数字集成电路——电路系统与设计 项目
数字集成电路-电路系统与设计
数字集成电路-电路系统与设计数字电路设计的抽象层次:器件->电路->门->模块->系统时钟偏差对全局信号都可能产⽣影响,是⾼性能⼤系统的设计关键。
集成电路的成本:固定成本+可变成本;固定成本可理解为研发成本,⾮重复的成本;可变成本可理解为⽣产制造(芯⽚成本和封测成本)过程中产⽣的成本,与良率也有关,控制芯⽚⾯积能够有效且直接的控制芯⽚成本。
⼀个门电路要想具有再⽣性,其VTC(电压传输特性)应当具有⼀个增益⼤于1的过渡区,以及增益⼩于1的合法区域,如下图:封装可按照封装材料,互连层数量,散热⽅式进⾏分类:封装材料:陶瓷封装、塑封(⾼分⼦聚合物)NMOS与PMOS,以增强型为例,NMOS VGS>Vth时导通,PMOS |VGS|>|Vth|时导通,且VGS<0。
CMOS反相器电压传输特性(VTC)推导:上式为CMOS上下管需要遵守的规则。
结合上式得到,下图为CMOS中上官PMOS部分不同栅极输⼊电压下,下管NMOS电流与输出电压的关系为了使NMOS和PMOS的传输特性能够符合上式DC成⽴,需要根据⼆者的V-I曲线找到交叉点,使其满⾜DC平衡找到上图中的DC平衡交叉点,并提取绘制得到CMOS的电压传输特性如下图,可以看出CMOS的电压传输特性具有再⽣性其中res表⽰呈电阻特性PMOS和NMOS的电流⽅向问题:源極的源是指載流⼦的起點;漏極的漏是指載流⼦的終點。
載流⼦從源極出發,穿過溝道,到達漏極,從外部看,載流⼦最終從漏極漏出去了。
顯然,NMOS和PMOS的載流⼦是不同的,因此導致了令⼈困惑的電流⽅向問題。
盯住載流⼦即可,別被電流⽅向迷惑。
可以簡單地認為,柵極和襯底間的電壓超過閾值後,漏極和源極就接通了,⽽電流⼤⼩則是由柵漏源三極間的電壓決定。
因為MOS是對稱結構,所以源極和漏極無區別且可互換。
關於D和S,也就是漏和源,其實是從⼯藝⾓度觀察的結果。
在MOS中,有兩種載流⼦,⼀種是電⼦,另⼀種是空⽳,標記為N和P。
集成电路设计与集成系统主要课程
集成电路设计与集成系统主要课程集成电路设计与集成系统是电子信息工程专业中的重要课程,旨在培养学生掌握集成电路设计和集成系统的基本理论和实践技能。
本文将从两个方面进行阐述,分别是集成电路设计和集成系统。
集成电路设计是指将多个电子元器件(如晶体管、电阻、电容等)集成在一个芯片上,形成一个完整的电路系统。
集成电路设计是电子工程中的核心技术之一,对电子设备的性能和功能起着至关重要的作用。
在集成电路设计的课程中,学生将学习到数字电路设计、模拟电路设计、射频电路设计等方面的知识。
在数字电路设计方面,学生将学习到数字电路的基本原理和设计方法,包括逻辑门电路的设计、组合逻辑电路的设计、时序逻辑电路的设计等。
通过学习这些内容,学生可以掌握数字电路设计的基本技能,为后续的集成电路设计打下坚实的基础。
在模拟电路设计方面,学生将学习到模拟电路的基本原理和设计方法,包括放大器电路的设计、滤波器电路的设计、功率放大器电路的设计等。
通过学习这些内容,学生可以了解模拟电路设计的基本原理和方法,并能够运用所学知识解决实际问题。
在射频电路设计方面,学生将学习到射频电路的基本原理和设计方法,包括射频放大器电路的设计、射频混频器电路的设计、射频滤波器电路的设计等。
通过学习这些内容,学生可以了解射频电路设计的基本原理和方法,并能够应用所学知识进行射频电路设计。
集成系统是由多个集成电路组成的一个完整的系统。
集成系统的设计是集成电路设计的延伸和拓展,旨在将多个集成电路组合成一个具有特定功能的系统。
在集成系统的课程中,学生将学习到系统级设计、系统级集成、系统级测试等方面的知识。
在系统级设计方面,学生将学习到系统级设计的基本原理和方法,包括需求分析、系统框架设计、系统接口设计等。
通过学习这些内容,学生可以掌握系统级设计的基本技能,为后续的集成系统设计打下坚实的基础。
在系统级集成方面,学生将学习到不同集成电路之间的连接和通信方式,包括串行通信、并行通信、总线通信等。
数字集成电路 电路系统与设计
数字集成电路电路系统与设计
数字集成电路是指将若干个数字电路组合在一起,形成一个完整
的电路系统的过程。
数字集成电路充分利用了数字电子技术的优势,
将不同的数字电路模块集成至一个芯片上,从而大大提高了电路系统
的性能和可靠性。
数字集成电路的设计需要遵循特定的规范和标准,包括电路功能
的设计、电路参数的计算和选取,以及电路布局和制造等方面。
同时,数字集成电路的设计需要充分考虑电路系统的稳定性、抗干扰能力、
低功耗、高可靠性等特点,以满足不同应用场景的需求。
数字集成电路常常应用于各种高精度、高复杂度数字系统中,包
括计算机、通信系统、音视频处理、自动化控制等领域。
在数字集成
电路的设计和制造中,还需要根据具体应用场景选择不同的设计方案
和制造工艺,以获得最优性能和可靠性。
【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
【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。
数字集成电路设计 pdf
数字集成电路设计一、引言数字集成电路设计是一个广泛且深入的领域,它涉及到多种基本元素和复杂系统的设计。
本文将深入探讨数字集成电路设计的主要方面,包括逻辑门设计、触发器设计、寄存器设计、计数器设计、移位器设计、比较器设计、译码器设计、编码器设计、存储器设计和数字系统集成。
二、逻辑门设计逻辑门是数字电路的基本组成单元,用于实现逻辑运算。
常见的逻辑门包括与门、或门、非门、与非门和或非门等。
在设计逻辑门时,需要考虑门的输入和输出电压阈值,以确保其正常工作和避免误操作。
三、触发器设计触发器是数字电路中用于存储二进制数的元件。
它有两个稳定状态,可以存储一位二进制数。
常见的触发器包括RS触发器、D触发器和JK触发器等。
在设计触发器时,需要考虑其工作原理和特性,以确保其正常工作和实现预期的功能。
四、寄存器设计寄存器是数字电路中用于存储多位二进制数的元件。
它由多个触发器组成,可以存储一组二进制数。
常见的寄存器包括移位寄存器和同步寄存器等。
在设计寄存器时,需要考虑其结构和时序特性,以确保其正常工作和实现预期的功能。
五、计数器设计计数器是数字电路中用于对事件进行计数的元件。
它可以对输入信号的脉冲个数进行计数,并输出计数值。
常见的计数器包括二进制计数器和十进制计数器等。
在设计计数器时,需要考虑其工作原理和特性,以确保其正常工作和实现预期的功能。
六、移位器设计移位器是数字电路中用于对二进制数进行移位的元件。
它可以对输入信号进行位移操作,并输出移位后的结果。
常见的移位器包括循环移位器和算术移位器等。
在设计移位器时,需要考虑其工作原理和特性,以确保其正常工作和实现预期的功能。
七、比较器设计比较器是数字电路中用于比较两个二进制数的元件。
它可以比较两个数的值,并输出比较结果。
常见的比较器包括并行比较器和串行比较器等。
在设计比较器时,需要考虑其工作原理和特性,以确保其正常工作和实现预期的功能。
八、译码器设计译码器是数字电路中用于将二进制数转换为另一种形式的元件。
数字集成电路 数字集成电路设计流程和设计方法PPT课件
pmos p2 (i2, il, b); pmos p3 (i3, i2, c); pmosp4 (il, vdd, b); pmos p5 (i2, il, c); pmos p6 (i3, i2, a); pmos p7 (co, vdd, i3); end module
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pmos p4 (i4, vdd, b); pmos p5 (i4, vdd, a); pmos p6 (co, vdd, en); pmos n6 (co, vss, en); end module
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2.2 设计描述
• 四、物理描述
•
电路的物理描述是用来定义在硅表面的物理实现,并由物理实现
数字集成电路设计总体上可分为
1.电路设计(前端设计)
电路设计是指根据对ASIC的要求或规范,从电路系统的行为描述开 始,直到设计出相应的电路图,对于数字系统来说就是设计出它的 逻辑图或逻辑网表
2.版图设计(后端设计)
版图设计就是根据逻辑网表进一步设计集成电路的物理版图,也就 是制造工艺所需的掩膜版的版图。
Verilog-HDL 描述进位算法描述
module carry (co,a,b,c); output co; input a,b,c;
wire #10 co=(a&b)|(a&c)|(b&c) end module
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2.2 设计描述
• 三、结构描述
•
结构描述规定了电路系统的结构,规定了元件之间的连接关系,
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2.1 设计流程
• 二、Top-Down设计
•
从电路行为到逻辑结构的转换是由逻辑综合这一步骤自动进行的。逻辑综合
数字集成电路--电路、系统与设计
数字集成电路是现代电子产品中不可或缺的一部分,它们广泛应用于计算机、手机、汽车、医疗设备等领域。
数字集成电路通过在芯片上集成大量的数字电子元件,实现了电子系统的高度集成和高速运算。
本文将从电路、系统与设计三个方面探讨数字集成电路的相关内容。
一、数字集成电路的电路结构数字集成电路的电路结构主要包括逻辑门、寄存器、计数器等基本元件。
其中,逻辑门是数字集成电路中最基本的构建元件,包括与门、或门、非门等,通过逻辑门的组合可以实现各种复杂的逻辑功能。
寄存器是用于存储数据的元件,通常由触发器构成;而计数器则可以实现计数和计时功能。
这些基本的电路结构构成了数字集成电路的基础,为实现各种数字系统提供了必要的支持。
二、数字集成电路与数字系统数字集成电路是数字系统的核心组成部分,数字系统是以数字信号为处理对象的系统。
数字系统通常包括输入输出接口、控制单元、运算器、存储器等部分,数字集成电路在其中充当着处理和控制信号的角色。
数字系统的设计需要充分考虑数字集成电路的特性,包括时序和逻辑的正确性、面积和功耗的优化等方面。
数字集成电路的发展也推动了数字系统的不断完善和创新,使得数字系统在各个领域得到了广泛的应用。
三、数字集成电路的设计方法数字集成电路的设计过程通常包括需求分析、总体设计、逻辑设计、电路设计、物理设计等阶段。
需求分析阶段需要充分了解数字系统的功能需求,并将其转化为具体的电路规格。
总体设计阶段需要根据需求分析的结果确定电路的整体结构和功能分配。
逻辑设计阶段是将总体设计转化为逻辑电路图,其中需要考虑逻辑函数、时序关系、并行性等问题。
电路设计阶段是将逻辑电路图转化为电路级电路图,包括门电路的选择和优化等。
物理设计阶段则是将电路级电路图转化为实际的版图设计,考虑布线、功耗、散热等问题。
在每个设计阶段都需要充分考虑电路的性能、面积、功耗等指标,以实现设计的最优化。
结语数字集成电路作为现代电子系统的关键组成部分,对于数字系统的功能和性能起着至关重要的作用。
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
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.。
数字集成电路:电路系统与设计(第二版)
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。
数字集成电路—电路、系统与设计
数字集成电路(IC)在当今的电子装置和系统中发挥着至关重要的作用。
这些电路的设计将大量电子组件集成到一个单一芯片上,提供高性能和紧凑的尺寸。
在本篇文章中,我们将探索数字IC设计的关键方面,侧重于电路,系统和设计方面。
我们探索数字IC的电路方面。
数字 IC由晶体管,电阻器,电容器等基本电子元件构建而成,这些电子元件相互连接,可以实现逻辑功能。
现代数字IC集成水平惊人,数十亿晶体管被包装成一个芯片。
这种密集的集成使得在很小的物理空间内可以执行复杂的功能,如微处理器,内存单元,以及通信接口。
数字IC还设计为高速运行,消耗最小功率。
实现高速运行需要仔细考虑信号传播延迟,交叉对讲,以及动力消散。
为了应对这些挑战,IC设计师采用了先进的电路设计技术,如管道衬线,时钟标注,以及动力标注,以优化数字电路的性能和能效。
转到系统方面,数字IC常是更大的电子系统的一部分,它们与其他组件如传感器、起动器和通信接口相互作用。
数字IC的设计必须考虑到系统层面的要求,包括与外部组件的接口,处理输入、输出信号,以及支持各种通信协议。
数字IC在系统层面设计中的一个有趣例子是汽车电子领域。
现代车辆配备了广泛的数字IC,控制发动机,传输,安全系统,以及信息娱乐等功能。
这些IC必须满足可靠性、性能和安全性的严格要求,同时与各种传感器和起动器接口。
汽车数字IC的设计不仅涉及电路层面的考虑,还涉及系统层面的方面,如故障耐受性,通信协议,以及实时操作。
让我们谈谈数字IC的设计方面。
IC设计开始于具体说明电路的功能,之后是建筑和逻辑设计,电路执行,以及验证。
设计过程涉及各种工具和技术,包括逻辑综合、地点和路线、时间分析和功能核查。
设计可制造性和可检验性是关键考虑因素,可确保能够大规模生产高产量的IC并测试其可靠性。
IC设计中一个有趣的例子是开发适用于加密货币开采的集成电路。
为此目的设计的ASIC高度优化,用于履行采矿所需的密码散列功能,与一般用途处理器相比,往往能达到更高的性能和能源效率。
数字集成电路——电路、系统与设计(第二版)Chapter10_timing
Contamination and Propagation Delays
Settle to a final value
Begin to change
Settle to a final value
Begin to change as the last time D begins to change
Settle to a final value as the last time D settles to a final value
Digital Integrated Circuits A Design Perspective
Chapter10: Timing Issues
EE141
Timing Issues
Timing Issues
10.1 Timing Classification
10.2 Synchronous Design
Datapath Structure with Feedback
- Routing the clock so that only negative skew occurs is not feasible
- Design of a low-skew clock network is essential
10.2.5 Clock Distribution Techniques
10.3 Summary
EE141
Timing Issues
Sequencing Methods
Registers
2-Phase Latches
EE141
Timing Issues
Timing Diagrams
数字集成电路--电路、系统与设计(第二版)复习资料
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。
数字电路与系统设计实验报告
数字电路与系统设计实验报告学院:班级:姓名:实验一基本逻辑门电路实验一、实验目的1、掌握TTL与非门、与或非门和异或门输入与输出之间的逻辑关系。
2、熟悉TTL中、小规模集成电路的外型、管脚和使用方法。
二、实验设备1、二输入四与非门74LS00 1片2、二输入四或非门74LS02 1片3、二输入四异或门74LS86 1片三、实验内容1、测试二输入四与非门74LS00一个与非门的输入和输出之间的逻辑关系。
2、测试二输入四或非门74LS02一个或非门的输入和输出之间的逻辑关系。
3、测试二输入四异或门74LS86一个异或门的输入和输出之间的逻辑关系。
四、实验方法1、将器件的引脚7与实验台的“地(GND)”连接,将器件的引脚14与实验台的十5V连接。
2、用实验台的电平开关输出作为被测器件的输入。
拨动开关,则改变器件的输入电平。
3、将被测器件的输出引脚与实验台上的电平指示灯(LED)连接。
指示灯亮表示输出低电平(逻辑为0),指示灯灭表示输出高电平(逻辑为1)。
五、实验过程1、测试74LS00逻辑关系(1)接线图(图中K1、K2接电平开关输出端,LED0是电平指示灯)(2)真值表2、测试74LS02逻辑关系(1)接线图(2)真值表3、测试74LS86逻辑关系接线图(1)接线图(2)真值表六、实验结论与体会实验是要求实践能力的。
在做实验的整个过程中,我们首先要学会独立思考,出现问题按照老师所给的步骤逐步检查,一般会检查处问题所在。
实在检查不出来,可以请老师和同学帮忙。
实验二逻辑门控制电路实验一、实验目的1、掌握基本逻辑门的功能及验证方法。
2、掌握逻辑门多余输入端的处理方法。
3、学习分析基本的逻辑门电路的工作原理。
二、实验设备1、基于CPLD的数字电路实验系统。
2、计算机。
三、实验内容1、用与非门和异或门安装给定的电路。
2、检验它的真值表,说明其功能。
四、实验方法按电路图在Quartus II上搭建电路,编译,下载到实验板上进行验证。
数字集成电路——电路、系统与设计
IC,这些微小但强大的芯片,是我们电子设备的无名英雄,从我们口袋里的光滑智能无线终端,到我们桌子上的强大的截肢者,甚至我们车上最先进的汽车系统。
当它到数字集成电路时,全部是创建顶尖的系统,来传递心跳的性能,而吸电就像一个花哨的鸡尾酒,永远,永远,投球在可靠性上。
这些电路是数据处理、信号处理和控制系统的摇滚巨星,使得我们技术精湛的世界开始运转。
但是,在所有的滑翔和魅力背后,工作上有大量的脑力。
设计数字集成电路就像开始一个令人惊叹的冒险,任务包括设定舞台有规格,通过模型化将人物带入生命,在模拟中通过脚步化,通过合成来伤害它们的存在,最后通过彻底的验证确保一切的平稳航行。
就像是数字交响乐的策划者,进行电路,系统和设计技术的和谐混合,在区块上创建最高效和可靠的集成电路。
这是一个疯狂的旅程,但有人必须做到这一点!设计数字集成电路需要使用不同的工具和方法来开发和改进数字系统。
首先要弄清楚数字系统需要做什么以及它需要多好的表现我们用维利洛格和VHDL等特殊语言创建模型并测试数字系统。
接下来,我们把模型变成逻辑门列表,我们努力确保设计符合所有要求。
我们用半导体制造来制造实际的电路。
这涉及到根据设计创建布局和建造电路。
数字集成电路领域是一个不断发展和动态的研究领域,其特点是设计方法、技术和应用方面不断取得进展。
随着数字系统继续在各种电子装置和系统中发挥重要作用,对数字集成电路设计专业人才的需求日益增加。
对这一领域感兴趣的个人必须在数字电路、系统和设计原则方面奠定坚实的基础,并随时了解数字集成电路技术的最新发展。
只要具备必要的知识和技能,就能够有助于创造创新的数字集成电路,推动技术进步,提高电子系统的性能。
数字集成电路设计流程
数字集成电路设计是一个复杂而系统性强的工程,通常包括以下几个主要步骤:1. 确定需求在设计数字集成电路之前,首先需要明确设计的功能和性能要求,包括输入输出接口、逻辑功能、时序要求等方面的设计需求。
2. 概念设计通过对需求进行分析和理解,进行电路结构和功能的初步设计,确定电路的整体架构和模块划分,制定初步的电路设计方案。
3. 逻辑设计根据概念设计的结果,进行逻辑电路设计,包括逻辑门的选择、逻辑电路的设计与优化等,确保电路满足功能需求。
4. 电气特性设计在逻辑设计的基础上,进行电气特性设计,包括时序分析、电气参数分析等,保证电路在电气特性上符合要求。
5. 物理布局设计进行物理布局设计,确定芯片内各功能块的布局位置,考虑信号线路长度、时延等因素,使得布局紧凑且方便布线。
6. 时序分析与优化进行时序分析,保证电路中的时序要求得到满足,并对电路进行时序优化,减少时序迟滞,提高电路的性能。
7. 电路仿真与验证通过电路仿真软件对设计的电路进行仿真验证,包括功能仿真、时序仿真等,确保设计的准确性和可靠性。
8. 物理布线设计根据物理布局设计结果进行布线设计,连接各功能块之间的信号线路,考虑信号传输的稳定性和功耗等因素。
9. 版图设计生成版图设计,包括器件的排列、连线规划等,生成最终的版图文件,为后续的制造加工做准备。
10. 设计规则检查(DRC)和布局VS电气规则检查(LVS)进行设计规则检查和布局与电气规则检查,确保设计符合制造工艺要求和电气规范。
11. 前期验证进行前期验证,包括功能验证、时序验证等,确保设计符合需求,并进行必要的调整和优化。
12. 准备生产完成设计验证后,准备将设计文件交付给芯片制造厂商进行生产加工,最终完成数字集成电路设计流程。
以上是数字集成电路设计的主要流程,每个步骤都非常重要,需要经过严格的设计和验证。
在实际设计过程中,还会涉及到许多细节和技术要点,需要设计工程师具备扎实的专业知识和经验。
数字集成电路——电路、系统与设计
数字集成电路——电路、系统与设计目录第一部分基本单元第1章引论1.1 历史回顾1.2 数字集成电路设计中的问题1.3 数字设计的质量评价1.4 小结1.5 进一步探讨第2章制造工艺2.1 引言2.2 CMOS集成电路的制造2.3 设计规则——设计者和工艺工程师之间的桥梁2.4 集成电路封装2.5 综述:工艺技术的发展趋势2.6 小结2.7 进一步探讨设计方法插入说明A——IC版图第3章器件3.1 引言3.2 二极管3.3 MOS(FET)晶体管3.4 关于工艺偏差3.5 综述:工艺尺寸缩小3.6 小结3.7 进一步探讨设计方法插入说明B——电路模拟第4章导线4.1 引言4.2 简介4.3 互连参数——电容、电阻和电感4.4 导线模型4.5 导线的SPICE模型4.6 小结4.7 进一步探讨第二部分电路设计第5章CMOS反相器5.1 引言5.2 静态CMOS反相器——直观综述5.3 CMOS反相器稳定性的评估——静态特性5.4 CMOS反相器的性能——动态特性5.5 功耗、能量和能量延时5.6 综述:工艺尺寸缩小及其对反相器衡量指标的影响5.7 小结本文由整理提供5.8 进一步探讨第6章CMOS组合逻辑门的设计6.1 引言6.2 静态CMOS设计6.3 动态CMOS设计6.4 设计综述6.5 小结6.6 进一步探讨设计方法插入说明C——如何模拟复杂的逻辑电路设计方法插入说明D——复合门的版图技术第7章时序逻辑电路设计7.1 引言7.2 静态锁存器和寄存器7.3 动态锁存器和寄存器7.4 其他寄存器类型7.5 流水线:优化时序电路的一种方法7.6 非双稳时序电路7.7 综述:时钟策略的选择7.8 小结7.9 进一步探讨第三部分系统设计第8章数字IC的实现策略8.1 引言8.2 从定制到半定制以及结构化阵列的设计方法8.3 定制电路设计8.4 以单元为基础的设计方法8.5 以阵列为基础的实现方法8.6 综述:未来的实现平台8.7 小结8.8 进一步探讨设计方法插入说明E——逻辑单元和时序单元的特性描述设计方法插入说明F——设计综合第9章互连问题9.1 引言9.2 电容寄生效应9.3 电阻寄生效应9.4 电感寄生效应9.5 高级互连技术9.6 综述:片上网络9.7 小结9.8 进一步探讨第10章数字电路中的时序问题10.1 引言10.2 数字系统的时序分类本文由整理提供10.3 同步设计——一个深入的考察10.4 自定时电路设计10.5 同步器和判断器10.6 采用锁相环进行时钟综合和同步10.7 综述:未来方向和展望10.8 小结10.9 进一步探讨设计方法插入说明G——设计验证第11章设计运算功能块11.1 引言11.2 数字处理器结构中的数据通路11.3 加法器11.4 乘法器11.5 移位器11.6 其他运算器11.7 数据通路结构中对功耗和速度的综合考虑11.8 综述:设计中的综合考虑11.9 小结11.10进一步探讨第12章存储器和阵列结构设计12.1 引言12.2 存储器内核12.3 存储器外围电路12.4 存储器的可靠性及成品率12.5 存储器中的功耗12.6 存储器设计的实例研究12.7 综述:半导体存储器的发展趋势与进展12.8 小结12.9 进一步探讨设计方法插入说明H——制造电路的验证和测试本文由整理提供。
数字集成电路--电路、系统与设计(第二版)复习资料
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。
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-Chapter 6 Designing
4
Chapter 6 Problem Set
VDD F G
A B
A
A B
A
Figure 6.6 Two-input complex logic gate.
11.
Design and simulate a circuit that generates an optimal differential signal as shown in Figure 6.7. Make sure the rise and fall times are equal.
2
VDD E 6 A A 6 B 6 C 6 D 6 F A B C D 4 4 4 4 E 1 A B C D E 4 4 4 4 E 1 6 F 6 B 6 C 6 D
Chapter 6 Problem SetVDD 6Circ来自it ACircuit B
Figure 6.2 Two static CMOS gates.
Digital Integrated Circuits - 2nd Ed
3
2.5 V
PMOS
M2 W/L = 0.5μm/0.25μm Vout Vin M1 W/L = 4μm/0.25μm NMOS Figure 6.4 Pseudo-NMOS inverter.
a. What is the output voltage if only one input is high? If all four inputs are high? b. What is the average static power consumption if, at any time, each input turns on with an (independent) probability of 0.5? 0.1? c. Compare your analytically obtained results to a SPICE simulation.
数字集成电路与系统设计
数字集成电路与系统设计是指基于数字电路技术和集成电路技术,设计和实现数字电路系统的过程。
它涵盖了从电路级到系统级的设计和实现,包括电路设计、逻辑设计、芯片设计、系统设计和验证等方面。
在数字集成电路与系统设计中,需要考虑以下几个方面:
电路设计:根据系统需求和功能要求,设计各种数字电路,包括逻辑门、寄存器、计数器、多路选择器等。
电路设计要考虑电路的功耗、时序要求、可靠性等因素。
逻辑设计:根据系统功能需求,将电路设计抽象成逻辑功能的表示,使用逻辑门和时序元件进行逻辑功能的实现。
逻辑设计要考虑时序关系、数据通路、控制信号等。
芯片设计:基于所需的电路和逻辑设计,进行芯片级的设计,包括电路布局、线路布线、电源分配、时钟设计等。
芯片设计要考虑电路的集成度、功耗、散热等因素。
系统设计:将多个数字电路组合成完整的系统,包括处理器、存储器、输入输出接口等。
系统设计要考虑系统的性能、功耗、可靠性、通信接口等。
验证与测试:对设计的数字电路和系统进行验证和测试,确保其功能正确和性能满足要求。
验证与测试包括功能验证、时序验证、功耗测试、可靠性测试等。
数字集成电路与系统设计是现代电子技术领域的重要组成部分,它广泛应用于计算机、通信、控制系统等领域,推动了数字技术的发展和应用。
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Digital Integrated Circuits – A Design Perspective 2/eJan M. Rabaey, Anantha Chandrakasan, Borivoje NikolićChapter 11 and 6Design Project: 32-bit Arithmetic Logic Unit (Phase 1)1.Designing a 32-bit atithmetic-logic unit – BackgroundArithmetic-logic units are the heart of any microprocessor. This semester, we will design the critical part of a 32-bit ALU.1.1.High level structureThe high-level block diagram of a high-performance ALU is shown in Figure 1.ALU’s have four major parts:•Arithmetic block: This block is used to perform arithmetic operations such as addition, subtraction and comparison. The core of the arithmetic block is an adder. In thearchitecture presented in Figure 1, the adder uses carry look-ahead and sum-select techniques (the blocks labeled CARRYGEN, SUMGEN and SUMSEL).•Logic block: This block is used to perform simple bitwise logic operations such as AND (masking), OR and XOR (the block labeled LU in Figure 1)•Multiplexers: These blocks are used to select the appropriate inputs for the arithmetic and logic blocks. Usually more than two buses arrive at the inputs of the ALU (9 buses inFigure 1, selected by 9:1 MUX’s). Sometimes these multiplexers are used to performsome simple logic operations. The 5:1 MUX is a programmable shifter: its inputs containshifted versions of the 9:1 MUX output. The 2:1 MUX can be programmed to invert oneof the operands (this can be used to execute a subtraction using just an adder).•Registers: these blocks are used to store the operands and the results. Usually, these registers are not part of the microprocessor’s register file (though not always the case).Note that there is a bus looping from the output back to the input of the ALU, allowing it touse the newly computed results as operands in the next cycle. This is usually a very long wire(1.6mm in this case) and therefore puts a significant load on the previous stages.2.Implementation and ConstraintsThe goal of this project is to design the carry-lookahead adder for an ALU to be used in a high-performance or mobile microprocessor with a particular set of optimization criteria. The project will be completed in TWO phases. In the first phase of the project, you will choose a circuit style, design the logic, and lay out basic cells for a 32-bit adder. You will also have to do some pencil-and-paper optimization in order to meet the stated design goals and constraints. The complete adder will be assembled and simulated in PHASE 2.Physical and electrical specifications and constraints:2.1. TECHNOLOGY: The design is to be implemented in a 0.25 µm CMOS process with 4 metal layers. The SPICE technology is in the g25.mod file.2.2. POWER SUPPLY: You are free to choose any supply voltage and logic swing up to 2.5 V. Make sure that you use the appropriate model when you perform any hand analysis.2.3. PERFORMANCE METRIC: The propagation delay for static CMOS design is defined as the time interval between the 50% transition point of the inputs and the 50% point of the worst-case output signal. Make sure you pick the worst-case condition and state EXPLICITLY in your report what that condition is. For dynamic designs, the propagation delay is defined in this case as the delays of the evaluate phase ONLY (at least in this phase of the project)!2.4. AREA: The area is defined as the smallest rectangular box that can be drawn around the design. Since the ALU must interface with the cache, all of the row-matched inputs must be accessible from the left side of the design, in row-address order. In the first phase of the project, you should make area estimations based on the total transistor width and the wiring complexity. An expression for prediction of the area will be provided on the web page.2.5. Each bit slice in the adder should accommodate 9 metal-5 busses and is 144λ (36 metal pitch) wide. Other circuits in the datapath set this constraint.2.5. NAMING CONVENTIONS: The input operands of the adder are named A<31:0> andB<31:0>. The output is SUM<32:0>, where SUM<32> is the carry out bit.2.6. REGISTERS: In this phase of the design, you do not need registers. The data flow from input to output should be combinational logic.2.7. CLOCKS: There should be no global clock since the design is combinational. If you choose to use dynamic logic, you are permitted a precharge/evaluate clock, but the result must become available after ONE evaluate phase (no pipelined logic). Remember that the load capacitance of the clock should be included in the energy analysis.2.8. V OH, V OL, NOISE MARGINS: You are free to choose your logic swing. The noise margins should be at least 10% of the voltage swing. Test this by computing the VTC between one of the inputs and the output signals (with the other outputs set to the appropriate values) for a staticdesign. For a dynamic circuit, apply an input signal with a 10% noise value added to the input and observe the outputs.2.9. RISE AND FALL TIMES: All input signals have rise and fall times of 50 ps. The rise and fall times of the output signals (10% to 90%) should not exceed 200ps.2.10. LOAD CAPACITANCE: Your adder is driving a 1.6mm long bus with 9 loads evenly distributed. Each capacitive load is equal to the adder input capacitance. Each wire in the bus is 4λ wide with 4λ spacing in M5.2.11. INPUT CAPACITANCE: Each input of the adder should not load the previous stage with more than 50fF (less is OK).The goal is to minimize the delay, power and area of the design. Your delay should not exceed 12 fanout-of-4 (FO4) inverter delays.In the first phase, you should make the following decisions:•Circuit family (complementary static CMOS, pass-transistor, dynamic).•Type of carry-lookahead tree to be used.Remember that your decisions on the logic level significantly affect the final delay, area and power. You should discuss your designs with the TAs.In this first phase, your design will consist of:•Pencil-and-paper design of the adder.•Identification of the critical path in the adder.•Hand optimization of the delay.•Schematics and layouts of all different cells that will be used.•Estimation of delay, area and power of your design, by simulating individual cells with proper loading.3.SimulationAnalyze the circuit by using either SPICE or IRSIM to simulate the design and prove that it functions correctly. You will need to determine the input pattern that causes the worst-case propagation delay or energy consumption by analyzing your circuit schematic.4.ReportThe quality of your report is as important as the quality of your design. One must sell the design by justifying all design decisions. Be sure to provide all relevant information and eliminate unnecessary material. Organization, conciseness, and completeness are of paramount importance. Do not repeat information we already know. Use the templates provided on the web page (Word and PDF formats). E-mail an electronic version of your report as a Word or PDF file. Make sure to fill in the cover-page and use the correct units. A report must be submitted at the end of each phase of the project.Your report should discuss your overall design philosophy and the important design decisions made at the logic and circuit level. Discuss why your approach increases the operating speed or helps to reduce energy or area, while meeting the performance specifications. Provide estimatesof your results and describe how you arrived at them. Describe the sizing methodology used in your design. Include schematics and highlight important elements and the critical paths.The total report should not contain more than four pages. You are not allowed to add any other sheets.The organization of the report should be based on the following outline:Cover page: Names, project title, decisions about logic family and the carry path, performance estimates.Page 1: Executive summary, overall design decisions: circuit style, drawing of the carry-lookahead tree, critical path, sizing, remarks, and motivationsPage 2-3: Schematics and layouts of all circuit cells. Simulation results for power and area. Remember, a good report is like a good layout: it should perform its function (convey information) in the smallest possible area with the least delay and energy (to the reader) possible. The quality of the report is an important (major) part of the grade!Design of an 32-bit Arithmetic Logic Unit – Phase II1. Physical design of a 32-bit adderIn the second phase of the project, you must realize the physical design of the complete 32-bit adder you proposed in phase 1. You should complete the schematic in SUE, and lay out the design using MAX, the layout editor you have become intimately familiar with throughout this semester, using the cells designed in the first phase.Your schematic and layout should be well organized, and easy to understand. Your layout must be free of design rule errors, and must include wells and sufficient contacts to the wells. Note that the latest version of MAX provides the easy generation of contacts, via’s, and wells through the via p-cell normally located in the right sub-window of the layout editor.As with any layout assignment, you will quickly discover that drawing a layout is much simpler if you plan things out ahead of time. It is much easier to have a general layout strategy than to just blindly draw objects on the screen. For example, it is important to plan out how you will distribute the supply and ground rails in your design. As discussed in phase 1, we suggest that you draw them horizontally in metal 1. A design that is very regular can easily be tiled and reused, saving you a lot of time.Plan ahead: develop a top-level plan of your design. Try to determine optimal routing and placement of wires. Use common sense in laying out your circuit and remember that long transistors must be built properly. Plan your cells accordingly. You can modify them slightly from phase 1 if they don’t fit your top-level floorplan.Use the same specifications for the height of the bit-slice as you had in phase 1. Arrange the bit slices accordingly, and connect the power and ground. Make your wiring plan: in which metal layers are you going to route long carries? Use metal layers 1-4 for all routing, metal 5 should be used only for routing the bus on the top level. Make sure that you properly strap the supply rails, and that you distribute the clocks if you are using dynamic logic.2. Updating ResultsMost likely, mapping your design into a physical implementation will cause some significant changes in the energy and delay of your circuit. However, the functional operation shouldn’t change! You must ensure that your layout and schematic are functionally equivalent by performing LVS (layout-versus-schematic) check. You must therefore perform both a full functional and performance analysis on your extracted layout.The goal of this phase of the project is to compare the results before and after physical design, not to improve on the design goals. Explain any major deviations from your results in Phase I. Try not to make any significant changes (i.e. adder architecture, etc.) to your original design of Phase I. You may make minor modifications to the circuit that do not change the underlying foundation of your design. If you find it absolutely necessary to alter a major part of your circuit (because of non-functionality or unacceptable results), a full motivation should be provided in the report.3. SimulationYou should simulate the netlist extracted from your layout, with signals and loading specified as in phase 1. You will report the delay, area and power of your extracted design and compare them to your estimates from phase 1.4. ReportYour report for this phase of the project serves to accomplish two things:1) You should discuss the overall layout strategy and how it is related to your original design goals.2) Compare your results in this phase to those that you obtained in the first phase of the project, including any changes you made to the design.The total report should not contain more than two pages. You are NOT allowed to add any other sheets, except for important plots. Use the following guidelines to govern your report content and length:•Page 1: Summary of the performance of your design, estimates from phase 1 and extracted data from phase 2.•Page 2: Executive summary, overall design decisions, floorplan, remarks, and motivations. •Page 3: Layout of your adder, with labeled terminals.Also, you should prepare the poster containing 9 slides to present to the class.。