只用一个1位二进制全加器为基本元件和一些辅助的时序电路,设计一个8位串行二进制全加器,要求: 1、能在8-9个时钟脉冲后完成8位二进制数(加数被加数的输入方式为并行)的加法运算,电路须考虑进位输入Cin和进位输出Cout;需要在Quartus II 13.0中使用的VHDL语言
时间: 2024-03-15 22:43:55 浏览: 105
下面是一个实现要求的8位串行二进制全加器的VHDL代码:
```vhdl
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
entity serial_adder is
Port ( clk : in STD_LOGIC;
reset : in STD_LOGIC;
A : in STD_LOGIC_VECTOR (7 downto 0);
B : in STD_LOGIC_VECTOR (7 downto 0);
Cin : in STD_LOGIC;
sum : out STD_LOGIC_VECTOR (7 downto 0);
Cout : out STD_LOGIC);
end serial_adder;
architecture Behavioral of serial_adder is
component one_bit_adder is
Port ( A : in STD_LOGIC;
B : in STD_LOGIC;
Cin : in STD_LOGIC;
S : out STD_LOGIC;
Cout : out STD_LOGIC);
end component;
signal Cin_i : std_logic := '0';
signal S_i : std_logic_vector(7 downto 0);
signal Cout_i : std_logic_vector(7 downto 0);
begin
process(clk, reset)
begin
if reset = '1' then
S_i <= (others => '0');
Cout_i <= (others => '0');
Cin_i <= '0';
sum <= (others => '0');
Cout <= '0';
elsif rising_edge(clk) then
if Cin_i = '1' then
S_i(0) <= not A(0) xor not B(0) xor '1';
Cout_i(0) <= (not A(0) and not B(0)) or (not A(0) and Cin_i) or (not B(0) and Cin_i);
else
S_i(0) <= A(0) xor B(0);
Cout_i(0) <= A(0) and B(0);
end if;
for i in 1 to 7 loop
S_i(i) <= S_i(i-1) xor A(i) xor B(i);
Cout_i(i) <= (S_i(i-1) and A(i)) or (S_i(i-1) and B(i)) or (A(i) and B(i));
end loop;
sum <= S_i;
Cout <= Cout_i(7);
Cin_i <= Cin;
end if;
end process;
one_bit: for i in 0 to 7 generate
one_bit_adder_inst: one_bit_adder port map(
A => A(i),
B => B(i),
Cin => Cin_i,
S => S_i(i),
Cout => Cout_i(i)
);
end generate one_bit;
end Behavioral;
```
其中,`one_bit_adder` 是一个1位全加器的VHDL代码,需要单独实现。在主要的 `serial_adder` 实体中,使用 `one_bit_adder` 实例化8次,分别对每个二进制位进行加法运算。同时,使用 `S_i` 和 `Cout_i` 分别存储每个二进制位的和和进位,最后将 `S_i` 输出为8位二进制数的和,将 `Cout_i(7)` 输出为进位输出 `Cout`。为了实现串行输入,需要在每个时钟上升沿时将进位输入 `Cin` 的值存储在 `Cin_i` 中,以便在下一次计算中使用。
需要注意的是,在 `one_bit_adder` 中,由于使用了异或门和与非门来实现异或和进位的计算,因此需要使用 `STD_LOGIC_UNSIGNED` 库来支持位级运算。
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