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As a memory segment is 64k does it mean that 64k are the total addresses in a segment if 64K is the total addresses in a segment then what is the sizebytes of one segment 64k?
Yes, if a memory segment is 64K, it means there are 64K total addresses in that segment. Since 1K (kilobyte) equals 1024 bytes, a 64K segment would have a total size of 64 * 1024 bytes, which equals 65,536 bytes. Therefore, the size of one segment is 65,536 bytes.
Why microcomputer system based on 8 bit microprocessor have 64K memory?
Even though the 8085 is an 8 bit microprocessor, it can address 64K memory, because it has a 16 bit address bus.
How many memory location can be addressed by microprocessor 8085?
The 8085 can address 216, or 65536 different memory locations.
Why in memory segmentation of 8086 it use 64Kb instead of 1Mb?
Each segment in the 8086/8088 is 64KB because that is how Intel designed the microprocessor. The offset address is 16 bits, making the allowable range 64KB. See the related questions link for a further discussion of segmented architecture.
What is .model small in 8086?
One code-segment. One data-segment. Thus neither code nor data may be greater than 64K
What is program segment prefix in microprocessor?
mujhe nahi aata
Why does not Motorola's microprocessor has segment register?
There is insufficient information in the question to properly answer it. Which Motorola microprocessor are you talking about? Please restate the question.
What is segment in microprocessor?
segment is for converting physical address to logical address , here on taking 8086 microprocessor as example, we have 20 address lines but it is capable of taking only 16 address lines.... so to convert that 20 into 16 segment is used....
What is the maximum memory address space that the processor can access directly if it is connected to a 16 bit memory considering a hypothetical microprocessor generating a 16-bit address?
If you assume that it has a 16-bit data bus, then it would be 128k so the microprocessor can access 2^16 points, which is 64k (from it being a 16bit address) 16bits = 2 bytes (memory) so through a 16 bit memory, it can access 2*64k, which is 128k alternatively, if its 8bit memory, 8bits=1byte 1*64k = 64k I'm no expert, and i was searching for the answer myself, hope this helped
What is a segment used for?
A segment is a chunk (segment) of memory that is 64Kb in size. Due to the design of the 8086/8088 there are 64K possible segments, ecah overlapping the next by 16 bytes, for a total addressibility of 1 Mb. In the instruction model, a segment is the locus of addresses that can be reached in one instruction, without stopping to load a new value into a segment register. It is also called a near, or 16 bit address.
How many segment can be directly addressed at a particular time by 8086 microprocessor?
There are four segment registers in the 8086/8088, Code Segment (CS), Stack Segment (SS), Data Segment (DS), and Extra Segment (ES). As a result, there are four segments that can be directly addressed at a particular time, i.e. without an extra instruction to reload a segment register.
Why 7 segment display is used in microprocessors?
A 7 segment display is an I/O device, and it is not necessarily used in microprocessors. The choice of I/O devices is a function of system design, not microprocessor design.
