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At any given time the 8086 works with four segments in this address space. How many bytes are contained in each segment?
In the 8086 microprocessor architecture, each segment can contain 64 kilobytes (KB) of data. Since 1 KB is equal to 1024 bytes, this means each segment can hold 65,536 bytes. The four segments typically used are the code segment, data segment, stack segment, and extra segment, allowing the processor to manage different types of information efficiently within its 1 MB addressable memory space.
How many bits are bit addressable in the internal RAM and what is the range of the address?
Of the 128-byte internal RAM of the 8051, only 16 bytes are bit-addressable. The rest must be accessed in byte format. The bit-addressable RAM locations are 20H to 2FH.
What is size of instruction queue of 8086?
6 bytes
What is the length of the instruction queue in 8086?
6 bytes
What is bit addressable memory in microcontroller 8051?
The bit addressable memory in 8051 is compose from 210 bits: - bit address space: 20H - 2FH bytes RAM = 00H - 7FH bits address; - SFR registers; The following addresses are NOT bit addressable, only 1-byte addressable: - 32 bytes RAM from 00H to 1FH (R0 - R7 registers in all four banks); - 80 bytes RAM general user from 30H to 7FH.
Which are internal bit addressable RAM locations of 8051?
The 16 bytes (128 bits) at internal RAM locations 0x20-0x2F are bit-addressable.
Which instruction in 8086 microprocessor is 6 bytes long?
mov
How many memory address does this number of address lines allow the 8086 to access directly?
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
How many memory addresses does this number of address lines allow the 8086 to access directly?
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
What is the size of the interrupt vector table in the 8086-8088 microprocessor?
There are 256 different interrupt vectors in the 8086/8088. Each vector is a far CS:IP address, which is four bytes. That makes the interrupt vector table 1,024 bytes.
Difference between bit addressable and byte addressable in micro controller?
Hey ! in bit addressable , one can reach the bits (PORT 1.0,PORT1.1,PORT1.2,..etc..) of the ports in the programming while byte addressable only allows the bytes(PORT 1) to be addressed. therefore,there is difference in programming instructions too.
How many bits would you need to address a 2M by 32 memory if the memory is byte-addressable?
To address a 2M by 32 memory in a byte-addressable system, you first need to determine the total number of bytes. Since 2M refers to 2 megawords and each word is 32 bits (or 4 bytes), the total size is 2M × 4 bytes = 8M bytes. To find the number of bits needed for addressing, calculate the logarithm base 2 of the total bytes: ( \log_2(8M) = \log_2(8 \times 2^{20}) = 3 + 20 = 23 ). Therefore, you would need 23 bits to address the memory.
