数据选择器的工作原理:数据选择器就是在数字信号的传输过程中,从一组数据中选出某一个来送到输出端,也叫多路开关(下图所示)
【比如输入了4个信号,我只要其中一个,那我就需要通过控制端进行选择】
以双四选一数据选择器进行说明,内部电路如下图:
以其中一个进行说明:
D 10 D 13 D_{10} ~D_{13} D10 D13 :数据输入端;
A 1 , A 0 A_{ 1} ,A_{ 0} A1,A0 :为选通地址输入端;
Y 1 Y _{1} Y1 :输出端;
S 1 S _{1} S1 :附加控制端。当 S 1 ’ = 0 , S 1 = 1 S_{ 1 }’=0,S_{ 1 }=1 S1’=0,S1=1时
备注:TG为传输门,以TG1为例子,当上边是高电平,下边是低电平,则TG1导通,信号可以通过
由电路写出输出端逻辑式:
从后面写起
Y 1 = S 1 X Y_{1}=S_{1}X Y1=S1X
X = A 1 ′ X 1 + A 1 X 2 X = A_{1}'X_{1}+A_{1}X_{2} X=A1′X1+A1X2
x 1 x_{1} x1有两种看法,一种是两个传输门只有一个能通,而前后两个非门又相互抵消,所以 X 1 = D 10 A 0 ′ + D 11 A 0 X_{1}=D_{10}A_{0}'+D_{11}A_{0} X1=D10A0′+D11A0
另外一种直接通过逻辑表达式化简:
X 1 = ( D 10 ′ A 0 ′ + D 11 ′ A 0 ) ′ X_{1}=(D_{10}'A_{0}'+D_{11}'A_{0})' X1=(D10′A0′+D11′A0)′
= ( D 10 ′ A 0 ′ ) ′ ( D 11 ′ A 0 ) ′ =(D_{10}'A_{0}')'(D_{11}'A_{0})' =(D10′A0′)′(D11′A0)′
= ( D 10 + A 0 ) ( D 11 + A 0 ′ ) =(D_{10}+A_{0})(D_{11}+A_{0}') =(D10+A0)(D11+A0′)
= D 10 D 11 + A 0 A 0 ′ + D 10 A 0 ′ + D 11 A 0 =D_{10}D_{11}+A_{0}A_{0}'+D_{10}A_{0}'+D_{11}A_{0} =D10D11+A0A0′+D10A0′+D11A0
= D 10 A 0 ′ + D 11 A 0 =D_{10}A_{0}'+D_{11}A_{0} =D10A0′+D11A0
【本处化简运用了摩根定律等常用公式】
同理:
X 2 = A 0 ′ D 12 + A 0 D 13 X_{2}=A_{0}'D_{12}+A_{0}D_{13} X2=A0′D12+A0D13
所以:
Y 1 = S 1 ( A 1 ′ ( D 10 A 0 ′ + D 11 A 0 ) + A 1 ( A 0 ′ D 12 + A 0 D 13 ) ) Y_{1}=S_{1}( A_{1}'(D_{10}A_{0}'+D_{11}A_{0})+A_{1}(A_{0}'D_{12}+A_{0}D_{13})) Y1=S1(A1′(D10A0′+D11A0)+A1(A0′D12+A0D13))
= S 1 ( D 10 A 0 ′ A 1 ′ + D 11 A 0 A 1 ′ + D 12 A 1 A 0 ′ + D 13 A 1 A 0 ) =S_{1}( D_{10}A_{0}'A_{1}'+D_{11}A_{0}A_{1}'+D_{12}A_{1}A_{0}'+D_{13}A_{1}A_{0}) =S1(D10A0′A1′+D11A0A1′+D12A1A0′+D13A1A0)
【本次分析采用从电路出发进行分析的方法,有别于之前从功能设计电路】
“八选一”需要3位地址线,所以 S S S充当第三位地址线,同一时间,两片的 S S S相反,只选择其中一片
所以输出端逻辑表达式为:
因为 S = 1 S=1 S=1时,四选一输出端的逻辑表达式为:
Y = D 10 A 0 ′ A 1 ′ + D 11 A 0 A 1 ′ + D 12 A 1 A 0 ′ + D 13 A 1 A 0 Y=D_{10}A_{0}'A_{1}'+D_{11}A_{0}A_{1}'+D_{12}A_{1}A_{0}'+D_{13}A_{1}A_{0} Y=D10A0′A1′+D11A0A1′+D12A1A0′+D13A1A0
所以我们需要将逻辑函数化成最小项之和的形式
Y = A B ′ C + A B ′ C ′ + A B C ′ + A B ′ C ′ + A ′ B ′ C ′ + A B C Y=AB'C+AB'C'+ABC'+AB'C'+A'B'C'+ABC Y=AB′C+AB′C′+ABC′+AB′C′+A′B′C′+ABC
假如我们选取 A , B A,B A,B为地址输入线,则再化简:
Y = 1 A B ′ + 1 A B + C ′ A ′ B ′ + 0 A ′ B Y=1AB'+1AB+C'A'B'+0A'B Y=1AB′+1AB+C′A′B′+0A′B
令 A 1 = A , A 0 = B A_{1}=A,A_{0}=B A1=A,A0=B,
则 D 11 = 1 , D 12 = 1 , D 10 = C ′ , D 12 = 0 D_{11}=1,D_{12}=1,D_{10}=C',D_{12}=0 D11=1,D12=1,D10=C′,D12=0
【王老师的课件是假设BC为地址输入线,道理相同】
八选一的设计道理相同,不做赘述
试用双4选1数据选择器74HC153构成全减器,设A为被减数,B为减数, C I C_{ I} CI 为低位的借位,D为差, C O C_{ O} CO 为向高位的借位
分析:三个输入变量,两个输出,一个输出一片四选一是够的,需要两片
【这里判断注意一下,不是4+1=5蛤!】
问题的关键就是写出逻辑表达式,其他就像上面一样啦