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Processor: Data Path Components CS 251 Fall 2019 CS 240 Spring 2020 Principles of Programming Languages Foundations of Computer Systems Ben Wood Ben Wood Sequential Logic 2 1 3 Latch: CC-BY Rberteig@flickr and State Instruction ALU


  1. λ Processor: Data Path Components CS 251 Fall 2019 CS 240 Spring 2020 Principles of Programming Languages Foundations of Computer Systems Ben Wood Ben Wood Sequential Logic 2 1 3 Latch: CC-BY Rberteig@flickr and State Instruction ALU Registers Fetch and Memory Decode Output depends on inputs and stored values . (vs. combinational logic: output depends only on inputs) Elements to store values: latches, flip-flops, registers, memory https://cs.wellesley.edu/~cs240/s20/ Sequential Logic 1 Sequential Logic 2 Bistable latches SR latch S R Q Q' Q (stable) Q' (stable) 0 0 0 1 0 1 0 0 1 0 1 0 Suppose we somehow get a 1 (or a 0?) on here. 1 0 ? ? 1 0 Q Q 0 1 ? ? 0 1 = Set Reset 0 S 0 R Q Q Q Q Sequential Logic 3 Sequential Logic 4

  2. SR latch D latch Q R R D Q Q S Data bit R Q R Q Q C S Q Q S Clock S if C = 0, then SR latch stores current value of Q. S R R Q if C = 1, then D flows to Q: Q Q if D = 0, then R = 1 and S = 0, Q = 0 if D = 1, then R = 0 and S = 1, Q = 1 Q S Sequential Logic 5 Sequential Logic 6 ex Time matters! Clocks Clock: free-running signal with fixed cycle time = clock period = T. D Clock frequency = 1 / clock period C Falling edge Q Rising edge Clock period A clock controls when to update a sequential logic element's state. Assume Q has an initial state of 0 Sequential Logic 7 Sequential Logic 8

  3. Synchronous systems D flip-flop with falling-edge trigger Inputs to state elements must be valid on active clock edge. leader follower E D Q Q L Q F D L D F D latch D latch Q C L Q L C F Q F State State element Combinational logic element 1 2 C Q next becomes Q now Can still read Q now follower stores E as Q Clock leader stores D as E Time Sequential Logic 9 Sequential Logic 10 ex Time matters! Reading and writing in the same cycle D D Q D Flip-Flop Assume Q is initially 0. Clock C Q C E Q Assume Q and E have an initial state of 0 Sequential Logic 11 Sequential Logic 12

  4. *Half a byte! A 1-nybble* register D flip-flop = one bit of storage (a 4-bit hardware storage cell) 0 D Q D Flip-Flop C Q D Q D Q 1 1 D Flip-Flop D Flip-Flop C Q C Q 0 D Q D Flip-Flop C Q 1 D Q D Flip-Flop Write C Q Clock Sequential Logic 13 Sequential Logic 14 Register file Read register number 1 Register 0 Register 1 M Read register Read ports . . . u Read data 1 r selector 1 x (data out) Read register Register n – 2 Read data 1 w r selector 2 Read ports Register n – 1 Why 2? Read data 2 Write register w selector Read register r number 2 Write data w Write? Write port M 0 = read u Read data 2 1 = write r = log 2 number of registers x w = bits in word Array of registers, with register selectors, write/read control, input port for writing data, output ports for reading data. Sequential Logic 15 Sequential Logic 16

  5. Write port (data in) RAM (Random Access Memory) write control clock Write C 0 Register 0 1 D . n -to-2 n . register number Register number . decoder C A B Register 1 n – 2 D n – 1 . . . C Register n – 2 D C Register n – 1 incoming data Register data D Similar to register file, except… Sequential Logic 17 Sequential Logic 18 16 x 4 RAM 4-bit address 1101 4 to 16 decoder data out Sequential Logic 19

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