Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 20 (Q.No. 6)
6.
An 8085 microprocessor based system uses a 4 k x 8 bit RAM whose address in AAOOH. The address of the last byte in this RAM is
Discussion:
3 comments Page 1 of 1.
Ahir said:
7 years ago
Last address ={ initial add in hexa + hexa equivalent of memory size -1}.
Raj said:
8 years ago
According to me, It is option C.
Jane said:
8 years ago
4k x 8 = 2^12 x 8 bits = 2^12 bytes,
Range = 0001 0000 0000 0000 = 2^12.
Last address =first address+ range = AA00 + 1000 = BA00.
Range = 0001 0000 0000 0000 = 2^12.
Last address =first address+ range = AA00 + 1000 = BA00.
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