《电子线路练习与单元测试(作业薄式)》参考答案.pdf 62页

'参考答案第1章半导体器件1．1半导体的基本特性练习题一、判断题１．×　２．×　３．√　４．√　５．√　６．×　７．×　８．×二、填空题１．导体绝缘体２．锗硅３．热敏性光敏性掺杂性４．热敏光敏５．等于小于大于６．电子电流空穴电流三、选择题１．Ｃ，Ｄ　　２．Ｂ　　３．Ｂ　　４．Ａ　　５．Ｃ四、简答题１．半导体具有掺杂性、热敏性和光敏性。掺杂性是指在纯净的半导体物质中适当地掺入微量杂质，则其导电能力将会成百万倍地增加。利用这一特性可制造出各种不同用途的半导体器件，如晶体二极管、晶体三极管等。热敏性是指半导体的电阻率会随温度升高而降低。利用半导体的热敏性可制成自动控制用的热敏元件（热敏电阻）。光敏性是指半导体受到光照时，其导电能力随之增强。利用半导体的光敏性可制成自动控制用的光电器件，如光电池、光电管和光敏电阻等。２．半导体可分为纯净半导体、Ｐ型半导体和Ｎ型半导体。纯净半导体又称为本征半导体，是指不含杂质的硅或锗晶体。Ｐ型半导体是在四价的本征半导体中掺入了三价元素，譬如极小量（一千万之一）的铟，合成的晶体。由于三价元素进入四价元素中，因此，晶体结构中就产生了１个空穴。由于少１个价电子，因此空穴带正电。Ｐ型的“Ｐ”正是取“Ｐｏｓｉｔｉｖｅ（正）”一词的第一个字母。Ｎ型半导体是把五价的元素，譬如砷，掺入四价的本征半导体中，因此，产生多余１个价电子，价电子显负电性。Ｎ是从“Ｎｅｇａｔｉｖｅ（负）”中取的第一个字母。３．这一说法不正确。Ｎ型半导体的价电子挣脱原子核的束缚就成为自由电子，而少了价电子的原子对应形成正离子。对于整个Ｎ型半导体来说，每产生一个自由电子，就对应形成一个正离子，自由电子与正离子数量相等，故成电中性。149 ·························································································································参考答案················1．2晶体二极管练习题一、判断题１．×　２．×　３．×　４．√　５．√　６．×二、填空题１．高低导通截止２．塑料负正３．击穿４．单向导电性最大整流电流最高反向工作电压反向饱和电流最高工作频率５．０．５　　０．７６．０．２　　０．３７．锗硅三、选择题１．Ｃ　　２．Ａ　　３．Ｂ　　４．Ｃ　　５．Ｄ　　６．Ｂ四、分析题１．IＤ＝０ｍＡ，VＤ＝１０Ｖ。２．VＡ＝５Ｖ，二极管导通，VＢ＝４．３Ｖ；VＡ＝－５Ｖ，二极管截止，VＢ＝０Ｖ。五、简答题测其正向电阻越小越好，反向电阻越大越好。1．3特殊二极管练习题一、判断题１．×　２．√　３．×　４．×　５．√　６．√二、填空题１．相同陡直２．Iｚｍａｘ～Iｚｍｉｎ３．０．７Ｖ　　７．５Ｖ４．几至几十毫安　　１．５～２．５Ｖ５．大小６．减小增大７．光电流８．光电二极管三、选择题１．Ｃ　　２．Ｄ　　３．Ａ　　４．Ｄ　　５．Ｂ　　６．Ｂ四、简答题１．稳定电压为７Ｖ，稳压二极管利用反向击穿时，通过管子的电流在很大范围内变化，而管子两端电压变化很小的特性来起稳压的作用。２．图１３２（ａ）中，VＺ＝１０．６Ｖ；图１３２（ｂ）中，VＺ＝６Ｖ。倡３．ＬＥＤ光源的主要有以下几个特点：150 ················参考答案·························································································································（１）低压电压：ＬＥＤ使用低压电源，供电电压在６～２４Ｖ之间，根据产品不同而异，所以它是一个比使用高压电源更安全的电源，特别适用于公共场所。（２）耗能低：消耗能量比相同光效的白炽灯减少８０％。（３）适用性强：每个单元ＬＥＤ小片是３～５ｍｍ的正方形，面积很小，所以可以制备成各种形状的器件，并且适合于易变的环境。（４）稳定性强：１０万小时，光衰为初始值的５０％。（５）颜色多变：改变电流可以变色，发光二极管方便地通过化学修饰方法，调整材料的能带结构和带隙，实现红黄绿兰橙多色发光。如小电流时为红色的ＬＥＤ，随着电流的增加，可以依次变为橙色，黄色，最后为绿色。1．4晶体三极管练习题一、判断题１．×　２．×　３．√　４．×　５．×　６．√二、填空题１．发射极基极集电极２．发射结加正向偏压集电结加反向偏压３．iＣ４．共发射极交流电流放大系数５．集电基发射三、选择题１．Ａ　　２．Ａ　　３．Ｂ　　４．Ｄ　　５．Ｃ　　６．Ｂ四、简答题１．３—三极管，Ｄ—ＮＰＮ型硅材料，Ｇ—高频小功率管，１００—序号，Ａ—规格号。２．采用金属封装，由于金属外壳能安装于散热器上，便于散热，所以适用于大功率场合使用。五、计算题１．（１）IＣ＝４．８ｍＡ。－（２）β＝２４。２．IＣ＝１．１８ｍＡ，IＥ＝１．２ｍＡ。1．5场效晶体管练习题一、判断题１．√　２．×　３．√　４．√　５．×二、填空题１．输入阻抗高噪声低热稳定性好耗电省２．栅源电压VＧＳ３．漏源漏极栅源４．栅源漏极漏源５．栅源漏极151 ·························································································································参考答案················６．刚开始截止三、选择题１．Ａ　　２．Ｂ　　３．Ａ　　４．Ｂ　　５．Ｃ　　６．Ｃ　　７．Ｃ　　８．Ｂ四、分析题１．（１）因为VＧＳ由负到正时，IＤ由大到小，所以它是绝缘栅Ｐ沟道耗尽型场效晶体管。（２）夹断电压VＧＳ（ｏｆｆ）是IＤ≈０时的VＧＳ值，由图中可知VＧＳ（ｏｆｆ）≈３Ｖ。（３）IＤＳＳ是VＧＳ＝０时的IＤ值，由图中可知IＤＳＳ≈－８ｍＡ。２．（１）由转移特性曲线可知，vＧＳ的电压为负极性，也就是结型场效晶体管的栅极ｇ接电源的负极，源极ｓ接电源的正极，且正常工作时应加反偏电压，因此，可判断与源极连接的是Ｎ型半导体，即Ｎ沟道结型场效晶体管。（２）由转移特性曲线可求得：vＧＳ＝０时，iＤ值是３ｍＡ，即IＤＳＳ＝３ｍＡ；iＤ＝０时的vＧＳ值为－４Ｖ，即VＧＳ（ｏｆｆ）＝－４Ｖ。第1章半导体器件单元测试卷一、判断题１．×　２．√　３．×　４．√　５．√　６．×　７．×　８．√　９．×　１０．×　１１．×１２．√　１３．√　１４．√　１５．×二、填空题１．ＰＮ２．小大３．正向反向单向导电４．定值　　０．７　　０．３５．负反向６．０．７Ｖ　　５Ｖ７．反向正向８．降低增大９．基极集电极１０．电流放大倍数下降到正常值的２／３以下１１．电压栅漏１２．绝缘栅结型ＮＰ１３．二极三极三、选择题１．Ｄ　　２．Ｄ　　３．Ｂ　　４．Ｃ　　５．Ｂ　　６．Ｃ　　７．Ｂ　　８．Ｃ　　９．Ｂ　　１０．Ｃ１１．Ａ１２．Ｄ１３．Ａ１４．Ｂ１５．Ｃ四、简答题１．ＮＰＮ型硅管，第一只脚为ｂ极，第二只脚为ｅ极，第三只脚为ｃ极。２．（１）VＧＳ（ｏｆｆ）＝－３Ｖ。（２）IＤＳＳ≈５ｍＡ。152 ················参考答案·························································································································（３）VＧＳ曲线簇的范围是VＧＳ≤０，因此，可判定为Ｎ沟道结型场效应管。３．（１）反向击穿后，稳压二极管的特性曲线比一般二极管陡直，反向电流急剧变化。（２）在反向击穿情况下，普通二极管会损坏，由于稳压二极管是特殊工艺制造的硅二极管，只要反向电流不超过极限电流，管子工作在击穿区并不会损坏，属可逆击穿。五、计算题１．VR＝６．３Ｖ，I＝６．３ｍＡ。２．图１６（ａ）中，V１截止，VＡＢ＝１２Ｖ；图１６（ｂ）中，V２导通，VＡＢ＝１４．７Ｖ。３．β＝８０。４．图１７（ａ）中，vＯ＝７Ｖ；图１７（ｂ）中，vＯ＝０．７Ｖ。第2章放大电路基础2．1三极管基本放大电路练习题一、判断题１．√　２．×　３．√　４．×　５．×　６．√　７畅√二、填空题１．将输入的微弱电信号放大成幅度足够大的输出信号２．放大输出波形失真３．交流信号未输入时，电路中的电压、电流都不变化４．IＢＱIＣＱVＣＥＱ５．截止失真６．饱和失真７．发射极８．基极三、选择题１．Ｄ　　２．Ｂ　　３．Ａ　　４．Ａ　　５．Ａ　　６．Ｃ四、作图题图Ａ２１153 ·························································································································参考答案················五、简答题输出电压波形属于截止失真。产生截止失真的原因是：IＢＱ偏小时，静态工作点偏低。三极管工作在截止区附近，在输入电压vｉ的负半周时，三极管的发射结将在一段时间内处于反向偏置，造成iＣ负半周、vｏ的正半周相应的波顶被削去。放大电路出现截止失真时，可适当减小偏置电阻Rｂ，将偏置电流IＢＱ增大，则可消除截止失真。2．2放大电路的分析方法练习题一、判断题１．√　２．×　３．√　４．√　５．×　６．×二、填空题１．放大倍数输入电阻输出电阻２．电压增益GV＝２０ｌｇAV３．大４．静态工作点放大倍数输入电阻输出电阻５．直流电流通过三、选择题１．Ｂ　　２．Ｂ　　３．Ｃ　　４．Ｄ　　５．Ｃ　　６．Ａ四、综合题（１）直流通路　　　　　　　　　　　　　　（３）交流通路图Ａ２２图Ａ２３（２）IＢＱ＝３７畅７μＡ，IＣＱ＝３畅８ｍＡ，VＣＥＱ＝－４畅４Ｖ。（４）AV＝－１３５，Rｉ＝０．８５ｋΩ，Rｏ＝２ｋΩ。（５）Rｂ＝２００ｋΩ。（６）此为ＰＮＰ型的三极管，图２１１（ｂ）所出现的失真为截止失真，调整Rｂ使之减少，通过增大IＢ，可消除失真。154 ················参考答案·························································································································2．3工作点稳定放大电路练习题一、判断题１．√　２．√　３．√　４．×　５．×二、填空题１．增大增大减低２．大些小些３．Rｂ２RｅCｅ４．稳定静态电流５．I２≥１０IＢＱVＢＱ≥３VＢＥＱ６．VＣＥＱIＢＱ（或IＣＱ）三、选择题１．Ａ　　２．Ｃ　　３．Ｂ　　４．Ｂ　　５．Ａ四、分析题该电路是分压式偏置放大电路，其电路原理图如图Ａ２４所示。稳定静态工作点的过程如下：温度t↓→集电极电流IＣＱ↓→发射极电流IＥＱ↓→发射极电位VＥＱ↓→发射结偏压VＢＥＱ↑→基极电流IＢＱ↑→集电极电流IＣＱ↑。五、计算题（１）IＥＱ＝３．３ｍＡ，IＢＱ＝６６μＡ，VＣＥＱ＝２．１Ｖ。（２）AV＝－１４２．５，Rｉ＝０．７ｋΩ，Rｏ＝２ｋΩ。（３）AV＝－７１。图Ａ２４2．4场效晶体管放大电路练习题一、判断题１．×　２．√　３．√　４．√　５．√二、填空题１．电压输入电压VＧＳ偏置分压栅极稳定静态工作点２．源极旁路电容消除Rｓ对交流信号的衰减作用隔直流耦合交流信号３．放大倍数４．小大５．分压式自偏压式三、选择题１．Ｄ　　２．Ａ　　３．Ｃ　　４．Ａ　　５．Ｃ四、作图题１．分压偏置放大电路155 ·························································································································参考答案················图Ａ２５　　２．自偏压放大电路图Ａ２６2．5多级放大电路练习题一、判断题１．√　２．×　３．√　４．×　５．√　６．√二、填空题１．耦合２．阻容耦合变压器耦合直接耦合３．信号源负载输入信号（信号源）４．幅频特性曲线156 ················参考答案·························································································································５．通频带６．上限频率下限频率７．fＨ－fＬ三、选择题１．Ｂ　　２．Ｄ　　３．Ｃ　　４．Ａ　　５．Ｃ四、分析题１畅（１）放大电路级间采用变压器耦合方式。（２）Ｖ２管组成的放大电路采用固定偏置，Ｖ３管组成的放大电路采用分压式偏置。（３）放大电路的结构框图如图Ａ２７所示。图Ａ２７２．fＬ＝５０Ｈｚ，fＨ＝１５００Ｈｚ，BW＝１４５０Ｈｚ。图Ａ２８第2章放大电路基础单元测试卷一、判断题１．×　２．×　３．√　４．√　５．×　６．√　７．×　８．×　９．√　１０．√　１１．×１２．×　１３．√　１４．×　１５．×二、填空题（30分，每个填空1分）１．静态工作点２．直流交流直流成分交流成分３．偏置电阻４．４０５．大容量电容短路短路电容开路６．放大倍数电压放大倍数电流放大倍数功率放大倍数７．大信号电流８．饱和截止９．分压式偏置电路集电极基极偏置电路157 ·························································································································参考答案················１０．降低１１．提高放大倍数１２．大些１３．交流直流１４．输入输入１５．集电极电阻Rｃ三、选择题１．Ｄ　　２．Ａ　　３．Ｃ　　４．Ｃ　　５．Ｂ　　６．Ｄ　　７．Ａ　　８．Ｂ　　９．Ｄ　　１０．Ｂ四、计算题１．（１）IＢＱ≈４０μＡ，IＣＱ≈２ｍＡ，VＣＥＱ≈６Ｖ。（２）AV≈－７８，Rｉ≈０．９６ｋΩ，Rｏ＝３ｋΩ。（３）Rｂ＝４５０ｋΩ。２．（１）IＣＱ≈３．５ｍＡ，IＢＱ≈１２０μＡ，VＣＥＱ≈３．２５Ｖ。（２）AV≈－２８，Rｉ≈０．５ｋΩ，Rｏ＝１ｋΩ。五、作图题１．原理电路图实物接线图图Ａ２９图Ａ２１０２．直流通路交流通路图Ａ２１１图Ａ２１２158 ················参考答案·························································································································３．温度t↓→集电极电流IＣＱ↓→三极管压降VＣＥＱ↑→基极电流IＢＱ↑→集电极电流IＣＱ↑六、分析题图２６（ａ）Rｃ两端的输出信号被C２交流短路，无放大电压输出。图２６（ｂ）发射结无正偏压，三极管处于截止状态。图２６（ｃ）电源极性接反，三极管无法正常工作。第3章常用放大器3．1集成运算放大器练习题一、判断题１．√　２．×　３．×　４．×　５．√　６．√　７．×　８．×二、填空题１．输入级中间级输出级偏置电路为各级电路提供静态工作点２．AＶＤ／AＶＣ３．AＶＤ＝∞　　rｉｄ＝∞　　rｏ＝０　　BW＝∞４．反相输入端电位零５．反相放大器同相放大器差分放大器６．加法运算电路减法运算电路７．加强对电源的滤波调整电路板的布线结构避免电路接线过长８．输入信号为零时，要使输出为零，在输入端所加的补偿电压三、选择题１．Ｃ　　２．Ｂ　　３．Ａ　　４．Ａ　　５．Ｄ四、作图题１．图Ａ３１vＩ１vＩ２２．该电路是加法运算电路，vＯ＝－Rｆ＋。R１R２159 ·························································································································参考答案················图Ａ３２３．图Ａ３３４．反相器如图Ａ３４（ａ）所示，电压跟随器如图Ａ３４（ｂ）所示。图Ａ３４3．2负反馈放大器练习题一、判断题１．×　２．√　３．√　４．√　５．√　６．√　７．×　８．×二、填空题１．基本放大电路反馈电路输入输出２．负反馈正反馈３．瞬时极性法４．静态工作点放大倍数５．串联负反馈电流负反馈电压负反馈三、选择题１．Ｃ　　２．Ｂ　　３．Ａ　　４．Ａ　　５．Ｃ四、分析题１．图３２１（ａ）中，Rｅ１为第一级的电流串联负反馈；Rｅ２、Cｅ２为第二级的直流电流串联负反160 ················参考答案·························································································································馈；Rｅ３为第三级的电流串联负反馈；Rｆ为第三级对第一级的电流串联负反馈。电路特点：输出电流稳定，输入电阻增大，输出电阻增大。图３２１（ｂ）中，Rｅ１为第一级的电流串联负反馈；Rｅ２、Cｅ２为第二级的直流电流串联负反馈；Rｆ为第二级对第一级的电压串联负反馈。电路特点：输出电压稳定，输入电阻增大，输出电阻降低。２．３２２图（ａ）中，R２为电压串联负反馈，图３２２（ｂ）中，R３、R４为电压并联负反馈。五、计算题AV＝６０，AVＦ＝２０，F＝０．０３３。3．3低频功率放大器练习题一、判断题１．√　２．×　３．×　４．×　５．√　６．√　７．×　８．√二、填空题１．甲类乙类甲乙类２．截止区边缘３．PＣＭV（ＢＲ）ＣＥＯIＣＭ散热４．特性相等相反轮流５．交越６．自激三、选择题１．Ｂ　　２．Ｄ　　３．Ａ　　４．Ａ　　５．Ｂ四、分析题（１）ＯＴＬ电路的工作方式为甲乙类方式。（２）RＰ２、Ｖ１与Ｖ２为Ｖ３、Ｖ４提供一静态偏置电压，使之处于微导通状态，可消除交越失真。Ｖ１与Ｖ２还具有温度补偿作用。（３）VＥ１＝７．５Ｖ。（４）调整电位器RＰ１，可使Ｖ３管射极电位VＥ１为７．５Ｖ；调整电位器RＰ２，可使Ｖ３、Ｖ４有合适的射极工作电流。（５）Pｏｍ≈３．５Ｗ。五、作图题图Ａ３５161 ·························································································································参考答案················3．4谐振放大器练习题一、判断题１．√　２．×　３．×　４．×　５．√二、填空题１．分散选频集中选频２．放大电路谐振回路选频放大高频放大中频放大３．单谐振双谐振通频带选择性４．集成放大电路集中滤波器陶瓷声表面波５．互感耦合电容耦合三、选择题１．Ｄ　　２．Ｂ　　３．Ｃ　　４．Ａ　　５．Ｃ四、简答题１．解决小信号谐振放大器通频带与选择性之间矛盾的途径主要有：（１）采用双调谐谐振回路。（２）采用既有较好选择性又有较宽通频带的谐振元器件，如声表面波滤波器。２．该谐振放大器使用的是三端陶瓷滤波器。工作原理：输入信号经０．０１μＦ电容C１耦合到Ｖ１管的基极进行放大，放大后的信号从Ｖ１管ｃ、ｅ极输出，经三端陶瓷滤波器进行选频滤波，然后送到Ｖ２管进行放大，输出信号通过０．０１μＦ的输出耦合电容C２输出。３．３４３图（ａ）为互感耦合双调谐放大器，两个调谐回路之间是靠线圈之间的电磁感应来实现信号的传递，耦合程度主要通过调整线圈的位置或调整电感的磁心位置进行调整；图３４３（ｂ）为电容耦合双调谐放大器。两个调谐回路之间的信号传递是靠电容的耦合来实现，耦合程度主要通过调整耦合电容的容量来调整。第3章常用放大器单元测试卷一、判断题１．√　２．√　３．√　４．×　５．√　６．×　７．×　８．×　９．√　１０．√　１１．×１２．√　１３．×　１４．×　１５畅√二、填空题１．集成运算放大器负反馈放大器功率放大器谐振放大器２．差模输入信号共模输入信号３．差模放大倍数４．共模差模５．放大抑制６．降低提高减小展宽输入电阻输出电阻７．串联并联８．１　　高小电流162 ················参考答案·························································································································９．双电源改为单电源直接耦合改为电容耦合１０．双峰状１１．体积小重量轻耐振动选频特性好三、选择题１．Ｂ　　２．Ｄ　　３．Ｄ　　４．Ａ　　５．Ｄ　　６．Ｂ　　７．Ａ　　８．Ｃ　　９．Ｄ　　１０．Ｃ四、计算题１．VＯ＝－３Ｖ。２．vＯ＝５．１Ｖ。五、分析题１．图３５（ａ）所示为电压并联负反馈，图３５（ｂ）所示为电流并联负反馈，图３５（ｃ）所示为电压串联负反馈，图３５（ｄ）所示为电压串联负反馈，图３５（ｅ）中，Rｅ１对第一级起电流串联负反馈，Rｅ２对第二级起直流电流串联负反馈，Rｅ３对第三级起电压串联负反馈，Rｆ是三级的电压串联负反馈。２．C４起隔直流、耦合输出信号的作用，另一个功能是充电后，能为Ｖ５管提供工作电源；C２和R２为自举元件，用于克服输出信号的半顶失真；RＰ１是电压并联负反馈的电阻，稳定静态工作点和提高输出信号的稳定性，调RＰ１可使输出端为VＣＣ／２；RＰ２可调整输出管的静态电流的大小；Ｖ１和Ｖ２为输出管提供偏置电压，并有温度补偿作用。３．（１）谐振频率调在需要放大的输入信号频率f０上。（２）可以减小三极管内部的结电容对LC调谐回路的影响，提高工作频率的稳定度，并使三极管的输出阻抗与LC回路匹配。（３）电容C是LC调谐回路的谐振电容，电容C１是电阻R２的交流信号旁路电容。（４）电阻R４用来降低放大器输出LC调谐回路的品质因数Q，它的作用是加宽放大器的通频带。六、作图题１．R１＝１００ｋΩ，R２＝３３．３ｋΩ，R３＝２５ｋΩ，R４＝１２．５ｋΩ。图Ａ３６163 ·························································································································参考答案················　　２．图Ａ３７第4章直流稳压电源4．1整流电路练习题一、判断题１．×　２．×　３．√　４．√　５．√　６．√二、填空题１．交流直流２．单向导电３．零无穷大４．将交流电网电压变换成整流电路要求的交流电压５．８．５Ｖ６．较高较大较高小三、选择题１．Ｄ　　２．Ｂ　　３．Ａ　　４．Ｃ　　５．Ｂ　　６．Ｃ四、作图题１．略　　２．略　　３．见图Ａ４１五、计算题１．VＬ＝１０．８Ｖ，VＲＭ≥１７Ｖ，IＦＭ＝１０８ｍＡ。２．（１）V２≈２２．２Ｖ。（２）VＲＭ≥３１畅４Ｖ，IＦＭ＝１Ａ。4．2滤波电路练习题一、判断题图Ａ４１１．×　２．√　３．√　４．×　５．√　６．√二、填空题１．交流成分164 ················参考答案·························································································································２．电感电容３．并串４．大大５．输出电压脉动大三、选择题１．Ｄ　　２．Ｂ　　３．Ａ　　４．Ｂ　　５．Ｄ四、作图题略五、计算题１．（１）V２＝１０Ｖ。（２）VＲＭ≥１４．１４Ｖ，IＦＭ≥１２０ｍＡ。２．（１）VＬ＝２４Ｖ。（２）滤波电容选择１５０～２５０μＦ，耐压应大于２８Ｖ。4．3稳压电路练习题一、判断题１．×　２．√　３．×　４．√　５．√　６．√二、填空题１．输入电压负载２．硅稳压二极管三极管集成稳压器３．电流压降电压４．电流调节作用５．固定可调６．输入输出公共７．大低８．开关状态三、选择题１．Ｃ　　２．Ｃ　　３．Ａ　　４．Ａ　　５．Ｂ　　６．Ａ四、计算题（１）VＺ＝５Ｖ；（２）VＬｍｉｎ＝８Ｖ，VＬｍａｘ＝１３．３Ｖ。五、简答题Ｗ７８０５使用时，输入电压vｉ的最大值不超过３０Ｖ，输入端１与输出端３之间的压降不小于３Ｖ。电解电容C１、C２具有滤波作用，起平滑电压作用，C３、C４具有抑制高频干扰信号的作用。第4章直流稳压电源单元测试卷一、判断题１．√　２．×　３．√　４．×　５．×　６．√　７．×　８．√　９．×　１０．×　１１．×165 ·························································································································参考答案················１２．√　１３．×　１４．√　１５．√二、填空题１．稳定直流电整流电路滤波电路稳压电路２．二极管半波整流桥式整流３．２４．１．５５．２６．电容电感复式７．漏电爆裂８．９　　１２９．输出直流电压１０．较小１１．三极管串联１２．正　　１２１３．输入输出调整１４．饱和截止小高三、选择题１．Ｂ　　２．Ｃ　　３．Ｂ　　４．Ｂ　　５．Ａ　　６．Ｄ　　７．Ｂ　　８．Ａ　　９．Ｄ　　１０．Ｃ四、简答题１．（１）二极管Ｖ１断路，此时全波整流变成半波整流，v２正半周波形无法送到RＬ上。（２）二极管Ｖ１短路，此时二极管Ｖ２或变压器二次侧线圈可能烧毁。（３）二极管Ｖ１接反，这样在v２负半周时，变压器二次侧输出直接加到两个导通的二极管Ｖ１、Ｖ２上，造成二次侧线圈或二极管Ｖ１、Ｖ２过流以至烧毁。（４）二极管Ｖ１、Ｖ２极性均接反，此时正常的整流通路均被切断，电路无输出电压，VＬ＝０。（５）将四个二极管Ｖ１～Ｖ４极性均接反。电路图略。２．（１）取样环节：R３、R４、RＰ；比较放大电路：Ｖ２，R１；基准电压：R２、ＶＺ；调整环节：Ｖ１、R１。（２）稳压过程：VＩ↑→VＬ↑→VＢ２↑→VＢＥ２↑→IＢ２↑→IＣ２↑→VＣ２↓VＬ↑←VＣＥ１↑←IＣ１↓←IＢ１↓←VＢＥ１↓五、计算题１．二极管：VＲＭ＞３５．４Ｖ，IＦＭ＞０．２５Ａ；滤波电容：C＝２５０～４１６μＦ，VC＞３５．４Ｖ。２．VＬｍａｘ＝２３．１Ｖ，VＬｍｉｎ＝１１．６Ｖ。六、作图题１．略。２．图Ａ４２166 ················参考答案·························································································································第5章正弦波振荡器5．1正弦波振荡器的基本知识练习题一、判断题１．√　２．√　３．×　４．×　５．×　６．×二、填空题１．能量转换直流电能２．放大电路选频电路反馈网络３．稳定小４．φ＝２nπ　　AVF＞１　　AVF＝１５．RCLC石英晶体６．扰动电压放大器放大选频网络选频７．三极管非线性区域放大倍数三、选择题１．Ｄ　　２．Ｄ　　３．Ａ　　４．Ｃ　　５．Ｂ四、作图题图Ａ５１五、简答题１．正弦波振荡器由放大电路、选频网络、正反馈网络三部分所组成。放大电路———振荡器的核心，将直流电源提供的能量转换成交流信号能量，补充振荡过程中的能量损耗，以获得连续的等幅正弦波。放大电路一般还具有稳幅功能，靠振荡管自身的非线性来稳幅。选频网络———从信号中选出所需的频率，使振荡器产生一个单一频率的正弦波。正反馈网络———将选频网络选出的所需频率信号送回到输入端放大。２．扩音机产生声音啸叫是由于音箱喇叭发出的声音直接或经反射后进入麦克风，这些反馈声经扩音机放大后又从音箱喇叭发出，如此反复，产生类似放大器正反馈引起的自激振荡。引起啸叫的原因很多，如室内环境、使用麦克风的数量以及麦克风和音箱的距离、指向性等等。可以用以下方法来消除扩音机的声音啸叫：（１）调整扩音机的音量电位器，将音量适当降低。（２）调整音箱位置，使之远离麦克风。（３）调整音箱的方向，使之不朝向麦克风说话方向。5．2RC振荡器练习题一、判断题１．×　２．√　３．×　４．√　５．√二、填空题１１１．RC串并联选频vｉ零３２πRC167 ·························································································································参考答案················２．频率调节方便波形失真小３．负温度系数４．电路停振振荡波形失真５．RC振荡器三、选择题１．Ｃ　　２．Ａ　　３．Ｄ　　４．Ｂ　　５．Ｃ　　６．Ｄ四、作图题电路的名称是RC桥式正弦波振荡电路。五、综合题（１）为了满足振幅平衡条件，应使同相放大器AV≥３，RＦ而AV＝１＋，因此，取RＦ＝２２ｋΩ。R１（２）振荡频率的调节范围是１０４８～７５３Ｈｚ。（３）调反馈电阻RＦ，适当将RＦ阻值调大。（４）调反馈电阻RＦ，适当将RＦ阻值调小。图Ａ５２5．3LC振荡器练习题一、判断题１．×　２．√　３．×　４．×　５．√二、填空题１．共基极变压器耦合式负载线圈选频网络发射基L２的匝数两个线圈之间距离振荡频率f０１２．电容三点式三个电极电容支路的三个点选频网络C１C２２πLC１＋C２三、选择题１．Ｃ　　２．Ｂ　　３．Ｄ　　４．Ａ　　５．Ｄ四、作图题图Ａ５３168 ················参考答案·························································································································五、计算题电路为改进型电容三点式振荡器，f０＝３５ｋＨｚ。5．4石英晶体振荡器练习题一、判断题１．√　２．×　３．×　４．√　５．√　６．√二、填空题１．电感电容－３－２２２．平板电容C０几皮法到几十皮法电感L　　１０～１０Ｈ电阻R　　１０Ω　－２－１电容C　　１０～１０ｐＦ１３．纯电阻R２πLC１４．电感CC０２πLC＋C０５．振荡频率稳定度高三、选择题１．Ｃ　　２．Ｄ　　３．Ａ　　４．Ｂ　　５．Ｃ　　６．Ａ四、简答题１．该电路为并联石英晶体振荡器。谐振时，石英晶振工作在电感状态，它与C２、C１电容组成并联谐振选频网络。２．图５４４（ａ）所示电路不能振荡。因为在并联石英晶体振荡电路中，石英晶体只有与谐振电容C１、C２并联才能构成选频网络。图５４４（ｂ）所示电路不能振荡。石英晶体支路构成的是负反馈，不满足振荡的相位平衡条件。图５４４（ｃ）所示电路不能振荡。三极管的集电极没有直流通路，无法正常放大信号。第5章正弦波振荡器单元测试卷一、判断题１．×　２．√　３．×　４．×　５．√　６．×　７．√　８．√　９．√　１０．×　１１．×１２．√　１３．×　１４．√　１５．×二、填空题１．等于相位相同２．选频３．接通电源瞬间产生的扰动电压４．文氏桥同相放大器RC串并联选频网络169 ·························································································································参考答案················５．３６．容易起振稳定输出信号幅度７．变压器耦合式三点式８．fｓ至fｐ频率范围之间fｓ至fｐ频率范围之外９．石英晶体改进式电容三点１０．输出波形是否正常反偏电压小于正常放大时的数值升高三、选择题１．Ｂ　　２．Ｂ　　３．Ｄ　　４．Ｄ　　５．Ｂ　　６．Ｄ　　７．Ｃ　　８．Ｂ　　９．Ｄ　　１０．Ａ四、简答题１．图５２（ａ）所示电路不能振荡，RC串并联选频网络与反相放大器组成的是负反馈，不满足振荡的相位条件。图５２（ｂ）所示电路不能振荡，不是正反馈。图５２（ｃ）所示电路不能振荡，L１将三极管的ｃ极和ｅ极直流短路，三极管不能正常工作。图５２（ｄ）所示电路不能振荡，不是正反馈。图５２（ｅ）所示电路能振荡，电容三点式振荡电路。图５２（ｆ）所示电路能振荡，并联型石英晶体振荡电路。２．①调整上偏置电阻，使电路的放大倍数达到最大；②更换β值较大的三极管；③将L３的匝数增加。３．（１）该电路为并联型石英晶体振荡电路。（２）石英晶体ＸＴ为选频元件，振荡器的振荡频率主要由ＸＴ的固有频率所决定，C２只能对振荡频率作微调。五、计算题１．f０＝３１．８３ｋＨｚ。２．该电路为电感三点式振荡电路。f０ｍａｘ＝４１１ｋＨｚ，f０ｍｉｎ＝１１２．６ｋＨｚ。第6章高频信号处理电路6．1调幅与检波练习题一、判断题１．×　２．√　３．×　４．√　５．√　６．×　７．×　８．√二、填空题１．调幅、调频、调相２．频率变换３．高电平低电平４．大信号放大调幅５．小信号调幅放大６．非线性特性LC谐振回路基极集电极发射极170 ················参考答案·························································································································７．普通调幅波８．抑制载波的双边带９．ＡＭ三、选择题１．Ｂ　　２．Ａ　　３．Ｂ　　４．Ｄ　　５．Ｃ　　６．Ａ四、简答题１．载波信号通过高频变压器Ｔω加到调制管Ｖ的基极，调制电压vΩ通过变压器ＴΩ加到调制管Ｖ的集电极回路中，与电源电压VＣＣ相串联，等效为集电极的时变电源电压。调制管Ｖ工作在过压状态，调谐于载波频率的集电极LC谐振回路上的输出信号即为所需的调幅波。２．（１）该电路属于二极管包络检波电路。（２）RＬ１、C２、C３构成RC—π型滤波器，对截去负半周的调图Ａ６１幅波进行滤波，即可取出调制信号———声音。（３）若检波二极管开路，收音机将收不到声音，因为检波二极管开路不仅检波电路无法工作，而且将使收音机的中频调制信号通路断开。（４）电位器Rｐ用于调整检波器输出电压大小，即调节收音机音量的大小。6．2调频与鉴频练习题一、判断题１．×　２．×　３．√　４．√　５．×二、填空题１．角频率偏移量调制电压灵敏度２．调制深度最大角频偏调制信号频率３．振幅频率４．直接间接５．从调频波中取出调制信号等幅调频波低频调制三、选择题１．Ｂ　　２．Ｃ　　３．Ｄ　　４．Ａ　　５．Ａ四、简答题调频电路的高频等效电路如图Ａ６２所示。图６２２中，R５、Rｐ为变容二极管偏压电阻，为变容二极管提供固定反向偏压；C５为隔直电容，亦为调制信号vΩ的耦合电容；调制信号加于变容二极管两端，对变容二极管进行调制作用，使其等效电容CＤ按一定规律变化。电路构成电感三点式振荡器，由于变容二极管的CＤ容量随调制信号vΩ而变化，因而使LC回路的参数随着变化，即振荡频率亦随调制信号vΩ而改变。所以，该电路输出的是调频信号。R１１为直流负反馈电阻；R１为集电极直流降压电阻；C６、C８为高频振荡图Ａ６２171 ·························································································································参考答案················信号提供通路，对低频调制信号起隔离作用。五、计算题３KｆVΩ２π×２５×１０×２３１．Δfｍ＝＝Ｈｚ＝５０×１０Ｈｚ＝５０ｋＨｚ２π２π３kｆVΩ２π×２５×１０×２mｆ＝＝３＝５０Ω２π×１０３Δfｍ７５×１０２．当调制信号频率为１００Ｈｚ时，mｆ＝＝＝７５０；当调制信号频率为１５ｋＨｚ时，mｆfΩ１００３Δfｍ７５×１０＝＝３＝５fΩ１５×１０6．3变频器练习题一、判断题１．×　２．√　３．√　４．√　５．√二、填空题１．下变频２．载波频率３．混频电路本机振荡选频电路４．三极管变频器输入信号电压vＳ三极管ｂｅ结的非线性特性LC并联谐振回路５．选频网络６．fＬ＋fＳfＬ－fＳ三、选择题１．Ｂ　　２．Ａ　　３．Ｃ　　４．Ｄ　　５．Ａ四、简答题１．就其功能而言，它们是一致的。不同的是变频包括了混频和本机振荡两部分．即电路中的非线性器件本身既能产生本振信号，又能实现频率变换，因此，变频器又称为自激混频器。而混频器则不同，电路中的非线性器件只进行频率变换，其本振信号是由另外的电路产生的，混频器又称为他激混频器。一般对工作性能要求不高的接收机采用自激混频器，要求高的则采用他激混频器。２．（１）电路的功能是将接收的高频信号下变频为中频信号，该电路称为三极管变频电路。（２）L５和C１、C１０组成调谐回路，用于调谐接收电台信号，经电感线圈L６耦合加到三极管Ｖ１的基极与三极管发射极。利用线圈L１和线圈L２的耦合形成正反馈，电路自激产生本机振荡信号，本机振荡信号的频率取决于线圈L２和电容C１６、C９、C５、C８构成的谐振电路。接收信号与本机振荡信号混合产生差频信号，即４６５ｋＨｚ的中频信号，L３、C１１和L４、C１４组成了双谐振回路作为中频选频网络，从输出的混频信号选出其中的４６５ｋＨｚ中频信号送到中频放大电路去。R１、R２为三极管偏置电阻，R４为射极稳定电阻。C４对R２起交流管路作用。第6章高频信号处理电路单元测试卷一、判断题１．√　２．×　３．×　４．×　５．√　６．√　７．×　８．×　９．×　１０．√　１１．√172 ················参考答案·························································································································１２．√　１３．×　１４．√　１５．√二、填空题１．低频调制高频载波信号的振幅振幅调制信号２．调制信号幅度载波信号的频率频率调制信号３．检波鉴频４．波形变换调频调幅波振幅变化５．输出信号电压输入调频波频率６．上变频７．三极管８．另一频率调制规律三、选择题１．Ａ　　２．Ｂ　　３．Ｃ　　４．Ｄ　　５．Ａ　　６．Ａ　　７．Ｄ　　８．Ｂ　　９．Ｄ　　１０．Ｂ四、简答题１．输入信号与输出信号比较，波形包络基本不变，主要的变化在于载频降低了。由此可断定图６２所示的功能电路为下变频电路。２．图６３（ａ）所示电路用于双边带调幅波的检波，vＡＭ是双边带调幅输入信号，vω是与被抑制的原载波同频同相的同步信号。加到二极管Ｖ１上的电压信号是（vＡＭ＋vω），加到二极管Ｖ２上的电压信号是（vＡＭ－vω），经过二极管Ｖ１、Ｖ２检波，从上半部检波器输出解调信号＋vΩ，从下半部检波器输出解调信号－vΩ，所以可从１、２端取出解调信号２vΩ。图６３（ｂ）所示电路用于调频波的鉴频，两个LC谐振回路分别调谐在f０１和f０２上，f０１高于调频波的中心频率fω，f０２低于调频波的中心频率fω，f０１和f０２对称分布fω两侧。调频信号在两个谐振回路产生的调幅—调频波分别为v１和v２，鉴频器的总输出电压为vｏ＝vｏ１－vｏ２，vｏ随f变化的规律就是合成鉴频特性。３．方框１的名称为混频电路，本机振荡的信号与接收的信号，在混频电路的非线性元件的作用下产生和频fＬ＋fＳ、差频fＬ－fＳ和其他频率mfＬ±nfＳ，再由混频电路中选频网络取出所需的频率信号。方框２的名称为检波电路，检波电路的功能是从中频调幅波中取出调制信号———声音信号。五、计算题１．fＬ＝fω＋fＩ＝１２００ｋＨｚ＋４６５ｋＨｚ＝１６６５ｋＨｚ。３３KｆVΩ２π×１０×３３KｆVΩ２π×１０×３２．Δfｍ＝＝Ｈｚ＝３×１０Ｈｚ，mｆ＝＝３＝３。２π２πΩ２π×１０１１３．f０１≈＝Ｈｚ＝１１２畅６ｋＨｚ－３－１２２πL１C１２×３畅１４１０×１０×２００×１０１１f０２≈＝Ｈｚ－３－１２２πL１（C１＋C２）２×３畅１４×１０×１０×（２００＋２００）×１０　＝７９．６ｋＨｚf０１＋f０２１１２畅６＋７９畅６fω＝＝ｋＨｚ＝９６畅１ｋＨｚ２２173 ·························································································································参考答案················枟电子线路枠模拟电路期中考试卷一、单项选择题１．Ｃ　　２．Ａ　　３．Ｂ　　４．Ｄ　　５．Ｃ　　６．Ｃ　　７．Ａ　　８．Ａ　　９．Ｄ　　１０．Ｂ二、多项选择题１．ＢＣＤ　　２．ＢＣＤ　　３．ＡＤ　　４．ＢＣ　　５．ＣＤ　　６．ＡＢＣＤ　　７．ＡＢＣＤ　　８．ＢＣ９．Ｄ１０．ＡＢＣＤ三、填空题１．加正向电压加反向电压　　０．７Ｖ　　０．３Ｖ２．放大区饱和区截止区３．基极集电极微小较大４．可调电阻区放大区击穿区５．IＢＱIＣＱVＣＥＱ６．３Ｖ７．电压串联负反馈８．０．７０７９．０　　∞１０．输出输入１１．截止四、分析计算题Rｂ２１２畅５１．（１）VＢＱ＝VＣＣ＝×１５Ｖ＝３ＶRｂ１＋Rｂ２５０＋１２畅５VＥＱ＝VＢＱ－VＢＥＱ＝３Ｖ－０．７Ｖ＝２．３ＶVＥＱ２畅３ＶIＣＱ≈IＥＱ＝＝≈３ｍＡRｅ０畅７５IＣＱ３ｍＡIＢＱ＝＝＝０．０３ｍＡβ１００VＣＱ＝VＣＣ－IＣＱ（Rｃ＋Rｅ）＝１５Ｖ－３×（２＋０．７５）Ｖ＝６．７５Ｖ２６２６（２）rｂｅ＝３００Ω＋（１＋β）Ω＝３００Ω＋１０１×Ω≈１１７５Ω≈１．１７ｋΩIＥ３Rｉ≈rｂｅ＝１．１７ｋΩR０＝RＣ＝２ｋΩR′Ｌ０畅８６AV＝－β＝－１００×≈－７３γｂｅ１畅１７174 ················参考答案·························································································································（３）去掉Cｅ后，电压放大倍数AV下降，输入电阻rｉ增大。vＩ１vＩ２vＩ３２．（１）vＯ＝－RＦ＋＋R１R２R３＝－２×（０畅２Ｖ＋０畅３Ｖ＋０Ｖ）＝－１ＶvＯ（２）vＩ３＝－＋vＩ１＋vＩ２２０畅３Ｖ＝－＋０畅２Ｖ－０畅４Ｖ２＝０畅０５Ｖ３．图Ｔ１６（ａ）的反馈元件是R２，反馈的类型是电压并联负反馈；图Ｔ１６（ｂ）的反馈元件是R，反馈的类型是电流串联负反馈；图Ｔ１６（ｃ）的交流反馈元件是Rｅ１，反馈的类型是电流串联负反馈。五、作图题１．图ＡＴ１１２．直流通路交流通路图ＡＴ１２175 ·························································································································参考答案················枟电子线路枠模拟电路期末考试卷一、单项选择题１．Ａ　　２．Ｂ　　３．Ｂ　　４．Ａ　　５．Ｃ　　６．Ａ　　７．Ｃ　　８．Ｄ　　９．Ｂ　　１０．Ｄ二、多项选择题１．ＡＣ　　２．Ｂ　　３．ＡＢ　　４．ＣＤ　　５．ＢＣＤ　　６．ＡＣ　　７．Ｂ　　８．ＢＣ９．ＡＣ　　１０．Ａ三、填空题１．集电极　　１２．单向导电３．１．２５Ｗ４．绝缘栅型结型低频跨导gｍ５．LC串联回路电感６．基准电压下V２的直流负载电阻V１的偏置电阻７．整流滤波稳压８．足够大高小散热要好９．高频调幅波调制二极管包络检波双边带调幅检波１０．变频四、作图题１．图ＡＴ２１　　　　　　　　　　　２．电路的名称是ＯＴＬ放大电路，对低频信号起功率放大和输出。图ＡＴ２２176 ················参考答案·························································································································３．自选作图题（１）下变频电路的框图和波形图（２）开关电源结构框图图ＡＴ２３五、分析计算题１．为了稳定输出电压，应引入电压负反馈。反馈电阻Rｆ应接在Ｖ２管的集电极ｃ２与Ｖ１管的发射极ｅ１间。为了稳定输出电流，应引入电流负反馈。反馈电阻Rｆ应接在Ｖ２管的发射极ｅ２与Ｖ１管的基极ｂ１间。２．（ａ）×，（ｂ）√，（ｃ）√。３．（１）VＬ＝１５Ｖ。（２）１脚为输入端，２脚为接地端，３脚为输出端。（３）三端稳压器的输入电压VＩ应选择比输出电压高２～３Ｖ，取VＩ＝１８Ｖ。VＩ１８（４）V２＝＝Ｖ＝１５Ｖ。１畅２１畅２（５）C１为整流电路的低频滤波电容，C２用于滤除稳压集成电路的输入高频干扰信号，C３为用于消除输出电压的波动，并具有消振作用。第7章数字电路基础7．1脉冲与数字信号练习题一、判断题１．×　２．√　３．√　４．×　５．×　６．√　７．×　８．√177 ·························································································································参考答案················二、填空题１．模拟数字连续不连续短暂２．逻辑关系逻辑３．Vｍ伏（Ｖ）tＷ４．DD＝T５．陡峭上升越快６．方波　　５０％三、选择题１．Ａ　　２．Ｃ　　３．Ｂ　　４．Ｄ　　５．Ｃ四、分析题１．正逻辑：D１D２D３D４D５D６D７D８D９D１０＝0101101001。负逻辑：D１D２D３D４D５D６D７D８D９D１０＝1010010110。２．（１）脉冲幅度Vｍ＝（垂直距离ｄｉｖ）×（挡位Ｖ／ｄｉｖ）＝５ｄｉｖ×２００ｍＶ／ｄｉｖ＝１Ｖ。（２）脉冲上升时间tｒ＝０畅９ｄｉｖ×１．００ｍｓ／ｄｉｖ＝０．９０ｍｓ；脉冲下降时间tｆ＝０．９ｄｉｖ×１．００ｍｓ／ｄｉｖ＝０．９０ｍｓ。（３）脉冲周期T＝７．６ｄｉｖ×１．００ｍｓ／ｄｉｖ＝７．６０ｍｓ；脉冲宽度tＷ＝３．８ｄｉｖ×１．００ｍｓ／ｄｉｖ＝３．８０ｍｓ。tＷ３畅８ｍｓ（４）占空比D＝＝×１００％＝５０％。T７畅６ｍｓ图Ａ７１7．2RC电路的应用练习题178 ················参考答案·························································································································一、判断题１．×　２．×　３．√　４．√　５．√二、填空题１．将矩形脉冲变换为尖脉冲２．电阻脉冲宽度３．将矩形脉冲变换为锯齿波４．电容远大于５．远小于远大于６．微分积分三、选择题１．Ａ　　２．Ｃ　　３．Ｃ　　４．Ｄ　　５．Ａ四、计算题１１．tＷ＝DT＝×３０μｓ＝１０μｓ。３３－１２图７２４（ａ）中，τ＝RC＝１００×１０Ω×２０００×１０Ｆ＝２００μｓ；由于τ远大于tＷ，不满足微分电路的条件，所以不是微分电路，而是耦合电路。３－１２图７２４（ｂ）中，τ＝RC＝２００×１０Ω×１００×１０Ｆ＝２０μｓ；由于τ未远大小于tＷ，不满足积分电路的条件，所以不是积分电路。－６１tＷ１０×１０２．图７２４（ａ）要满足微分电路条件，要求τ＝RC≤tＷ，即C≤＝３Ｆ＝２０ｐＦ。５５R５×１００×１０－６３tＷ３×１０×１０３．图７２４（ｂ）要满足积分电路条件，要求τ＝RC≥３tＷ，即C≥＝３Ｆ＝１５０ｐＦ。R２００×１０7．3数制与码制练习题一、判断题１．×　２．√　３．√　４．×　５．√　６．√　７．×　８．×二、填空题１．10逢二进一借一当二２．（10010）２　　（00011000）８４２１３．（４３）１０４．（３５）１０５．逢十六进一６．（Ａ）１６　　（Ｃ）１６　　（Ｆ）１６７．２　　０　　余数８．８４２１码　　５４２１码余３码三、选择题１．Ｃ　　２．Ａ　　３．Ｄ　　４．Ａ　　５．Ｂ四、简答题１．在计算机中，广泛采用的是由0和1两个基本数码组成的二进制数，而不使用人们习惯的十进179 ·························································································································参考答案················制数，原因如下：（１）二进制数在物理上最容易实现。例如，可以只用高、低两个电平表示1和0，也可以用脉冲的有无或者脉冲的正负极性表示它们。（２）二进制数用来表示的二进制数的编码、计数、加减运算规则最为简单。（３）二进制数的两个数码1和0正好与逻辑命题的两个值“是”和“否”或“真”和“假”相对应，为计算机实现逻辑运算和程序中的逻辑判断提供了便利的条件。２．在数字系统中，各种数据要转换为二进制代码才能进行处理，而人们习惯于使用十进制数，所以在数字系统的输入输出中仍采用十进制数，这样就产生了用４位二进制数表示１位十进制数的方法，这种用于表示十进制数的二进制代码称为二十进制代码（ＢｉｎａｒｙＣｏｄｅｄＤｅｃｉｍａｌ），简称为ＢＣＤ码。它具有二进制数的形式以满足数字系统的要求，又具有十进制数的特点（只有１０种有效状态）。五、计算题１．（11011）２＝（２７）１０。２．（５６）１０＝（111000）２。３．（100111000）８４２１＝（１３８）１０。7．4逻辑门电路基础练习题一、判断题１．×　２．×　３．√　４．×　５．√　６．√二、填空题１．高较强低大２．Ｐ型和Ｎ型绝缘栅场效晶体管３．与门或门非门４．＋５．５Ｖ　　－０．５Ｖ５．与门非门６．0011三、选择题１畅Ａ　　２．Ｃ　　３．Ａ　　４．Ｂ　　５．Ｄ　　６．Ａ四、作图题１．　　　　　　　　　　　　　　　　　　　　　　２畅图Ａ７２图Ａ７３180 ················参考答案·························································································································7．5逻辑代数运算法则及逻辑函数化简练习题一、判断题１．×　２．√　３．×　４．√　５．×　６．√二、填空题１．逻辑功能电路简化元器件可靠性２．或项数每个与项３．BC４．A·B·D５．A６．A三、选择题１．Ｂ　　２．Ｄ　　３．Ａ　　４．Ｃ　　５．Ａ　　６．Ｂ四、化简题　　１．Y＝AB＋ABC（D＋E）２．Y＝AB＋AC＋BC＝AB［1＋C（D＋E）］＝AB＋C（A＋B）＝AB＝AB＋ABC＝AB＋C五、证明题１．证明：　AB＋AC＋BC＝AB＋C（A＋B）＝AB＋ABC＝AB＋C２．证明：　AB＋BD＋AD＋DC＝AB（1＋D）＋BD＋AD＋DC＝AB＋ABD＋BD＋AD＋DC＝AB＋D（AB＋B＋A＋C）＝AB＋D（A＋B＋A＋C）＝AB＋D３．证明：　ACD＋ACD＋AD＋BC＋BC＝AD（C＋C）＋AD＋B（C＋C）＝AD＋AD＋B＝D（A＋A）＋B＝B＋D181 ·························································································································参考答案················第7章数字电路基础单元测试卷一、判断题１．√　２．√　３．√　４．√　５．×　６．×　７．√　８．×　９．×　１０．√　　１１．√１２．√　１３畅×　１４．√　１５．√二、填空题１．脉冲信号２．不连续突变矩形３．陡峭下降越慢４．逻辑电路５．断开闭合开关６．４　　１７．（0011）２８．４．７５～１８．０Ｖ９．A１０．逻辑函数表达式真值表逻辑图１１．B＋C＋D１２．1A三、选择题１．Ｂ　　２．Ｂ　　３．Ｂ　　４．Ｄ　　５．Ｃ　　６．Ｃ　　７．Ｄ　　８．Ｄ９．Ｂ　　１０．Ａ四、分析题（共20分，第1小题12分，第2小题8分）１．图Ａ７４（１）脉冲波形的幅度Vｍ＝４ｄｉｖ×１０ｍＶ／ｄｉｖ＝４０ｍＶ。（２）脉冲上升时间tｒ＝１．８ｄｉｖ×２．００ｍｓ／ｄｉｖ＝３．６０ｍｓ。182 ················参考答案·························································································································脉冲下降时间tｆ＝１．８ｄｉｖ×２．００ｍｓ／ｄｉｖ＝３．６０ｍｓ。（３）脉冲周期T＝４．９ｄｉｖ×２．００ｍｓ／ｄｉｖ＝９．８０ｍｓ。２．图７５（ａ）Y＝AB＋AB；图７５（ｂ）Y＝ABCD。五、计算题１．（10101）２＝（２１）１０。２．（２８）１０＝（11100）２。３．（01111001）８４２１＝（７９）１０。－３１tＷ１０×１０－６４．要满足微分电路条件，要求τ＝RC≤tＷ，即C≤＝３Ｆ＝１×１０Ｆ＝１μＦ。５５R５×２×１０六、化简题１．Y＝AB＋ABC（D＋E）＝AB［1＋C（D＋E）］＝AB２．Y＝AB＋AC＋BCD＋A＝A（B＋1）＋AC＋BCD＝A＋AC＋BCD＝A＋C＋BCD＝A＋C七、作图题１．　　　　　　　　　　　　　　　　　　　２畅图Ａ７５图Ａ７６第8章组合逻辑电路8．1组合逻辑电路的基本知识练习题一、判断题１．×　２．×　３．√　４．√　５．×　６．×　７．×　８．√183 ·························································································································参考答案················二、填空题１．逻辑门电路２．该时刻的输入信号原来状态３．逻辑函数表达式化简和变换真值表逻辑功能４．真值表逻辑函数表达式化简逻辑电路图三、选择题１．Ｂ　　２．Ｃ　　３．Ｄ　　４．Ｄ　　５．Ａ四、分析题１．图８１１（ａ），Y１＝AB（A＋B）＝A＋B；图８１１（ｂ），Y２＝AB＋AB＝AB＋AB。２．（１）S＝AB＋ABC＝AB（２）输入输出ABSC0000011010101101　　（３）该电路能对输入端的A、B进行相加，S满足二进制数的加法运算，用于表示输出的和。C满足“逢二进一”的规则，用于表示进位数。这种加法运算只考虑了两个加数本身，而没有考虑由低位来的进位，所以称为半加器。8．2编码器练习题一、判断题１．×　２．√　３．√　４．√　５．×　６．×　７．×二、填空题１．将特定意义的数字、文字、符号信息等２．完成编码操作３．被编码的信号相对应的代码４．ＢＣＤ代码十进制的数字０～９５．同时输入优先级别最高６．１６　　６三、选择题１．Ａ　　２．Ｃ　　３．Ｃ　　４．Ｄ　　５．Ｃ　　６．Ｄ四、分析题Y１＝I１＋I３，Y２＝I２＋I３。184 ················参考答案·························································································································输入输出I３I２I１I０Y２Y１000100001001010010100011I０、I１、I２、I３表示４路输入，可以代表十进制的０、１、２、３，输出是对应的二进制码00、01、10、11，故该电路是２位二进制编码器。五、作图题１．　　　　　　　　　　　　　　　　２．图Ａ８１图Ａ８２8．3译码器练习题一、判断题１．×　２．×　３．√　４．×　５．√　６．√二、填空题１．编码２．ＢＣＤ码十进制码３．二进制码与二进制码对应的信息４．８　　１５．ＢＣＤ码６．100７．３三、选择题１．Ｄ　　２．Ａ　　３．Ａ　　４．Ｂ　　５．Ｂ四、分析题１．Y０＝ABY１＝ABY２＝ABY３＝AB185 ·························································································································参考答案················输入输出ABY３Y２Y１Y０001110011101101011110111　　２．（１）数码管显示的是数字“６”。（２）A３A２A１A０＝0101。（３）数码管正常应显示数字“８”，显示“０”则表明g发光段缺失，出现缺段的通常原因是：数码管g脚连接不良、或数码管内部相应的g段发光二极管损坏。8．4数据选择器及数据分配器练习题一、判断题１．√　２．√　３．√　４．×　５．×　６．√二、填空题１．传输总线２．有选择不同的输出端３．单刀八掷开关４．分配５．数据输入数据输出数据地址输入输出６．D３７．０　　不工作８．８　　２三、选择题１．Ｃ　　２．Ｄ　　３．Ｂ　　４．Ａ　　５．Ｂ四、分析题１．Y＝ABC＋ABC＋ABC。２．设置A２A１A０＝110。３．设置A２A１A０＝011。４．F１＝Y４·Y７＝A２A１A０·A２A１A０＝ABC·ABC＝ABC＋ABCF２＝Y２·Y３·Y４＝A２A１A０·A２A１A０·A２A１A０＝ABC＋ABC＋ABC＝AB＋ABC186 ················参考答案·························································································································第8章组合逻辑电路单元测试卷一、判断题１．√　２．√　３．√　４．×　５．×　６．√　７．√　８．√　９．×　１０．×　１１．√１２．√　１３．×　１４．√　１５．×二、填空题１．编码器译码器数据选择器数据分配器２．二进制代码３．４　　１４．通用译码器显示译码器５．译码禁止状态６．共阴极共阳极７．0不工作低电平８．４９．１６　　10101111１０．D３三、选择题１．Ａ　　２．Ｂ　　３．Ａ　　４．Ｄ　　５．Ｃ　　６．Ａ　　７．Ｃ　　８．Ａ　　９．Ｃ　　１０．Ｄ四、分析题１．Y＝ABC·AC＝ABC＋AC＝AB＋AC。Y＝1的ABC的变量取值应为100、110、111。２．（１）Y＝AB＋AB。　　（２）真值标表如右表所示。ABY（３）两个输入量A、B同为1或同为0时，001输出为1，否则为0，所以该电路的功能是用来010判断输入信号是否相同，称其为“一致判别100电路”。111　　３．输入输出A１A０Y00D０01D１10D２11D３图Ａ８３187 ·························································································································参考答案················４．设置７４ＬＳ１５１的A２A１A０＝101，７４ＬＳ１３８的A２A１A０＝001。五、作图题１．图Ａ８４２．图Ａ８５３．先将逻辑函数Y＝AB＋BC＋AC化简为Y＝AB＋BC，再画出对应的电路图（图Ａ８６）。图Ａ８６４．先将逻辑函数Y＝（A＋B）AB化简为Y＝AB，再画出对应的电路图（图Ａ８７）。图Ａ８７第9章触发集成器9．1RS触发器练习题一、判断题１．√　２．√　３．×　４．√　５．√　６．×二、填空题１．R＝S＝0２．置1置0保持３．与非或非相同４．两个与非门５．输出节拍６．基本RS触发器７．SRQQ188 ················参考答案·························································································································８．时钟脉冲CP三、选择题１．Ａ　　２．Ｃ　　３．Ｂ　　４．Ｃ　　５．Ｄ　　６．Ｂ四、作图题１．图Ａ９１２．图Ａ９２３．图Ａ９３9．2触发器的几种常用触发方式练习题一、判断题１．×　２．√　３．√　４．√　５．√　６．×189 ·························································································································参考答案················二、填空题１．CP为高电平２．同步触发上升沿触发下降沿触发主从触发３．同步主从边沿４．主从非５．陡峭６．陡峭三、选择题１．Ａ　　２．Ｂ　　３．Ｃ　　４．Ｄ　　５．Ａ四、作图题１．（ａ）同步RS触发器　　（ｂ）上升沿RS触发器　　（ｃ）下降沿RS触发器　　（ｄ）主从RS触发器图Ａ９４２．图Ａ９５３．图Ａ９６190 ················参考答案·························································································································9．3JK触发器练习题一、判断题１．×　２．√　３．√　４．×　５．×　６．×　７．√　８．×二、填空题１畅JKQQ２．置0３．置1４．00５．翻转６．主触发器从触发器非门７．11三、选择题１．Ｂ　　２．Ａ　　３．Ｄ　　４．Ａ　　５．Ｃ四、作图题１．图Ａ９７２．图Ａ９８３．图Ａ９９191 ·························································································································参考答案················9．4D触发器练习题一、判断题１．√　２．×　３．√　４．√　５．√　６．×二、填空题１．置0置1２．DSＤRＤQQ３．置0４．置1５．高低６．非门三、选择题１．Ｂ　　２．Ｄ　　３．Ｃ　　４．Ｂ　　５．Ａ四、作图题１．图Ａ９１０２．图Ａ９１１３．图Ａ９１２４．图Ａ９１３192 ················参考答案·························································································································9．5T触发器练习题一、判断题１．√　２．×　３．√　４．√　５．×二、填空题１．翻转保持２．TSＤRＤQQ３．翻转４．保持不变５．低高三、选择题１．Ｃ　　２．Ｂ　　３．Ｂ　　４．Ａ　　５．Ｄ　　６．Ａ四、作图题１．图Ａ９１４２．图Ａ９１５３．输出脉冲的频率为２５０Ｈｚ，Q的波形如图Ａ９１６所示：图Ａ９１６193 ·························································································································参考答案················第9章集成触发器单元测试卷一、判断题１．√　２．×　３．×　４．√　５．√　６．√　７．×　８．√　９．×　１０．×二、填空题１．２　　触发信号２．下降上升主从３．10４．1５．同步６．CP脉冲置1置0７．会随之产生多次变化８．下降上升９．10１０．T触发器D触发器１１．状态不定三、选择题１．Ｃ　　２．Ｂ　　３．Ｂ　　４．Ａ　　５．Ａ　　６．Ｄ　　７．Ｄ　　８．Ｂ　　９．Ｃ　　１０．Ｄ四、分析题１．触发器可根据触发方式和逻辑功能来分类。根据触发方式不同，即信号的输入方式以及触发器状态随输入信号变化的规律不同，触发器可分为无时钟脉冲控制的直接触发器、有时钟脉冲控制的同步触发器、主从触发器和边沿触发器。按逻辑功能又可分为RS触发器、JK触发器、D触发器、T触发器等几类型。２．以与非门RS触发器为例，当R＝0、S＝0时，触发器两个输出都为1，不再是互补关系．且在输入低电平信号同时变为高电平后，触发揣的状态不能确定，此时称为触发器的不定状态。在正常工作时，不允许输入端R和S同时力0，即要求输入信号遵守S＋R＝1的约束条件。可通过控制S、R输入信号或选其他无约束条件的触发器，如JK触发器。３．７４ＬＳ７６属于JK型的触发器，集成电路内包含２个独立触发器。五、作图题１．图Ａ９１７194 ················参考答案·························································································································２．图Ａ９１８３．图Ａ９１９４．图Ａ９２０195 ·························································································································参考答案················第10章时序逻辑电路10．1寄存器练习题一、判断题１．√　２．√　３．√　４．×　５．×　６．√二、填空题１．电路原来状态２．记忆电路３．移位４．置0、置1５．数码移位６．并行并行７．寄存器８．单拍接收式输入端数码输出端总清零低三、选择题１．Ｃ　　２．Ｃ　　３．Ｂ　　４．Ａ　　５．Ｄ四、作图题在CP作用下的Q２、Q１、Q０的波形如图Ａ１０１所示。图Ａ１０１五、分析题（１）需４个脉冲配合。（２）各触发器状态为0010。（３）要从G４、G３、G２、G１门并行输出所存数码，应使读出控制端A＝1。10．2计数器练习题一、判断题１．√　２．×　３．×　４．√　５．√　６．√196 ················参考答案·························································································································二、填空题１．同时２．不同时３．二进制十进制N进制４．同步异步５．加法减法６．４７．７　　１５三、选择题１．Ｃ　　２．Ａ　　３．Ｃ　　４．Ｄ　　５．Ｂ四、作图题１．图Ａ１０２２．图Ａ１０３３．（１）构成２位二进制异步递增计数器，触发器输出端对应于各CP脉冲的状态见表Ａ１０１。（２）连接线ａ断路。表A101CPQ１Q００00１01２10３11197 ·························································································································参考答案················４00第10章时序逻辑电路单元测试卷一、判断题１．×　２．×　３．×　４．√　５．×　６．×　７．×　８．√　９．√　１０．√二、填空题１．存储数码和信息２．组合逻辑电路时序逻辑电路３．双拍接收式单拍接收式４．触发器门电路５．单向双向６．３　　３７．计数器８．触发器门电路９．４　　１５１０．加法１１．减法１２．10101111三、选择题１．Ｃ　　２．Ｄ　　３．Ｂ　　４．Ｂ　　５．Ｄ　　６．Ｂ　　７．Ｄ　　８．Ｃ　　９．Ｃ　　１０．Ｂ四、简答题１．计数器出现进位不正常的故障时，首先，检查低位触发器有无进位信号输出；其次，检查高位触发器的J、K输入端的接线有无错误及有无开路；第三，检查高位触发器元件是否损坏及相关的连接线是否开路。２．（１）７４ＬＳ１６０为十进制计数集成电路。（２）CR是清零端，将CR置于低电平，计数器实现清零；Q０～Q３为８４２１ＢＣＤ码的４位数码输出端；D０～D３为置数输入端；CTＴ、CTＰ是计数控制端，全为高电平时为计数状态，若其中有一个是低电平，则处于保持数据的状态。３．计数器和寄存器都由触发器组成，确有很多相似之处，但有两点不同之处。一是输入不同：计数器仅有CP输入；而寄存器不仅有CP输入，而且还有数据输入。二是使用的触发器不同：计数器选用的触发器具有翻转功能，如T触发器或JK触发器（设置为翻转功能），一般不用D触发器，因为D触发器不具有翻转功能；而寄存器通常使用D触发器比较多，也有用RS触发器来构成寄存器的。五、作图题１．３位单拍接收式数码寄存器如图Ａ１０４所示。198 ················参考答案·························································································································图Ａ１０４199 ·························································································································参考答案················　　２．异步十进制加法计数器如图Ａ１０５所示。图Ａ１０５３．该寄存器为移位寄存器，画出的工作波形如图Ａ１０６所示。图Ａ１０６４．各触发器的时钟CP０＝CP，CP１＝Q０，CP下跳沿到达时，触发器翻转。显然，电路实现了２位二进制异步加法计数器的功能，图Ａ１０７是其工作波形图。图Ａ１０７第11章脉冲波形的产生与变换11．1多谐振荡器练习题一、判断题１．×　２．√　３．√　４．×　５．×二、填空题１．矩形脉冲２．低电平高电平稳定３．１．４RC电源电压４．非门200 ················参考答案·························································································································５．不是很高６．石英晶体７．８５０Ω～２ｋΩ　　１０～１００ｋΩ三、选择题１．Ｂ．　　２．Ａ　　３．Ｃ　　４．Ｃ　　５．Ｄ四、作图题１．将非门与阻容元件连接成RC多谐振荡器，如图Ａ１１１所示。２．用集成电路７４ＬＳ００来组成的RC多谐振荡器的实物图如图Ａ１１２所示。图Ａ１１１图Ａ１１２五、计算题３－１２１．T≈１．４RC＝１．４×１×１０Ω×２２００×１０Ｆ＝３．０８μｓ，１f０＝≈３２５ｋＨｚ。T１T２．T＝≈５μｓ，　R＝≈１．６２ｋΩ。f０１．４C11．2单稳态触发器练习题一、判断题１．√　２．×　３．√　４．×　５．×　６．×二、填空题１．稳态暂稳态２．不可重复触发可重复触发３．稳态暂稳态充放电４．整形处理延时控制定时控制５．０．７RC６．暂稳态的时间对应缩短三、选择题１．Ｃ　　２．Ｄ　　３．Ｂ　　４．Ｃ　　５．Ａ四、作图题１．将图１１２３中的元件连接成单稳态触发器。201 ·························································································································参考答案················图Ａ１１３２．单稳态触发器的输出波形如图Ａ１１４所示。图Ａ１１４11．3施密特触发器练习题一、判断题１．√　２．×　３．×　４．√　５．×　６．√　７．×　８．×二、填空题１．输入触发信号２．上限触发电平VＴＨ下限触发电平VＴＬ３．VＴＨ－VＴＬ４．波形变换整形处理幅度鉴别５．ＣＭＯＳＴＴＬ施密特反相器施密特与非门６．单稳态触发器施密特触发器三、选择题１．Ｄ　　２．Ｄ　　３．Ａ　　４．Ｃ　　５．Ｂ　　６．Ａ四、作图题如图Ａ１１５所示。五、计算题R１＋R２VＴＨ＝VＴ＝３．６Ｖ，R２R１＋R２R１VＴＬ＝VＴ－VＣＣ＝１．２Ｖ，R２R２202 ················参考答案·························································································································图Ａ１１５R１ΔVＴ＝VＤＤ＝２．４ＶR２11．4555时基电路及其应用练习题一、判断题１．√　２．×　３．√　４．√　５．√　６．×二、填空题１．电压比较器电阻分压器基本RS触发器输出缓冲器开关管２．８Ｖ　　４Ｖ３．低高４．触发输入５．锯齿波矩形波相同６．３～１８Ｖ７．多谐施密特三、选择题１．Ａ　　２．Ｄ　　３．Ｂ　　４．Ｃ　　５．Ｄ四、作图题vＯ的波形如图Ａ１１６所示。图Ａ１１６五、计算题１．f＝３４０Ｈｚ；D＝８５．７％。203 ·························································································································参考答案················２．回差电压越大，施密特触发器的抗干扰能力越强，但灵敏度越低。当VＤＤ＝１２Ｖ时，VＴＨ＝８Ｖ，VＴＬ＝４Ｖ，ΔVＴ＝４Ｖ。３．该电路为５５５单稳态触发器，tＷ＝１．１ｍｓ。第11章脉冲波形的产生与变换单元测试卷一、判断题１．×　２．×　３．√　４．×　５．√　６．√　７．×　８．×　９．√１０．√二、填空题１．脉冲信号源２．RC定时元件３．门电路石英晶体４．石英晶体的串联谐振频率无关５．单稳态触发器施密特触发器６．暂稳态时间tＷ７．上限触发电平VＴＨ下限触发电平VＴＬ８．数字模拟９．ＴＴＬ型单时基电路ＣＭＯＳ型单时基电路１０．多谐振荡器单稳态触发器施密特触发器１１．输入触发信号１２．VＴＨ三、选择题１．Ｃ　　２．Ａ　　３．Ｄ　　４．Ｂ　　５．Ａ　　６．Ｂ　　７．Ａ　　８．Ｃ　　９．Ｃ　　１０．Ｂ四、简答题１．图１１２所示是石英晶体多谐振荡器的电路，电路中石英晶体的作用是选频，它只让频率等于其串联谐振频率的信号通过并形成正反馈。这样，该电路的振荡频率就等于石英晶体的串联谐振频率。２．图１１３所示为施密特触发器的电路。当光线强时，光敏三极管Ｖ导通，输入电压VＩ的值增大，当大于VＴＨ时，电路状态翻转，输出为低电平，发光二极管ＬＥＤ暗。当光线弱时，光敏三极管Ｖ截止，输入电压VＩ的值变小，当低于VＴＬ时，电路状态就发生翻转，输出为高电平，发光二极管ＬＥＤ亮。五、作图题１．用５５５时基电路接成施密特触发器、单稳态电路、多谐振荡器如图Ａ１１７（ａ）、（ｂ）、（ｃ）所示。２．单稳态触发器的输出波形如图Ａ１１８所示。六、计算题１．tＷ＝０．７RC＝１０５μｓ。204 ················参考答案·························································································································图Ａ１１７图Ａ１１８－６tＷ１５０×１０２．由tＷ＝１．１RC，可得R＝＝－１２Ω＝２２ｋΩ。１．１Ｃ１．１×６２００×１０３．该电路是５５５时基电路组成的多谐振荡器。T＝０．７（R１＋２R２）C１＝１４３．６μｓ，１f＝＝６．９６ｋＨｚ。T枟电子线路枠数字电路期中考试卷一、单项选择题１．Ａ　　２．Ｂ　　３．Ｃ　　４．Ｂ　　５．Ａ　　６．Ｂ　　７．Ｃ　　８．Ｄ　　９．Ａ　　１０．Ｄ二、多项选择题１．ＢＤ　　２．ＢＤ　　３．ＡＢ　　４．ＡＢＣ　　５．ＡＤ　　６．ＢＤ　　７．ＡＣＤ　　８．ＡＢＣ９．ＢＤ　　１０．ＡＣ三、填空题１．开关２．十进制二进制３．逢二进一幂４．01５．1011001６．0101７．逻辑图真值表逻辑函数式８．编码器译码器数据选择器数据分配器９．高电平状态低电平状态高阻状态EN控制三态门的工作状态205 ·························································································································参考答案················１０．３１１．８１２．７　　阳四、分析计算题１．Y＝ABC＋ABC＋ABC＋ABC。２．Y＝A磑B。３．Y＝（A＋B）（A＋B）＝AB＋AB。４．（１）D端输入信号送至输出端Y６。（２）设置A２A１A０＝101。５．（１）数码管显示的是数字“７”。（２）A３A２A１A０＝0010。五、作图题１．图ＡＴ３１２．Y１＝ABC，输入全高，输出才高；Y２＝A＋B＋C，输入全低，输出才低。图ＡＴ３２３．表AT31输入输出I３I２I１I０Y１Y０000100001001010010100011图ＡＴ３３206 ················参考答案·························································································································枟电子线路枠数字电路期末考试卷一、单项选择题１．Ｂ　　２．Ｄ　　３．Ｃ　　４．Ｂ　　５．Ａ　　６．Ｄ　　７．Ｄ　　８．Ｂ　　９．Ｃ　　１０．Ａ二、多项选择题１．ＡＢ　　２．ＢＤ　　３．ＡＢＤ　　４．ＢＣ　　５．ＡＤ　　６．Ｃ　　７．ＣＤ　　８．ＡＢＣ９．ＢＣ　　１０．ＡＢＤ三、填空题１．５Ｖ２．七段数码显示器显示十进制数字及部分字母３．原来的状态４．Q５．相反６．整形　　１．１RC７．RS触发器JK触发器D触发器T触发器８．0状态1状态９．置1置0１０．1１１畅计数器１２．寄存器１３．４１４．３　　７　　１５１５．数码移位四、分析计算题１．（１）电路名称为５５５多谐振荡器，功能是自激产生脉冲信号。１（２）f１＝０．７（R１＋２RＰ＋２R２）C１＝－３Ｈｚ≈６畅２ｋＨｚ，０．７×（１＋２×１０＋２×１）×０．０１×１０１１f２＝＝－３Ｈｚ≈１１ｋＨｚ。０畅７（R１＋RＰ＋２R２）C０畅７×（１＋１０＋２×１）×０畅０１×１０２．tＷ＝０．７RC＝７０μｓ。五、作图题１．２．３．电路实现D触发器的功能。207ＡＴ４４ ·························································································································参考答案················图ＡＴ４１图ＡＴ４２图ＡＴ４３４．图ＡＴ４５５．208 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