数字集成电路分析与设计 第四章答案
《数字电子技术基础》第四章习题答案
第四章 集 成 触 发 器 4.1R d S d Q Q不定4.2 （1CP=1时如下表)(2) 特性方程Q n+1=D(3)该电路为锁存器（时钟型D 触发器）。
CP=0时，不接收D 的数据；CP=1时，把数据锁存。
(但该电路有空翻)4.3 (1)、C=0时该电路属于组合电路；C=1时是时序电路。
(2)、C=0时Q=A B +; C=1时Q n+1=B Q BQ nn+= (3)、输出Q 的波形如下图。
A B C Q4.4CP D Q 1Q 2图4.54.5 DQ QCPT4.6 Q 1n 1+=1 Q 2n 1+=Q 2n Q n 13+=Q n 3 Q Q 4n 14n+=Q1CP Q2Q3Q44.7 1、CP 作用下的输出Q 1 Q 2和Z 的波形如下图； 2、Z 对CP 三分频。
DQ QCPQ1DQ QQ2ZRd CP Q1Q2Z14.8由Q D J Q KQ J Q KQ n 1n n n n +==+=⋅得D 触发器转换为J-K 触发器的逻辑图如下面的左图；而将J-K 触发器转换为D 触发器的逻辑图如下面的右图CPD Q QJKQ QDQ QJ KCP4.9CP B CA4.10CP X Q1Q2Z4.11 1、555定时器构成多谐振荡器 2、u c, u o 1, u o 2的波形u c u o 1u o 2t t t 1.67V3.33V3、u o 1的频率f 1=1074501316..H z ⨯⨯≈ u o 2的频率f 2=158H z4、如果在555定时器的第5脚接入4V 的电压源，则u o 1的频率变为1113001071501232....H z ⨯⨯+⨯⨯≈4.12 图(a)是由555定时器构成的单稳态触发电路。
1、工作原理（略）；2、暂稳态维持时间t w =1.1RC=10ms(C 改为1μF)；3、u c 和u o 的波形如下图：u ou ct t tu i (ms)(ms)(ms)5 10 25 30 45 503.33V4、若u i 的低电平维持时间为15m s ，要求暂稳态维持时间t w 不变，可加入微分电路4.13由555定时器构成的施密特触发器如图(a)所示 1、电路的电压传输特性曲线如左下图； 2、u o 的波形如右下图；3、为使电路能识别出u i 中的第二个尖峰，应降低555定时器5脚的电压至3V 左右。
《数字电路-分析与设计》1--10章习题及解答(部分)_北京理工大学出版社
第五章习题5-1 图题5-1所示为由或非门组成的基本R-S 锁存器。
试分析该电路，即写出它的状态转换表、状态转换方程、状态图、驱动转换表和驱动方程，并画出它的逻辑符号，说明S 、R 是高有效还是低有效。
解：状态转换表：状态转换驱动表5-2 试写出主从式R-S 触发器的状态转换表、状态转换方程、状态图、驱动转换表和驱动方程，注意约束条件。
解：与R-S 锁存器类似，但翻转时刻不同。
5-3 试画出图5.3.1所示D 型锁存器的时序图。
解：G=0时保持，G=1时Q=D 。
图题5-1 或非门组成的基本R-S 锁存器S R状态转换方程：Q n+1Q n+1=S+RQ n状态转换图： S =Q n+1R=Q n+1 状态转换驱动方程： 逻辑符号： 输入高有效 G D Q图题5-3 D 型锁存器的时序图5-4试用各种描述方法描述D锁存器：状态转换表、状态转换方程、时序图、状态转换驱动表、驱动方程和状态转换图。
5-5锁存器与触发器有何异同？5-6试描述主从式RS触发器，即画出其功能转换表，写出状态方程，画出状态表，画出逻辑符号。
5-7试描述JK、D、T和T'触发器的功能，即画出它们的逻辑符号、状态转换表、状态转换图，时序图，状态转换驱动表，写出它们的状态方程。
5-8试分析图5.7.1(a) 所示电路中虚线内电路Q’与输入之间的关系。
5-9试分析图5.7.1(b)所示电路的功能，并画出其功能表。
5-10试用状态方程法完成下列触发器功能转换：JK→D, D→T, T→D, JK→T, JK→T’, D→T’。
解：JK→D：Q n+1=JQ+KQ，D：Q n+1=D=DQ+DQ。
令两个状态方程相等：D=DQ+DQ =JQ+KQ。
对比Q、Q的系数有：J=D，K=D逻辑图略。
5-11试用驱动表法完成下列触发器功能转换：JK→D, D→T, T→D, JK→T, JK→T’, D→T’。
解：略。
5-12用一个T触发器和一个2-1多路选择器构成一个JK触发器。
数字电子技术基础第四章习题及参考答案
数字电子技术基础第四章习题及参考答案第四章习题1．分析图4-1中所示的同步时序逻辑电路，要求：（1）写出驱动方程、输出方程、状态方程；（2）画出状态转换图，并说出电路功能。
CPY图4-12．由D触发器组成的时序逻辑电路如图4-2所示，在图中所示的CP脉冲及D作用下，画出Q0、Q1的波形。
设触发器的初始状态为Q0=0，Q1=0。
D图4-23．试分析图4-3所示同步时序逻辑电路，要求：写出驱动方程、状态方程，列出状态真值表，画出状态图。
CP图4-34．一同步时序逻辑电路如图4-4所示，设各触发器的起始状态均为0态。
（1）作出电路的状态转换表；（2）画出电路的状态图；（3）画出CP作用下Q0、Q1、Q2的波形图；（4）说明电路的逻辑功能。
图4-45．试画出如图4-5所示电路在CP波形作用下的输出波形Q1及Q0，并说明它的功能（假设初态Q0Q1=00）。
CPQ1Q0CP图4-56．分析如图4-6所示同步时序逻辑电路的功能，写出分析过程。
Y图4-67．分析图4-7所示电路的逻辑功能。
（1）写出驱动方程、状态方程；（2）作出状态转移表、状态转移图；（3）指出电路的逻辑功能，并说明能否自启动；（4）画出在时钟作用下的各触发器输出波形。
CP图4-78．时序逻辑电路分析。
电路如图4-8所示：（1）列出方程式、状态表；（2）画出状态图、时序图。
并说明电路的功能。
1C图4-89．试分析图4-9下面时序逻辑电路：（1）写出该电路的驱动方程，状态方程和输出方程；（2）画出Q1Q0的状态转换图；（3）根据状态图分析其功能；1B图4-910.分析如图4-10所示同步时序逻辑电路，具体要求：写出它的激励方程组、状态方程组和输出方程，画出状态图并描述功能。
1Z图4-1011．已知某同步时序逻辑电路如图4-11所示，试：（1）分析电路的状态转移图，并要求给出详细分析过程。
（2）电路逻辑功能是什么，能否自启动？（3）若计数脉冲f CP频率等于700Hz，从Q2端输出时的脉冲频率是多少?CP图4-1112．分析图4-12所示同步时序逻辑电路，写出它的激励方程组、状态方程组，并画出状态转换图。
数字电路与数字电子专业技术课后答案第四章
第四章 逻辑函数及其符号简化1.列出下述问题的真值表，并写出逻辑表达式：(1) 有A 、B 、C 三个输入信号，如果三个输入信号中出现奇数个1时，输出信号F=1，其余情况下，输出 F= 0.(2) 有A 、B 、C 三个输入信号，当三个输入信号不一致时，输出信号F=1，其余情况下，输出为0.(3) 列出输入三变量表决器的真值表•解：(1 )(1) F=AB+ A B1⑵ F= AB+ A C(3) F= (A+B+C) (A+B+ C ) (A+ B +C) (A+ B +C ) 解： (1) AB = 00 或 AB=11 时 F=1(2) ABC110 或 111 或 001,或 011 时 F=1 (3) ABC = 100 或 101 或 110 或 111 时 F=1 3. 用真值表证明下列等式.(1) A+BC = (A+B) (A+C)(2) A BC+A B C+AB C = BC ABC +AC ABC +AB ABC (3) A B + BC + AC =ABC+ A B C ⑷ AB+BC+AC=(A+B)(B+C)(A+C)A B C F0 0 0 00 01 10 10 10 11 0 1 0 0 11 01(2 )1 B 10 0 1 10 10 10 11 11 00 11 0 1 1 (3 )1 1 0 1 A 1B C F0 00 00 01 00 10 00 11 1 1 0 0 01 011F= A B C+ A B C +A B C +ABCF= (A+B+C) ( A + B +C )F= A BC+A B C+AB C +ABC ,F 的值为“ 12.对下列函数指出变量取哪些组值时(5) ABC+ A + B + C=1证：(1 )A B C A+BC(A+B)(A+C)0 0 0 0 00 0 1 0 00 1 0 0 00 1 1 1 11 0 0 1 11 0 1 1 11 1 0 1 1A B C ABC + ABC + ABC —BCABC + ACA B C + ABABC0 0 0 0 00 0 1 0 00 1 0 0 00 1 1 1 11 0 0 0 01 0 1 1 11 1 0 1 11 1 1 0 0(5 )A B C AB+ BC3 AC ABC + A"B"C—0 0 0 1 10 0 1 0 00 1 0 0 00 1 1 0 01 0 0 0 01 0 1 0 01 1 0 0 01 1 1 1 1ABC AB+BC+AC (A+B)(B+C)(A+C)0 0 00 0 10 1 00 1 11 0 01 0 11 1 0 11ABC ABC + A + B + C0 0 0 10 0 1 10 1 0 10 1 1 11 0 0 11 0 1 1(5 )4. 直接写出下列函数的对偶式F'及反演式F的函数表达式(1) F=[A B(C+D)][B C D+B(C +D)](2) F= A BC + ( A +B C ) (A+C)⑶F= AB+ CD + BC + D + CE + D + E(4)F=C+AB?AB+ D解：(1) F'= [ A +B+CD]+[(B+ C+ D)?B+C D]]F = [A+ B + C D ]+[( B +C+D) ?( B +CD ]](2) F'= (A+ B + C)?[A?(B + C)AC]F = (A+ B<C)?[A?(B + C)+ A C]⑶F、=C?(A + B)+(A + B)?DF = C?(A + B) + (A + B)?D5. 若已知x+y = x+z，问y = z吗？为什么？解：y不一定等于z,因为若x=1时,若y=0,z=1,或y=1,z=0,则x+y = x+z = 1,逻辑或的特点，有一个为1则为1。
3篇4章习题解答浙大版集成电路课后答案
第四章 功率变换电路题 一双电源互补对称电路如图题所示，设已知V CC =12V ，R L =16Ω，v I 为正弦波。
求：(1)在三极管的饱和压降V CES 可以忽略不计的条件下，负载上可能得到的最大输出功率Pom=?；(2)每个管子允许的管耗P Cm 至少应为多少?(3)每个管子的耐压｜V (BR)CEO ｜应大于多少? 图题解：(1) 负载上可能得到的最大输出电压幅度V om =12V (W 5.416212222=⨯==L om om R V P ) (2) (W)9.02.0(max)==om CM p P ∴CM P ≥(3) CEO BR V )(≥24V题 在图题所示的OTL 功放电路中，设R L =8Ω，管子的饱和压降｜ＶCES ｜可以忽略不计。
若要求最大不失真输出功率(不考虑交越失真)为9W ，则电源电压V CC 至少应为多大?(已知v i 为正弦电压。
)图题解：W 982)21(2)21(22(max)=⨯==CC L CC om V R V P V CC =24(V)∴电源电压V CC 至少24V题 OTL 放大电路如图题所示，设T 1、T 2特性完全对称，v i 为正弦电压，V CC =10V ，R L =16Ω。
试回答下列问题：(1)静态时，电容C 2两端的电压应是多少?调整哪个电阻能满足这一要求?(2)动态时，若输出电压波形出现交越失真，应调整哪个电阻?如何调整?(3)若R 1=R 3=Ω，T 1、T 2管的β=50,｜V BE ｜=,Pcm=200mW,假设D 1、D 2、R 2中任意一个开路，将会产生什么后果?图题解：(1) 静态时，电容C 2两端的电压应为5V 。
调整R 1、R 3，可调整上、下两部分电路的对称性，从而使C 2两端电压为5V 。
(2) 若出现交越失真，应调大R 2，使b １b 2间电压增大，提供较大的静态电流。
(3) 若D 1、D 2、R 2中任意一个开路，则(mA)58.322121=-==R V V I I BE CCB B I C1=I C2=βI B1=179(mA)P C =I C1·V CE =I Ｃ１·5V=895(mW)＞Pcm,∴功率管会烧坏。
数字集成电路--电路、系统与设计(第二版)课后练习题-第四章 导线-Chapter 4 The Wire
1Chapter 4 Problem SetChapter 4Problems1.[M, None, 4.x] Figure 0.1 shows a clock-distribution network. Each segment of the clock net-work (between the nodes) is 5 mm long, 3 μm wide, and is implemented in polysilicon. Ateach of the terminal nodes (such as R ) resides a load capacitance of 100 fF.a.Determine the average current of the clock driver, given a voltage swing on the clock linesof 5 V and a maximum delay of 5 nsec between clock source and destination node R . Forthis part, you may ignore the resistance and inductance of the networkb.Unfortunately the resistance of the polysilicon cannot be ignored. Assume that eachstraight segment of the network can be modeled as a Π-network. Draw the equivalent cir-cuit and annotate the values of resistors and capacitors.c.Determine the dominant time-constant of the clock response at node R .2.[C, SPICE, 4.x] You are designing a clock distribution network in which it is critical to mini-mize skew between local clocks (CLK 1, CLK 2, and CLK 3). You have extracted the RC net-work of F igure 0.2, which models the routing parasitics of your clock line. Initially, you notice that the path to CLK 3 is shorter than to CLK 1 or CLK 2. In order to compensate for this imbalance, you insert a transmission gate in the path of CLK 3 to eliminate the skew.a.Write expressions for the time-constants associated with nodes CLK 1,CLK 2 and CLK 3.Assume the transmission gate can be modeled as a resistance R 3.b.If R 1 = R 2 = R 4 = R 5 = R and C 1 = C 2 = C 3 = C 4 = C 5 = C , what value of R 3 is required to balance the delays to CLK 1, CLK 2, and CLK 3?c.For R =750Ω and C =200fF, what (W /L )’s are required in the transmission gate to elimi-nate skew? Determine the value of the propagation delay.d.Simulate the network using SPICE, and compare the obtained results with the manually obtained numbers.3.[M, None, 4.x]Consider a CMOS inverter followed by a wire of length L . Assume that in thereference design, inverter and wire contribute equally to the total propagation delay t pref . Youmay assume that the transistors are velocity-saturated. The wire is scaled in line with the idealwire scaling model . Assume initially that the wire is a local wire .a.Determine the new (total) propagation delay as a a function of t p ref , assuming that technol-ogy and supply voltage scale with a factor 2. Consider only first-order effects.b.Perform the same analysis, assuming now that the wire scales a global wire , and the wire length scales inversely proportional to the technology.Figure 0.1Clock-distribution network.SR2Chapter 4 Problem Setc.Repeat b, but assume now that the wire is scaled along the constant resistance model. You may ignore the effect of the fringing capacitance.d.Repeat b, but assume that the new technology uses a better wiring material that reduces the resistivity by half, and a dielectric with a 25% smaller permittivity.e.Discuss the energy dissipation of part a. as a function of the energy dissipation of the orig-inal design E ref .f.Determine for each of the statements below if it is true, false, or undefined, and explain in one line your answer. - When driving a small fan-out, increasing the driver transistor sizes raises the short-circuit power dissipation. - Reducing the supply voltage, while keeping the threshold voltage constant decreases the short-circuit power dissipation.- Moving to Copper wires on a chip will enable us to build faster adders.- Making a wire wider helps to reduce its RC delay.- Going to dielectrics with a lower permittivity will make RC wire delay more impor-tant.4.[M, None, 4.x] A two-stage buffer is used to drive a metal wire of 1 cm. The first inverter is of minimum size with an input capacitance Ci=10 fF and an internal propagation delay t p0=50 ps and load dependent delay of 5ps/fF. The width of the metal wire is 3.6 μm. The sheet resis-tance of the metal is 0.08 Ω/, the capacitance value is 0.03 fF/μm 2and the fringing field capacitance is 0.04fF/μm.a.What is the propagation delay of the metal wire?pute the optimal size of the second inverter. What is the minimum delay through the buffer?c.If the input to the first inverter has 25% chance of making a 0-to-1 transition, and the whole chip is running at 20MHz with a 2.5 supply voltage, then what’s the power con-sumed by the metal wire?5.[M, None, 4.x]To connect a processor to an external memory an off -chip connection is neces-sary. The copper wire on the board is 15 cm long and acts as a transmission line with a charac-teristic impedance of 100Ω.(See F igure 0.3). The memory input pins present a very highimpedance which can be considered infinite. The bus driver is a CMOS inverter consisting ofvery large devices: (50/0.25) for the NMOS and (150/0.25) for the PMOS, where all sizes areClock CLK 1CLK 2CLK 3R 1R 2R 5R 4R 3Model as:Figure 0.2RC clock-distribution network.driver C 1C 3C 4C 5C 2Digital Integrated Circuits - 2nd Ed3 in μm. The minimum size device, (0.25/0.25) for NMOS and (0.75/0.25) for PMOS, has theon resistance 35 kΩ.a.Determine the time it takes for a change in the signal to propagate from source to destina-tion (time of flight). The wire inductance per unit length equals 75*10-8 H/m.b.Determine how long it will take the output signal to stay within 10% of its final value. Youcan model the driver as a voltage source with the driving device acting as a series resis-tance. Assume a supply and step voltage of 2.5V. Hint: draw the lattice diagram for thetransmission line.c.Resize the dimensions of the driver to minimize the total delay.L=15cmMemoryZ=100ΩFigure 0.3The driver, the connecting copper wire and thememory block being accessed.6.[M, None, 4.x] A two stage buffer is used to drive a metal wire of 1 cm. The first inverter is aminimum size with an input capacitance C i=10 fF and a propagation delay t p0=175 ps whenloaded with an identical gate. The width of the metal wire is 3.6 μm. The sheet resistance ofthe metal is 0.08 Ω/, the capacitance value is 0.03 fF/μm2 and the fringing field capacitanceis 0.04 fF/μm.a.What is the propagation delay of the metal wire?pute the optimal size of the second inverter. What is the minimum delay through thebuffer?7.[M, None, 4.x] For the RC tree given in Figure 0.4 calculate the Elmore delay from node A tonode B using the values for the resistors and capacitors given in the below in Table 0.1.Figure 0.4RC tree for calculating the delay4Chapter 4 Problem SetTable 0.1Values of the components in the RC tree of Figure 0.4Resistor Value(Ω)Capacitor Value(fF)R10.25C1250R20.25C2750R30.50C3250R4100C4250R50.25C51000R6 1.00C6250R70.75C7500R81000C82508.[M, SPICE, 4.x] In this problem the various wire models and their respective accuracies willbe studied.pute the 0%-50% delay of a 500um x 0.5um wire with resistance of 0.08 Ω/,witharea capacitance of 30aF/um2, and fringing capacitance of 40aF/um. Assume the driverhas a 100Ω resistance and negligible output capacitance.•Using a lumped model for the wire.•Using a PI model for the wire, and the Elmore equations to find tau. (see Chapter 4, figure4.26).•Using the distributed RC line equations from Chapter 4, section 4.4.4.pare your results in part a. using spice (be sure to include the source resistance). Foreach simulation, measure the 0%-50% time for the output•First, simulate a step input to a lumped R-C circuit.•Next, simulate a step input to your wire as a PI model.•Unfortunately, our version of SPICE does not support the distributed RC model as described in your book (Chapter 4, section 4.5.1). Instead, simulate a step input to yourwire using a PI3 distributed RC model.9.[M, None, 4.x] A standard CMOS inverter drives an aluminum wire on the first metal layer.Assume Rn=4kΩ, Rp=6kΩ. Also, assume that the output capacitance of the inverter is negli-gible in comparison with the wire capacitance. The wire is .5um wide, and the resistivity is0.08 Ω/..a.What is the "critical length" of the wire?b.What is the equivalent capacitance of a wire of this length? (For your capacitance calcula-tions, use Table 4.2 of your book , assume there’s field oxide underneath and nothingabove the aluminum wire)Digital Integrated Circuits - 2nd Ed510.[M, None, 4.x] A 10cm long lossless transmission line on a PC board (relative dielectric con-stant = 9, relative permeability = 1) with characteristic impedance of 50Ω is driven by a 2.5Vpulse coming from a source with 150Ω resistance.a.If the load resistance is infinite, determine the time it takes for a change at the source toreach the load (time of flight).Now a 200Ω load is attached at the end of the transmission line.b.What is the voltage at the load at t = 3ns?c.Draw lattice diagram and sketch the voltage at the load as a function of time. Determinehow long does it take for the output to be within 1 percent of its final value.11.[C, SPICE, 4.x] Assume V DD =1.5V . Also, use short-channel transistor models forhand analy-sis.a.The Figure 0.5 shows an output driver feeding a 0.2 pF effective fan-out of CMOS gates through a transmission line. Size the two transistors of the driver to optimize the delay.Sketch waveforms of V S and V L , assuming a square wave input. Label critical voltages and times.b.Size down the transistors by m times (m is to be treated as a parameter). Derive a first order expression for the time it takes for V L to settle down within 10% of its final voltage pare the obtained result with the case where no inductance is associated with the wire.Please draw the waveforms of V L for both cases, and comment.e the transistors as in part a). Suppose C L is changed to 20pF. Sketch waveforms of V S and V L , assuming a square wave input. Label critical voltages and instants.d.Assume now that the transmission line is lossy. Perform Hspice simulation for three cases:R=100 Ω/cm; R=2.5 Ω/cm; R=0.5 Ω/cm. Get the waveforms of V S , V L and the middle point of the line. Discuss the results.12.[M, None, 4.x] Consider an isolated 2mm long and 1μm wide M1(Metal1)wire over a silicon substrate driven by an inverter that has zero resistance and parasitic output capccitance. How will the wire delay change for the following cases? Explain your reasoning in each case.a.If the wire width is doubled.b.If the wire length is halved.c.If the wire thickness is doubled.d.If thickness of the oxide between the M1 and the substrate is doubled.13.[E, None, 4.x] In an ideal scaling model, where all dimensions and voltages scale with a fac-tor of S >1 :L=350nH/m 10cm C=150pF/m inV DDV DD V S V LC L =0.2pF Figure 0.5Transmission line between two inverters6Chapter 4 Problem Seta.How does the delay of an inverter scale?b.If a chip is scaled from one technology to another where all wire dimensions,including thevertical one and spacing, scale with a factor of S, how does the wire delayscale? How doesthe overall operating frequency of a chip scale?c.Repeat b) for the case where everything scales, except the vertical dimension of wires (itstays constant).。
数字设计原理与实践第四章答案
=W X Y Z Z+W X X Y Z
+W W X Y Z +W X Y Y Z
0
习题4.6(b)
F = A B +A B C D+A B D E+A B C E+A B C E
F X Y Z = X,Y,Z (1， 2, 4, 7) FD = X,Y,Z (0,3,5,6) X,Y,Z (1,2, 4, 7) =F
所以是自对偶的
习题4.47
（e）F’(A,B,…,Z)=FD(A’,B’,…,Z’) FD(A,B,…,Z)=F’(A’,B’,…,Z’) P135 当为1的变量数大于3个时，
00 01
11
10 0
0 0
0 0
0
习题4.24
(X+Y)(X'+Z)=XX'+XZ+X'Y+YZ = XZ+X'Y+YZ （由T11） =XZ+X'Y 证毕 习题4.25
N输入与门可以由N-1个2输入的与来实现。 对于N输入与非门是不可以由N-1个2输入的 与非门来实现的。可举反例来证明。
F=A B A B+A B
习题4.36
A
B
F
0 0 1 1
0 1 0 1
1 0 0 1
F=A
B A B+A B
习题4.39
两输入的与非门可以构成完全集； 由题可知，2 输入的与门，或门，反相器可 以构成完全集，所以只要证明 2 输入的与门， 或门，反相器可以由与非门来表示， AB=((AB)')'=((AB)'·1)' A+B=((A+B)')'=(A'·B')' =((A·A)'·(B·B)')' A'=(A·A)'
数字集成电路课后习题1-4章作业解析
VOH = VGG − VT = VDD − VT 0 + γ
(
(V
SB
+ 2 φf − 2 φf
))
= VGG − VT 0 − γ VOH + 2 φ f + γ 2 φ f = 1.6 − 0.4 − 0.2 VOH + 0.88 + 0.2 0.88 = 1.388 − 0.2 VOH + 0.88
QOX 6 ×1011 ×1.6 ×10−19 0.06 V = = COX 1.6 ×10−6 VT0 =−0.99 − (−0.88) − (−0.188) − 0.060 =+0.018 V
计算 PMOS 器件的阈值电压： kT N D 3 ×1017 = φFn = ln 0.026 ln = 0.44 V q ni 1.4 ×1010
VOL 2 (0.1×10−4 )(8 ×106 )(1.6 − VOL − 0.4) 2 1 270 1.2 0.4 V − − ( ) OL = 0.1 VOL 2 (1.6 − VOL − 0.4) + 0.6 1 + 0.6
+0.99 V φGC = φFn − φG ( gate ) = 0.44 + 0.55 = QB 0 3 ×10−7 = = +0.188 V QB 0 = 3 ×10−7 C / cm 2 COX 1.6 ×10−6 QOX 6 ×1011 ×1.6 ×10−19 = = 0.06 V COX 1.6 ×10−6 VT0 =0.99 − (+0.88) − (+0.188) − 0.060 =−0.138 V
VOH = 1.11V 由此可知，VGG 实际要大于 1.6 V，接近 1.7 V，才能使 VOH 达到 1.2 V。 计算 VOL 时忽略体效应， ∴
数字电子技术基础(第四版)课后习题答案_第四章
第4章触发器[题4.1]画出图P4.1所示由与非门组成的基本RS触发器输出端Q、Q的电压波形，输入端S、R的电压波形如图中所示。
图P4.1[解]见图A4.1图A4.1[题4.2]画出图P4.2由或非门组成的基本R-S触发器输出端Q、Q的电压波形，输出入端S D，R D的电压波形如图中所示。
图P4.2[解]见图A4.2[题4.3]试分析图P4.3所示电路的逻辑功能，列出真值表写出逻辑函数式。
图P4.3 [解]由真值表得逻辑函数式01=+=+SR Q R S Q nn[题4.4]图P4.4所示为一个防抖动输出的开关电路。
当拨动开关S 时，由于开关触点接触瞬间发生振颤，D S 和D R 的电压波形如图中所示，试画出Q 、Q 端对应的电压波形。
图P4.4[解]见图A4.4图A4.4[题4.5]在图P4.5电路中，若CP 、S 、R 的电压波形如图中所示，试画出Q 和Q 端与之对应的电压波形。
假定触发器的初始状态为Q =0。
图P4.5[解]见图A4.5图A4.5[题4.6]若将同步RS触发器的Q与R、Q与S相连如图P4.6所示，试画出在CP信号作用下Q和Q端的电压波形。
己知CP信号的宽度t w = 4 t Pd 。
t Pd为门电路的平均传输延迟时间，假定t Pd≈t PHL≈t PLH，设触发器的初始状态为Q=0。
图P4.6图A4.6[解]见图A4.6[题4.7]若主从结构RS触发器各输入端的电压波形如图P4.7中所给出，试画Q、Q端对应的电压波形。
设触发器的初始状态为Q=0。
图P4.7[解] 见图A4.7图A4.7R各输入端的电压波形如图P4.8所示，[题4.8]若主从结构RS触发器的CP、S、R、DS。
试画出Q、Q端对应的电压波形。
1D图P4.8[解] 见图A4.8图A4.8[题4.9]已知主从结构JK触发器输入端J、K和CP的电压波形如图P4.9所示，试画出Q、Q端对应的电压波形。
设触发器的初始状态为Q = 0。
数字电路第四章答案
数字电路第四章答案【篇一：数字电路答案第四章时序逻辑电路2】p=1，输入信号d被封锁，锁存器的输出状态保持不变；当锁存命令cp=0，锁存器输出q?d，q=d；当锁存命令cp出现上升沿，输入信号d被封锁。
根据上述分析，画出锁存器输出q及 q的波形如习题4.3图（c）所示。
习题4.4 习题图4.4是作用于某主从jk触发器cp、j、k、 rd及 sd 端的信号波形图，试绘出q端的波形图。
解：主从jk触发器的 rd、且为低有效。
只有当rd?sd?1 sd端为异步清零和复位端，时，在cp下降沿的作用下，j、k决定输出q状态的变化。
q端的波形如习题4.4图所示。
习题4.5 习题4.5图（a）是由一个主从jk触发器及三个非门构成的“冲息电路”，习题4.5图（b）是时钟cp的波形，假定触发器及各个门的平均延迟时间都是10ns，试绘出输出f的波形。
cpf cp100ns10nsq（a）f30ns10ns（b）（c）习题4.5图解：由习题4.5图（a）所示的电路连接可知：sd?j?k?1，rd?f。
当rd?1时，在cp下降沿的作用下，且经过10 ns，状态q发生翻转，再经过30ns，f发生状态的改变，f?q。
rd?0时，经过10ns，状态q=0。
根据上述对电路功能的分析，得到q和f的波形如习题4.5图（c）所示。
习题4.6 习题4.6图（a）是一个1检出电路，图（b）是cp及j端的输入波形图，试绘出 rd端及q端的波形图（注：触发器是主从触发器，分析时序逻辑图时，要注意cp=1时主触发器的存储作用）。
cpj（a）qd（c）cp j（b）习题图解：分析习题4.6图（a）的电路连接：sd?1,k?0,rd?cp?q；分段分析习题4.6图（b）所示cp及j端信号波形。
（1）cp=1时，设q端初态为0，则rd?1。
j信号出现一次1信号，即一次变化的干扰，且k=0，此时q端状态不会改变；（2）cp下降沿到来，q端状态变为1，rd?cp，此时cp=0，异步清零信号无效；（3）cp出现上升沿，产生异步清零信号，使q由1变为0，在很短的时间里 rd又恢复到1；（4）同理，在第2个cp=1期间，由于j信号出现1信号，在cp下降沿以及上升沿到来后，电路q端和 rd端的变化与（2）、（3）过程的分析相同，其波形如习题4.6图（c）所示。
数字集成电路分析和设计第四章答案
P4.1. Problem should refer to Figure P4.2.a. All inverters but the CMOS inverter consume static power then the output is high.Notice that in the first three inverters when the input is high, there is always a directconnection from V DD to G ND .b. None of the static inverters consumes power when the input is low because there is nopath from V DD to G ND .c. All inverters but the saturated enhancement inverter has a V OH of 1.2 V.d. Only the CMOS inverter has a V OL of 0 V.e. Except for the CMOS inverter, all the other inverte rs’ functionality depend on therelative sizes of the transistors.P4.2. Problem should refer to Figure P4.1a. Resistive loadb. Saturated-enhancement loadIterate to produce:To compute V OL we can ignore body effect and equate currents:Solve for 0.03OL V V ≈c. Linear-enhancement loadIterate to produce:This tells us that V GG should have been above 1.6V <closer to 1.7 V>.To compute V OL we can ignore body effect and equate currents. Note that the load issaturated even though we call it a linear-enhancement load. The driver is alsosaturated due to the device sizes used.Solve for 0.69V OL V ≈d. CMOSP4.3. For this problem, you are required to use the formulae:We already know that V OH =1.2 V and V OL =0 V. For V S use:Next V IL and V IH are estimated as follows:ThereforeWhen we cut the size of the PMOS device in half, the VTC shifts to the left. So V IL , V S , and V IH will all shift to the left. The recalculation of the switching threshold produces V S =0.566V. We can compute V IL to be roughly 0.533V and V IH to be roughly 0.667V.ThereforeP4.4. Similar approach as in P4.3. Run SPICE to check results.P4.5. First, set up the equation.Now solve for χ.This implies that a very large <W/L>P is needed to reach the desired value. It also reveals the limitations of the models. SPICE would be needed to obtain an acceptable solution if the switching threshold of 0.9V is truly desired.P4.6. SPICEP4.7. The advantages of the pseudo-PMOS is that it can reach a V OH of V DD while the pseudo-NMOS V OH can never reach that value. Additionally, the pseudo-NMOS’s V OH dependson the relative sizings of the inverters.The disadvantage is the dual of its advantage. The pseudo-PMOS inverter can never reach a V OL of 0 V. In addition, the pseudo-PMOS device will have to be approximately twice as large as a pseudo-NMOS device with comparable characteristics. This is due to the unequal mobility of holes and electrons. The pseudo-PMOS’s NMOS pull -down device is twice as strong as the pseudo-NMOS’s PMOS pull -up device, that means that the pseudo-PMOS’s PMOS wi ll have to be bigger than the NMOS device in a pseudo-NMOS.P4.8. a> Circuit is a buffer with degraded outputs.Output swing calculation:When IN DD V V =, output voltage is OH DD TN V V V =-. Since the source of NMOS transistor is not connected to substrate <ground>, we must take into account body effect.When 0IN V V =, output voltage is ||OL TP V V =. Since the source of PMOS transistor is not connected to substrate <V DD >, we must take into account body effect.Therefore the output swing is DD TN V V - to ||TP V with full accounting for body effect.b> Assume that the input is at 0 and the output is at |V TP |. As the input is increased, the output will stay constant until the NMOS device turns on. That will occur at V IN =|V TP |+V TN . The upper transistor behaves as a source follower and will pull the output along as the input rises until the output reaches V DD -V TN . However, as the input is reduced in value the output stays at its highvalue until the PMOS device turns on. This occurs at V IN=V DD-< |V TP|+V TN>. Then the PMOS device acts as a source follower and the output drops linearly to |V TP| as the input is reduced.c> The gain of the circuit is close to unity but slightly below this value. The circuit has poor noise rejection properties as it lacks the regenerative properties <this is a consequence of low gain>.d> SPICE run.P4.9.Resistive Load inverter:Saturated Enhancement Load inverter <ignoring body-effect>:Linear Enhancement Load inverter <ignoring body-effect>:The linear enhancement load inverter requires the largest pull-down device since it has the strongest pull up device. The resistive load inverter is next and the saturated enhancement load requires the smallest pull-down device.P4.10.We will illustrate the process and estimate the solutions for this problem.We already know that V OH=1.2 V and V OL=0 V. For V S use:Next V IL and V IH are estimated as follows:We can compute V IL to be roughly 0.533V.We can compute V IH to be roughly 0.667V.When we double the size of the PMOS device, the VTC shifts to the right. So V IL, V S, and V IH will all shift to the right. The recalculation of the switching threshold produces V S=0.6V.We can compute V IL to be roughly 0.55V and V IH to be roughly 0.65V.P4.11.The peak current would occur when both devices are in saturation and when V out=V in=V S.We can easily compute V S as:P4.12.As the required V OL becomes smaller, the W D/W L ratio becomes larger.P4.13.SPICEP4.14.The expression for the switching threshold of a CMOS inverter is:Solving for χ.Now solving for the ratio of sizes.Solving for χ.Now solving for the ratio of sizes.In the first case <0.6S DD V V >, the PMOS is much larger than the NMOS, so t PLH issmaller and t PHL is larger. The reverse is true for the second case.P4.15 <a> It does not have the regenerative property since the gain is less than one.<b> The last inverter would have an output of about 0.8V.<c> It is not possible to define the noise margin for this gate. Even a properinput eventually produces the incorrect output.P4.16 Both gates would work as a tristate buffer. However, as we shall find out in Chapter 7, the second one is prone to charge-sharing. That is, when the output is high and the EN signal is low, if the input goes high, the output may drop slightly in value due to loss of charge to the adjacent internal node.。
数字集成电路设计 第四章导线.ppt
导线. 17
合肥工业大学应用物理系
接触电阻(contact resistance)
• 布线层之间的转接将给导线带来额外的电阻 – 尽可能地使信号线保持在同一层上并避免过多的接触或通孔 – 使接触孔较大可以降低接触电阻(电流集聚在实际中将限制接触孔 的最大尺寸)
• 典型接触电阻，RC, (最小尺寸) – 金属或多晶至n+、p+以及金属至多晶为 5 ~ 20 – 通孔(金属至金属接触)为1 ~ 5
例4.1 金属导线电容
考虑一条布置在第一层铝上的10cm长，1m宽的铝线，计算总的电容值。
平面(平行板)电容： ( 0.1×106m2 )×30aF/m2 = 3pF
边缘电容：
2×( 0.1×106m )×40aF/m = 8pF
总电容：
11pF
现假设第二条导线布置在第一条旁边，它们之间只相隔最小允许的距离， 计算其耦合电容。
Capacitance-only
注意：这些附加的电路元件并不处在实际的单个点上，而是分布在导 线的整个长度上
导线. 6
合肥工业大学应用物理系
寄生简化
• 电感的影响可以忽略 – 如果导线的电阻很大(例如截面很小的长铝导线的情形) – 外加信号的上升和下降时间很慢
• 采用只含电容的模型 – 当导线很短，导线的截面很大时 – 当所采用的互连材料电阻率很低时
D2 C1R1 C2 R1 R2
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1 r2
2
Vin
c1
c2
ri-1 i-1 ri
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ci-1
ci
rN
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Di C1R1 C2R1 R2 ... Ci R1 R2 ... Ri
(完整word版)数电1-10章自测题及答案(2)
第一章绪论一、填空题1、根据集成度的不同，数字集成电路分位以下四类：小规模集成电路、中规模集成电路、大规模集成电路、超大规模集成电路。
2、二进制数是以2为基数的计数体制,十六体制数是以16为基数的计数体制。
3、二进制数只有0和1两个数码，其计数的基数是2，加法运算的进位规则为逢二进一。
4、十进制数转换为二进制数的方法是：整数部分用除2取余法，小数部分用乘2取整法，十进制数23。
75对应的二进制数为10111.11。
5、二进制数转换为十进制数的方法是各位加权系数之和，二进制数10110011对应的十进制数为179。
6、用8421BCD码表示十进制时，则每位十进制数可用四位二进制代码表示，其位权值从高位到低位依次为8、4、2、1。
7、十进制数25的二进制数是11001，其对应的8421BCD码是00100101。
8、负数补码和反码的关系式是:补码=反码+1。
9、二进制数+1100101的原码为01100101，反码为01100101，补码为01100101。
-1100101的原码为11100101，反码为10011010,补码为10011011。
10、负数-35的二进制数是—100011，反码是1011100，补码是1011101。
二、判断题1、二进制数有0~9是个数码,进位关系为逢十进一。
（）2、格雷码为无权码,8421BCD码为有权码。
（√）3、一个n位的二进制数，最高位的权值是2^n+1. （√）4、十进制数证书转换为二进制数的方法是采用“除2取余法”. (√）5、二进制数转换为十进制数的方法是各位加权系之和。
（√)6、对于二进制数负数，补码和反码相同。
（)7、有时也将模拟电路称为逻辑电路。
（）8、对于二进制数正数,原码、反码和补码都相同. (√）9、十进制数45的8421BCD码是101101。
（）10、余3BCD码是用3位二进制数表示一位十进制数. ( ）三、选择题1、在二进制技术系统中,每个变量的取值为（A ）A、0和1B、0～7C、0~10D、0~F2、二进制权值为（B ）A、10的幂B、2的幂C、8的幂D、16的幂3、连续变化的量称为( B )A、数字量B、模拟量C、二进制量D、16进制量4、十进制数386的8421BCD码为（B ）A、0011 0111 0110B、0011 1000 0110C、1000 1000 0110D、0100 1000 01105、在下列数中，不是余3BCD码的是( C ）A、1011B、0111C、0010D、10016、十进制数的权值为（D ）A、2的幂B、8的幂C、16的幂D、10的幂7、负二进制数的补码等于（D )A、原码B、反码C、原码加1D、反码加18、算术运算的基础是（A ）A、加法运算B、减法运算C、乘法运算D、除法运算9、二进制数-1011的补码是（D ）A、00100B、00101C、10100D、1010110、二进制数最高有效位（MSB）的含义是( A )A 、最大权值B 、最小权值C 、主要有效位D 、中间权值第二章 逻辑代数基础一、填空题1、逻辑代数中三种最基本的逻辑运算是与运算、或运算、非运算。
数字集成电路设计第四章习题
1. 如下图所示时钟数, 根据下表中提供的电容电阻数据, 计算从节点A到节点B的Elmore 延时。
图计算延时的RC树
表Values of the components in the RC tree
Resistor Value( ) Capacitor Value(fF)
R1 0.25 C1 250
R2 0.25 C2 750
R3 0.50 C3 250
R4 100 C4 250
R5 0.25 C5 1000
R6 1.00 C6 250
R7 0.75 C7 500
R8 1000 C8 250
3等分并插入2个传播延时为100ps的反相器，计算在这种情况下各层上整个导线的传播延时。
3．设计一个时钟分布网络，在各个时钟之间的最小偏差是很关键的问题，从一个时钟网络中抽象出如下图所示的RC网络，最初CLK3比CLK1和CLK2的路径更短，为了补偿这一不平衡，在CLK3的路径中插入一个传输门。
1)写出节点CLK3、CLK1和CLK2的时间常数，假设传输门用R3模拟；
2)如果R1=R2=R4=R5=R，C1=C2=C3=C4=C5=C，R3为多大时可以平衡；
3)当R=750Ω，C=200fF，传输门有多大的W/L比可以消除偏差；。
《数字电路-分析与设计》第四章习题及解答3(部分) 北京理工大学出版社
4-42设计一个具有多输出函数的组合网络。
该网络有两个输入信号X 1和X 0，两个控制信号C 1和C 0，以及两个输出函数F 1和F 0。
控制信号对输出函数的影响由下表所示：例如：当C 1=1且C 0=0时，F 1(X 1，X 0，C 1，C 0)=0而F 0(X 1，X 0，C 1，C 0)=X 0。
请选择合适的SSI 或者MSI 实现此组合逻辑网络。
解：01101011010101100),,,(C C X C C C C X C C C C X X F ⋅+⋅+⋅+⋅=011011C C X C C X ⋅+⋅= 01C X =01001001010101000),,,(C C X C C X C C C C C C X X F ⋅+⋅+⋅+⋅=10C X =用SSI 实现组合网络：用MSI 实现组合网络：4-43试分析图题4-43所示电路的竞争冒险现象。
画出在A =B =0的情况下，C由“0”变为“1”、再由“1”变为“0”时的各级波形。
设门电路的传输延迟时间为t pd 。
说明在什么情况下会产生毛刺，应如何消除。
X 1C 0F 1(X 1,X 0,C 1,C 0) X 0 C 1F 0(X 1,X 0,C 1,C 0)1 X 0“0“0X 1(X 1,X 0,C 1,C 0)1 X 0“0“0X 0(X 1,X 0,C 1,C 0)解：函数的逻辑表达式如下：C B C A F +++= )C B )(C A (++=当A =B =0时，0=⋅=C C F 。
所以C 变化时，输出将产生正尖峰脉冲（“1”型）冒险。
A =B =0时的电路图如下：A =B =0且C 变化时的波形图如下：加“选通”信号可消除冒险（“毛刺”）现象，如下图所示：图题 4-43C “0”C “0”CF CEC。
数字电子技术第四章习题答案
第四章习题答案
4.1 Y=A’B’C’+A’BC+AB’C+ABC’
A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 Y 1 0 0 1 0 1 1 0
三变量奇偶检测电 路，当输入便两种 有偶数个1 有偶数个1时，输出 否则为0 为1，否则为0。
《数字电子技术基础》第五版 数字电子技术基础》
4.12 用3线-8线译码器实现多输出逻辑函数。 线译码器实现多输出逻辑函数。 线 线译码器实现多输出逻辑函数
' ' Y1 = AC = AB ' C + ABC = m5 + m7 = ( m5 m7 )' = (Y5'Y7' )' ' ' ' ' Y2 = A ' B ' C + AB ' C '+ BC = ( m1m3 m4 m7 )' = (Y1'Y3'Y4'Y7' )' ' ' ' Y3 = B ' C '+ ABC ' = ( m0 m4 m6 )' = (Y0'Y4'Y6' )'
Y’2(2) 1 1 0 1 1
Y’2(1) 1 1 1 0 1
D2 1 1 1 1 0
设片1优先级别 设片 优先级别 最低， 优先 最低，片4优先 级别最高。 级别最高。 输出5位数为原 输出 位数为原 码D4D3D2D1D0
《数字电子技术基础》第五版 数字电子技术基础》
《数字电路-分析与设计》第四章习题及解答2(部分) 北京理工大学出版社
4-15试分析图题4-15所示各电路的逻辑功能。
列出真值表，写出函数表达式。
解： （a ）加中间变量如右图所示：.)()(;;;32413121B A B A B A B A B AG G G B A B AB B G G B A AB A G A G B A G ⊕=+=+++=⋅=+=+=⋅=+=+=⋅=⋅=∴;B A )B A B A (A G A F =+⋅=⋅=41A B A G F =⊕==42⊙B ； ;)(43B A B A B A B B G F =+=⋅=F 1、F 2和F 3的真值表如右所示：由F 1、F 2和F 3的逻辑表达式知，这是一位比较器。
（b ）加中间变量如右图所示： ;;;13121B A B AB B G G B A AB A G A G B A G +=+=⋅=+=+=⋅=⋅= ∴.;12321AB G F B A B A B A G G F ==⊕=+=+=F 1和F 2的真值表如右所示：由F 1和F 2的逻辑表达式知，这是一位半加器。
F 1是和，F 2是进位。
12 （b ）23（a ）112（b ）2 3（a ）14-16图题4-16是一个多功能逻辑运算电路，图中S3、S2、S1、S0为控制输入端。
试列表说明该电路在S3、S2、S1、S0的各种取值组合下F与A、B的逻辑关系。
解：由图写出F关于变量S3、S2、S1、S0、A、B的函数表达式：ABSSBSBAABSF++⊕+=123可以看出，以7与8号之间为分界线，上、下位置对称的函数F互为补函数。
4-19试分析图题4-19所示电路的逻辑功能。
列出真值表，写出函数表达式。
图题4-16F10图题4-19F1F2（a）F（b）解：由图（a ）知：∑∑=+++=⋅⋅⋅==++=⋅⋅=)7,4,2,1(;)6,5,3(01201201201201201201201220120120120120120121m A A A A A A A A A A A A A A A A A A A A A A A A F m A A A A A A A A A A A A A A A A A A FF 1和F 2的真值表如右所示：由图（b ）知：∑=+++++=+++=⋅+⋅+⊕⋅+⋅=)14,12,6,5,3,2(0)(m D C AB D ABC D BC A D C B A D C B A CD B A D AB D BC A D C B A C B AAB B A D C B A C B A F F 的真值表如右所示： 4-21写出图题4-21所示逻辑电路输出函数的最小项之和式与最大项之积式。
(完整版)数字电子技术第四章答案
习题44－1 分析图P4－1所示的各组合电路，写出输出函数表达式，列出真值表，说明电路的逻辑功能。
解：图（a ）：1F AB =；2F A B =e ；3F AB = 真值表如下表所示： A B 1F2F3F0 0 0 1 0 0 1 0 0 1 1 0 1 0 0 111其功能为一位比较器。
A>B 时，11F =；A=B 时，21F =；A<B 时，31F = 图（b ）：12F AB AB F AB =+=； 真值表如下表所示： A B 1F2F功能：一位半加器，1F 为本位和，2F 为进位。
图（c ）：1(0,3,5,6)(1,2,4,7)F M m ==∑∏2(0,1,2,4)(3,5,6,7)F M m ==∑∏真值表如下表所示：功能：一位全加器，1F 为本位和，2F 为本位向高位的进位。
图（d ）：1F AB =；2F A B =e ；3F AB =功能：为一位比较器，A<B 时，1F ＝1；A=B 时，2F ＝1；A>B 时，3F ＝14－2 分析图P4－2所示的组合电路，写出输出函数表达式，列出真值表，指出该电路完成的逻辑功能。
解：该电路的输出逻辑函数表达式为：100101102103F A A x A A x A A x A A x =+++因此该电路是一个四选一数据选择器，其真值表如下表所示：1A0AF0 0 0x 0 1 1x 1 0 2x 1 13x4－3 图P4－3是一个受M 控制的代码转换电路，当M ＝1时，完成4为二进制码至格雷码的转换；当M ＝0时，完成4为格雷码至二进制的转换。
试分别写出0Y ,1Y ,2Y ,3Y 的逻辑函数的表达式，并列出真值表，说明该电路的工作原理。
解：该电路的输入为3x 2x 1x 0x ，输出为3Y 2Y 1Y 0Y 。
真值表如下： 3x2x1x0x3Y2Y1Y0YM=10 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 1 0 0 1 1 0 0 1 0 0 1 0 0 0 1 1 0 0 1 0 1 0 1 1 1 0 1 1 0 0 1 0 1 01 1 0 0 1 0 0 M=0 1 0 0 0 1 1 1 1 1 0 0 1 1 1 1 0 1 0 1 0 1 1 0 0 1 0 1 1 1 1 0 1 1 1 0 0 1 0 0 0 1 1 0 1 1 0 0 1 1 1 1 0 1 0 1 1 111111由此可得：1M =当时，33232121010Y x Y x x Y x x Y x x =⎧⎪=⊕⎪⎨=⊕⎪⎪=⊕⎩ 完成二进制至格雷码的转换。
《数字电路与系统设计》第4章习题答案
4.1分析图4.1电路的逻辑功能解：（1）推导输出表达式（略）(2) 列真值表（略）4.6 试设计一个将8421BCD 码转换成余3码的电路。
解： 电路图略。
4.7 在双轨输入条件下用最少与非门设计下列组合电路： 解：略4.8 在双轨输入信号下，用最少或非门设计题4.7的组合电路。
解：将表达式化简为最简或与式：(1)F=(A+C)(⎺A+B+⎺C)= A+C+⎺A+B+⎺C(2)F=(C+⎺D)(B+D)(A+⎺B+C)= C+⎺D+B+D+A+⎺B+C(3)F=(⎺A+⎺C)(⎺A+⎺B+⎺D)(A+B+⎺D)= ⎺A+⎺C+⎺A+⎺B+⎺D+A+B+⎺D(4)F=(A+B+C)(⎺A+⎺B+⎺C)= A+B+C+⎺A+⎺B+⎺C 4.9 已知输入波形A 、B 、C 、D ，如图P4.4所示。
采用与非门设计产生输出波形如F 的组合电路。
解： F=A ⎺C+⎺BC+C ⎺D 电路图略4.10 电话室对3种电话编码控制，按紧急次序排列优先权高低是：火警电话、急救电话、普通电话，分别编码为11，10，01。
试设计该编码电路。
解：略4.11 试将2/4译码器扩展成4/16译码器 解：A 3A 2A 1 A 0⎺Y 0⎺Y 1⎺Y 2⎺Y 3 ⎺Y 4 ⎺Y 5⎺Y 6⎺Y 7 ⎺Y 8⎺Y 9⎺Y 10⎺Y 11 ⎺ Y 12⎺Y 13⎺Y 14⎺Y 15A 1 ⎺EN ⎺Y 3A 0 2/4 ⎺Y 2译码器 ⎺Y 1⎺Y 0⎺EN A 1 2/4(1)A 0 ⎺Y 0⎺Y 1⎺Y 2⎺Y 3⎺EN A 1 2/4(2) A 0 ⎺Y 0⎺Y 1⎺Y 2⎺Y 3 ⎺EN A 1 2/4(3) A 0 ⎺Y 0⎺Y 1⎺Y 2⎺Y 3 ⎺EN A 1 2/4(4) A 0 ⎺Y 0⎺Y 1⎺Y 2⎺Y 34.12试用74138设计一个多输出组合网络，它的输入是4位二进制码ABCD，输出为：F1：ABCD是4的倍数。
数字电路与逻辑设计第四章答案PPT课件
转了就不会翻回来。但主从SR触发器在CLK＝1期
间S、R多次改变时主触发器会多次翻转。
主从JK触发器芯片74HC72简介
置位端
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J1 J2 J3 Q GND
74HC72
VCC PR CLK K3 K2 K1 Q
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CHAPTER 4P4.1. Problem should refer to Figure P4.2.a. All inverters but the CMOS inverter consume static power then the output is high. Notice that in the first three inverters when the input is high, there is always a direct connection from V DD to G ND .b. None of the static inverters consumes power when the input is low because there is no path from V DD to G ND .c. All inverters but the saturated enhancement inverter has a V OH of 1.2 V.d. Only the CMOS inverter has a V OL of 0 V.e. Except for the CMOS inverter, all the other invert ers’ functionality depend on the relative sizes of the transistors. P4.2. Problem should refer to Figure P4.1a. Resistive loadOH DD V V =()()()()()()()6310.1111.20.055V 1260101010 1.20.4DD DDOL W L L DD T N OX L DD T V V V kR V V C R V V μ-==+-+-==+⨯-b.Saturated-enhancement load()001.20.40.988OH DD T DD T DD T V V V V V V V γγγ=-=-+=--+=--=-Iterate to produce:0.733V OH V =To compute V OL we can ignore body effect and equate currents:()22()2()1N OL L sat DD OL TL IDD TI OL I DD OL TL CN L OLCN IV W V V V WV V V L V V V E L V E L μν⎡⎤--∴--=⎢⎥--+⎛⎫⎣⎦+⎪⎝⎭()2262(270)(0.2)(810/)(1.20.4)1.2 1.20.40.12(1.20.4)0.610.6OL OL OL OL OL cmV m cm s V V s V V V μ⎡⎤⨯---∴--=⎢⎥--+⎛⎫⎣⎦+ ⎪⎝⎭Solve for 0.03OL V V ≈ c.Linear-enhancement load()001.60.41.388OH GG T DD T GG T V V V V V V V γγγ=-=-+=--=--=-Iterate to produce:1.11V OH V =This tells us that V GG should have been above 1.6V (closer to 1.7 V).To compute V OL we can ignore body effect and equate currents. Note that the load issaturated even though we call it a linear-enhancement load. The driver is also saturated due to the device sizes used.22()()()()I sat DD TI L sat GG OL TL DD TI CN I GG OL TL CN LW V V W V V V V V E L V V V E L νν---∴=-+--+ 6262(0.1)(810/)(1.60.4)(0.1)(810/)(1.20.4)(1.20.4)0.6(1.60.4)0.6OL OL m cm s V m cm s V μμ⨯--⨯-∴=-+--+ Solve for 0.69V OL V ≈ d. CMOSOH DD V V = 0V OL V =P4.3. For this problem, you are required to use the formulae:2(/)()1(/)out DD TP N P TN IL N P V V V k k V V k k --+=+ 2(/)()1(/)out TN P N DD TP IH P N V V k k V V V k k ++-=+We already know that V OH =1.2 V and V OL =0 V. For V Suse:()4,16:10.80.410.611N P S W W X V Vλλ======+==+Next V IL and V IH are estimated as follows:2(/)()2 1.20.4(1)(0.4)2 1.20.551(/)1(1)2out DD TP N P TN out out IL N P V V V k k V V V V V k k --+---+-====++ 2(/)()20.4(1)(1.20.4)2 1.20.651(/)1(1)2out TN P N DD TP out out IH P N V V k k V V V V V V k k ++-++-+====++Therefore0.5500.551.20.650.55L H NM V NM V V=-==-=When we cut the size of the PMOS device in half, the VTC shifts to the left. So V IL , V S , and V IHwill all shift to the left. The recalculation of the switching threshold produces V S =0.566V. We can compute V IL to be roughly 0.533V and V IH to be roughly 0.667V. Therefore0.53300.5331.20.6670.533L H NM V NM V V=-==-=P4.4. Similar approach as in P4.3. Run SPICE to check results. P4.5. First, set up the equation.231DD TP TN S DD V V V V V χχ-+==+Now solve for χ.()()()223312331233113322331.20.401.20.4DD TP TN DD DD DD TP DD TN DD TP DD TN DD TP DD TNV V V V V V V V V V V V V V V V V χχχχχχ-+=+-=--=---===--This implies that a very large (W/L)P is needed to reach the desired value. It also reveals the limitations of the models. SPICE would be needed to obtain an acceptable solution if the switching threshold of 0.9V is truly desired. P4.6. SPICEP4.7. The advantages of the pseudo-PMOS is that it can reach a V OH of V DD while the pseudo-NMOS V OH can never reach that value. Additionally, the pseudo-NMOS’s V OH depends on the relative sizings of the inverters.The disadvantage is the dual of its advantage. The pseudo-PMOS inverter can never reach a V OL of 0 V. In addition, the pseudo-PMOS device will have to be approximately twice as large as a pseudo-NMOS device with comparable characteristics. This is due to the unequal mobility of holes and electrons. The pseudo-PMOS’s NMOS pull -down device is twice as strong as the pseudo-NMOS’s PMOS pull -up device, that means that the pseudo-PMOS’s PMOS will have to be bigger than the NMOS device in a pseudo -NMOS.P4.8. a) Circuit is a buffer with degraded outputs.Output swing calculation:When IN DD V V =, output voltage is OH DD TN V V V =-. Since the source of NMOS transistor is not connected to substrate (ground), we must take into account body effect.When 0IN V V =, output voltage is ||OL TP V V =. Since the source of PMOS transistor is not connected to substrate (V DD ), we must take into account body effect.Therefore the output swing is DD TN V V - to ||TP V with full accounting for body effect.b) Assume that the input is at 0 and the output is at |V TP |. As the input is increased, the output will stay constant until the NMOS device turns on. That will occur at V IN =|V TP |+V TN . The upper transistor behaves as a source follower and will pull the output along as the input rises until the output reaches V DD -V TN . However, as the input is reduced in value the output stays at its high value until the PMOS device turns on. This occurs at V IN =V DD -( |V TP |+V TN ). Then the PMOS device acts as a source follower and the output drops linearly to |V TP | as the input is reduced.V OUTV INDD TP TN V OH =V DD -V V OL =|Vc) The gain of the circuit is close to unity but slightly below this value. The circuit has poor noise rejection properties as it lacks the regenerative properties (this is a consequence of low gain).d) SPICE run.P4.9. Resistive Load inverter:2[2()]1DD OL N n ox OH T OL OL L N OL C V V W C V V V V R L V E L μ-=--⎛⎫+ ⎪⎝⎭ 621.20.1(270)(1.610)[2(1.20.4)0.10.1]0.1100.110.60.2N N W k W mμ--⨯=--⎛⎫+ ⎪⎝⎭∴= Saturated Enhancement Load inverter (ignoring body-effect):()22()2()1N ox out L sat ox DD out TL Iin TI out I DD out TL CN L outCN I C V W C V V V W V V V L V V V E L V E L μν⎡⎤----=⎢⎥--+⎛⎫⎣⎦+⎪⎝⎭6422(270)(1.610)0.1(10)(8)(1.6)(1.20.10.4)[2(1.20.4)0.10.1]0.10.1(1.20.10.4)0.610.60.1I N W W mμ--⨯----=--+⎛⎫+ ⎪⎝⎭∴= Linear Enhancement Load inverter (ignoring body-effect):()22()2()1N ox out L sat ox DD out TL Iin TI out I DD out TL CN L outCN I C V W C V V V W V V V L V V V E L V E L μν⎡⎤----=⎢⎥--+⎛⎫⎣⎦+⎪⎝⎭6422(270)(1.610)0.1(10)(8)(1.6)(1.60.10.4)[2(1.20.4)0.10.1]0.10.1(1.60.10.4)0.610.60.6I N W W mμ--⨯----=--+⎛⎫+ ⎪⎝⎭∴= The linear enhancement load inverter requires the largest pull-down device since it has the strongest pull up device. The resistive load inverter is next and the saturated enhancement load requires the smallest pull-down device.P4.10. We will illustrate the process and estimate the solutions for this problem. We already know that V OH =1.2 V and V OL =0 V. For V Suse:()0.4,0.8:1.410.80.41.410.5661 1.41N P S W um W um X V ======+==+Next V IL and V IH are estimated as follows:2(/)()1(/)out DD TP N P TN IL N P V V V k k V V k k --+=+2 1.20.4(2)(0.4)20.61(2)3out out IL V V V ---+-==+ We can compute V IL to be roughly 0.533V.2(/)()1(/)out TN P N DD TP IH P N V V k k V V V k k ++-=+20.4(2)(1.20.4)221(2)3out out IH V V V ++-+==+We can compute V IH to be roughly 0.667V.When we double the size of the PMOS device, the VTC shifts to the right. So V IL , V S , and V IHwill all shift to the right. The recalculation of the switching threshold produces V S =0.6V. We can compute V IL to be roughly 0.55V and V IH to be roughly 0.65V.P4.11. The peak current would occur when both devices are in saturation and when V out =V in =V S .We can easily compute V S as:||1DD TP TNS V V V V χχ-+=+χ==1.4=1.2|0.4|(1.4)(0.4)0.571 1.4S V V -+==+242()(0.4)(10)(8)(1.6)(0.570.4)20()(0.570.4)0.6sat ox S T DSS T C W C V V I A V V E L νμ---==≈-+-+ P4.12. As the required V OL becomes smaller, the W D /W L ratio becomes larger. P4.13. SPICEP4.14. The expression for the switching threshold of a CMOS inverter is:||1DD TP TNS V V V V χχ-+=+a. 0.6S DD V V = Solving for χ.()||0.610.60.6||0.60.4||0.60.4||DD TP TNDD DD DD DD TP TN DD TN DD TP DD TN DD TP V V V V V V V V V V V V V V V V V χχχχχχχ-+=++=-+-=--=- ()()0.41.20.40.4||0.250.60.61.20.4DD TP DD TN V V V V χ--===--Now solving for the ratio of sizes.χ==0.2564PNWW=∴=b.0.4S DDV V=Solving for χ.()||0.410.40.4||0.40.6||0.40.6||DD TP TNDDDD DD DD TP TNDD TN DD TPDD TN DD TPV V VVV V V V VV V V VV V V Vχχχχχχχ-+=++=-+-=--=-()()0.61.20.40.6||40.40.41.20.4DD TPDD TNV VV Vχ--===--Now solving for the ratio of sizes.χ==42PNWW==In the first case (0.6S DDV V=), the PMOS is much larger than the NMOS, so t PLH is smaller and t PHL is larger. The reverse is true for the second case.P4.15 (a) It does not have the regenerative property since the gain is less than one.(b) The last inverter would have an output of about 0.8V.(c) It is not possible to define the noise margin for this gate. Even a properinput eventually produces the incorrect output.P4.16 Both gates would work as a tristate buffer. However, as we shall find out in Chapter 7, the second one is prone to charge-sharing. That is, when the output is high and the EN signal is low, if the input goes high, the output may drop slightly in value due to loss of charge to the adjacent internal node.。