Question
Download Solution PDFAn 8085-microprocessor based system uses a 4K × 8 bit RAM whose starting address is AA00H. The address of the last byte in this RAM is:
This question was previously asked in
ISRO (VSSC) Technical Assistant Electrical 2017 Official Paper
Answer (Detailed Solution Below)
Option 3 : B9FFH
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Detailed Solution
Download Solution PDFSpecification of RAM = 4K × 8 bit
Size of the RAM = 22 × 210 × 8 bit
= 212 × 8 bit
So 12 address lines will be engaged.
Number of address locations = 0FFFH
Starting address = AA00H
So, final address = (AA00H + 0FFFH) = B9FFH
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