0自我检测题1DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD2DDDT6.2DDDDCMOS或非DDDDDDDDDDDDDaDbdDeDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD1DDDRdDDDD2DDDRDDDD3DDDCDDDD4DDDVDDDDDDDDDDD5DDDDDDDDDDDDDDDDDDDDDDDDDDvlDDDDDDDDDDDDDDDD1DDD2DDD3DDDDDDDDDDDDDDDDDDDDDaPDDDDDDDDDcDDDDDDDDDvO3DDD4DDD5DDvlDDDvODDDDDT6.3DDDDDDDDDDDDDDDDDbPDDDDDDDdDDDDDDDDQaQv+IVI1廿一节-一■_—―QbQv：tvI0vOh^AAAAA-0tQcQQdDDT6.12ADDDDDDDDBDDDDDDDDCDDDDDDDDDDDDDDDDDDADDDDDBDDDDDCDDDDDDDDDDDDDDADDDDDDBDDDDDDCDDDDDDDDDDDDDADDDDDBDDDDDDCDTODDDDDDDDDDADDDDDDBDDDDDDCDTODDDDDDDDDCd10DDDDDDDDDDDDDDDADDDDVDDBDDDDDDT6.10DDDDDDDDDDDDDDDDDDDADDDDBDDDDDDCDDDDDD12DD555DDDDDDDDDDT6.12DDDDDDDDDDDADDDDDDBDDDDDDCDDDDDDDSRODDVDR1T]DISCv-TT-VI2-R2UvxCC-T■RLL748655535二二O.OlyF—i4DDDDDDDDDDDDDD5DDDDDDDDDDDDDDDDDDDDDDDDDDD6DDDDDDDDDDDDDDDD7DDDDDDDDDDD8DDDDDDDDDDDDDD9DDCMOSDDDDDDDDDDDDDDDDDA.0.7RCB.RCC.1.1RCD.2RCG2DT6.10555DDDDDDDDDDDDDDDDDDDDDDDDDDDDRDCODD习题CDDDDDDDT6.30tD3DVT+2DDR2CDDDDvCDDDDDvODDDDDDDDDvCvO1DDDDDP6.1ODDG』GDDCMOSODD12D1DDDDDDDDD2DDDDDDDDDD3DDDDDDDDDDDDDDD1DDDDDDDDDDDDDDDD2DDDDDRV=■+VD1——DT-2DDR2DDDD=V•呂DDR22DDP6.2DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDP6.2DDvORvIvO21JO112O1DP6.1C=F3.由CC40106构成的电路如图P6.3(a)所示，图P6.3(b)为CC40106的电压传输特性曲线，图P6.3(c)中的输入V］高电平脉宽和低电平脉宽均大于时间常数RC。要求画出vI作用下的vA、vO1和vO2波形。图P6.3解：波形图如下：vOvIa)b)tttc)占空比可调电路PR4DDDP6.4DDDDDDD1DDDDDDDDDD2DDDDDDDDDDDDDDD5DDDDDDDDDDDDDDDv°DCMOS或非DDDDDDDDDDDvDDvDvDDDDDDIO1O2tDDDDDWGGvIvO2DP6.4rnu±12V=TT=0.7RCW555DDDDDDDDDDDDDDDDDVDDvDDP6.5DDDDDv/VI6v/VO120ms6．1）构成电路的名称；（2）已知输入信号波形V[,画出电路中vO的波形（标明vO波形的脉冲宽度）;图P6.6解：（1）555组成的单稳态触发器。（2）vI、vO波形如图所示。输出脉冲宽度由下式求得：T尸RCln3=100X103X3.3X10-6X1.1=363（ms）WvO解：vO0.0Mrtt图P6.5ttvIVT+V7.图P6.7(a)所示为由555定时器构成的心率失常报警电路。经放大后的心电信号v如图P6.7(b)所示,v的峰值V=4V。IIm(1)分别说出555定时器I和555定时器II所构成单元电路的名称；(2)对应分别画出A、B、D三点波形；(3)说明心率失常报警的工作原理。a)(b)图P6.7解：(1)电路I为施密特触发器，电路II为可重触发单稳态触发器;(2)AUUU8.由集成定时器7555构成的电路如图P6.8所示，请回答下列问题。4tR+6VvC/VJLRr6.22k765553I1vv'CI225-I-O.OlyF1）构成电路的名称；（2）画出电路中vC、vO的波形（标明各波形电压幅度，vO波形周期）。1100kC宁1PF解：取C=0.1pF,则R+2R=—”1x10-3-=14.3kQ12Cln20.1x10-6x0.7取R2=5.1kQ,则R]=4.1kQ,可用一只3.3kQ的固定电阻和一只2kQ的精密电位器组成。原理图如图所示。vO图P6.9P6.11Da叮vlDvODDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDODDlkHzU10DDDP6.10DDDDDD555DDDDDDDDDDDDDaDbDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD1DDD555DDDDDDDDDD1DDDDDDD2DDDDDDDDD555DDDDDDDDDDDQDDDDDDDDDDDDDDDDDDDDDD555DDDDDDDDDDQDDDDDDDDDDDDDDD11D4DDDDDDDDD74161DDDDDDDDD74LS121DDDDP6.11DaDDDDDDD1DDD74161DDDDDDDDDDDD2DDD74LS121DDDDDDDDDTDDWD3DDCPDDDDDD1msDDDDP6.11DbDDDDDDDDDDDDvO1P6.12DDDD1DDD555DDDDDDDDDDDDDDADBDCDDDDDDDDDD2DDDDDDDDZDDDDDDDDDDDDDDDDO12叮555DDDDDDMQ24DDDDDDDDIDDDDDDDDDDD3DDDDDDDDDDDDDDDD640HzDDDDCDDDD103CPpFu0.1pF解：（1）T管截止时才能起振。因此AB=11或C=1时即可起振。2）[UWWUW0.7(R]+2R2)640x0.7x(1+20)Q3VCAB图P6.123)T=0.7C(R1+2R2)
