Digital Electronics - The 8051 Microcontroller - Discussion
Discussion Forum : The 8051 Microcontroller - General Questions (Q.No. 17)
17.
The last 96 locations in the internal data memory are reserved for general-purpose data storage and stack.
Discussion:
7 comments Page 1 of 1.
Pankaj said:
10 years ago
Its 80 bytes.
Internal data memory of 128 bytes is divided as,
1. Four register banks, each of 8 bytes. i.e 8*4 = 32 bytes.
2. Bit addressable memory of 16 bytes.
3. Remaining for general purpose data storage and stack, i.e 4.
128 - (32+16) = 80 bytes.
Internal data memory of 128 bytes is divided as,
1. Four register banks, each of 8 bytes. i.e 8*4 = 32 bytes.
2. Bit addressable memory of 16 bytes.
3. Remaining for general purpose data storage and stack, i.e 4.
128 - (32+16) = 80 bytes.
(1)
Vishal said:
1 decade ago
Because if we start from 00h then first 32 locations out of 128are allocated for register banks, then 16 locations for bit-addressable area so now left 80 (128-32-16=80) all this remaining locations are assigned for general purpose.
Khan said:
10 years ago
32 locations out of 128 are allocated for 4 register banks, than 16 locations for bit-addressable area. So now 80 (128 - 32 - 16 = 80) remaining locations are assigned for general purpose.
Manjunatha CR said:
8 months ago
Total 128 bytes =128 location.
32 bytes = 4 Register bank.
16 bytes = Bit addressable.
80 byes = general purpose.
32 bytes = 4 Register bank.
16 bytes = Bit addressable.
80 byes = general purpose.
Dhruvin said:
1 decade ago
30 to 7F are allocated as general purpose locations.
So 7F-30 = 79(d)....means 0-79...total 80 locations.
So 7F-30 = 79(d)....means 0-79...total 80 locations.
Manjusha Bhosale said:
10 years ago
Yes, its 80 locations.
Amith said:
1 decade ago
its 80
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