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1、1DigitalLogicDesignandApplicationLecture#21CounterUESTC,Spring201325.DecodingBinary-CounterStatesThedecoderoutputsmaycontain“glitch”onstatetransitionswheretwoormorecounterbitschange.——functionhazard3DecoderWaveformsThedecoderoutputsmaycontain“glitch”onstatetransitionswheretwoor
2、morecounterbitschange.——functionhazard012345670124Glitch-freeoutputs8-bitregisterRegisteredoutputsdelayedbyoneclocktick.56.CascadingCounter(synchronous)CLOCKRESET_LLOAD_LCNTEND0D1D2D3Q4Q5Q6Q774x16374x16374x162BCDcounter74x162个位十位计数范围:0~25599两个芯片同步工作Free-running6RCO=QD∙QC∙QB∙QARC
3、O-in→111111QAQBQCQD0Counting↓Hold↓低片计数器每计满一个周期后,高片计一个数。76.CascadingCounter(asynchronous)CLOCKRESET_LLOAD_LCNTEND0D1D2D3Q4Q5Q6Q774x16374x163思考:利用低位的进位控制高位的时钟行不行?74x16374x163CNTEN——两个芯片异步工作。Free-running87.Modulo-mCounterUseSSIdevices——statemachinedesignusinggatesandflip-flopsUseMSI
4、counter——realizeamodulo-mcounterusingann-bitbinarycounterConsidertwocasesm<2nm>2nUsingclearorloadinputsCascadingCounter((a)synchronous)9Designamodulo-11counterusinga74x163——m<2nClearafterCount=(1010)B=(10)D0,1,2,3,4,5,6,7,8,9,10S0S1S2S3S4S12S11S10S9S8S7S6S5S13S14S15CLR_Lasserte
5、dNote:modulo-11↔ClearafterCount=(10)D→0,1,2,3,…1010Quiz:如果是74x161(异步清零)可以这样连接吗?——利用1011状态异步清零,会出现“毛刺”Designamodulo-11counterusinga74x163——m<2nClearafterCount=(1010)B=(10)D0,1,2,3,4,5,6,7,8,9,10,0,1,2,3,…0101不可,1010状态将会以“毛刺”形态出现,造成少计一个数的行为。11074X16111Designamodulo-11counterusinga7
6、4x163Load5(0101)2afterCount=155,6,7,8,9,10,11,12,13,14,15S0S1S2S3S4S12S11S10S9S8S7S6S5S13S14S15——m<2nLD_Lasserted→5,6,…1512Designamodulo-11counterusinga74x163Load0101(5)afterCount=155,6,7,8,9,10,11,12,13,14,15,5,6,…——m<2n1010清零法需要额外设计进位输出端,置数法不需要Quiz:usestates3~13tobuildamodulo-1
7、1counter?13Usinga74x163asaexcess-3counterS0S1S2S3S4S12S11S10S9S8S7S6S5S13S14S15S30011S121100→Countingfrom3to12LD_Lasserted141100Usinga74x163asaexcess-3counterCountingfrom3to12001115CLKQ0Q1Q2Q3COUNT34567891011123Q310,dutycycle=50%16Modulo-mCounter(m>2n)Cascade,thenclearorloadDesi
8、gnamodulo-193counterusing74x163Cascadetw