 Hello and welcome to the series of video lectures on the subject microprocessor for secondary IT students. I am Dr. Srisail Sharadkarjbar and in this video lecture we are going to study architecture of 80286 microprocessor. At the end of this session you will be able to write key features of 80286 microprocessor. You will also be able to describe architecture of 80286 microprocessor. 80286 microprocessor can be seen as an advanced version of 8086 microprocessor with memory management and protection abilities, until 80286 is a high performance 16-bit microprocessor unit. By 16-bit we mean that this microprocessor is able to handle 16-bit data at any instant of time. This microprocessor provides multitasking and multi-user environment. Its performance is almost 6 times faster than 8086 microprocessor. It has a dedicated memory manager and similar to 8086 it has a compatible numerical data processor 80287 for computationally intensive tasks. Let us see the features of 80286 microprocessor. 80286 microprocessor is designed for multi-user and multitasking systems. This microprocessor can be operated at 3 operating clock speeds namely 12.5 MHz, 10 MHz and 8 MHz. This microprocessor has got 24-bit address bus. Hence it can address 2 REST to 24 that is 16 MB of physical memory. This microprocessor has got one special operating mode and in that mode it can address 1 GB of virtual memory per task. This microprocessor is available in 68-pin IC package. It has got 2 modes of operation, real address mode and protected virtual address mode. Unlike 8086 this microprocessor has got 4 separate processing units. The 8086 has 2 separate processing units namely bus interface unit and execution unit whereas this microprocessor has got 4 separate processing units which act parallely. The first unit is bus unit, the second unit is instruction unit, the third unit is address unit and fourth unit is execution unit. Through memory management unit it can address a virtual memory space of 1 GB. It has got 8 general purpose registers namely AX, BX, CX, DX, source index, destination index, base pointer and stack pointer. It has got 4 segment registers namely code segment register, data segment register, extra segment register and stack segment register. Now pause the video for 2 minutes and write down the answer of the following question. I hope you have written the answer. The correct answer is 1 GB of virtual memory can be addressed by 80286 microprocessor. The diagram shows the architectural block diagram of 80286 microprocessor. As mentioned earlier the architecture of 80286 microprocessor has got 4 separate processing units namely address unit, bus unit, instruction unit and execution unit. And one thing which is worth noting here is that all these 4 units work parallely that is their execution is simultaneous in nature. Now let us see each processing unit in detail. Before that let us understand what is the job of a CPU. The job of CPU is to fetch, decode and execute the instructions which are stored in the physical memory or random access memory. Now let us consider the address unit. This unit computes the physical addresses of instructions and data that CPU wants access to. In real mode address unit computes address using segment base and offset address similar to 8086 and similar to 8086 case the address in case of real mode is 20 bit in nature. In protected mode it behaves as a complete memory management unit. In virtual mode the processor uses all 24 lines to address up to 16 megabyte of physical memory and using the concepts of mapping of physical memory to the secondary memory and the descriptor tables it is able to address 1 gigabyte of virtual memory space per task. As you can see here in this processing unit there are various blocks. The physical address adder gives the 20 bit or 24 bit physical address depending on the mode which is operating. The input to this physical address adder are this offset and the segment base. This offset address is the output of this offset adder. The segment limit checker checks whether the current offset address does not exceed the segment size that is 64 kilobyte. So using segment base and offset address this physical address adder gives the output of either 20 bit physical address or 24 bit physical address depending on the mode in which this processor is operating. The next unit is bus unit. The bus unit performs same operation as that of bus interface unit of 8086 microprocessor. This unit acts as an interface to the outside world for this microprocessor. This unit performs all the memory and IO read and write operations. It prefixes the instruction bytes and store these in the queue. It also controls the transfer of data to and from processor extension devices such as 802 87 Mac co-processor. As you can see here this block contains address latches which latches this 20 or 24 bit physical address onto this address latches. It also generates some control signals such as bus high enable, memory or IO read operation. It has got 24 address lines A0 to A23. This preface unit works for the purpose of storing the instructions into the 6 byte preface queue beforehand. This processor extension interface is useful for interfacing the Mac co-processor with this microprocessor. The bus control logic generates the control signals. Whatever data that is input from the random access memory is available onto these D0 to DWIFT in data lines. The next unit is instruction unit. The instruction unit decodes 3 preface instructions and stores them in a queue. From the queue these preface instructions are taken by the execution unit. As you can see here there is an instruction decoder which decodes the instructions which are input from the 6 byte preface queue. These 3 decoded instructions are stored in this queue and these decoded instructions are then given to this execution unit. Let us see the execution unit. The execution unit is exactly the same as the execution unit of 8086 microprocessor. This unit executes the instructions which it receives from the instruction unit with the help of its 16-bit registers. In real mode this unit has the same set of registers as that of 8086 except one 16-bit register the machine status word is there. As you can see here in the execution unit there are mainly 3 blocks arithmetic and logic unit control unit which receives some control signals from the external world and some general purpose registers. So depending on the instructions the output may be stored in the general purpose register or it may be given to the outside world. This is the reference. Thank you very much.