异步十六进制加法器.ewb

Multisim2001软件的仿真电路实例261例

Electronics Workbench Circuit File
Version: 5
Charset: ANSI
Description: 
"         对于同步计数器，由于各个状态之间的转换均用同一个CP控制，在CP跳变时刻同时决定各个输出状态是否改变。所以，在分析、设计的过程中，仅仅将CP作为状态转换的时间因素来考虑。而在异步时序电路的分析和设计中，首先应考虑的是该级触发器的CP是否有跳变。触发器有了其CP的跳变才能考虑其输出状态的变化，如若没有其CP合适跳变到达，触发器只能处在保持状态。该图是由4级JK触发器构成，各级触发器的激励输入J、K均为1，为计数触发器（T`）型结构。只要CP下跳，其状态就要翻转。电路是异步的，只有CP1有计数输入脉冲CP控制，后三级的CP时钟信号均为前一级的输出端Q。只有在前级的输出由状态1变为状态0时，后级的状态才会改变。触发器的状态方程为：" 
"   			        Q4^n+1=[Q4^n]`	Q3下跳沿有效" 
"			        Q3^n+1=[Q3^n]`	Q2下跳沿有效" 
"			        Q2^n+1=[Q2^n]`	Q1下跳沿有效" 
"			        Q1^n+1=[Q1^n]`	CP下跳沿有效" 
"         如果将 Q4^nQ3^nQ2^nQ1^n=0000作为初始状态，CP1（CP）下跳使Q1翻转，即 Q1^n+1=1。由于Q1是由0变1，使CP2（Q1）上跳，因此 Q2^n+1= Q2^n=0保持不变。Q2不变化使CP3、CP4都没有下跳，Q3、Q4都不变化。所以 Q4^nQ3^nQ2^nQ1^n的下一状态为0001。当 Q4^nQ3^nQ2^nQ1^n=0001，CP1（CP）脉冲使Q1发生翻转Q1^n+1=0。CP2（Q1）由1变为0的下跳使  Q2^n+1=[Q2^n]`=1，CP3（Q2）的上跳使Q3和Q4仍保持原状态不变，则Q4^nQ3^nQ2^nQ1^n的下一状态为0010，其余状态转换可以类推。当电路为 Q4^nQ3^nQ2^nQ1^n=1111状态时，再有一个CP来，其下跳使Q1^n+1翻转为0，Q1下跳使Q2^n+1翻转为0，Q2下跳又使Q3^n+1翻转为0，Q3下跳又使Q4^n+1翻转为0，电路回到 Q4^nQ3^nQ2^nQ1^n=0000的初始状态，完成一次状态的循环，电路实现了二进制加法计数。异步二进制加法计数器的特点是电路结构简单，但速度慢。随着位数的增加，计数器从受时钟触发到稳定状态的建立，时延也大大增加。" 
"        4位异步二进制计数器状态转移表：" 
"           序			原状态				次状态" 
"          号		Q4	Q3	Q2	Q1	Q4	Q3	Q2	Q1" 
"          0		 0	 0	 0	 0	 0	 0	 0	 1	 " 
"          1		 0	 0	 0	 1	 0	 0	 1	 0" 
"          2		 0	 0	 1	 0	 0	 0	 1	 1" 
"          3		 0	 0	 1	 1	 0	 1	 0	 0" 
"          4	 	 0	 1	 0	 0	 0	 1	 0	 1" 
"          5	 	 0	 1	 0	 1	 0	 1	 1	 0" 
"          6	 	 0	 1	 1	 0	 0	 1	 1	 1" 
"          7		 0	 1	 1	 1	 1	 0	 0	 0" 
"          8		 1	 0	 0	 0	 1	 0	 0	 1	 " 
"          9		 1	 0	 0	 1	 1	 0	 1	 0" 
"          10		 1	 0	 1	 0	 1	 0	 1	 1" 
"          11		 1	 0	 1	 1	 1	 1	 0	 0" 
"          12	 	 1	 1	 0	 0	 1	 1	 0	 1" 
"          13	 	 1	 1	 0	 1	 1	 1	 1	 0" 
"          14	 	 1	 1	 1	 0	 1	 1	 1	 1" 
"          15		 1	 1	 1	 1	 0	 0	 0	 0  " 

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