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
Verilog HDL数字设计与综合(第二版) 第四章课后习题答案
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1.模块的基本组成部分有哪些?哪几个部分必须出现?答:模块的基本组成部分包括:关键字module ,模块名、端口列表、端口声明和可选的参数声明,模块内部还有5个基本组成部分是:变量声明、数据流语句、低层模块实例、行为语句块以及任务和函数。
在模块的所有组成部分中,只有module 、模块名和endmodule 必须出现,其他部分都是可选的,用户可以根据设计的需要随意选用。
2.一个不与外部环境交互的模块是否有端口?模块定义中是否有端口列表?答:严格意义上讲,一个模块如果与外界没有任何交互,那么自然是不需要有而且是不可能有端口的。
在没有端口的情况下,端口列表自然也是不存在的。
3.一个4位并行移位寄存器的I/O 引脚如下图所示。
写出模块shift_reg 的定义,只需写出端口列表和端口定义,不必写出模块的内部结构。
答:模块代码如下:module shift_reg(clock,reg_in,reg_out);input clock;input [3:0] reg_in;output [3:0] reg_out;//********XXXXXXX**************endmodule4.定义一个顶层模块stimulus ,在其中声明reg 变量REG_IN (4位)和CLK (1位)以及wire变量REG_OUT (4位)。
在其中调用(实例引用)模块shift_reg ,实例名为sr1,使用顺序端口连接。
答:模块代码如下module stimulus();reg CLOCK;reg [3:0] REG_IN;wire [3:0] REG_OUT;//*******connect in order************shift_reg sr1(CLOCK,REG_IN,REG_OUT);endmodulereg_in [3:0] clockreg_out [3:0]32Verilog HDL数字设计与综合(第二版)5.将上题的端口连接方法改为命名连接。
数字集成电路习题答案PPT课件
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的最小尺寸为(W=0.375m,L=0.25m,即W/L=0.375/0.25)
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DB Ck RBk k 1 R1C1 R1C2 (R1 R3 )C3 R1C4 (R1 R3 )C5 (R1 R3 R6 )C6 (R1 R3 )C7 (R1 R3 R6 R8 )C8 0.25* 250 0.25*750 (0.25 0.5) * 250 0.25* 250 (0.25 0.5) *1000 (0.25 0.5 1) * 250 (0.25 0.5) *500 (0.25 0.5 11000) * 250 62.5 187.5 187.5 62.5 750 437.5 375 250437.5 0.2525(ns)
【精品】数字集成电路电路、系统与设计第二版课后练习题第六章CMOS组合逻辑门的设计
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【精品】数字集成电路--电路、系统与设计(第二版)课后练习题第六章CMOS组合逻辑门的设计第六章 CMOS组合逻辑门的设计1.为什么CMOS电路逻辑门的输入端和输出端都要连接到电源电压?CMOS电路采用了MOSFET(金属氧化物半导体场效应管)作为开关元件,其中N沟道MOSFET(NMOS)和P沟道MOSFET(PMOS)分别用于实现逻辑门的输入和输出。
NMOS和PMOS都需要连接到电源电压,以使其能够正常工作。
输入端连接到电源电压可以确保信号在逻辑门中正常传递,输出端连接到电源电压可以确保输出信号的正确性和稳定性。
2.为什么在CMOS逻辑门中要使用两个互补的MOSFET?CMOS逻辑门中使用两个互补的MOSFET是为了实现高度抗干扰的逻辑功能。
其中,NMOS和PMOS分别用于实现逻辑门的输入和输出。
NMOS和PMOS的工作原理互补,即当NMOS导通时,PMOS截止,当PMOS导通时,NMOS截止。
这样的设计可以在逻辑门的输出上提供高电平和低电平的稳定性,从而提高逻辑门的抗干扰能力。
3.CMOS逻辑门的输入电压范围是多少?CMOS逻辑门的输入电压范围通常是在0V至电源电压之间,即在低电平和高电平之间。
在CMOS逻辑门中,低电平通常定义为输入电压小于0.3Vdd(电源电压的30%),而高电平通常定义为输入电压大于0.7Vdd(电源电压的70%)。
4.如何设计一个基本的CMOS逻辑门?一个基本的CMOS逻辑门可以由一个NMOS和一个PMOS组成。
其中,NMOS的源极连接到地,栅极连接到逻辑门的输入,漏极连接到PMOS的漏极;PMOS的源极连接到电源电压,栅极连接到逻辑门的输入,漏极连接到输出。
这样的设计可以实现逻辑门的基本功能。
5.如何提高CMOS逻辑门的速度?可以采取以下方法来提高CMOS逻辑门的速度:•减小晶体管的尺寸:缩小晶体管的尺寸可以减小晶体管的电容和电阻,从而提高逻辑门的响应速度。
•优化电源电压:增加电源电压可以提高晶体管的驱动能力,从而加快逻辑门的开关速度。
chap4-6
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第四章数字集成电路的基本单元电路-动态CMOS电路动态逻辑电路的特点静态电路:靠管子稳定的导通、截止来保持输出状态动态电路:靠电容来保存信息V DDV V V outΦDD AMMP2P1A BC LB Y =A .B M MN2V out动态电路的优点AN1B:相对NMOS 电路:动态电路可降低功耗,无比电路电路:用动态电路简化电路提高速度相对CMOS 电路:用动态电路简化电路,提高速度—预充求值动态CMOS 电路的构成Φ=0,预充;Φ=1,求值V DDV ΦV outA C LoutA M1B存在的问题:Φ=0,A =B =1,V V 解决了预充过程OH 小于DD下拉支路导通问题outV outΦΦ富NMOS 动态电路Φ=0,预充;Φ=1,求值富PMOS 动态电路Φ=1,预充;Φ=0,求值下降时间影响速度上升时间影响速度YCBNΦY AB C=+Y AB C=+富NMOS 电路实现富PMOS 电路实现—预充求值电路中的电荷分享问题M1V V out (0) =V (0) =0V DD ()1()M1C B1Φf L DD L V V C V C C V C )(1+=出现电荷分享的条件:时LDDL DD L f C C C C V /111+=+=Φ=0时,A =0;Φ=1时,A =1;B 始终为0。
电荷分享过程中的节点电平变化M1V outΦ极端情况:C L =C 1, 则V f =V DD /2一般情况:般情况:C L >C 11C V V V V =−−()outDD DD TN LCMOS 管电容的耦合作用对电荷分享的影响V DDV outC C GDC V A AALC C GSV 1AC LC C GS GD V 1V out 11—预充求值电路的级连举例A=B=1,C=0M P1V outΦCV1M N1V2不能用富富NMOS注意:NMOS与富NMOS(或富PMOS与富PMOS)电路直接级连。
数字集成电路--电路、系统与设计(第二版)课后练习题第六.
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数字集成电路--电路、系统与设计(第⼆版)课后练习题第六.Digital Integrated Circuits - 2nd Ed 11 DESIGN PROJECT Design, lay out, and simulate a CMOS four-input XOR gate in the standard 0.25 micron CMOS process. You can choose any logic circuit style, and you are free to choose how many stages of logic to use: you could use one large logic gate or a combination of smaller logic gates. The supply voltage is set at 2.5 V! Your circuit must drive an external 20 fF load in addition to whatever internal parasitics are present in your circuit. The primary design objective is to minimize the propagation delay of the worst-case transition for your circuit. The secondary objective is to minimize the area of the layout. At the very worst, your design must have a propagation delay of no more than 0.5 ns and occupy an area of no more than 500 square microns, but the faster and smaller your circuit, the better. Be aware that, when using dynamic logic, the precharge time should be made part of the delay. The design will be graded on themagnitude of A × tp2, the product of the area of your design and the square of the delay for the worst-case transition.。
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
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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).。
数字集成电路:电路系统与设计(第二版)
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数字集成电路:电路系统与设计(第二版)简介《数字集成电路:电路系统与设计(第二版)》是一本介绍数字集成电路的基本原理和设计方法的教材。
本书的内容覆盖了数字电路的基础知识、逻辑门电路、组合逻辑电路、时序逻辑电路、存储器和程序控制电路等方面。
通过学习本书,读者可以了解数字集成电路的概念、设计方法和实际应用。
目录1.数字电路基础知识 1.1 数字电路的基本概念 1.2 二进制系统与数制转换 1.3 逻辑运算与布尔代数2.逻辑门电路 2.1 与门、或门、非门 2.2 与非门、或非门、异或门 2.3 多输入门电路的设计方法3.组合逻辑电路 3.1 组合逻辑电路的基本原理 3.2 组合逻辑电路的设计方法 3.3 编码器和译码器4.时序逻辑电路 4.1 时序逻辑电路的基本原理 4.2 同步时序电路的设计方法 4.3 异步时序电路的设计方法5.存储器电路 5.1 存储器的基本概念 5.2 可读写存储器的设计方法 5.3 只读存储器的设计方法6.程序控制电路 6.1 程序控制电路的基本概念 6.2 程序控制电路的设计方法 6.3 微程序控制器的设计方法内容概述1. 数字电路基础知识本章主要介绍数字电路的基本概念,包括数字电路与模拟电路的区别、数字信号的表示方法以及数制转换等内容。
此外,还介绍了数字电路中常用的逻辑运算和布尔代数的基本原理。
2. 逻辑门电路逻辑门电路是数字电路中的基本组成单元,本章主要介绍了与门、或门、非门以及与非门、或非门、异或门等逻辑门的基本原理和组成。
此外,还介绍了多输入门电路的设计方法,以及逻辑门电路在数字电路设计中的应用。
3. 组合逻辑电路组合逻辑电路是由逻辑门电路组成的,本章主要介绍了组合逻辑电路的基本原理和设计方法。
此外,还介绍了编码器和译码器的原理和应用,以及在数字电路设计中的实际应用场景。
4. 时序逻辑电路时序逻辑电路是在组合逻辑电路的基础上引入了时序元件并进行时序控制的电路。
本章主要介绍了时序逻辑电路的基本原理和设计方法,包括同步时序电路和异步时序电路的设计。
电路与电子学基础第二版第四章答案
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电路与电子学基本第二版第四章答案4.1 解:用万用表测量二极管的正向直流电阻,选择量程越大,通二极管的电流就减小,由二极管的伏安特性曲线可知,电流急剧减小时,电压减小的很慢,所以测量出来的电阻值会大副增大。
4.2 (a) D 导通,U ab =12V(b) D 1,D 2截止,U ab =0V 改为:D 1导通,D 2截止,U ab =0V (c) D 1截止,D 2导通,U ab =-12V 4.3 (a) U=-5V I=K10)5(5--?=1mA(b) U=-5V I=0A (c) U=3V I=K 1)5(3--=8mA (d) U=8V I=K61212--=4mA?4.4 图4.5 (a) 因为30V>D 1z 的稳定电压6V ,所以D 1z 导通,D 2z 稳压,故U 0=0.7+9=9.7V (b) 因为30V> D 1z + D 2z 的稳定电压,所以D 1z 与 D 2z 都起稳压作用,故U 0=6+9=15V (c) 因为30V> D 1z 的稳定电压6V ,所以D 1z 稳定,D 2z 导通,故U 0=6+0.7=6.7V (d) 因为30V> D 1z 的正向导通电压,所以D 1z 导通,D 2z 截止,故U 0=0.7V (e) 因为30V>9V ,30V>6V ,故D 1z 起稳压作用,D 2z 截止,故U 0=6V 4.6 ① 因为12V>6V ,所以D z 稳压,故U 0=6V ,I z =K2620--K26=4mA<5mA ,稳压效果差② 因为5V<6V ,所以D z 不稳压,有计算可知,D z 视为开路,所以U 0=5V ,I z =0mA4.7 图4.8 ⑴此晶体管类型为PNP 型,1为集电极,2为基极,3为发射极。
⑵β=03.02.1=404.9 (a)饱和区(b)放大区 (c)放大区 (d)截止区 (e)放大区 (f)截止(g)临界饱和 (h)损坏4.10 β=50,I CEO =0,U CEO BR )(=25V ,P CM =504.11 选择β=50的管子,因为虽然β=200的放大系数大,但其I CEO 也较大,在考虑温度影响的情况下,应选用温度稳定性好的,故选β=50的管子。
(完整版)二篇4章2浙大版集成电路课后答案
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题2.4.18 试用负边沿JK 触发器和“与-或-非”门构成一个四位数码并行寄存和一个四位数码串行输入右移移位寄存器。
解:令C 是并行寄存数据和实现右向移位操作的控制端,其用JK 触发器构成的框图如图所示:令C=1并行存数,C=0时为右移串入后,得出各组合电路的逻辑函数,现以1J 3和1K 3函数为例,列出真值表,求出函数式,其它式子也照此类推。
输 入 输 出 C Q 2 D 3 1J 3 1K 30 0 0 0 1 0 0 1 0 1 0 1 0 1 0 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 0 1 1 1 11 0233311Q C CD K J +==122211Q C CD K J +==011111Q C CD K J +==SRD C CD K J +==00011由四个函数式画出的电路图如图所示:题2.4.19 图题2.4.19是一个实现串行加法的电路图,被加数11011及加数10111已分别存入二个五位被加数和加数移位寄存器中。
试分析并画出在六个时钟脉冲作用下全加器输出S i 端、进位触发器Q 端以及和数移位寄存器中左边第一位寄存单元的输出波形(要求时间一一对应)。
1D 2D 3DFF3 FF2 FF1 FF0 SR D 1J C1 1K 组合 逻辑电路 1JC1 1K 组合 逻辑电路1J C1 1K 组合 逻辑电路 1J C1 1K 组合 逻辑电路 CP0D0Q 1Q 2Q 3Q C D 1 D S1J QC11K11J Q C11K11J Q C11K11J Q C11K1CPC Q 3Q 2Q 1Q 0D 3 D 2 D 0≥1 &≥1 &≥1 &≥1 &图题2.4.19解:解该题时,注意全加器是一个合逻辑电路,而移位寄存器和触发器是一个时序电路,要注意时序关系。
其波形如图:题2.4.20 (1)试分析图题2.4.20(a)、(b)所示计数器的模是多少?采用什么编码进行计数?(2)若计数脉冲频率f CP 为700Hz 时,从Q 2端、Q 0端输出时的频率各为多少?图题2.4.20解:分析计数器电路有多种方法,列表法:以CP 为顺序,依次列出触发器的初态、输入,和次态,可以得出结论。
数字电路与系统设计课后习题答案
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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
解:F=AB+BC+AC
或:F=AB+AC+BC
(4) F=ACD+BC+BD+AB+AC+BC
解:F=AD+C+B
(5) F=AC+BC+B(AC+AC)
解:F=AC+BC
2.12用卡诺图把下列函数化简为最简与或式
(1)F(A,B,C)=m(0,1,2,4,5,7)
解:F=B+AC+AC
图略
(2)F(A,B,C,D)=m(0,2,5,6,7,9,10,14,15)
(3)F(ABC)=∏M(1,3,4,5,7)
2.10试写出下列各函数表达式F的F和F的最小项表达式。
(1)F=ABCD+ACD+BCD
(2)F=AB+AB+BC
解:(1)F=∑m(0,1,2,3,5,6,7,8,9,10,13,14)
F'=∑m(1,2,5,6,7,8,9,10,12,13,14,15)
数字集成电路--电路、系统与设计(第二版)复习资料
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第一章 数字集成电路介绍第一个晶体管,Bell 实验室,1947第一个集成电路,Jack Kilby ,德州仪器,1958 摩尔定律:1965年,Gordon Moore 预言单个芯片上晶体管的数目每18到24个月翻一番。
(随时间呈指数增长)抽象层次:器件、电路、门、功能模块和系统 抽象即在每一个设计层次上,一个复杂模块的内部细节可以被抽象化并用一个黑匣子或模型来代替。
这一模型含有用来在下一层次上处理这一模块所需要的所有信息。
固定成本(非重复性费用)与销售量无关;设计所花费的时间和人工;受设计复杂性、设计技术难度以及设计人员产出率的影响;对于小批量产品,起主导作用。
可变成本 (重复性费用)与产品的产量成正比;直接用于制造产品的费用;包括产品所用部件的成本、组装费用以及测试费用。
每个集成电路的成本=每个集成电路的可变成本+固定成本/产量。
可变成本=(芯片成本+芯片测试成本+封装成本)/最终测试的成品率。
一个门对噪声的灵敏度是由噪声容限NM L (低电平噪声容限)和NM H (高电平噪声容限)来度量的。
为使一个数字电路能工作,噪声容限应当大于零,并且越大越好。
NM H = V OH - V IH NM L = V IL - V OL 再生性保证一个受干扰的信号在通过若干逻辑级后逐渐收敛回到额定电平中的一个。
一个门的VTC 应当具有一个增益绝对值大于1的过渡区(即不确定区),该过渡区以两个有效的区域为界,合法区域的增益应当小于1。
理想数字门 特性:在过渡区有无限大的增益;门的阈值位于逻辑摆幅的中点;高电平和低电平噪声容限均等于这一摆幅的一半;输入和输出阻抗分别为无穷大和零。
传播延时、上升和下降时间的定义传播延时tp 定义了它对输入端信号变化的响应有多快。
它表示一个信号通过一个门时所经历的延时,定义为输入和输出波形的50%翻转点之间的时间。
上升和下降时间定义为在波形的10%和90%之间。
对于给定的工艺和门的拓扑结构,功耗和延时的乘积一般为一常数。
最新数字集成电路--电路、系统与设计(第二版)复习资料
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精品文档第一章数字集成电路介绍第一个晶体管,Bell实验室,1947第一个集成电路,Jack Kilby ,德州仪器,1958 摩尔定律:1965年,Gordon Moore预言单个芯片上晶体管的数目每18到24个月翻一番。
(随时间呈指数增长)抽象层次:器件、电路、门、功能模块和系统抽象即在每一个设计层次上,一个复杂模块的内部细节可以被抽象化并用一个黑匣子或模型来代替。
这一模型含有用来在下一层次上处理这一模块所需要的所有信息。
固定成本(非重复性费用)与销售量无关;设计所花费的时间和人工;受设计复杂性、设计技术难度以及设计人员产出率的影响;对于小批量产品,起主导作用。
可变成本(重复性费用)与产品的产量成正比;直接用于制造产品的费用;包括产品所用部件的成本、组装费用以及测试费用。
每个集成电路的成本=每个集成电路的可变成本+固定成本/产量。
可变成本=(芯片成本+芯片测试成本+ 封装成本)/最终测试的成品率。
一个门对噪声的灵敏度是由噪声容限NM (低电平噪声容限)和NM (高电平噪声容限)来度量的。
为使一个数字电路能工作,噪声容限应当大于零,并且越大越好。
NM = V°H - V IH NM L = V lL - V OL 再生性保证一个受干扰的信号在通过若干逻辑级后逐渐收敛回到额定电平中的一个。
一个门的VTC应当具有一个增益绝对值大于1的过渡区(即不确定区),该过渡区以两个有效的区域为界,合法区域的增益应当小于1。
理想数字门特性:在过渡区有无限大的增益;门的阈值位于逻辑摆幅的中点;高电平和低电平噪声容限均等于这一摆幅的一半;输入和输出阻抗分别为无穷大和零。
传播延时、上升和下降时间的定义传播延时tp定义了它对输入端信号变化的响应有多快。
它表示一个信号通过一个门时所经历的延时,定义为输入和输出波形的50%翻转点之间的时间。
上升和下降时间定义为在波形的10%和90%之间。
对于给定的工艺和门的拓扑结构,功耗和延时的乘积一般为一常数。
数字集成电路分析和设计第四章答案
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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.。
《数字电路与系统设计》第4章习题答案
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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。
数字集成电路设计第四章习题
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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;。
(参考资料)数字集成电路课后习题1-4章作业解析
![(参考资料)数字集成电路课后习题1-4章作业解析](https://img.taocdn.com/s3/m/602679ec0912a21615792911.png)
QB0 =−3×10−7 C / cm××1100−−76
=−0.188 V
= QOX COX
6= ×1011.16××11.60−×610−19
0.06 V
VT0 =−0.99 − (−0.88) − (−0.188) − 0.060 =+0.018 V
计算 PMOS 器件的阈值电压:
VGS −VT + EC L
(1.2 − 0.4)(6)(0.2) 1.2 − = 0.4 + (6)(0.2)
0.48V
VDS = 0.2V
∴ VDS < VDSAT
d. 饱和
VGS>VT,VD > VG 肯定工作在饱和区。对于长沟道器件,如果满足这个关系 就工作在饱和区。而发生速度饱和的短沟道器件的 VDSAT 比长沟道器件的要 小,如果电压偏置能使长沟道器件饱和,那么肯定能使速度饱和的短沟道器件 饱和。
VGS = VG −VS = 1.2 −1.1 = 0.1V V=T V= T 0 0.4V ∴ VGS < VT
c. 线性
VGS = VG −VS = 1.2 − 0 = 1.2V V=T V= T 0 0.4V ∴ VGS > VT
不在饱和区的判断依据:
= VDSAT
(= VGS −VT ) EC L
(N 型)
对于(b)中的 PMOS 器件:
NI
= − QI q
= − (1.6
×10−6 )(1.24 1.6 ×10−19
−
0.4)
= 8.4 ×1012 ions / cm2
(P 型)
d) 从上面的计算可以看到,NMOS 用 N 型多晶硅栅和 PMOS 用 P 型多晶硅栅算得的阈值 电压比较小,在沟道区使用与衬底相同的离子掺杂即可调整到期望值(NMOS:P 型注 入;PMOS:N 型注入)。如果我们在 MOS 管的栅极中采用跟衬底相同类型的离子注 入,得到的阈值电压很大,偏离期望值很多,调整起来比较困难。另外,源极和漏极 的制作过程采用自对准工艺,如果栅极的注入类型和源漏一致,一步即可完成离子注 入,简化了器件制作的工艺流程。
数字电路与逻辑设计第四章答案PPT课件
![数字电路与逻辑设计第四章答案PPT课件](https://img.taocdn.com/s3/m/f92c18ddf12d2af90342e6a3.png)
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
NC CLR
J1 J2 J3 Q GND
74HC72
VCC PR CLK K3 K2 K1 Q
Q*= 1
若Q=1, Q¢=0
在CLK的
S主=0
在CLK=1时,主 触发器翻转为
R主=1
“0”,即 Q*主=
,即Q*=
0,
Q*¢
=
0 1
Q*= Q ¢
脉冲触发方式的动作特点:
1.分两步动作:第一步在CLK=1时,主触发器受输 入信号控制,从触发器保持原态;第二步在CLK到 达后,从触发器按主触发器状态翻转,故触发器输 出状态只改变一次 。CLK 到达时从触发器的状态 不一定能按此刻输入信号的状态来确定,而必须考 虑整个CLK=1输入信号的变化过程。
X
K=Q
J
K
Q
4.3.2 边沿JK触发器
由于主从JK触发器存在一次变化问题,所以抗干扰能力差。为
了提高触发器工作的可靠性,希望触发器的次态(新态)仅决 定于CLK的下降沿(或上升沿)到达触时发刻脉的冲输入信号的状态, 与CLK的其它时刻的信号无关。这样下出降现沿了翻各种边沿触发器。
转
&
J
CP 1
1
状态转换图
SD S
Q
RD R
Q
输入
SD
RD
0
0
0
1
1
0
1
1
输
数字集成电路--电路、系统与设计(第二版)课后练习题 第五章 CMOS反相器
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C H A P T E R5T H E C M O S I N V E R T E R Quantification of integrity,performance,and energy metrics of an inverterOptimization of an inverter design5.1Exercises and Design Problems5.2The Static CMOS Inverter—An IntuitivePerspective5.3Evaluating the Robustness of the CMOSInverter:The Static Behavior5.3.1Switching Threshold5.3.2Noise Margins5.3.3Robustness Revisited5.4Performance of CMOS Inverter:The DynamicBehavior5.4.1Computing the Capacitances5.4.2Propagation Delay:First-OrderAnalysis5.4.3Propagation Delay from a DesignPerspective5.5Power,Energy,and Energy-Delay5.5.1Dynamic Power Consumption5.5.2Static Consumption5.5.3Putting It All Together5.5.4Analyzing Power Consumption UsingSPICE5.6Perspective:Technology Scaling and itsImpact on the Inverter Metrics180Section 5.1Exercises and Design Problems 1815.1Exercises and Design Problems1.[M,SPICE,3.3.2]The layout of a static CMOS inverter is given in Figure 5.1.(λ=0.125µm).a.Determine the sizes of the NMOS and PMOS transistors.b.Plot the VTC (using HSPICE)and derive its parameters (V OH ,V OL ,V M ,V IH ,and V IL ).c.Is the VTC affected when the output of the gates is connected to the inputs of 4similargates?.d.Resize the inverter to achieve a switching threshold of approximately 0.75V .Do not lay-out the new inverter,use HSPICE for your simulations.How are the noise margins affected by this modification?2.Figure 5.2shows a piecewise linear approximation for the VTC.The transition region isapproximated by a straight line with a slope equal to the inverter gain at V M .The intersectionof this line with the V OH and the V OL lines defines V IH and V IL .a.The noise margins of a CMOS inverter are highly dependent on the sizing ratio,r =k p /k n ,of the NMOS and PMOS e HSPICE with V Tn =|V Tp |to determine the valueof r that results in equal noise margins?Give a qualitative explanation.b.Section 5.3.2of the text uses this piecewise linear approximation to derive simplifiedexpressions for NM H and NM L in terms of the inverter gain.The derivation of the gain isbased on the assumption that both the NMOS and the PMOS devices are velocity saturatedat V M .For what range of r is this assumption valid?What is the resulting range of V M ?c.Derive expressions for the inverter gain at V M for the cases when the sizing ratio is justabove and just below the limits of the range where both devices are velocity saturated.What are the operating regions of the NMOS and the PMOS for each case?Consider theeffect of channel-length modulation by using the following expression for the small-signalresistance in the saturation region:r o,sat =1/(λI D ).Figure 5.1CMOS inverter layout.InOutGND V DD =2.5V.Poly Metal1NMOSPMOSPolyMetal12λ182THE CMOS INVERTER Chapter 53.[M,SPICE,3.3.2]Figure 5.3shows an NMOS inverter with resistive load.a.Qualitatively discuss why this circuit behaves as an inverter.b.Find V OH and V OL calculate V IH and V IL .c.Find NM L and NM H ,and plot the VTC using HSPICE.d.Compute the average power dissipation for:(i)V in =0V and (ii)V in =2.5Ve HSPICE to sketch the VTCs for R L =37k,75k,and 150k on a single graph.ment on the relationship between the critical VTC voltages (i.e.,V OL ,V OH ,V IL ,V IH )and the load resistance,R L .g.Do high or low impedance loads seem to produce more ideal inverter characteristics?4.[E,None,3.3.3]For the inverter of Figure 5.3and an output load of 3pF:a.Calculate t plh ,t phl ,and t p .b.Are the rising and falling delays equal?Why or why not?pute the static and dynamic power dissipation assuming the gate is clocked as fast as possible.5.The next figure shows two implementations of MOS inverters.The first inverter uses onlyNMOS transistors.V OH V OL inV outFigure 5.2A different approach to derive V IL and V IH .V outV in M 1W/L =1.5/0.5+2.5VFigure 5.3Resistive-load inverterR L =75k ΩSection 5.1Exercises and Design Problems183a.Calculate V OH ,V OL ,V M for each case.e HSPICE to obtain the two VTCs.You must assume certain values for the source/drain areas and perimeters since there is no layout.For our scalable CMOS process,λ =0.125μm,and the source/drain extensions are 5λfor the PMOS;for the NMOS the source/drain contact regions are 5λx5λ.c.Find V IH ,V IL ,NM L and NM H for each inverter and comment on the results.How can you increase the noise margins and reduce the undefined region?ment on the differences in the VTCs,robustness and regeneration of each inverter.6.Consider the following NMOS inverter.Assume that the bulk terminals of all NMOS deviceare connected to GND.Assume that the input IN has a 0V to 2.5V swing.a.Set up the equation(s)to compute the voltage on node x .Assume γ=0.5.b.What are the modes of operation of device M2?Assume γ=0.c.What is the value on the output node OUT for the case when IN =0V?Assume γ=0.d.Assuming γ=0,derive an expression for the switching threshold (V M )of the inverter.Recall that the switching threshold is the point where V IN =V OUT .Assume that the devicesizes for M1,M2and M3are (W/L)1,(W/L)2,and (W/L)3respectively.What are the limitson the switching threshold?For this,consider two cases:i)(W/L)1>>(W/L)2V DD =2.5V V IN V OUTV DD =2.5V V IN V OUT M 2M 1M 4M 3W/L=0.375/0.25W/L=0.75/0.25W/L=0.375/0.25W/L=0.75/0.25Figure 5.4Inverter ImplementationsV DD =2.5V OUTM1IN M2M3V DD =2.5Vx184THE CMOS INVERTER Chapter 5ii)(W/L)2>>(W/L)17.Consider the circuit in Figure 5.5.Device M1is a standard NMOS device.Device M2has allthe same properties as M1,except that its device threshold voltage is negative and has a valueof -0.4V.Assume that all the current equations and inequality equations (to determine themode of operation)for the depletion device M2are the same as a regular NMOS.Assume thatthe input IN has a 0V to 2.5V swing.a.Device M2has its gate terminal connected to its source terminal.If V IN =0V ,what is the output voltage?In steady state,what is the mode of operation of device M2for this input?pute the output voltage for V IN =2.5V .You may assume that V OUT is small to simplify your calculation.In steady state,what is the mode of operation of device M2for this input?c.Assuming Pr (IN =0)=0.3,what is the static power dissipation of this circuit?8.[M,None,3.3.3]An NMOS transistor is used to charge a large capacitor,as shown in Figure5.6.a.Determine the t pLH of this circuit,assuming an ideal step from 0to 2.5V at the input node.b.Assume that a resistor R S of 5k Ωis used to discharge the capacitance to ground.Deter-mine t pHL .c.Determine how much energy is taken from the supply during the charging of the capacitor.How much of this is dissipated in M1.How much is dissipated in the pull-down resistanceduring discharge?How does this change when R S is reduced to 1k Ω.d.The NMOS transistor is replaced by a PMOS device,sized so that k p is equal to the k n ofthe original NMOS.Will the resulting structure be faster?Explain why or why not.9.The circuit in Figure 5.7is known as the source follower configuration.It achieves a DC levelshift between the input and the output.The value of this shift is determined by the current I 0.Assume x d =0,γ=0.4,2|φf |=0.6V ,V T 0=0.43V ,k n ’=115μA/V 2and λ=0.V DD =2.5VOUTM1(4μm/1μm)IN M2(2μm/1μm),V Tn =-0.4VFigure 5.5A depletion load NMOSinverterV DD =2.5VOutFigure 5.6Circuit diagram with annotated W/L ratios=5pFSection 5.1Exercises and Design Problems 185a.Suppose we want the nominal level shift between V i and V o to be 0.6V in the circuit in Figure 5.7(a).Neglecting the backgate effect,calculate the width of M2to provide this level shift (Hint:first relate V i to V o in terms of I o ).b.Now assume that an ideal current source replaces M2(Figure 5.7(b)).The NMOS transis-tor M1experiences a shift in V T due to the backgate effect.Find V T as a function of V o for V o ranging from 0to 2.5V with 0.5V intervals.Plot V T vs.V oc.Plot V o vs.V i as V o varies from 0to 2.5V with 0.5V intervals.Plot two curves:one neglecting the body effect and one accounting for it.How does the body effect influence the operation of the level converter?d.At V o (with body effect)=2.5V,find V o (ideal)and thus determine the maximum error introduced by the body effect.10.For this problem assume:V DD =2.5V ,W P /L =1.25/0.25,W N /L =0.375/0.25,L =L eff =0.25μm (i.e.x d =0μm),C L =C inv-gate ,k n ’=115μA/V 2,k p ’=-30μA/V 2,V tn0=|V tp0|=0.4V,λ =0V -1, γ=0.4,2|φf |=0.6V ,and t ox =e the HSPICE model parameters for parasitic capacitance given below (i.e.C gd0,C j ,C jsw ),and assume that V SB =0V for all problems except part (e).Figure 5.7NMOS source follower configuration V DD =2.5V V iV oV DD =2.5VV i V oV bias =(a)(b)I o1um/0.25um M1186THE CMOS INVERTER Chapter 5##Parasitic Capacitance Parameters (F/m)##NMOS:CGDO=3.11x10-10,CGSO=3.11x10-10,CJ=2.02x10-3,CJSW=2.75x10-10PMOS:CGDO=2.68x10-10,CGSO=2.68x10-10,CJ=1.93x10-3,CJSW=2.23x10-10a.What is the V m for this inverter?b.What is the effective load capacitance C Leff of this inverter?(include parasitic capacitance,refer to the text for K eq and m .)Hint:You must assume certain values for the source/drain areas and perimeters since there is no layout.For our scalable CMOS process,λ =0.125μm,and the source/drain extensions are 5λfor the PMOS;for the NMOS the source/drain contact regions are 5λx5λ.c.Calculate t PHL ,t PLH assuming the result of (b)is ‘C Leff =6.5fF’.(Assume an ideal step input,i.e.t rise =t fall =0.Do this part by computing the average current used to charge/dis-charge C Leff .)d.Find (W p /W n )such that t PHL =t PLH .e.Suppose we increase the width of the transistors to reduce the t PHL ,t PLH .Do we get a pro-portional decrease in the delay times?Justify your answer.f.Suppose V SB =1V,what is the value of V tn ,V tp ,V m ?How does this qualitatively affect C Leff ?ing Hspice answer the following questions.a.Simulate the circuit in Problem 10and measure t P and the average power for input V in :pulse(0V DD 5n 0.1n 0.1n 9n 20n),as V DD varies from 1V -2.5V with a 0.25V interval.[t P =(t PHL +t PLH )/2].Using this data,plot ‘t P vs.V DD ’,and ‘Power vs.V DD ’.Specify AS,AD,PS,PD in your spice deck,and manually add C L =6.5fF.Set V SB =0Vfor this problem.b.For Vdd equal to 2.5V determine the maximum fan-out of identical inverters this gate candrive before its delay becomes larger than 2ns.c.Simulate the same circuit for a set of ‘pulse’inputs with rise and fall times of t in_rise,fall =1ns,2ns,5ns,10ns,20ns.For each input,measure (1)the rise and fall times t out_rise andV DD =2.5VV IN V OUTC L =C inv-gateL =L P =L N =0.25μmV SB-+(W p /W n =1.25/0.375)Figure 5.8CMOS inverter with capacitiveSection 5.1Exercises and Design Problems 187t out_fall of the inverter output,(2)the total energy lost E total ,and (3)the energy lost due to short circuit current E short .Using this data,prepare a plot of (1)(t out_rise +t out_fall )/2vs.t in_rise,fall ,(2)E total vs.t in_rise,fall ,(3)E short vs.t in_rise,fall and (4)E short /E total vs.t in_rise,fall.d.Provide simple explanations for:(i)Why the slope for (1)is less than 1?(ii)Why E short increases with t in_rise,fall ?(iii)Why E total increases with t in_rise,fall ?12.Consider the low swing driver of Figure 5.9:a.What is the voltage swing on the output node (V out )?Assume γ=0.b.Estimate (i)the energy drawn from the supply and (ii)energy dissipated for a 0V to 2.5V transition at the input.Assume that the rise and fall times at the input are 0.Repeat the analysis for a 2.5V to 0V transition at the input.pute t pLH (i.e.the time to transition from V OL to (V OH +V OL )/2).Assume the input rise time to be 0.V OL is the output voltage with the input at 0V and V OH is the output volt-age with the input at 2.5V .pute V OH taking into account body effect.Assume γ =0.5V 1/2for both NMOS and PMOS.13.Consider the following low swing driver consisting of NMOS devices M1and M2.Assumean NWELL implementation.Assume that the inputs IN and IN have a 0V to 2.5V swing andthat V IN =0V when V IN =2.5V and vice-versa.Also assume that there is no skew between INand IN (i.e.,the inverter delay to derive IN from IN is zero).a.What voltage is the bulk terminal of M2connected to?V in V out V DD =2.5V W L 3μm 0.25μm =p 2.5V0V C L =100fFW L 1.5μm 0.25μm=n Figure 5.9Low Swing DriverV LOW =0.5VOutM1ININ M225μm/0.25μm 25μm/0.25μmC L =1pFFigure 5.10Low Swing Driver188THE CMOS INVERTER Chapter 5b.What is the voltage swing on the output node as the inputs swing from 0V to 2.5V .Showthe low value and the high value.c.Assume that the inputs IN and IN have zero rise and fall times.Assume a zero skewbetween IN and IN.Determine the low to high propagation delay for charging the outputnode measured from the the 50%point of the input to the 50%point of the output.Assumethat the total load capacitance is 1pF,including the transistor parasitics.d.Assume that,instead of the 1pF load,the low swing driver drives a non-linear capacitor,whose capacitance vs.voltage is plotted pute the energy drawn from the lowsupply for charging up the load capacitor.Ignore the parasitic capacitance of the driver cir-cuit itself.14.The inverter below operates with V DD =0.4V and is composed of |V t |=0.5V devices.Thedevices have identical I 0and n.a.Calculate the switching threshold (V M )of this inverter.b.Calculate V IL and V IH of the inverter.15.Sizing a chain of inverters.a.In order to drive a large capacitance (C L =20pF)from a minimum size gate (with inputcapacitance C i =10fF),you decide to introduce a two-staged buffer as shown in Figure5.12.Assume that the propagation delay of a minimum size inverter is 70ps.Also assumeV DD =0.4VV IN V OUTFigure 5.11Inverter in Weak Inversion RegimeSection 5.1Exercises and Design Problems 189that the input capacitance of a gate is proportional to its size.Determine the sizing of thetwo additional buffer stages that will minimize the propagation delay.b.If you could add any number of stages to achieve the minimum delay,how many stages would you insert?What is the propagation delay in this case?c.Describe the advantages and disadvantages of the methods shown in (a)and (b).d.Determine a closed form expression for the power consumption in the circuit.Consider only gate capacitances in your analysis.What is the power consumption for a supply volt-age of 2.5V and an activity factor of 1?16.[M,None,3.3.5]Consider scaling a CMOS technology by S >1.In order to maintain compat-ibility with existing system components,you decide to use constant voltage scaling.a.In traditional constant voltage scaling,transistor widths scale inversely with S,W ∝1/S.To avoid the power increases associated with constant voltage scaling,however,youdecide to change the scaling factor for W .What should this new scaling factor be to main-tain approximately constant power.Assume long-channel devices (i.e.,neglect velocitysaturation).b.How does delay scale under this new methodology?c.Assuming short-channel devices (i.e.,velocity saturation),how would transistor widthshave to scale to maintain the constant power requirement?1InAdded Buffer StageOUTC L =20pF C i =10fF‘1’is the minimum size inverter.??Figure 5.12Buffer insertion for driving large loads.190THE CMOS INVERTER Chapter5DESIGN PROBLEMUsing the0.25μm CMOS introduced in Chapter2,design a static CMOSinverter that meets the following requirements:1.Matched pull-up and pull-down times(i.e.,t pHL=t pLH).2.t p=5nsec(±0.1nsec).The load capacitance connected to the output is equal to4pF.Notice that thiscapacitance is substantially larger than the internal capacitances of the gate.Determine the W and L of the transistors.To reduce the parasitics,useminimal lengths(L=0.25μm)for all transistors.Verify and optimize the designusing SPICE after proposing a first design using manual -pute also the energy consumed per transition.If you have a layout editor(suchas MAGIC)available,perform the physical design,extract the real circuitparameters,and compare the simulated results with the ones obtained earlier.。
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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).。