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EE3A1 Computer Hardware and Digital Design Lecture 2 Introduction to VHDL
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Introduction We’ll look at some simple design in VHDL These will be the examples used in lab 1
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A simple example NAND gate 1 st thing to do: Say what it looks like to the rest of the system List of inputs and outputs Port map
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Port map: inputs and outputs Uppercase is a VHDL keyword Lower case is name I have chosen Mode can be IN or OUT ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC ); END;
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Port map: inputs and outputs Type STD_LOGIC can be ‘0’, ‘1’, ‘X’ or ‘U’. ‘X’ means unknown ‘U’ means uninitialized ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC ); END;
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Architecture ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC ); END; Now we know how many inputs and outputs Next we say how outputs derive values from inputs: Architecture
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Architecture ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC ); END; ARCHITECTURE simple OF nandgate IS BEGIN c <= a NAND b; END;
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Simulate it Check our design by giving it inputs Simulator says what the output would do Output is 0 when both inputs are 1: It works
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Multiple architectures We can give more than 1 architecture Distinguish them by different names ARCHITECTURE simple OF nandgate IS BEGIN c <= a NAND b; END; ARCHITECTURE complicated OF nandgate IS BEGIN c <= NOT ( a AND b ); END;
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BEGIN and END statements C: Block defined by { and } for (i=1; i<=n; i++) { a[i]=i; b[i]=a[i]*a[i]; } A block is a group of statements that should be treated as one
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BEGIN and END statements VHDL: Block defined by BEGIN and END FOR i IN ( 1 TO N ) LOOP BEGIN a(i) = i; b(i) = a(i) * a(i); END LOOP; A block is a group of statements that should be treated as one
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Semicolons C: for (i=1; i<=n; i++); /* WRONG semicolon */ {; /* ALSO WRONG semicolon */ a[i]=i; b[i]=a[i]*a[i]; } ; indicates end of statement. Statements that "open up" a block don't take semicolons:
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Semicolons VHDL: FOR i IN ( 1 TO N ) LOOP; --WRONG semicolon BEGIN; --ALSO WRONG semicolon a(i) = i; b(i) = a(i) * a(i); END LOOP; ; indicates end of statement. Statements that "open up" a block don't take semicolons:
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ENTITY doesn’t need a BEGIN ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC ); END; ARCHITECTURE simple OF nandgate IS BEGIN c <= a NAND b; END;
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Style issues VHDL is not case sensitive. In (black and white) books, VHDL keywords are normally in one particular case In editors, VHDL keywords are normally in one particular colour
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Style issues These are the same: ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC); END; entity NANDGATE is port ( A, B: in std_logic; C: out std_logic); end; entity nandgate is port ( a, b: in std_logic; c: out std_logic); end;
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Spaces, indents and line breaks Have no effect on code meaning Used to enhance clarity These are the same: ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC); END; ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC); END;
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Annotating END statements ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC); END; ARCHITECTURE simple OF nandgate IS BEGIN c <= a NAND b; END; It’s good style to say what you are ENDing
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Annotating END statements Often helps avoid bugs or confusion Usually optional ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC); END; ARCHITECTURE simple OF nandgate IS BEGIN c <= a NAND b; END ARCHITECTURE simple; It’s good style to say what you are ENDing
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Comments Comments are introduced by two dashes -- This is a comment
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Libraries void main () { printf(“My first C program”); } ERROR: printf not found
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Libraries #include void main () { printf(“My first C program”); } Works OK
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The IEEE library ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC); END ENTITY nandgate; ARCHITECTURE simple OF nandgate IS BEGIN c <= a NAND b; END ARCHITECTURE simple; u Error message “Cannot recognise type STD_LOGIC”.
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Opening libraries Definition of STD_LOGIC is held in library Must open library in order to get at definitions Main library is called IEEE LIBRARY IEEE; USE IEEE.XXXX.YYYY XXXX is sub-library (package) YYYY is feature that you want to use.
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Opening libraries Definition of STD_LOGIC is held in library Must open library in order to get at definitions Main library is called IEEE LIBRARY IEEE; USE IEEE.XXXX.ALL If you want to open all features of the package
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Using STD_LOGIC LIBRARY ieee; USE ieee.std_logic_1164.all; ENTITY nandgate IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC ); END ENTITY nandgate; ARCHITECTURE simple OF nandgate IS BEGIN c <= a NAND b; END ARCHITECTURE simple;
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Questions What is the block diagram for this entity declaration? ENTITY question1 IS PORT ( a: IN STD_LOGIC; b: OUT STD_LOGIC; c: IN STD_LOGIC; d: OUT STD_LOGIC ); END ENTITY question1; What is wrong with this architecture (2 problems)? ARCHITECTURE simple OF question1 IS BEGIN b <= a NAND c; END ARCHITECTURE complicated;
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What is the block diagram for this entity declaration? ENTITY question1 IS PORT ( a: IN STD_LOGIC; b: OUT STD_LOGIC; c: IN STD_LOGIC; d: OUT STD_LOGIC ); END ENTITY question1;
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Questions What is wrong with this architecture? ARCHITECTURE simple OF question1 IS BEGIN b <= a NAND c; END ARCHITECTURE complicated; Output d is never assigned a value Mismatch in architecture names
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Example 2 Arithmetic Logic Unit Heart of a microprocessor 1 st part of the assignment To build ALU we need u Conditionals u Signals that are many bits wide u Arithmetic on signals
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Conditionals LIBRARY ieee; USE ieee.std_logic_1164.ALL;
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Conditionals LIBRARY ieee; USE ieee.std_logic_1164.ALL; ENTITY equals IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC); END ENTITY equals;
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Conditionals LIBRARY ieee; USE ieee.std_logic_1164.ALL; ENTITY equals IS PORT ( a, b: IN STD_LOGIC; c: OUT STD_LOGIC); END ENTITY equals; ARCHITECTURE number1 OF equals IS BEGIN c <= '1' WHEN a=b ELSE '0'; END ARCHITECTURE number1;
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Signals that are more than 1 bit STD_LOGIC_VECTOR(0 TO 3) Four members a(0), a(1), a(2) and a(3). Each is of type STD_LOGIC.
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Example of 4-bit device LIBRARY ieee; USE ieee.std_logic_1164.ALL; ENTITY orgate IS PORT ( a, b: IN STD_LOGIC_VECTOR(0 TO 3); c: OUT STD_LOGIC_VECTOR(0 TO 3)); END ENTITY orgate;
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Example of 4-bit device ARCHITECTURE number1 OF orgate IS BEGIN C(0) <= a(0) OR b(0); C(1) <= a(1) OR b(1); C(2) <= a(2) OR b(2); C(3) <= a(3) OR b(3); END ARCHITECTURE number1;
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Example of 4-bit device ARCHITECTURE number2 OF orgate IS BEGIN c <= a OR b; END ARCHITECTURE number2; The compiler can figure out that a,b,c are 4 bits wide
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Example of 4-bit device ARCHITECTURE number3 OF orgate IS BEGIN C(0 TO 3) <= a(0 TO 3) OR b(0 TO 3); END ARCHITECTURE number3; Making the loop explicit
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STD_LOGIC_VECTOR values Single bit STD_LOGIC assignment: a <= ‘1’;--single quotes 4-bit STD_LOGIC_VECTOR assignment: a <= “1110”;-- double quotes.
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Direction of numbering Can number upwards or downwards Both represent the same decimal number u 14 if it’s signed; -2 if it’s unsigned
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Direction of numbering In digital logic design bit 0 is usually lsb So in VHDL index usually counts downwards
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Arithmetic on STD_LOGIC_VECTOR s Comparator g b ELSE ‘0’; Signed or unsigned? u 1111 is +15 or –1? u 1110 is +14 or –2?
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Arithmetic on STD_LOGIC_VECTOR s VHDL has two different versions of +,-,>,< etc. u STD_LOGIC_SIGNED u STD_LOGIC_UNSIGNED Import one or the other
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Arithmetic on STD_LOGIC_VECTOR s LIBRARY ieee; USE ieee.std_logic_1164.ALL; USE ieee.std_logic_signed.ALL; ENTITY comp IS PORT ( a, b: IN STD_LOGIC_VECTOR(3 DOWNTO 0); g: OUT STD_LOGIC); END ENTITY comp; ARCHITECTURE simple OF comp IS BEGIN g b ELSE ‘0’; END ARCHITECTURE simple; 1111 will be interpreted as -1
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Arithmetic on STD_LOGIC_VECTOR s LIBRARY ieee; USE ieee.std_logic_1164.ALL; USE ieee.std_logic_unsigned.ALL; ENTITY comp IS PORT ( a, b: IN STD_LOGIC_VECTOR(3 DOWNTO 0); g: OUT STD_LOGIC); END ENTITY comp; ARCHITECTURE simple OF comp IS BEGIN g b ELSE ‘0’; END ARCHITECTURE simple; 1111 will be interpreted as 15
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ALU example OpcodeOperation 00a + b 01a – b 10a and b 11a or b 16 bit ALU Operation is selected by opcode
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ALU example LIBRARY ieee; USE ieee.std_logic_1164.ALL; USE ieee.std_logic_signed.ALL; ENTITY alu IS PORT ( a, b: IN STD_LOGIC_VECTOR(15 DOWNTO 0); opcode: IN STD_LOGIC_VECTOR(1 DOWNTO 0); c: OUT STD_LOGIC_VECTOR(15 DOWNTO 0) ); END ENTITY alu;
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ALU example ARCHITECTURE simple OF alu IS BEGIN c <= a + b WHEN opcode=”00” ELSE a - b WHEN opcode=”01” ELSE a OR b WHEN opcode=”10” ELSE a AND b WHEN opcode=”11”; END ARCHITECTURE simple;
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Simulate the ALU a = “0000000001110111”, 0077H. b = “0000000000000001”, 0001H. As opcode changes “00”,”01”,”10”,”11”, (0,1,2,3) output gets a+b, then a-b, then a OR b then a AND c.
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Synthesise the example It works as expected Synthesise to gate level description Much quicker and easier than doing it the old fashioned way More flexible too
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Summary Intro to VHDL Basic examples for use in lab 1
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