数电研讨---医用生产线设计报告.docx

PROJECT REPORT实验课程： 数字电路EDA实验实验名称： 医用生产线【Requirement】Be able to preset the number of tablets per bottle, for example ,fifty tablets each bottle. Every box contains twenty-four bottles, stop count until 18 Use software to design the simulation, Quartus II will be better. 【Principle and framework given by teacher】【Our system framework】【1.0】【2.0】【Modular design and simulation】Keyboard prcessor【Function introduction】 As project requirement, we can optional set keyboard size and what is the key arrangement. However , in consideration of engineering application, we chose 4*4 keyboard , as indicated below.Just as the picture illustrated above, We define “A” are the first counter enable pin .When the A pin is high , corresponding counter accept 8-bit number as its period. So as B and C.D is the reset put which clear the number of register. *and # are use as back-up. Basic on this keyboard framework we can easily achieve our program use VHDL.At first ,we divide the keyboard into a 4 row and 4 column and then judge its value using case sentence.【Programming】LIBRARY IEEE;USE IEEE.STD_LOGIC_1164.ALL;ENTITY KEYBOARD IS PORT(CLK: IN STD_LOGIC;ROW: IN STD_LOGIC_VECTOR(3 DOWNTO 0); /行向量COL: IN STD_LOGIC_VECTOR(3 DOWNTO 0); /列向量DATA: OUT STD_LOGIC_VECTOR(3 DOWNTO 0);F:OUT STD_LOGIC; /给寄存器判断是否有输入EN1: OUT STD_LOGIC;EN2: OUT STD_LOGIC;EN3: OUT STD_LOGIC);END ENTITY KEYBOARD;ARCHITECTURE RTL OF KEYBOARD ISSIGNAL MID: STD_LOGIC_VECTOR(7 DOWNTO 0);SIGNAL NUM: STD_LOGIC_VECTOR(3 DOWNTO 0);BEGINPROCESS(CLK,ROW,COL)BEGINMID NUM NUM NUM NUM NUM NUM NUM NUM NUM NUM NUM=1010;EN1=0;EN2=0;EN3 NUM=1111;EN1=1;EN2=0;EN3 NUM=1111;EN1=0;EN2=1;EN3 NUM=1111;EN1=0;EN2=0;EN3 NUM=1111;/1111的时候表示数据无效END CASE;END IF;DATA=NUM;F=(NUM(0) AND NUM(1) AND NUM(2) AND NUM(3);END PROCESS;END RTL;【Simulation waveform】Register【Function introduction】 This register has 4-bit input and 8-bit output, and a input named “carry” is also add to this register . When carry equal to “1”, put the 4-bit input to high output pin, when the other, the 4-bit input will be put into low output pin.【programming】LIBRARY IEEE;USE IEEE.STD_LOGIC_1164.ALL;ENTITY REGISTER_DIY IS PORT(F: IN STD_LOGIC;CLK: IN STD_LOGIC;CARRY : IN STD_LOGIC;RIN: IN STD_LOGIC_VECTOR(3 DOWNTO 0);OUT_LOW: OUT STD_LOGIC_VECTOR(3 DOWNTO 0);OUT_HIGH: OUT STD_LOGIC_VECTOR(3 DOWNTO 0);END REGISTER_DIY;ARCHITECTURE RTL OF REGISTER_DIY IS SIGNAL F_IN: STD_LOGIC_VECTOR(1 DOWNTO 0);BEGIN PROCESS(CLK,F,CARRY)BEGIN F_IN OUT_LOW OUT_HIGH NULL;END CASE;END IF;END PROCESS;END RTL;【Simulation waveform】BCD to binary convertorAt the very beginning, we should ask a question to ourselves. Why we need a BCD to binary converter? We use the keyboard to put in the number of pills per bottle which is from 0 to 99, it is BCD, but the counter only counts the binary number, so we need to convert the BCD to binary numbers.【Function introduction】At First, we divide the BCD into high bit and low bit, then use two registers to store them and convert high bit and low bit respectively. When we convert the high bit, we assume the low bit is zero, then convert the BCD with low bit zero to binary numbers. For example,0010(high bit) 00100000(BCD) 00010100(binary)Then we convert the low bit, similarly, we assume the high bit is zero, and then convert the BCD with high bit zero to binary numbers. For example,0010(low bit) 00000010(BCD) 00000010(binary)Finally, we need an adder to add the two binary numbers(high bit and low bit) to get the final binary numbers. Then we have finished BCD to binary conversion.00100010(BCD) 00010100(high bit)+00000010(low bit) 00010110(binary number)【Programming】【BCD to binary (high-bit)】library IEEE; use IEEE.std_logic_1164.all; entity BCD_binary_high is port ( CLK : IN STD_LOGIC; BCD_high: in bit_vector(3 downto 0); binary_high:out bit_vector(7 downto 0); end BCD_binary_high; architecture RTL of BCD_binary_high is begin process(BCD_high,CLK) begin IF CLKEVENT AND CLK=1 THEN case BCD_high is when 0000 = binary_high binary_high binary_high binary_high binary_high binary_high binary_high binary_high binary_high binary_high binary_high binary_low binary_low binary_low binary_low binary_low binary_low binary_low binary_low binary_low binary_low binary_low=01000000; end case; end process; end RTL;【Simulation waveform】Figure 1 BCD to binary converter(high-bit)Figure 2BCD to binary converter(low-bit)Adder【Function introduction】we need an adder to add the two binary numbers(high bit and low bit) to get the final binary numbers. Then we have finished BCD to binary conversion.【Programming】library IEEE; use IEEE.std_logic_1164.all; use IEEE.std_logic_unsigned.all; entity Unsigned_adder is port ( A,B: in std_logic_vector(7 downto 0); S: out std_logic_vector (7 downto 0); end Unsigned_adder ; architecture RTL of Unsigned_adder is begin S=A + B; end RTL;【Simulation waveform】Simulate successfully achieved the function of adder, but it has some delay and interference. I think it may caused by Race and Hazard.Counter【Function introduction】Set D and Q as the signal variable .D resource from the key board .Q is the internal signal of counter ,it circulates upon D.Once D has been settle ,Q is upcount for D times.When Q run for D times, Theres a OUTPUT pulse through Cd .One circulation end.【Programming】library ieee; use ieee.std_logic_1164.all; use ieee.std_logic_unsigned.all; entity bc3 is port(CLK:in std_logic; D:IN std_logic_vector(7 downto 0); Q:buffer std_logic_vector(7 downto 0); Cd:out std_logic;EN:in std_logic); end; architecture ONE of bc3 is begin process(CLK) begin if CLKevent and CLK=1and EN=0then Q=D; if QD-1 then Q=Q+1; else Q=00000000; end if