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The von Neumann Model – Chapter 4

The von Neumann Model – Chapter 4. COMP 2620. The LC-3 as a von Neumann Machine. The LC-3 as a von Neumann Machine. Note the two kinds of arrowheads in the diagram: Filled in – control signals for the processing of data elements

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The von Neumann Model – Chapter 4

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  1. The von Neumann Model – Chapter 4 COMP 2620 Dr. James Money COMP 2620

  2. The LC-3 as a von Neumann Machine

  3. The LC-3 as a von Neumann Machine • Note the two kinds of arrowheads in the diagram: • Filled in – control signals for the processing of data elements • Not Filled in – denotes data elements and the flow of data along the path • The ALU has two input data elements and one output • The ALUK line controls the operation performed by the ALU

  4. The LC-3 as a von Neumann Machine • Note the lines with the slash through them and the number next to it • This indicates the number of bits transmitted along the data and/or control path • Same notation applies to both data and control paths

  5. Memory • Memory contains the storage elements along with the MAR for addressing the memory elements • The MDR holds the contents of a memory location to/from storage • MAR is 16 bits, so the address space is 216 • The MDR is 16 bits indicating each memory location is 16 bits

  6. Input/Output • There are two: • Keyboard • Monitor • For the keyboard, there are two registers: • KBDR – keyboard data register which has the ASCII code if the keys struck • KBSR – keyboard status register for status about the keys struck

  7. Input/Output • The monitor has also two registers: • DDR – ASCII code of the character to be displayed on the screen • DSR – the associated status information for the character to be printed on the screen • We will talk more about these in detail in Chapter 8

  8. Processing Unit • This consists of the • ALU – arithmetic and logic unit • Eight registers • R0,R1,…,R7 • Use to store temporary values • Also used as operands for operations

  9. Processing Unit • The ALU only has three operations: • Addition • Bitwise AND • Bitwise NOT • We will have to spend some time implementing the other functions to achieve full assembly language

  10. Control Unit • This contains the parts to control flow of the program executing • It contains the finite state machine, which directs all activity • Processing is carried out step-by-step and not clock cycle by clock cycle • There is the CLK input to the finite state machine which specifies how long each clock cycle lasts

  11. Control Unit • The Instruction Register (IR) is input to finite state machine and has the current instruction • This is input to the finite state machine since it determines what activities must be carried out • The Program Counter (PC) keeps track of the next instruction to be executed

  12. Control Unit • Note all the outputs from the finite state machine are controls • These control processing in various parts of the computer • For example, the ALUK of 2 bits controls which operation is performed by the ALU • Another is the GateALU which controls whether the output of the ALU is provided to the processor during the current clock cycle.

  13. Instruction Processing • The most basic unit of processing is the instruction • There are two parts: • Opcode – what the instruction does • Operands – the parameters to the opcode, for example • Registers • Constants

  14. Instruction Processing • Each instruction is 16 bits, or one word on the LC-3 • The bits are numbered left to right from [15] to [0]. • Bits [15:12] contain the opcode • There are at most 24 = 16 opcodes • Bits [11:0] are operands

  15. Instruction Cycle • Each instruction is handled in a systematic way through a sequence of steps call the instruction cycle • Each step is called a phase • There are six phases to the cycle

  16. Instruction Cycle • The six phases of the instruction cycle are: • Fetch • Decode • Evaluate Address • Fetch Operands • Execute • Store Result

  17. FETCH • Fetch obtains the next instruction from memory and loads it into the IR of the control unit • We first use the PC to find the address of the next instruction

  18. FETCH • The FETCH stage takes multiple steps: • MAR is loaded with the value of PC and increment the PC • Memory is interrogated which results in the instruction being placed in the MDR • The IR is loaded with the contents of the MDR • Note the PC must be incremented at the same time

  19. FETCH • Note that the first step takes 1 machine cycle • The second one can take multiple machine cycles • Step 3 takes on machine cycle

  20. DECODE • This step looks at the highest 4 bits to determine what to do • A 4-to-16 decoder decides which of the 16 opcodes is to be processed • Input is IR[15:12] • The output line is the one that corresponds to the opcode

  21. EVALUATE ADDRESS • This phase reads the memory address needed to process the instruction • An example is the LDR • Causes a value stored in memory to be loaded into a register • The memory location is in the form base+offset • This final memory location is being evaluated at this step

  22. FETCH OPERANDS • This phase obtains the source operands • For the LDR example, the MAR is loaded with the address of the EVALUATE ADDRESS phase and reading memory with data in the MDR • For ADD, this would be obtaining the values for the source operands

  23. Execute • In this phase, the instruction is actually executed • For ADD, this is the step of performing the addition in the ALU

  24. STORE RESULT • The last phase is to write the result to the correct location • This may involve writing to memory or registers

  25. Instruction Cycle • Once STORE RESULT is done, the control unit starts again with a new machine instruction • It begins with FETCH and repeats • The PC already has the location of the next instruction to execute • Continues until processing order breaks

  26. Example: LC-3 ADD Instruction • LC-3 has 16-bit instructions. • Each instruction has a four-bit opcode, bits [15:12]. • LC-3 has eight registers (R0-R7) for temporary storage. • Sources and destination of ADD are registers. “Add the contents of R2 to the contents of R6,and store the result in R6.”

  27. Example: LC-3 LDR Instruction • Load instruction -- reads data from memory • Base + offset mode: • add offset to base register -- result is memory address • load from memory address into destination register “Add the value 6 to the contents of R3 to form amemory address. Load the contents of that memory location to R2.”

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