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Register Organization of 8086
8086 has a great set of registers containing special purpose and general purpose registers. All the 8086 resisters are 16-bit registers. The general purpose registers, may be utilized as either16-bit registers8 or -bit registers. The general purpose registers are either used for holding variables, data and intermediate results temporarily or for the other purposes like as counter or for storing offset address for some specific addressing modes etc. The special purpose registers are utilized as pointers, segment registers, offset storage registers for particular addressing modes or asindex registers. We will categorize the register set into 4 groups that are: General data Registers, Segment Registers, Pointer ad Index Registers, Flag Registers.
Software Interrupts Software interrupts are the result of an INT instruction in an executed program. It may be assumed as a programmer triggered event that immediately stops e
Addressing mode of 8086 : Addressing mode specify a way of locating operands or data. Depending on the data types used the memory addressing modes and in the instruction ,
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Flowchart for the sequence of 8251 Whether the control, mode or sync character register is selected depends on the accessing sequence. A flowchart of the sequencing is given i
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Typical link to modems for synchronous and asynchronous transmissions are shown in Figure. With regard to the synchronous connections it is consider that the timing is controlled
1 st Generation Microprocessor : At the end of the 70s a group of engineers developed a chip is able to processing data. This chip was called processor chip. Big processors w
ROR : Rotate Right without Carry: This instruction rotates the contents of destination operand to the bit-wise right either by one or by the count specified in register CL, exclud
8254 Programmable Timer A diagram of Intel's 8254 interval event/timer counter is given in Figure. The 8254 consists of 3 identical counting circuits, per of which has GATE and
Program : Write a program to perform a one byte BCD addition. Solution : It is consider that the operands are in BCD form, but the CPU considers it as hexadecimal and acco
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