Block Diagram 8086 Microprocessor Pdf
Contents.Description Features and performance The 80186 series was generally intended for, as with external memory. Therefore, to reduce the number of required, it included features such as, generator, channels, and external lines.The initial clock rate of the 80186 was 6, but due to more hardware available for the to use, especially for address calculation, many individual instructions ran faster than on an 8086 at the same clock frequency. For instance, the common register+immediate was significantly faster than on the 8086, especially when a memory location was both (one of the) operand(s) and the destination. Multiply and divide also showed great improvement being several times as fast as on the original 8086 and multi-bit shifts were done almost four times as quickly as in the 8086.A few new instructions were introduced with the 80186 (referred to as the 8086-2 instruction set in some ): enter/ leave (replacing several instructions when handling stack frames), pusha/ popa (push/pop all general registers), bound (check array index against bounds), and ins/ outs (input/output of string).
A useful immediate mode was added for the push, imul, and multi-bit shift instructions. These instructions were also included in the contemporary 80286 and in successor chips. (The instruction set of the 80286 is superset of the 80186's, plus new instructions for.). Of Intel 80C186.The (redesigned) version, 80C186, introduced, a power-save mode, and a direct interface to the or 80187 floating point numeric.Uses In personal computers The 80186 would have been a natural successor to the in personal computers.
Thus the size of the data bus is 16-bit as it can carry 16-bit data at a time. The architecture of 8086 microprocessor, is very much different from that of 8085 microprocessor.We have already discussed the introduction to the microprocessor. So, lets now proceed further and understand the architecture and working of 8086 microprocessor. Data segment register: It holds the address of the data segment. The data segment stores the data in the memory whose address is present in this 16-bit register.4.
Extra segment register: Here the starting address of the extra segment is present. The brogues don t shout me down. This register basically contains the address of the string data.It is to be noteworthy that the physical address of the instruction is achieved by combining the segment address with that of the offset address.6-byte pre-fetch queue: This queue is used in 8086 in order to perform pipelining.
As at the time of decoding and execution of the instruction in EU, the BIU fetches the sequential upcoming instructions and stores it in this queue.The size of this queue is 6-byte. This means at maximum a 6-byte instruction can be stored in this queue. The queue exhibits FIFO behaviour., first in first out. Execution Unit (EU)The Execution Unit (EU) performs the decoding and execution of the instructions that are being fetched from the desired memory location.Control Unit:Like the timing and control unit in 8085 microprocessor, the control unit in 8086 microprocessor produces control signal after decoding the opcode to inform the general purpose register to release the value stored in it. And it also signals the ALU to perform the desired operation.ALU:The arithmetic and logic unit carries out the logical tasks according to the signal generated by the CU. The result of the operation is stored in the desired register.Flag:Like in 8085, here also the flag register holds the status of the result generated by the ALU. It has several flags that show the different conditions of the result.Operand:It is a temporary register and is used by the processor to hold the temporary values at the time of operation.The reason behind two separate sections for BIU and EU in the architecture of 8086 is to perform fetching and decoding-executing simultaneously.
Working of 8086 MicroprocessorIn the previous section, we have discussed the operation of various sections of the BIU and EU. Now in this section, we will have a look at the overall processing cycle of 8086 microprocessor.So, basically, when an instruction is to be fetched from the memory, then firstly its physical address must be calculated and this is done at the BIU.The physical address of an instruction is given as:PA = Segment address Χ 10 + OffsetFor example: Suppose the segment address is 2000 H and the offset address is 4356 H. So, the generated physical address is 24356 H.Here, the code segment register provides the base address of the code segment which is combined with the offset address.The code segment contains the instructions. Each time an instruction is fetched the offset address inside the code segment gets incremented.So, once the physical address of an instruction is calculated by the BIU of the processor, it sends the memory location by the address bus to the memory.Further, the desired instruction at that memory location which is present in the form of the opcode is fetched by the microprocessor through the data bus.Suppose the instruction is ADD BL, CL. But, inside the memory, it will be in the form of an opcode. So, this opcode is sent to the control unit.The control unit decodes the opcode and generates control signals that inform the BL and CL register to release the value stored in it. Also, it signals the ALU to perform the ADD operation on that particular data.It is noteworthy that in any instruction, like ADD BL, CL.
Block Diagram 8086 Microprocessor Pdf Software
BL denotes the destination of the result of the add operation.This clearly shows that whatever, the operation is performed its result must be stored in the first register i.e., BL for this particular example.Let us take another example: Consider an instruction, ADD CL, 05H.This means that the operand which is 05H is to be added with the data present in the CL register and is stored in that particular register i.e., CL.In such condition, the operand is not provided to the control unit as only the opcode is required to be decoded by the CU. Hence the operand is directly provided to the ALU.Also, the status of this result is stored in the flag register. So, whenever, ALU carries out an operation, it simultaneously generates the result as well as its status.It is to be noteworthy that in BIU, pipelining fails whenever there is branching in the instruction. This is because generally instructions are present in a sequential manner.
But, sometimes the instructions are required to be executed unsequentially.However, in the queue, the instructions are stored sequentially. So, in case there exist a need for any random instruction to be decoded. The opcode stored in the queue will become invalid and must be cleared at that particular time.So, this is all about the block diagram and working of 8086 microprocessor.