📄 异步二进制加法器.ewb
字号:
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3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2进制加法计数。异步2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 "
EncryptionType: 2
UsingVectorGraphics: 0
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