VHDL - Get Now PowerPoint by linzhengnd


									Lecture 21: VHDL,
 Multiplication, I/O

   Soon Tee Teoh
      CS 147
• VHDL and Verilog HDL are the two most widely-
  used hardware description languages.
• VHDL stands for VHSIC Hardware Description
• VHSIC stands for Very High Speed Integrated
• Describe a logic circuit by function, data flow
  behavior and/or structure
• Structure and behavior are complementary ways
  of describing a system.
  – However, same behavior can be accomplished by
    different structure
              VHDL Blocks
• ENTITY block: Describes the interface for the
  design. Defines the input and output logic
  signals of the circuit.
• ARCHITECTURE block: describes the internal
  operation of the design.
• VHDL allows user to specify
  – Components of a circuit
  – Interconnection between components
• Many packages also allow user to input state
  diagram and translate to VHDL code
    Simulation and Synthesis
• Simulation tests if the logic works
• Synthesis takes into account timing
  considerations. User specifies the
  propagation delay of a component.
ENTITY block example
    entity half-adder is port (
              a, b: in bit;
              sum, carry: out bit);
    end half_adder

    entity ALU32 is port (
             A, B: in bit_vector( 31 downto 0);
             F: out bit_vector( 31 downto 0);
             FS: in bit_vector ( 3 downto 0);
             V, C, N, Z: out bit);
    end ALU32
ARCHITECTURE block example

    architecture half_adder_arch of half_adder is
                       sum <= (a xor b) after 5 ns;
                       carry <= (a and b) after 5 ns;
             end half_adder_arch;
  Putting components together (1):
              Example from Weijun Zhang (UCR)

First, define two components, OR_GATE and AND_GATE.

entity OR_GATE is                    entity AND_GATE is
port(    X:     in std_logic;        port(    A:    in std_logic;
         Y:     in std_logic;                 B:    in std_logic;
         F2:    out std_logic                 F1:   out std_logic
);                                   );
end OR_GATE;                         end AND_GATE;

architecture behv of OR_GATE is      architecture behv of AND_GATE is
begin                                begin
process(X,Y)                         process(A,B)
begin                                begin
         F2 <= X or Y;                        F1 <= A and B;
end process;                         end process;
end behv;                            end behv;
       Putting components together (1):
                       Example from Weijun Zhang (UCR)
Then, define a new entity, comb_ckt, that uses one instance of OR_GATE and one instance of AND_GATE.

entity comb_ckt is                     architecture struct of comb_ckt is
port(    input1: in std_logic;           component AND_GATE is
         input2: in std_logic;           port( A:         in std_logic;
         input3: in std_logic;                   B:       in std_logic;
         output: out std_logic                  F1:       out std_logic
);                                       );
end comb_ckt;                            end component;
                                         component OR_GATE is
                                         port( X:         in std_logic;
                                                Y:        in std_logic;
                                                F2:       out std_logic
                                         end component;
                                         signal wire: std_logic;
                                          Gate1: AND_GATE port map (A=>input1, B=>input2, F1=>wire);
                                          Gate2: OR_GATE port map (X=>wire, Y=>input3, F2=>output);
                                       end struct;
                        architecture TB of CKT_TB is

We want to test         component comb_ckt is
comb_ckt, which we      port(       input1: in std_logic;
built in the previous               input2: in std_logic;
page. So, we load it                input3: in std_logic;
into our test-bench.                output: out std_logic
                        end component;

                        -- declare all I/O ports from unit under test as signals.
                        -- signals are usually declared within architecture

                        signal T_input1, T_input2, T_input3, T_output: std_logic;


                          U_UT: comb_ckt port map (T_input1,T_input2,T_input3,T_output);


                                      variable err_cnt: integer := 0;

                                                                                           continue …

                             -- Test case 1
                                      T_input1 <= '0';
                                      T_input2 <= '0';
Set the inputs                        T_input3 <= '0';
                                      wait for 10 ns;
                                      assert (T_output=((T_input1 or T_input2) and T_input3))
In the waveform                       report "Failed Case1!" severity error;
simulation, watch the                 if (T_output/=((T_input1 or T_input2) and T_input3)) then
signals T_input1,                         err_cnt := err_cnt +1;
T_input2, T_input3 and                end if;
                                     -- Test case 2
                                     T_input1 <= '1';
                                     T_input2 <= '1';
                                     T_input3 <= '1';
                                     wait for 10 ns;
                                     assert (T_output=((T_input1 or T_input2) and T_input3))
                                     report "Failed Case1!" severity error;
                                     if (T_output/=((T_input1 or T_input2) and T_input3)) then
                                         err_cnt := err_cnt +1;
                                     end if;
                           end process;
                         end TB;
             How to Simulate
•   Load the circuits into a VHDL simulator.
•   Example of VHDL simulator: Symphony EDA
•   Compile the VHDL files.
•   Run the simulation.
           Assume multiplying unsigned numbers

                        1010110                          Multiplier
Product           1011100101110

          Same method as multiplication in decimal:
          Shift multiplicand to the left.
          Whenever there is a 1 bit in the multiplier,
          add the shifted multiplicand to the product.
                          Multiplication Circuit
                         Zero Fill More Sig                         32
                    64                             32                                         6

64-bit left shift register                     32-bit right shift register                            Down counter register


                                                        64                                                              6

                64-bit adder
                                                                                                                 Zero Detect
64                                                      All 0s

              64-bit register
                                       Write enable                 Bit0, the least significant bit

                                              The initializer begins by putting the Multiplicand in the
                                              left shift register, the Multiplier in the right shift register,
                                              the number 32 in the down counter register, and the
                    Product                   number 0 in the Product register.                                   Done
• Program and data need to be input.
• Results from computation need to be
  recorded or displayed.
• Examples: Keyboard, monitor, printer,
  hard disk, CD-ROM, network, speakers,
        Memory-mapped I/O
• Common data, address and control buses
  for both memory and I/O.
• Interface units and memory have distinct
• CPU read and write from interface units as
  if they were assigned memory addresses.
  Can use same memory read/write
    Communication between I/O
        Device and CPU
• Common methods:
  – Polling
  – Interrupt-driven
  – DMA
        Direct Memory Access (DMA)
• Sometimes need to transfer data between I/O device and memory.
   – e.g. transfer data between memory and hard disk
• Don’t involve the CPU so that the CPU can simultaneously execute
  other instructions.
• Method:
   – DMA Controller sends signal to CPU to request control of Address and
     Data buses.
   – CPU disconnects from Address and Data buses, and sends signal to
     DMA Controller to go ahead
   – DMA Controller transfers data to/from memory using the Address and
     Data buses. (At this time, CPU can execute any instructions that does
     not involve memory read/write)
   – When DMA Controller is done, DMA Controller signals to CPU
   – CPU regains control of Address and Data buses, and can now
     communicate with the memory.
• Problem with DMA: The stale data problem or coherency problem:
   – When device writes to memory directly, cache data is no longer valid
   – In write-back cache, when CPU writes to cache, data in memory
     accessed be device is invalid
                  Hard Disk
• A hard disk is used to
  store data. It is
  intermediate-speed,         platter
  nonvolatile and writable.
• A hard disk has many
• Each platter can be
  magnetized on one or
  both surfaces.
• There is one read/write
  head per recording
                         Hard Disk
• Each platter is divided into concentric tracks.
• The set of tracks at the same distance from the disk center on all
  platter surfaces is called a cylinder.
• Each track is divided into sectors containing a fixed number of bytes.
• In modern, high-capacity disks, there are more sectors in the longer
  outer tracks. Also, some sectors are reserved to replace defective



                 Hard Disk Timing
• The read/write head is used to access data. The set of heads are
  mounted on actuators that move the heads radially over the disk.
• To access data on the hard disk, several steps need to be
• First, we need to move the heads from the current cylinder to the
  desired cylinder. The time taken to do this is called the seek time.
• Second, we need to rotate the disk to reach the desired sector. The
  time taken to do this is called the rotational delay.
• Third, some time is needed by the disk controller to access the data.
  This time is called the controller time.
• Finally, the data is read. The rate at which data is read is called the
  disk transfer rate.
                Hard Disk Timing
• Typical times:
   – Seek time: 10 ms
   – Rotational delay: 6 ms
   – Controller time: negligible
• The total time required to locate a word on the disk is
  called the disk access time.
   – Disk access time = Seek time + Rotational delay + Controller
• Typical disk transfer rate: 3 MB/s
• Exercise:
   – The processor wants to load 1 MB of contiguous data from disk
     to memory. Suppose that the seek time is 10 ms, rotation delay
     is 6 ms, controller time is 0 ms, and the disk transfer rate is 3
     MB/s. How much time does it take?
          RAID (Redundant Array of
            Independent Disks)
•   Reliability and availability of data
•   What happens when hard disk fails? Data is lost.
•   Solution: Use RAID
•   Use array of hard disks instead of only one.
•   RAID has several levels.
•   RAID Level 1: Mirroring.
    –   Simply keep a duplicate of each disk.
    –   Example: 8 disks + 8 duplicate disks
    –   When one disk fails, repair by copying from duplicate disk
    –   How to detect disk failure: There is usually hardware on disk
        itself to detect that it has failed.
                      RAID Level 3
• RAID Level 3 is bit-interleaved parity.
   – Example: 8 disks + 1 parity disk
       • 8 disks contains data. A bit in the parity disk is set to 1 if the number
         of 1 bits in the same position in the 8 disks is odd. The parity bit is
         set to 0 otherwise.
       • When disk fails, the parity disk together with the 7 surviving disks
         can be used to reconstruct the failed disk.
       • Out of the 9 disks, only one can fail at a time. If two or more disks
         fail at the same time, the failure is not recoverable.
• How to protect against simultaneous multiple disk
   – RAID 6 can handle up to 2 simultaneous failures. It uses a P+Q
     (Reed-Solomon) coding scheme.
   – Otherwise, we can also use a 2D array of disks, with a parity
     disk for each row, and a parity disk for each column.

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