit RAM whose starting address is AAOOH. The address of last byte in this RAM is
- a)OFFFH
- b)1000 H
- c)B9 FFH
- d)BAOOH
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
An 8085 Microprocessor based system uses a 4 k x 8 bit RAM whose start...
4K x 8 = 212 x 8 bit it has 12 bits So address of last - byte will be AA 00 + 0 FFF = B9FF
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An 8085 Microprocessor based system uses a 4 k x 8 bit RAM whose start...
Answer:
Given information:
- Starting address of 4 k x 8 bit RAM = AAOOH
- RAM size = 4 k x 8 bits
To find:
- Address of last byte in the RAM
Solution:
- Size of RAM = 4 k x 8 bits = 4 x 1024 x 8 bits = 32,768 bits
- Each byte contains 8 bits
- Number of bytes in RAM = 32,768 bits / 8 bits per byte = 4,096 bytes
- Starting address of RAM = AAOOH = 1010 1010 0000 0000 2
- Last address of RAM = Starting address + Size of RAM - 1
- Last address of RAM = 1010 1010 0000 0000 2 + 4,096 - 1
- Last address of RAM = 1010 1010 1111 1111 2 = AAFH
- Therefore, the address of the last byte in the RAM = B9 FFH
Hence, the correct option is C) B9 FFH.
Given information:
- Starting address of 4 k x 8 bit RAM = AAOOH
- RAM size = 4 k x 8 bits
To find:
- Address of last byte in the RAM
Solution:
- Size of RAM = 4 k x 8 bits = 4 x 1024 x 8 bits = 32,768 bits
- Each byte contains 8 bits
- Number of bytes in RAM = 32,768 bits / 8 bits per byte = 4,096 bytes
- Starting address of RAM = AAOOH = 1010 1010 0000 0000 2
- Last address of RAM = Starting address + Size of RAM - 1
- Last address of RAM = 1010 1010 0000 0000 2 + 4,096 - 1
- Last address of RAM = 1010 1010 1111 1111 2 = AAFH
- Therefore, the address of the last byte in the RAM = B9 FFH
Hence, the correct option is C) B9 FFH.
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Solutions for An 8085 Microprocessor based system uses a 4 k x 8 bit RAM whose starting address is AAOOH. The address of last byte in this RAM isa)OFFFHb)1000 Hc)B9 FFHd)BAOOHCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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