How Intel's 8086 microprocessor access in 16-bit word? Explain with example.
The 8086 is a l6-bit microprocessor
chip designed by Intel All internal registers, as well as internal and external
data buses, are 16-bits wide, which firmly established the “16-bit
microprocessor” identity of the 8086. A 20-bit external address bus provides a
1 MB physical address space ( = 1,048,576). This address space is addressed
by means of internal memory "segmentation". The data bus is
multiplexed with the address bus in order to fit all of the control lines into
a standard 40-pin dual inline package. It provides a 16-bit I/O address bus,
supporting 64 KB of separate I/O space. The maximum linear address space is
limited to 64 KB, simply because internal address/index registers are only
16-bits wide. Programming over 64 KB memory boundaries involves adjusting the
segment registers (see below); this difficulty existed until the 80386
architecture introduced wider (32-bit) registers (the memory management
hardware in the 80286 did not help in this regard, as its registers are still
only 16-bits wide).
An electric circuit is made from different electrical components such as transistors, resistors, capacitors and diodes, that are connected to each other in different ways. These components have different behaviors. how to use a multimeter to test voltage of live wires
ReplyDelete