数字集成电路设计与分析
cmos数字集成电路设计流程
CMOS数字集成电路设计流程一、介绍CMOS数字集成电路设计是现代电子工程中的重要分支之一，涉及到数字逻辑、电子设计自动化、半导体器件物理和工艺等多个领域。
在数字集成电路的设计流程中，工程师需要进行功能分析、设计规划、逻辑综合、电路布局、版图设计、物理验证和后仿真等多个环节。
本文将就CMOS数字集成电路设计流程的各个环节进行详细介绍。
二、功能分析在进行CMOS数字集成电路设计之前，工程师需要首先完成功能分析。
在功能分析阶段，工程师需要明确电路的功能需求，包括各种逻辑门、寄存器、存储器等组件的功能与接口要求。
还需要对设计的电路进行规模估计，明确设计的规模和复杂度，为后续的设计规划和逻辑综合提供依据。
三、设计规划在完成功能分析之后，工程师需要进行设计规划。
设计规划阶段需要明确设计的总体结构、数据传输路径、时钟和控制信号的分配等。
还需要进行功耗和面积的预估，并确定设计的性能指标和约束条件等。
四、逻辑综合逻辑综合是数字集成电路设计的重要环节之一。
在逻辑综合过程中，工程师需要将设计的功能描述转换为门级网表，然后进行优化，包括面积优化、功耗优化、时序优化等。
逻辑综合的结果将是门级网表，为后续的电路布局和版图设计提供基础。
五、电路布局电路布局是数字集成电路设计的关键环节之一。
在电路布局过程中，工程师需要将逻辑综合的门级网表映射到物理结构上，并进行布线和布局设计。
电路布局需要考虑电路的面积、功耗、时序等多个方面的优化，并确保电路的稳定性和可靠性。
六、版图设计版图设计是数字集成电路设计中的重要环节之一。
在版图设计过程中，工程师需要将电路布局转换为实际的版图，并进行细化设计，包括晶体管布局、金属线路设计、接口电路设计等。
版图设计需要满足工艺规则和制约条件，确保设计的可制造性和可测试性。
七、物理验证物理验证是数字集成电路设计中不可或缺的一环。
在物理验证过程中，工程师需要进行电路的各种仿真和验证工作，包括静态时序分析、动态时序分析、功耗分析、布局抽取等。
数字集成电路分析与设计 第二章答案
CHAPTER 2P2.1. a) The solution for the NMOS case is based on Example 2.4: The equation for V T0 is: 02BT FB F OXQ V V C φ=-- Calculate each individual component.1710()1362OX 077200611196310ln 0.026ln 0.44 V 1.4100.440.550.99 V 4 3.510 F/cm1.610 F/cm 310310/0.188 V 1.610610 1.6100.1.610i FpA GC Fp G gate OXB B OX OX OX n kT q NC Q Q C cmC Q C φφφφεε-------⨯==-=-⨯=-=--=-==⨯=⨯⨯=⨯==⨯⨯⨯⨯==⨯TO 06 V V 0.99(0.88)(0.188)0.0600.018 V=------=+ For the PMOS device:1710()77200611196TO 310ln 0.026ln 0.44 V 1.4100.440.550.99 V 310310/0.188 V1.610610 1.6100.06 V 1.610V 0.99(0.88)(0.188)0.0600.138 D Fn i GC Fn G gate B B OX OX OX N kT q n Q Q C cmC Q C φφφφ-----⨯===⨯=-=+=+⨯=⨯==⨯⨯⨯⨯==⨯=---=-Vb) The magnitude of V T0 would be higher. Since the device is PMOS this means that V T0 islowered. Since the only thing that’s been changed is the doping of the gate, only G φ changes. The new V T0 then becomes:00.110.880.1880.6 1.24V T V =----=-c) Since V T0 will be adjusted with implanted charge (Q I ):60.40.0180.382(1.610)(0.382)IOXIOXI Q C Q V C Q V -=-==⨯To calculate the threshold implant level N I :I I I I qN Q Q N q==For the NMOS device from part(a):6122190.610 3.8210/1.610I I Q N ions cm q --⨯=-=-=⨯⨯ (p-type) For the PMOS device from part(a):612219(1.610)(0.40.138)2.6210/1.610I I Q N ions cm q --⨯-=-=-=⨯⨯ (n-type) For the PMOS device from part(b):612219(1.610)(1.240.4)8.410/1.610I I Q N ions cm q --⨯-=-=-=⨯⨯ (p-type)d) The advantage of having the gate doping be n + for NMOS and p + for PMOS could be seen from analysis above. Doping the gates in such a way leads to devices with lower threshold voltages, but enables the implant adjustment with the same kind of impurities that used in the bulk (p-type for NMOS and n-type for PMOS). If we were to use the same kind of doping in gate as in the body (i.e. n + for PMOS and p + for NMOS) that would lead to higher un-implanted threshold voltages. Adjusting them to the required lower threshold voltage would necessitate implantation of the impurities of the opposite type near the oxide-Si interface. This is not desirable. Also, the doping of the poly gate can be carried out at the same time as the source and drain and therefore does not require an extra step.P2.2. First, convert ox t to units of cm:810100cm222210cm 10ox t -=⨯=ÅÅNow, using the mobility equation:()()20 1.8568130/V70cm0.8114102210pep nGS T ox cm V s V V t μμθ--==≈⎛⎫⎛⎫-+ ⎪⎪+ ⎪⎝⎭⎝⎭P2.3. a) For each transistor, derive the region of operation. In our case, for 0V,0.4V GS V =, thetransistor is in the cutoff region and there is no current. For 0.8V,1.2V GS V =, firstcalculate the saturation voltage Dsat V using:()GS T C DSAT GS T C V V E L V V V E L-=-+For our transistors, this would be:Next, we derive the IV characteristics using the linear and saturation current equations,we get the graphs shown below.IV Characteristic of NMOS01020304050607000.20.40.60.811.2Volts (V)C u r r e n t (u A )IV Characteristic of PMOSVolts (V)C u r r e n t (u A )To plot DS I vs. GS V , first identify the region of operation of the transistor. For GS T V V <, the transistor is in the cutoff region, and there is negligible current. For GS T V V > and GS DS V V ≤, the transistor is in the saturation region and saturation current expression should be used. The graphis shown below. Clearly, it is closer to the linear model.Ids vs. Vgs of NMOS010********607000.20.40.60.811.21.4Vgs (V)I d s (V )P2.4. For each transistor, first determine if the transistor is in cutoff by checking to see if V GS isless than or greater than V T . V T may have to be recalculated if the source of the transistor isn’t grounded. If V GS is less than V T , then it is in cutoff, otherwise, it is in either triode or saturation.To determine if it is in the triode saturation region, check to see if V DS is less than or greater than V DSAT . If V DS is less than V DSAT , then it is in triode, otherwise, it is in saturation. a. Cutoff00.200.2V0.4V GS G S T T GS TV V V V V V V =-=-===∴<b. Cutoff01.2 1.20V0.4V GS G S T T GS TV V V V V V V =-=-===∴<c. Linear01.20 1.2V0.4V GS G S T T GS TV V V V V V V =-=-===∴>The transistor is not in the cutoff region.()()()()()()1.20.460.20.48V 1.20.460.20.2V GS T C DSATGS T C DS DS DSATV V E L V V V E L V V V --===-+-+=∴<d. Saturation: In this case, because D G V V > the transistor is in the saturation region. To see this, recognize that in a long-channel transistor if D G V V >, the transistor is in saturation. Since the saturation drain voltage Dsat V is smaller in a velocity-saturated transistor than in a long-channel transistor, if the long-channel saturation region equation produces a saturated transistor, than the velocity-saturated saturation region equation will also.P2.5. In both cases, the first step it to calculate the maximum value of X V given G V . If thevoltage at the drain is higher than this maximum value, then ,max X X V V =, otherwise,X D V V =. The maximum value of X V is G T V V - but 0T T V V ≠ because of body effect andwe consider its effect.(),max 0001.20.40.988X G T G T G T G T V V V V V V V V V γγγγ=-=-+=--=--+=--=-There are two ways to calculate this, either through iteration or through substitution. Iteration:For the iteration method, we need a starting value for V X,max . A good starting value would be 0 1.20.40.8V G T V V -=-=. We plug this value on the RHS of the equation, calculate a new V X,max and repeat until we reach a satisfactory converged value.Old Vx,max New Vx,max 0.800 0.728 0.728 0.734 0.734 0.734In this, only three iterations are needed to reach 0.734V. Substitution:The term makes things a bit tricky, we get around this by making the following substitution:2,max 2,max 0.880.88X X x V V x =+∴=-Therefore:,max 220.9880.880.98800.2 1.87X V x x x =--=-=+-2,max 1.27, 1.470.880.733,1.28X x V x ===-=-= We use the first value since second value is above V DD . a. Since ,max D X V V >, ,max 0.733V X X V V ==. b. Since ,max D X V V <, ,max 0.6V X X V V ==. P2.6.a. Initially, when 0V in V =, the transistor is in the cutoff region and 0V X V =. Thisvalue is constant until V in exceeds V t 0. From then, X in T V V V =- and body effect must be taken into account. This trend continues until 0.7V X D V V ==, and the value of V inat that point must be calculated. From then on, 0.7V X D V V ==. To plot V X in the second region, we first derive an expression for V X vs. V in.(),max 0000.40.212X G T G T in T in T in in V V V V V V V V V V V γγγγ=-=-+=---=--=--=--Substituting:2,max2,max 0.880.88X X x V V x =+∴=-Therefore:,max 220.2120.880.21200.20.66X in in in V V x V x x V =---=--=+--220.880.88XxV x====-=-⎝⎭Since this is a quadratic function, there will be two graphs of V X. Only one of thesegraphs intersects with V X in the first region. In this case, plug 0.4inV= and see which one gives 0V. In our case, it would be the ‘+’ version of the quadratic.To see where region 3 begins, we simply isolate V in:()()()22220.880.2 2.710.2 2.71440.2 2.711.16V4XinVV=-⎝⎭-+-==+-==The final graph is shown in Figure 错误！未找到引用源。
数字集成电路设计方法、流程
数字集成电路设计方法、流程数字集成电路设计是指将数字电路功能进行逻辑设计、电路设计和物理布局设计，最终实现数字电路在集成电路芯片上的实现。
数字集成电路设计方法包括：1.设计需求分析：对于待设计的数字电路，首先需要了解设计需求。
明确电路所需的功能、性能指标、工作条件等，以确定电路设计的目标和约束条件。
2.逻辑设计：通过使用硬件描述语言（HDL）或者可视化设计工具，设计数字电路的功能逻辑。
在逻辑设计中，使用逻辑门、寄存器、计数器、状态机等基本逻辑单元，以及组合逻辑和时序逻辑的方法，实现所需功能。
3.电路设计：根据逻辑设计的结果，进行电路级设计。
包括选择和设计适当的电路模型、搭建电路拓扑、设计功耗、提高抗噪声性能等。
在电路设计中，需要考虑电源电压、电路延迟、功耗、抗干扰性能等因素。
4.物理布局设计：根据电路设计的结果，进行芯片级物理布局设计。
将电路中的逻辑单元和电路模块进行排布，设计电路的物理连接，并确定芯片的尺寸、引脚位置等。
物理布局设计需要考虑电路的功耗、面积、信号干扰等因素。
5.时序分析：对于复杂的数字电路，在设计过程中需要进行时序分析，以确保电路在各种工作条件下都能正常工作。
时序分析包括时钟分析、延迟分析、时序约束等。
6.仿真验证：在设计完成后，通过仿真验证电路的功能和性能。
使用仿真工具对电路进行功能仿真、逻辑仿真和时序仿真，验证设计的正确性。
7.物理设计：在完成电路设计和仿真验证后，进行物理设计，包括版图设计、布线、进行负载和信号完整性分析，以及完成设计规则检查。
8.集成电路硅掩模制作：根据物理设计结果，生成集成电路的掩模文件。
掩模文件是制造集成电路所需的制作工艺图。
9.集成电路制造：根据掩模文件进行集成电路的制造。
制造过程包括光刻、蚀刻、沉积、离子注入等工艺。
10.设计验证和测试：在集成电路制造完成后，进行设计验证和测试，确保电路的功能和性能符合设计要求。
数字集成电路设计的流程可以总结为需求分析、逻辑设计、电路设计、物理布局设计、时序分析、仿真验证、物理设计、硅掩模制作、集成电路制造、设计验证和测试等步骤。
数字集成电路设计实验报告
数字集成电路设计实验报告
摘要：
本实验旨在设计一个数字集成电路，实现特定功能。
本报告将介绍实验目的、背景和理论知识、设计方法、实验步骤、结果分析和讨论以及实验总结。
1.实验目的：
设计一个数字集成电路，实现特定功能，并通过实验验证设计的正确性和可行性。
2.背景和理论知识：
简要介绍数字集成电路的基本概念和原理，并介绍与本实验相关的理论知识，包括逻辑门、布尔代数、时序电路等。
3.设计方法：
本部分将详细介绍实验中采用的设计方法，包括采用的逻辑门类型、布尔代数的转换方法、时序电路的设计方法等。
4.实验步骤：
本部分将详细描述实验的具体步骤，包括电路图的绘制、器件的选择和布局、逻辑设计的步骤、时序电路的设计方法、电路的仿真等。
5.结果分析和讨论：
本部分将对实验结果进行分析和讨论，比较设计与实际结果的差异，分析可能的原因，并讨论实验的局限性和改进方向。
6.实验总结：
总结实验过程中的收获和经验，评估实验的结果和设计的可行性，并提出对未来工作的展望和建议。
通过对数字集成电路设计实验的详细介绍和分析，本报告旨在提供一份完整的实验报告，帮助读者理解实验过程和结果，并为今后的设计工作提供参考。
集成电路设计与实现技术分析
集成电路设计与实现技术分析随着计算机和信息技术的迅猛发展，集成电路设计与实现技术成为了现代电子工程研究领域的一大重点。
集成电路是一种将数十亿个晶体管、二极管、电容等元件集成在一起的现代电子器件，可以实现各种复杂的电路功能，并具有功耗低、速度快、尺寸小等优点。
本文将从集成电路设计的基本原理、设计流程、常用工具和实现技术等方面进行分析。
一、集成电路设计的基本原理集成电路设计的基本原理是将一个大的电路功能模块进行分解，然后将各个分解出的模块进行功能设计和物理细节设计，最后将各个模块组合在一起形成完整的电路。
具体来说，集成电路设计的基本原理包括以下五个方面：1. 电路设计原理。
在集成电路设计中，需要根据需要设计出各种电路模块，包括模拟电路、数字电路、混合电路等。
针对不同类型的电路，需要采取不同的设计方法和设计流程。
2. 设计目标和指标。
在集成电路设计时，需要根据实际需要确定设计目标和指标，包括性能、功耗、可靠性、成本等，以确保设计效果和实际应用效果相符合。
3. 物理细节设计。
在集成电路设计时，需要考虑到电子元器件的物理特性，包括电阻、电容、电感、晶体管等，并根据实际情况进行物理细节设计，包括原理图设计、布局设计等。
4. 制造和加工工艺。
在集成电路设计时，需要考虑到制造和加工工艺的要求，包括工艺流程、工艺控制、工艺变量等，以确保集成电路可以成功制造和完好运行。
5. 整个电路的优化设计。
在集成电路设计中，需要对整个电路进行优化设计，包括优化模型、线路优化、布局优化等，以确保电路的各项指标达到最优化的设计效果。
二、集成电路设计的流程分析集成电路设计的流程通常包括电路分解、电路设计分析、电路综合和验证等四个步骤。
具体流程如下：1. 电路分解。
将大的电路模块分解成若干个小模块，然后进行单独设计和测试。
2. 电路设计分析。
对单个模块进行电路设计，包括原理图设计、和布局设计等。
3. 电路综合。
将各个模块根据指定的规则进行组合，形成完整的电路。
电路中的数字集成电路设计与分析
电路中的数字集成电路设计与分析数字集成电路（Digital Integrated Circuit，简称DIC）是现代电子电路中的重要组成部分。
它们基于数字信号处理和逻辑运算，被广泛应用于计算机、通信、控制系统等领域。
本文将分析数字集成电路的设计原理和技术，并探讨其在电路中的应用。
一、数字集成电路的基本原理1.1 数字电路和模拟电路的区别数字电路是一种使用二进制数表示信息的电路，通过处理离散的数字信号进行逻辑运算；而模拟电路则是通过处理连续的模拟信号进行运算。
数字电路具有精确性高、噪声干扰小等优点，适合用于逻辑运算和信号处理。
1.2 数字集成电路的分类数字集成电路根据功能和结构可以分为多种类型，包括时序电路、组合电路和存储电路等。
其中时序电路用于时钟信号控制的功能电路，组合电路用于逻辑运算的功能电路，存储电路用于存储信息的功能电路。
二、数字集成电路的设计过程2.1 设计规划在进行数字集成电路设计之前，需要明确设计目标，包括功能需求、性能指标和设计约束等。
同时，还需对设计流程和设计工具进行规划，确保设计过程的有效性和可行性。
2.2 逻辑设计逻辑设计是数字集成电路设计的核心环节，通过逻辑门、触发器等基本模块的组合和连接，实现设计目标的功能和逻辑运算。
逻辑设计需要使用专业的设计语言和工具，如VHDL、Verilog等。
2.3 电路图设计电路图设计是将逻辑设计转化为具体的电路图的过程，包括将逻辑门、触发器等模块转化为相应的元件和连线。
在电路图设计中，需要考虑电路的布局和连接方式，以满足电路的性能指标和工艺要求。
2.4 仿真和验证仿真和验证是数字集成电路设计的重要环节，通过软件仿真和硬件验证，验证设计的正确性和稳定性。
仿真和验证过程需要使用仿真工具和测试设备，确保设计结果符合预期。
2.5 物理设计和布局物理设计和布局是将电路图设计转化为真实芯片的过程。
在物理设计中，需要考虑芯片的几何结构、层次布局和连线规划等。
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数字集成电路分析与设计一、课程基本情况课程编号40260103开课单位微纳电子学系课程名称中文名称数字集成电路分析与设计英文名称Digital Integrated Circuit Analysis and Design教学目的与重点教学目的：1）让学生掌握数字集成电路的工作原理与分析方法2）让学生掌握数字集成电路与系统的设计流程和基本方法3）培养学生实际设计数字集成电路与系统的能力教学重点：1） CMOS反相器的特性，数字集成电路分析与设计的关键问题2）组合逻辑链的性能优化3）互连线的延时模型与分析4）同步时序电路的分析和设计5）数据通路运算单元的分析与设计6）存储器的工作原理的理解与分析课程负责人刘雷波吴行军课程类型□文化素质课□公共基础课□学科基础课□专业基础课■专业课□其它教学方式■讲授为主□实验/实践为主□专题讨论为主□案例教学为主□自学为主□其它授课语言■中文口中文+英文（英文授课＞50%）□英文□其他外语学分学时学分 3 总学时48考核方式及成绩评定标准作业：15%,课程设计：15%,期中考试（闭卷）：30%,期末考试（闭卷）：40%教材及主要参考书中文外文教材数字集成电路一电路、系统与设计（第二版）,JanM.Rabaey等著，周润德等译，电子工业出版社。
Jan M. Rabaey etc. “Digital Integrated Circuits , A Design Perspective (Second Edition)", Prentice Hall , 2003.主要参考书CMOS数字集成电路一分析与设计（第3版）,Sung-Mo Kang等著，王志功等译，清华大学出版社（影Sung-Mo Kang, Yusuf Leblebici,"CMOS Digital IntegratedCircuits-Analysis and Design(ThirdEdition)".三、课程主要教学内容9.4高级互连技术9. 5综述9.6总结第10章存储器（6学时）（教材第12章）10.1分类10.2结构10.3内核--- 存储单元和阵列10.4外围电路10.5可靠性10.6总结。
数字集成电路分析与设计 第三章答案
CHAPTER 3P3.1. The general approach for the first two parameters is to figure out which variables shouldremain constant, so that when you have two currents, you can divide them, and every variable but the ones you want to calculate remain. In this case, since the long-channel transistor is in saturation for all values of V GS and V DS , only one equation needs to be considered:()()2112DS N OX GS T DS W I C V V V Lμλ=-+ For the last two parameters, now that you have enough values, you can just choose oneset of numbers to compute their final values.a. The threshold voltage, V T0, can be found by choosing two sets of numbers with the same V DS ’s but with different V GS ’s. In this case, the first two values in the table can be used.()()()()()()211122222201022001121121.2 1.210000.82800.8DS N OX GS T DS DS N OX GS T DS T DS T DS T T W I C V V V L W I C V V V LV I V I V V μλμλ=-+=-+-⎛⎫-===⎪--⎝⎭ 00.35V T V ∴=b. The channel modulation parameter, λ, can be found by choosing two sets of numberswith the same V GS ’s but with different V DS ’s. In this case, the second and third values in the table can be used.()()221 1.225010.8247DS DS I I λλ+==+ -10.04V λ∴=c. The electron mobility, µn , can now be calculated by looking at any of the first three sets of numbers, but first, let’s calculate C OX .631062-31m 10μm22?.210μm1m 10 0.0351 1.610/2.210OX OX t C F cm--=⨯⨯===⨯Now calculate the mobility by using the first set of numbers.()()()()()()()()()()()()22111021262101111 1.21 1.222210002cm 348V-s 1.610(4.75)1.20.3510.04 1.21DS N OX GS T DS N OX T DS N OX GS T DS W W I C V V V C V L LA I W C V V V L μλμλμμλ-=-+=-+===⨯-+-+d. The body effect coefficient gamma, γ, can be calculated by using the last set of numbers since it is the only one that has a V SB greater than 0V.()()()()244124414411221 1.20.468VDS N OX GS T DS DS GS T N OX DS GS T T GS W I C V V V LI V V W C V LV V V V μλμλ=-+-=+-==-==12000.6VT T T T V V V V γγγ=+-====P3.2. The key to this question is to identify the transistor’s region of operation so that gatecapacitance may be assigned appropriately, and the primary capacitor that will dischargedat a rate of V It C ∂∂= by the current source may be identified. Then, because the nodes arechanging, the next region of operation must be identified. This process continues until the transistor reaches steady state behavior. Region 1:Since 0V GS V = the transistor is in the cutoff region. The gate capacitance is allocated to GB C . Since no current will flow through the transistor, all current will come from the source capacitor and the drain node remains unchanged.68-151010V V 6.67100.6671510s nsSB V I I t C C -∆⨯====⨯=∆⨯ The source capacitor will discharge until 1.1V GS T V V == when the transistor enters thesaturation region. This would require that the source node would be at 3.3 1.1 2.2V S G GS V V V =-=-=.()15961510 3.3 2.2 1.6510s 1.65ns 1010C t V I ---⨯∆=∆=-=⨯=⨯ Region 2:The transistor turns on and is in saturation. The current is provided from the capacitor atthe drain node, while the source node remains fairly constant. The capacitance at the drain node is the same as the source node so the rate of change is given by:68-151010V V 6.67100.6671510s nsSB V I I t C C -∆⨯====⨯=∆⨯ Since the transistor is now in the saturation region, GS V can be computed based on thecurrent flowing through the device.()22 1.1 1.37V 3.3 1.37 1.93VGS T GST S G GS kW I V V LV V V V V =-==+==-=-=This is where the source node settles. This means that most of the current is discharged through the transistor until the drain voltage reaches a value that puts the transistor at the edge of saturation.3.3 1.1 2.2VDS GS TD G T V V V V V V =-=-=-=If we assume that all the current comes from the transistor, and the source node remains fixed, the drain node will then discharge at a rate equal to that of the source node in the first region. Region 3:The transistor is now in the linear region the gate capacitance is distributed equally to both GS C and GD C . and both capacitors will discharge at approximately the same rate.-151510V0.28621510510nsV I A t C μ-∆===∆⨯⨯+⨯The graph is shown below.00.511.522.533.5024681012Time (ns)V o l t a g e (V )P3.3. The gate and drain are connected together so that DS GS V V = which will cause thetransistor to remain in saturation. This is a dc measurement so capacitances are not required. Connect the bulk to ground and run SPICE. P3.4. Run SPICE. P3.5. Run SPICE. P3.6. Run SPICE. P3.7. Run SPICE.P3.8. First, let’s look at the various parameters and identify how they affect V T .∙ L – Shorter lengths result in a lower threshold voltage due to DIBL. ∙ W – Narrow width can increase the threshold voltage.∙ V SB – Larger source-bulk voltages (in magnitude) result in a higher threshold voltage. ∙ V DS –Larger drain-source voltages (in magnitude) result in a lower threshold voltage due to DIBL. The transistor with the lowest threshold voltage has the shortest channel, larger width, smallest source-bulk voltage and largest drain-source voltage. This would be the first transistor listed.The transistor with the highest threshold voltage has the longest channel, smallest width,largest source-bulk voltage and smallest drain-source voltage. This would be the last transistor listed. P3.9. Run SPICE.P3.10. Run SPICE. The mobility degradation at high temperatures reduces I on and the increasemobile carriers at high temperatures increase I off . P3.11. The issues that prompted the switch from Al to Cu are resistance and electromigration.Copper wires have lower resistances and are less susceptible to electromigration problems. Copper on the other hand, reacts with the oxygen in SiO 2 and requires cladding around the wires to prevent this reaction.For low-k dielectrics, the target value future technologies is 2.High-k dielectrics are being developed as the gate-insulator material of MOSFET’s. This is because the current insulator material, SiO 2, can not be scaled any longer due to tunneling effects.P3.12. Self-aligned poly gates are fabricated by depositing oxide and poly before the source anddrain regions are implanted. Self-aligned silicides (salicides) are deposited on top of the source and drain regions using the spacers on the sides of the poly gate. P3.13. To compute the length, simply use the wire resistance equation and solve for L .LR TWRTWL ρρ==First convert the units of ρ to terms of μm. Aluminum:2.7μΩρ=cm 6Ω10μΩ⨯610μm100cm ⨯()()()0.027Ωμm1000.812963μm 2.96mm0.027RTWL ρ=====Copper:1.7μΩρ=cm 6Ω10μΩ⨯610μm100cm ⨯()()()0.017Ωμm1000.814706μm 4.71mm0.017RTWL ρ=====P3.14. Generally, the capacitance equation in terms of permittivity constants and spacing is:k C WL tε=a. 4k = ()()()()230048.8510 3.541100SiO k k C WL TL t S S Sεε-====b. 2k = ()()()()30028.8510 1.771100k k C WL TL t S SSεε-====The plots are shown below.Capacitance vs. Spacing01234567800.511.522.533.544.555.5Spacing (um)C a p a c i t a n c e (f F)。
数字集成电路分析与设计 第七章答案
CHAPTER 7P7.1. Assume that all nodes start at 0V. The first row outputs will be at DD T V V -. Since thesenodes are also the gate nodes of the second row of transistors, their source nodes will be at 2DD T V V -. Likewise, the last row of transistors have voltages of 3DD T V V -. However, this value is below 0V so we leave them at 0V.1.2V1.2V0.73V 0.73V 0.73V0.33V0.33V0.33V0V0V0VP7.2. (a)(b)(c)(d)P7.3. (a) First calculate V Q .()01.80.51.15Q DD T DD T V V V V V Vγ=-=-+=--=Since this is slightly below 1.3V (voltage at which the PMOS turns on), we assume that the PMOS is slightly on. Since the PMOS’s V GS is quite low (because Q is high) and its V DS is quite high (because Q is low), the transistor is very likely in saturation. Similarly for the NMOS, because its V GS is high and its V DS is low, it’s likely in the linear region. Equating the two currents:()()()()()()()()22,,222211DSNDSN CN NQ Q CN NSDP sat DSN linV N N OX GSN T DSN P sat OX GSP T V GSP T CP PN V N N OX Q T Q P sat OX DD Q T V DD Q T CP PE L N I I W C V V V W v C V V V V E L L W C V V V W v C V V V V V V E L L μμ=---=-++----=--++For simplicity we shall assume that 11Q CN NV E L +≈ and220QV ≈.()()()2N N OX Q T QP sat OX DD Q T DD Q T CP P NW C V V V W v C V V V V V V E L L μ---≈--+Solve to produce:0.0080V Q V ≈When the CLK goes low, the intermediate output suffers from clock feedthough. To calculate the effects of clock feedthrough, let us first compute the capacitances involved. The capacitance from the clock signal to Q is:(.2/)(.2)0.0.4fF GS OL C C fF um um ===The capacitance from the Q to ground is:()()()(),310.2320.2 1.4fF Q DN IN inv d g C C C C W C W =+=+=+=The capacitive feedthrough equation is:()210.04 1.80.05V 0.04 1.41.150.05 1.1VGS CLK Q GS Q Q Q Q C V V C C V V V -∆∆===-++=+∆=-=To get the new value of Q V , first determine the determine the regions of operation of the transistors in the inverter by calculating V S . Then, once again, use the currentequations to determine Q V .Since the new voltage of V Q is still greater than the switching voltage, the transistors are in the same regions:()()()()()()2000460.4100.2810P N sat OX DD Q T Q N N OX Q T DD Q T CP P OXW L v C V V V V W C V V V V V E L C μ---≈---+⨯⨯≈()()()21.8 1.10.50.2270OX C --()()0.016V1.10.5 1.8 1.10.5 4.8≈---+(b) In this case 1.8Q DD V V V == and 0Q V =. Clock feedthrough has no effect since the transmission gate CLK signals cancel each other out.()()()()()()()(),3151515315(23)312.5102100.2110(2)0.23(210)(0.2)312.5101100.2257.532.5pass pass inv d inv eqn g eff g eqn d t R C R C R C W C W C W R C Wps ps ps----=+=+++⎡⎤=⨯⨯+⨯+⨯+⎣⎦⨯⨯=+=P7.4.a. Out A BC =+BBOutb. Out AB BC C =++Outc. ()Out A B C AB ABC AB =+++=+BBOutd. ()()1Out A B C AB ABC AB AB C AB A B =+++=+=+==+OutP7.5.a. ()Out A B C =+b. ()()Out A B C D E =+++ P7.6.a. Out A BC =+c bclkclkV DDb. Out AB BCC =++a bclkclkV DDc.()Out A B C AB ABC AB =+++=+V DDd.()()()Out A B C AB A B C A B AB=+++=+++=+aclkclkV DDP7.7.Assuming that one of the transistors in each transmission gate is being driven by a min-sized inverter:a.()()()()122333passinvRC R R RLERC R R+====b.()()()()()()()()313133313133AAinvCCinvRRC RLERC R RRRC RLERC R R========()()()()339333BBinvRRC RLERC R R====P7.8.a. Out A sel B sel =⋅+⋅b.R inv 6.25k ΩC inv,diff 1.2fF C pass,gate 0.8fF C pass,diff0.8fFR pass 6.25k ΩCpass,diff0.8fFCpass,gate0.8fFfC inv,gate2.4f fFCpass,diff0.8fFc. ()()(),,,,,,2A C inv inv diff pass gate pass diff inv pass inv gate pass gate pass diff t R C C C R R fC C C -=++++++ d. (),,inv inv LOADC out inv diff LOAD inv inv diffR R C t fC C R C f f-=+=+ e.()()()()(),,,,,,,,220inv inv diff pass gate pass diff inv pass inv gate pass gate pass diff inv LOADinv inv diff inv LOAD inv pass inv gate t R C C C R R fC C C R C R C fR C dtR R C df f f =++++++++=+-===3.2=P7.9. In both of these cases, the logical effort is the same due to the fact that the longest pathfrom output to ground is three transistors long. Assume that the CLK arrives ahead of the signals. Then,12()26663R R LE R λλ+== P7.10. We will use 0.18um technology and the node names below:W=4W=4OutFor the two inverter inputs:()()()3230.2 1.2fF inv g C C W ===For the pass gate inputs:()0.4fF pass g C C W ==At node x:()(3)(2) 1.4x eff eff g C C W C W C W fF =++=At node y:()2((2))(2)2y eff g eff C C W C W C W fF =++=At node Out:()((2))(2) 1.2out eff g eff C C W C W C W fF =++=The shortest path is through the one of the G ND input nodes to the output:()()()()min 212.5 1.4212.5 1.247.5x out t RC RC k fF k fF ps =+=+=The longest path is through one of the inverters to the output.()()()()()()max 2312.5 1.4212.52312.5 1.2112.5sx y out t RC RC RC k fF k fF k fF p =++=++=P7.11. At 0t =: DD F V =0X =?Y =.When the a goes high the first time, the voltage at X would be computed using the charge-sharing formula:()101.21V 210F DDX X F C V V C C ===++But because the maximum allowable voltage at node x is 0.734V, set 0.734V X V = Then recomputed V F :()()()()10 1.220.734 1.05V 10F DD X X F F C V C V V C --===When Phi goes down, F DD V V = and V X and V Y remains the same. The next time the Phi goes up, all the internal nodes are 0. When Phi goes down, F DD V V = and V X and V Y remains at 0. P7.12.P7.13.a. The input settings that give you the worst-case charge sharing are any of 1a c e === and both of 0b d ==. Essentially, what you are doing it trying to create the greatest amount of parasitic capacitances without creating a path to G ND .b. Assuming that transistors share nodes to reduce capacitance.()()()()()()()12*11125(3)(5) 5.2fF 333190.2 1.8fF 5.2 1.8 1.34V 5.2 1.8g d g d C C W C W C W C C W W W C V V C C =++==++=====++ The actual voltage would be larger than this since the internal node cannot rise above V DD -V T .c. This circuit fails if the worse case voltage falls below the switching voltage which can be computed to be V S =0.92V. Therefore, the circuit will operate properly. P7.14. Both of these circuits act as latches. When EN is on, there is a path from the output toeither V DD or G ND . The first latch is better than the second because the second latch suffers from charge sharing. When EN is off, there is no path from the output to either of the sources, if IN is switching it is possible for whatever charge that is held on OUT to be shared with the internal nodes between the two NMOS’s or the two PMOS’s. Therefore, the second one is not as good as the first one. P7.15.a.OUT OL X DD TV V V V V ==-b. First, let ’s find the required change in voltage:()()2OUT DD OLX DD T DD T TV V V V V V V V V ∆=-∆=+--=Now, let’s set up the clock feedthrough equation and solve for C b :22b OUT X b XX X T Xb OUT X DD OL TC V V C C V C V C C V V V V V ∆∆=+∆==∆-∆--。
vlsi数字集成电路一般设计流程
vlsi数字集成电路一般设计流程VLSI数字集成电路一般设计流程数字集成电路（VLSI）是现代电子技术领域的重要组成部分，广泛应用于计算机、通信、消费电子等领域。
VLSI数字集成电路的设计流程是一个系统性的过程，涉及到从需求分析到电路设计、验证、布局布线等多个环节。
本文将介绍VLSI数字集成电路的一般设计流程。
一、需求分析需求分析是VLSI数字集成电路设计的第一步，主要目的是明确设计要求和功能需求。
在需求分析阶段，设计团队与客户或项目经理进行沟通，了解项目的背景、功能要求、性能指标等。
同时，还需要考虑电路的功耗、面积、可靠性等因素，以确定设计的整体目标。
二、框架设计在框架设计阶段，设计团队根据需求分析的结果，确定整个电路的结构和功能模块。
框架设计需要考虑各个模块之间的连接方式、数据传输方式、时序要求等。
同时，还需要确定使用的逻辑门、存储器、寄存器等基本元件，并进行初步的电路图设计。
三、逻辑设计逻辑设计是VLSI数字集成电路设计的核心环节，主要目的是将框架设计的功能模块转化为逻辑电路。
在逻辑设计阶段，设计团队使用硬件描述语言（如Verilog、VHDL）进行电路的建模和描述，利用逻辑门、时序电路等元件进行电路的逻辑实现。
四、验证验证是确保电路设计正确性的重要环节。
在验证阶段，设计团队需要使用仿真工具对电路进行功能仿真，并设计测试用例进行验证。
通过仿真和测试，可以发现电路设计中的错误或潜在问题，并对其进行修复和优化。
五、布局布线布局布线是将逻辑电路转化为物理电路的过程。
在布局布线阶段，设计团队将逻辑电路转化为实际的布局图，确定各个元件的位置和相互之间的连线关系。
同时，还需要考虑电路的面积、功耗、信号延迟等因素，并进行布线优化。
六、物理验证物理验证是检验布局布线结果的环节。
在物理验证阶段，设计团队对布局布线后的电路进行电气规则检查（DRC）和电磁规则检查（ERC），以确保电路的物理完整性和可靠性。
根据验证结果，可以对布局布线进行调整和优化。
数字集成电路分析与设计深亚微米工艺第三版课程设计
数字集成电路分析与设计深亚微米工艺第三版课程设计一、概述本文主要介绍数字集成电路分析与设计深亚微米工艺第三版课程设计。
本设计主要涉及数字集成电路设计的各个方面，包括数字逻辑设计、计算机组成原理、数字信号处理等。
本设计旨在深入探究数字电路和集成电路的设计和工艺细节，从而提高学生的专业技能和实践能力。
二、设计内容本次设计主要分为以下几个部分：1. 数字逻辑电路设计在本部分中，学生需要根据题目要求，设计数字逻辑电路的电路图和真值表，同时需要手动编写数字逻辑电路的代码，并利用VHDL语言进行编程实现。
本部分要求学生熟练掌握数字逻辑电路的设计方法和VHDL语言的编程技巧。
2. 计算机组成原理在本部分中，学生需要设计一个基于FPGA的计算机组成原理的电路图和真值表，并利用VHDL语言进行编程实现。
本部分要求学生深入理解计算机组成原理的设计思想，并熟练掌握FPGA电路设计和VHDL编程的技巧。
3. 数字信号处理在本部分中，学生需要设计一个数字信号处理的电路图和真值表，并利用Python语言进行编程实现。
本部分要求学生掌握数字信号处理的基本原理和算法，以及Python语言的编程技巧。
三、课程目标通过本次课程设计，学生应该达到以下目标：1. 掌握数字电路和集成电路的设计和工艺细节本设计涉及数字电路和集成电路的多个方面，要求学生深入理解电路设计和工艺细节，从而能够熟练掌握数字电路和集成电路的设计方法和实现流程。
2. 提高学生的专业技能和实践能力本设计要求学生进行实际的电路设计和编程实现，从而加深对数字电路和集成电路的理解和掌握。
通过实践，学生能够提高自己的专业技能和实践能力，为将来的工作打下坚实的基础。
3. 培养学生的团队合作和创新能力本设计要求学生分组进行合作，通过协作和交流，提高团队合作和创新能力。
学生需要思考如何在电路设计和编程实现中，发挥个人和团队的优势，提高工作效率。
四、总结数字集成电路分析与设计深亚微米工艺第三版课程设计，旨在提高学生的数字电路和集成电路设计能力，同时培养学生的实际操作能力和团队合作能力。
数字集成电路设计与系统分析答案
懂得1、Please illustrate the meaning of its voltage transfer characteristic to a logic gate, and describe the static behaviors showed in the voltage transfer characteristic curves.The electrical function of a gate is best expressed by its voltage transfer characteristic (VTC),which plots the output voltage as a function of the input voltage Vout=f(Vin).The high and low nominal voltage Voh and Vol;The gate or switching threshold voltage Vm,that is define as Vm=f(Vm)(The gate threshold voltage presents the midpoint of the switching characteristics,which is obtained when the output of a gate is short circuited to the input);The high and low input voltage Vih and Vil are defined by the point where the gain (=dVout/dVin)of the VTC equals -12、Please draw the voltage transfer characteristic curve of the inverter and label the static operation points in the VTC.3、Please describe the definition of noise margin and its physical significance(物理意义), then draw the figure of definition of noise margins.The noise margins represent the levels of noise that can be sustained(所允许的) when gates are cascaded. A measure of the sensitivity of a gate to noise is given by the noise margins NML(noise margin low) and NMH(noise margin high), which quantize the size of the legal “0” and “1”, respectively, and set a fixed maximum threshold on the noise value4、Please describe the meaning of the regenerative property and the conditions of a gate with regenerative property.A gate with regenerative property ensures that a disturbed signal converges back to a nominal voltage level after passing through a number of logical stages. The VTC should have a transient region (or undefined region) with a gain greater than 1 in absolute value, bordered by the two legal zones, where the gain should be less than 1 in absolute value5、What are the definitions of the fan-out and fan-in properties?The number that can be driven is termed the fan-out of circuit, that denotes the number of load gates N that are connected to the output of the driving gate. The fan-in of a gate is defined as the number of independent input nodes to the gate.6、How to describe the performance of a digital IC? Please illustrate the parameters used to characterize the transient performance of a logic family, and draw the associated figure of the definition of these parP ropagation delay time and rise/fall time can be used to characterize the transient performance of a logic family .Propagation delay time of a gate expresses the delay experienced by a signal when passing through a gate,which represent how quickly the gate responds to the changes at its inputs.Rise/fall time express how fast a signal transits between the different levels. Propagation delay time is defined as the period between the 50%transition points of the input and output signals.Rise/fall time is defined as the period between the 10% and 90% points of the total voltage transition at the output waveforms.1、Illustrate the basic structure and simple operation principle of MOS transistor.Four terminals:source, drain, gate, body; Vertical Structure: gate electrode, insulator, semiconductor substrate; Horizontal Structure: source region, channel region, drain region2、Illustrate the basic function of each terminal of MOS device, and describe the general terminal connections of NMOS and PMOS transistor, respectively.The source and the drain are the electrodes conducting the current. The gate electrode is thecontrolling terminal. The function of the body is secondaryIn NMOS devices, the source is defined as the n+ region which has a lower potential(电势) than the other n+ region, the drain. The source is the terminal with the higher potential in PMOS devices, The body is generally connected to a DC supply that is identical for all devices of the same type (GND for NMOS, VDD for PMOS).3、What does the transition (or input) characteristic of MOS transistor mean? And what conclusions we can find from the characteristic curve?It describes the relationship between the gate-source voltage and the drain-source current with the certain drain-source voltage .When the gate-source voltage is less than the threshold voltage, the conducting current is zero, that is, the NMOS transistor is in cutoff operation. When is larger than, the NMOS transistor is on.4、What does the current-voltage (or output) characteristic of MOS transistor mean? And what conclusions we can find from the I-V characteristic curve?.It describes the relationship between the drain-source voltage and the drain-source current with a certain gate-source voltageVgs > Vt , 0<VDS <VGS -VT : Linear modeThe inversion layer forms a continuous current path between the source and the drain.A drain current proportional to Vds will flow from the drain to the source through the conducting channel. The channel region acts as a voltage-controlled linear resister.5、Describe the operation modes of NMOS and PMOS transistors respectively, and define the corresponding ideal current equations.1、Explain the channel-length modulation, sub-threshold conduction, short-channel effect and narrow-channel effect. And illustrate their corresponding chief impacts on the device.This simple current equation prescribes a linear drain-bias dependence for the current in MOS transistors, determined by the empirical model parameter λ, called the channel-length modulation coefficientOne typical condition, which is due to the two-dimensional nature of channel current flow, is the sub-threshold conduction in small-geometry MOS transistors.As a working definition, a MOS transistor is called a short-channel device if its channel length is on the same order of magnitude as the depletion region thicknesses of the source and drain junctions.The short-channel effects that arise in this case are attributed to two physical phenomena: the limitations imposed on electron drift characteristics in the channel; the modification of the threshold voltage due to the shortening channel lengthMOS transistor that have channel widths on the same order of magnitude as the maxium depletion region thickness are defined as narrow channel devices.For MOSFET with small channel widths,the actual threshold voltage increases as a result of this extra depletion charge of the fringe depletion region.This fact is called narrow channel effect.2、Describe the three main components of the load capacitanceCL, when a logic gate is driving other fan-out gates. And sketch the capacitance model of NMOS transistor.Gate capacitances (of other inputs connected to out)Diffusion(or junction) capacitances (of drain/source regions)Routing capacitances (output to other inputs)1,Describe the basic structure and operation of a static CMOS inverter. Then draw theassociated transistor schematicThis structure consists of an enhancement-type NMOS transistor and an enhancement-type PMOS transistor, operating in complementary mode. So this configuration is called Complementary MOS (CMOS). The gate terminals of the PMOS and NMOS transistors are connected to form the inverter input. The drain terminals of the PMOS and NMOS transistors are connected to form the inverter output. The source and the substrate of the NMOS transistor are connected to the ground, while the source and body of PMOS transistor are connected to VDD The circuit topology is complementary push-pull in the sense that: For high input the NMOS transistor drives (pulls down) the output node while the PMOS transistor acts as the load, and for low input the PMOS transistor drives (pulls up) the output node while the NMOS transistor acts as the load.When the input is at VDD: The NMOS is on (conducting) while the PMOS is off (cut-off). A direct path exists between Vout and the ground node, resulting in a steady-state value of 0V at the output. When the input is at ground:The NMOS is off while the PMOS is on. A direct path exists between VDD and Vout, yielding a high output voltage (equal to VDD).Static CMOS logic:structure:The static CMOS style is really an extension of the static CMOS inverter to multiple inputs. A logic function in static CMOS must be implemented in both NMOS and PMOS transistors. It is the combination of the pull-up network(PUN) and the pull-down network(PDN). Each input always connects to PUN and PDN simultaneously. The function of the PUN is to provide a connection between the output and VDD anytime the output of the logic gate is meant to be 1 (based on the inputs). The function of the PDN is to connect the output to VSS when the output of the logic gate is meant to be 0.Opreation: The pull-down net should be “on” when the pull-up net is “off” and vice versa. For any given input combination, the output is connected either to VDD or to ground via a low-resistance path. A DC current path between the VDD and ground is not established for any of the input combinations. With the complementary nature of NMOS and PMOS, the pull-up or the pull-down is “on” alternately to implement the logic operation.Discuss the main problems for high fan-in static CMOS gates and the associated techniques for fast complex gates.tpHL = 0.69 Reqn(C1+2C2+3C3+4CL); Propagation delay deteriorates(恶化) rapidly as a function of fan-in quadratically in the worst case, Gates with a fan-in greater than 4 become excessively slow and must be avoided.tPLH increases linearly due to the linearly increasing value of the diffusion capacitance;tPHL increase quadratically due to the simultaneous increase the resistance and internal capacitance in serial part.Transistor sizing: as long as fan-out capacitance dominatesProgressive transistor sizing: This approach reduces the dominant resistance, while keeping the increase in capacitance within boundsTransfer gate:Configuration:The source and drain nodes serve as inputs and outputs, while the gate node serves as the control input, the body node is connected to the power/ground Operation: For NMOS transfer gate,it turns on while the gate control terminal goes high, and the input signal will be delivered to the output node; it turns off while the gate control terminal goes low, and the output node will be impedance.CMOS transmission gate:Configuration: The CMOS transmission gate consists of one NMOS and one PMOS transistor, with the source and drain connected in parallel; The gate voltages appliedto these two transistors are also set to be complementary signals. The substrate terminal of the NMOS transistor is connected to ground and the substrate terminal of the PMOS transistor is connected to Vdd.Operation: If the control signal C is logic-high (equal to Vdd), then both transistors are turned on and provide a low-resistance current path between the input and output nodes. If the control signal C is logic-low, then both transistors will be off, and the path between the input and output nodes will be in the high-impedance state. The weakness of one device is overcome by the strength of the other device, whether the output is transmitting a high or low value. This is a clear advantage of the CMOS transfer gate over the single transistor counterpart.DCVLS:Operation: Assume now that, for a given set of inputs, PDN1 conducts while PDN2 does not, and that Out and out are initially high and low, respectively. Turning on PDN1: Causes Out to be pulled down (below VDD−|VTP |); Out is in a high impedance state, as M2 and PDN2 are both turned off. At the point M2 turns on and starts charging out非to VDD — eventually turning off M1; This in turn enables Out to discharge all the way to GND.XOR/XNOR: When the signals A and B have the same values, there is one conducting path either AB or A非B非; Then the output F is pulled down;At the same time, the other pull-down paths connected to the F非are both turned off. When F is pulled down below VDD−|VTP |, M2 t urns on and starts charging F非to VDD —eventually turning off M1 and pulling down F to Gnd. When the signals A and B have the different values, there is one conducting path either AB非or A非B; Then the output F非is pulled down; At the same time, the other pull-down paths connected to the F are both turned off. When F非is pulled down below VDD−|VTP |, M1 turns on and starts charging F to VDD —eventually turning off M2 and pulling down F非to Gnd.Precharge-Evaluate dynamic CMOS:Operation: Precharge (when the clock signal Φ= 0)：The PMOS precharge transistor MP is conducting while the complementary NMOS transistor MN is off. The output load capacitance is precharged to VDD by MP, then VOH=VDD；The input voltages have no influence yet upon the output level since the complementary NMOS transistor MN is off. Evaluate (when the clock signal Φ=1)：The precharge transistor MP turns off while the NMOS evaluate transistor MN turns on. The output node voltage may now remain at the logic-high level or drop to a logic low, depending on the input voltage levels: If the input signals create a conducting path between the output node and the ground, PDN is on, and the output capacitance will discharge toward VOL=0;Otherwise, when PDN is off, the output voltage remains at VOH= VDD.Domino dynamic CMOS logic:When Φ=0, during precharge: The output of the n-type dynamic gate is charged up to VDD, and the output of the inverter is set to 0. When Φ=1, during evaluation: The dynamic gate conditionally discharges, and there are two possibilities: The output node of the dynamic CMOS stage is either discharged to a low level through the NMOS circuitry (1 to 0 transition), or it remains high. Consequently, the inverter output voltage can also make at most one transition during the evaluation phase, from 0 to 1.TSPC dynamic CMOS logic:Configuration:If one constrains a NORA stage to have only n-precharge gates, and not static gates, then a p-channel transistor can be eliminated from the clocked latch; The dynamic circuit technique to be presented in that it uses only one-phase clock signal, so no clock skew problem exists. The NORA design style can be simplified so that a single clock is sufficient. For the doubled n-C2MOS latch, when φ= 1, the latch is in the transparent evaluate mode and corresponds to 2 cascaded inverters (non-inverting); For the doubled n-C2MOS latch, when φ= 0, both inverters are disabled (hold mode) -- only the pull-up network is still active.Pipelined NORA dynamic CMOS system:Configuration: Consists of an np-CMOS logic sequence and a clocked CMOS output buffer; A pipelined system can be constructed by simply cascading alternating φ-section and φ -section, meaning that evaluation occurs during active φ and φ respectively;Operation:φ=0, during hold mode :N block performs the precharge operation and pulls node Out1 up to VDD through the p-type device Mp1, while p block performs the discharge operation and pulls the node Out2 down to zero through the n-type device Mn2; The clocked CMOS latch will not be in operation and the previous output voltage will be stored on the output load capacitor CL. φ=1, during evaluate mode:All cascaded NMOS and PMOS blocks evaluate output levels one after the other, and then the signal Out2 will be inversed to the output node by the clocked CMOS latch in operation;Operation Mode: Evaluate―Hold: All logic stages perform the precharge-discharge operation when the clock is high, and all stages evaluate output levels when the clock is low. Therefore, wewill call this circuit a section, meaning that evaluation occurs during active .Clocked CMOS dynamic circuit:Basic Structure:A pair of PMOS and NMOS transistors controlled by the complementary clock signals are cascaded in the pullup and pulldown paths of the static CMOS gate, respectively, then a CMOS logic gate can be synchronized with a clock. Operation: φ=1, during evaluation mode：The transistors Mp1 and Mp2 are both turned on, then this gate can evaluate normally as a CMOS inverter to generate the logic output In非; φ=0 , during hold mode: Both transistors Mp1 and Mp2 are off, decoupling the output from the input. The CMOS circuit cannot conduct and evaluate, then the output Q retains its previous value stored on the output capacitor CL.Sequential logic:Virtually all useful systems require storage of state information, leading to another class of circuits called sequential logic circuits. In these circuits, the output not only depends upon the current values of the inputs, but also upon preceding output values. In other words, a sequential circuit remembers some of the past history of the system; A sequential circuit consists of a combinational circuit and a memory block in the feedback loop.Combination logic:In all logic circuits described so far, the output is directly related to the input. Typically, there are no feedback loops between the output and the input in these circuits (also classified as non-regenerative circuits), so the outputs are always a logical combination of the inputs. As a class, these circuits are known as combinational logic circuits. Combinational logic circuits, described earlier, have the property that the output of a logic block is only a function of the current input values, assuming that enough time has elapsed for the logic gates to settle. Static storage:preserve state as long as the power is on;are built using positive feedback or regeneration with an intentional connection between the output and the input;useful when updates are infrequent (clock gating)Dynamic storage:store state on parasitic capacitors;only hold state for short periods of time (milliseconds);require periodic refresh to annihilate charge leakage;usually simpler, so higher speed and lower power;useful in datapath circuits that require high performance levels and are periodically clockedLatch: level sensitive circuit that passes inputs to Q when the clock is high (or low);input sampledon the falling edge of the clock is held stable when clock is low (or high)Register or Flip-flops (edge-triggered): edge sensitive circuits that only sample the inputs on a clock transitionpositive edge-triggered: 0- 1negative edge-triggered: 1 -0built using latches (e.g., master-slave flip-flops)。
数字集成电路分析与设计 第五章答案
CHAPTER 5P5.1. For each problem, restate each Boolean equation into a form such that it can be translatedinto the p and n-complex of a CMOS gate.a. ()()Out ABC BD ABC BD A B C B D =+=+=+++b. ()()()Out AB AC BC AB AC BC A B A C B C =++=++=+++c. ()()Out A B CD A AB C D A A B CD A A B CD A =+++=++=+++=++AbVddVddAb BbAAbVddP5.2.AP5.3. First, convert the equation into its p and n-complex.()()()()()()()()()()()Out A B C BC AB AB C BC AB AB C BC AB AB C BC AB AB C BC AB AB C B C =⊕+=++=++=+=++=+++VddP5.4. The truth table is given below in terms of voltages. The function is F A B =The worse case V OH is V DD and the worse case V OL is 0V.P5.5. The first circuit is a NOR gate while the second is a NAND gate. The V OL and V OHcalculated are for the worst-case scenario. To find this, assume only one transistor turns on, this just reduces to a pseudo-NMOS/PMOS inverter, so the other transistors are not important.a. The V OL for the pseudo-NMOS (in 0.18μm) is:()()()2,1N N OXNSAT OX P GSP TPP SATOL W C L N DD TN GSP TP CP PDD TN SAT P N OX v C W V V I V k V V V V E L V V v W L C μ-==--+-=()2DD TP N N OX V V W C μ-()()()()()20.1DD TP CP P DD TN SAT P N DD TPDDN N DD TP CP P DD TN V V E L V V v W L V V V W V V E L V V μ-+--==-+-()()()()()()()()()()()()226440.18100.2100.210 1.80.50.14μm=1.40.11.8270 1.80.5240.2 1.80.5SAT P N DD TPN DD N DD TP CP P DD TN v W L V V W V V V E L V V μλ---=-+-⨯⨯⨯-==-+-Since the minimum width is 2λ, we make that the width. The V OH for the pseudo-PMOS (in 0.18μm) is:()()()()()()2221SDPSDP CP PN P V P OX P SGP TP SDP SAT OX N GSN TN V GSN TN CN N N E L SAT OX I sat I lin C W V V V v C W V V V V E L L v C μ=---=-++()2P OX N DD TN DD TN CN NC W V V V V E L μ-=-+()()()()()()2201DD OH DD OH CP PV V P DD TPDDOH V V P E L W V V VV L ------+()()()()()()20.1824620.184.8(70) 1.80.50.180.2(10)(810)1.80.51.80.5 1.21P P W L ---⨯-=-++4.2P W λ≈The pseudo-PMOS circuit will have bigger devices than the pseudo-NMOS.P5.6. The steps to solving this question are the same as the pseudo-NMOS question in Chapter4.a. For V OH , recognize that GS T V V >= for operation so the output can only be as high asDD T V V -. Since 0SB V ≠, body effect must be taken into account and the full equationis:()()()001.20.40.2OH DD T DDT V V V V Vγγ=-+=-+=-+ Iteration produces V OH =0.73V.b. For V OL , we must first recognize that the worst-case V OL occurs when only one of the pull-down transistors is on. Next we identify the regions of operation of the transistors. In this case, the pull-up transistor is always in saturation and the pull-down is most likely in the linear region since it will have a high input (high V GS ) and a low output (low V DS ). Then, we equate the two currents together and solve for V OL :()()()()()()()()221222222211111224620.61(1)(270)1.20.4(0.13)(10)(810)1.20.42(1.20.42)0.61DS DS CN OL OLV N OX GS T DS sat OX GS T V GS T CN E LV OL OL V OL I sat I lin W C V V V W v C V V V V E LL V V V μ-=---=-++--⨯--=--++Using a programmable calculator or a spreadsheet program, V OL = 0.205V. The dc current with the output low is:()()()()2222222260.20520.2050.61(1)(270)(1.610)1.20.4(0.205)146.5DS DS CN V N OX GS T DS DS V ELW C V V V I L Aμμ---=+⨯--=+=The power with the output low is:(46.5)(1.2)55.8DS DD P I V A V W μμ===P5.7. See Example 5.2 which is based on the NAND gate. This question is the same except thatit addresses the NOR gate.With both inputs tied together, 88N P W W λλ==2χ=== ()()1.80.520.50.77V 112DD TP TNS V V V V χχ-+-+===++In the SPICE solution, the reason why the results vary for input A and B is due to body-effect.P5.8. The solution is shown below. Notice that there is no relevance with the lengths andwidths of the transistors when it comes to V OH , although they the do matter when calculating V OL.01.80.50.3 2.51Vout GG T GG out T V V V V V V γ=-=++=++=P5.9. For t PLH , we need to size the pull-up PMOS appropriately.()()()()15120.70.720.70.73010010845010PLH eqp LOAD p SQLOAD PLHLt RC R C WL W R C k t λλ--====Ω⨯=⨯For V OL :()()()()()()()()()()()()()2246660.1220.10.63 4.210810 1.610 1.20.4 1.08mA1.20.4240.1(270)(1.610)1.20.40.11138.577377232(3OLOL CN P sat OX GS T P GS T CP V N N OX OL TN OLN P V N N E LNN NW v C V V I sat V V E LW C V V V W I sat L L W W W stack L μλλλ---⨯⨯⨯--===-+-+--⨯--==++===⨯=2)155(2)W stack λ=P5.10. The circuit is shown below:()()()()()()()()31512315120.720.70.7301075106350100.720.70.712.510751026.6275010PLH EQP LOAD PP EQPLOAD PLHPHL EQN LOAD NN EQNLOAD PHLLt RC R C W L W R C t Lt RC R C W L W R C t λλλλλ----====⨯⨯=⨯====⨯⨯=≈⨯Because the number of transistors in series is more than one, we must multiply the widths by the appropriate number. Here, all the NMOS transistors will have a width of 54λ. The PMOS transistors will have widths of 126λ and 190λ, respectively.P5.11. We estimate the dc power and dynamic switching power for this problem.a. The circuit’s dc power can be computed by computing the dc current when the output is low. This is given by I DS =550uA/um x 0.1um=55uA. Then P DC =66uW when the output is low.b. Its dynamic power can be calculated by simply using the equation 2dyn DD P CV f α=. Therefore, P dyn =(50fF)(V DD -V TN )(V DD )(100MHz)=4.4uW.P5.12. The pseudo-NMOS inverter has static current when the output is low. We can estimate itas:()()()()()()()()224660.110810 1.610 1.20.425.6A 1.20.4240.1P sat OX GS T P GS T CP W v C V V I sat V V E Lμ--⨯⨯⨯--===-+-+Then the average static power is P stat =(25.6uA)(1.2)/2 =15.4uW.The dynamic power is dyn DD swing avg P CV V f ==(50fF)(1.2)(1.1)f avg assuming that V OL is 0.1V.For the CMOS inverter, the static power is almost zero: P stat =I sub V DD . It is far less than the pseudo-NMOS case. The dynamic power dyn DD swing avg P CV V f ==(50fF)(1.2)2f avg is slightly larger than the pseudo-NMOS case.VVINCMOS InverterV V INPseudo-NMOSP5.13. Model development to compute αsc .P5.14. The energy delivered by the voltage source is:()()200202DDDDV C sourceDD DD L L DDCL DDV CDDcap C LC L C C LdvE i t V dt V C dt C V dvC V dt dv V E i t v dt C v dt C v dv C dt∞∞∞∞========⎰⎰⎰⎰⎰⎰As can be seen, only half the energy is stored in the capacitor. The other half was dissipated as heat through the resistor.P5.15. The average dynamic power does not depend on temperature if the frequency stays thesame. However, the short-circuit current will increase as temperature increases. In addition, the subthreshold current increases as temperature increases. So the overall power dissipation will be higher. P5.16. The circuit is shown below. The delay should incorporate both Q and Qb settling in400ps. All NMOS and PMOS devices are the same size in both NAND gates.QQW()()()()()()()()15331220.70.70.70.720.71001030100.1212.5100.10.72400101μm N P P PHL PLH UP LOAD DOWN LOAD LOAD eqp eqn P N LOAD eqp eqn LOAD eqp eqn PL Lt t t R C R C C R R W W C R L R L WC R L R L W t --⎛⎫=+=+=+ ⎪⎝⎭+=++==≈P5.17. The small glitch in J propagates through the flop even though it is small. This is due tothe fact that the JK-flop of Figure 5.20 has the 1’s catching problem. P5.18. The small glitch in J does not propagate through the flop since the edge-triggeredconfiguration does not have a 1’s catching problem.P5.19. The positive-edge triggered FF is as follows:QQDS(a) With CK=D=0 and S=R=1, the outputs are(b) Now CK=0。
集成电路设计技术研究及其应用分析
集成电路设计技术研究及其应用分析在现代信息技术的发展中，集成电路被广泛应用于各种设备，从小型智能手机到大型超级计算机，都要依靠集成电路技术支撑。
由于这项领域的竞争越来越激烈，人们对于其研究和应用方面也更加关注。
本文将重点探讨集成电路设计技术的研究和其应用分析。
一、集成电路设计技术1. 单片集成电路单片集成电路是指在一个芯片上集成多种电子器件及其相应电路，其中最为广泛应用的是微处理器和数字信号处理器。
与传统电路相比，单片集成电路的电路复杂度更高，但体积更小。
其加工制造过程采用先进的光刻技术，能达到微米级别，从而制造出功耗更低、速度更快、规模更大的芯片。
2. 复杂性可编程集成电路随着电子技术的飞速发展，人们对集成电路性能和功能的要求也越来越高。
为了满足市场的需求，复杂性可编程集成电路(CPLD)因此应运而生。
该技术可以让设计人员在现有硬件上烧录自己的程序，快速完成特定电路的设计。
这种集成电路具有结构灵活、性能可调等特点，被广泛应用于系统级设计和硬件设计中。
3. 器件级集成电路器件级集成电路（ASIC）是一种特别定制的芯片，根据特定的需求，设计人员可以将电路设计成为单独定制的芯片。
ASIC主要应用于高速、高性能、高频率和低功耗等领域。
它可以包含数百万条逻辑门，能够实现极端的芯片性能，并广泛应用于通信、计算机、汽车、军工等领域。
二、集成电路设计应用分析1. 通信系统集成电路技术与通信系统密不可分，通信系统的构建离不开高性能的DSP和MCU等器件。
通过使用单片集成电路，可以将所有的信号处理器件集成到一起，提高了系统集成度。
在通信系统中，ASIC集成电路可以实现各种通信协议的实时处理和优化，使系统性能得到进一步提升。
2. 芯片生产在生产集成电路芯片的过程中，CPLD技术可以用于快速制造带有特定功能的通用芯片，从而降低生产成本。
而ASIC芯片则可以根据不同的需求，进行高度定制化的设计，生产出专用芯片，从而满足某些特定领域的需求。
数字集成电路设计与分析
数字集成电路设计与分析数字集成电路（Digital Integrated Circuit，简称DIC）是一种用于处理和传输数字信号的电路。
它由许多晶体管、二极管和其他电子元件组成，通过将信号转换为离散的数字形式来进行处理。
在现代科技和信息技术的推动下，数字集成电路已经广泛应用于计算机、通信、嵌入式系统等领域。
一、数字集成电路的设计原理数字集成电路的设计原理源于二进制逻辑电路的概念。
二进制逻辑电路利用布尔代数的运算规律，通过逻辑门的组合和连接来实现各种逻辑功能。
数字集成电路是在此基础上进一步发展而来。
数字集成电路的设计需要考虑以下几个方面：1. 逻辑功能：根据需求确定数字电路所需实现的逻辑功能，如加法器、乘法器、状态机等。
2. 硬件资源：根据逻辑功能确定所需的晶体管、电阻、电容等硬件资源，并进行布局和布线设计。
3. 时序与时钟：考虑电路中各元件的时序关系，确定时钟频率和时序控制策略。
4. 电源和接口：设计电源供应和与外部系统的接口电路，确保数字集成电路的正常工作和与外界的通信。
二、数字集成电路的分析方法数字集成电路的分析是为了验证其设计是否符合预期功能、时序要求和性能指标。
以下是常用的数字集成电路分析方法：1. 逻辑仿真：通过电路仿真软件，将输入信号应用到数字集成电路模型中，观察输出信号是否满足预期逻辑功能。
逻辑仿真可以帮助发现设计中的逻辑错误和时序问题。
2. 时序分析：通过时序分析工具，分析数字集成电路中各个时序路径的延迟和时钟频率。
时序分析可以帮助确定电路是否满足时序要求，避免出现时序冲突或时序违规的问题。
3. 功耗分析：通过电路仿真和电路特性提取工具，分析数字集成电路的功耗消耗和功耗分布。
功耗分析可以帮助优化电路的功耗性能，减少能源消耗。
4. 供电噪声分析：通过电磁仿真和噪声分析工具，分析数字集成电路中的供电噪声问题。
供电噪声分析可以帮助解决电路中的电源干扰和信号完整性问题。
5. 仿真验证：通过数字集成电路芯片级仿真和电路板级仿真，验证数字集成电路的功能和性能。
集成电路设计工具与分析方法
集成电路设计工具与分析方法现代技术的进步改变了人们的生活方式，从家电到计算机、移动设备再到云计算，更为底层的一种技术则是“集成电路”。
集成电路是通过把各种电子元器件（如电容、电阻等）集成在单一的晶片上面，从而制造出具备特定功能的芯片。
在如此快节奏的时代里，集成电路飞速发展成为我们日常生活中必不可少的一部分。
如今，集成电路的设计越来越依赖计算机软件工具，这些工具使设计者能够更精确地设计和模拟电路功能，并在更短的时间内更快地实现产品上市。
本文将重点探讨在集成电路设计和分析方面常用的软件工具和方法。
一. 集成电路设计工具1. 电路设计自动化系统（EDA）EDA系统在集成电路设计过程中非常重要。
它是一个完整的计算机仿真设计平台，能够将电路的功能验证和设计注释相结合。
常用的EDA系统软件包包括Cadence、Mentor Graphics和Synopsys等，它们提供了从电路原理到过程设计的各个方面的支持。
EDA系统能够将模型库、原理图、模拟器和自动生成器等多种功能集成在一起，提供了完整的电路设计解决方案。
2. 模拟器模拟器是集成电路仿真过程的核心部分。
电路仿真是为了验证电路功能正常性以及检查它们是否符合最初的规格说明。
模拟器的作用是使用不同的仿真方法对电路进行检查，以找到设计缺陷和错误。
软件工具包括SPICE模拟器和分析器等。
3. 模型库模型库是电路仿真所使用的重要资源。
它包括本身的原理图、芯片规格和仿真器等。
模型库通常由芯片制造商提供，并根据芯片规格和电路的功能进行分类。
使用模型库可以大大加快电路仿真的速度，同时也可以确保仿真结果真实可靠。
芯片制造商如Intel和TSMC都提供了很多常见的元器件和芯片的模型库，供工程师使用。
二. 集成电路分析方法1. SPICE仿真SPICE仿真是最常用的一种集成电路分析方法。
SPICE 是“电流，电压和功率模拟器的通称”。
它是模拟不同电路的特性来查找问题和验证电路功能的基本工具，其仿真结果可以帮助电路设计人员改进设计，后期的市场产品制造和测试也可以极大地受益。
数字集成电路设计与分析
问答：Point out design objects in the figure such as :design, cell, reference, port, pin, net, then write a command to set 5 to net ADesign: topReference: ADD DFFCell: U1 U2Port: A B clk sumPin: A B D QNet: A B SINSet_load 5 [get_nets A]why do we not choose to operate all our digital circuits at these low supply voltages?答：1）不加区分地降低电源电压虽然对减少能耗能正面影响，但它绝对会使门的延时加大2）一旦电源电压和本征电压（阈值电压）变得可比拟，DC特性对器件参数（如晶体管阈值）的变化就变得越来越敏感3）降低电源电压意味着减少信号摆幅。
虽然这通常可以帮助减少系统的内部噪声（如串扰引起的噪声），但它也使设计对并不减少的外部噪声源更加敏感）问道题：1. CMOS静态电路中，上拉网络为什么用PMOS,下拉网络为什么用NMOS管2. 什么是亚阈值电流，当减少VT时，VGS =0时的亚阈值电流是增加还是减少？3. 什么是速度饱和效应4. CMOS电压越低，功耗就越少？是不是数字电路电源电压越低越好，为什么？5. 如何减少门的传输延迟？ P2036. CMOS电路中有哪些类型的功耗？7. 什么是衬垫偏置效应。
8. gate-to-channel capacitance CGC,包括哪些部分VirSim有哪几类窗口3-6. Given the data in Table 0.1 for a short channel NMOS transistor withVDSAT = 0.6 V and k′=100 µA/V2, calculate VT0, γ, λ, 2|φf|, and W / L:解答:对于短沟道器件：在选择公式的时候，首先要确定工作区域，表格中的所有VDS均大于VDSAT，所以不可能工作在线性区域。
数字集成电路设计实验分析报告
哈尔滨理工大学数字集成电路设计实验报告学院：应用科学学院专业班级：电科12 - 1班学号： 32姓名：周龙指导教师：刘倩2015年5月20日实验一、反相器版图设计1.实验目的1）、熟悉mos晶体管版图结构及绘制步骤；2）、熟悉反相器版图结构及版图仿真；2. 实验内容1）绘制PMOS布局图；2）绘制NMOS布局图；3）绘制反相器布局图并仿真；3. 实验步骤1、绘制PMOS布局图：(1) 绘制N Well图层；(2) 绘制Active图层； (3) 绘制P Select图层； (4) 绘制Poly 图层； (5) 绘制Active Contact图层；(6) 绘制Metal1图层； (7) 设计规则检查；(8) 检查错误； (9) 修改错误； (10)截面观察；2、绘制NMOS布局图：(1) 新增NMOS组件；(2) 编辑NMOS组件；(3) 设计导览；3、绘制反相器布局图：(1) 取代设定；(2) 编辑组件；(3) 坐标设定；(4) 复制组件；(5) 引用nmos组件；(6) 引用pmos组件；(7) 设计规则检查；(8) 新增PMOS基板节点组件；(9) 编辑PMOS基板节点组件；(10) 新增NMOS基板接触点； (11) 编辑NMOS基板节点组件；(12) 引用Basecontactp组件；(13) 引用Basecontactn组件；(14) 连接闸极Poly；(15) 连接汲极；(16) 绘制电源线；(17) 标出Vdd 与GND节点；(18) 连接电源与接触点；(19) 加入输入端口；(20) 加入输出端口；(21) 更改组件名称；(22) 将布局图转化成T-Spice文件；(23) T-Spice模拟；4. 实验结果nmos版图pmos版图反相器的版图反相器的spice文件反相器的仿真曲线5.实验结论通过对仿真曲线的分析，当输入为高电平时，输出为低电平；当输入为低电平时，输出为高电平。
数字集成电路分析与设计深亚微米工艺第三版教学设计
数字集成电路分析与设计深亚微米工艺第三版教学设计教学背景随着信息技术的迅猛发展，数字集成电路的应用范围日益扩大，数字化程度逐渐加深。
数字集成电路分析与设计是电子信息工程专业中重要的基础课程之一，对于帮助学生掌握数字电路基本理论、设计方法和基础技能，具有重要意义。
深亚微米工艺是半导体工艺中一种重要的先进工艺，具有工作频率高、功耗低、稳定性好等优点，得到广泛应用。
如何将深亚微米工艺与数字集成电路分析与设计课程融合，达到更好的实践效果，是本次教学设计的核心问题。
教学目标•掌握数字电路的基本理论和设计方法。
•熟悉深亚微米工艺，了解其应用场景及特点。
•能够通过实验和设计，独立完成数字电路设计并进行仿真验证，达到对数字电路设计的初步掌握。
教学内容和方法教学内容第一部分：数字电路基础1.数字电路基本概念2.组合逻辑电路设计及仿真3.时序逻辑电路设计及仿真第二部分：深亚微米工艺1.深亚微米工艺简介2.晶体管和CMOS电路设计及仿真3.反演放大电路设计及仿真教学方法第一部分：数字电路基础1.理论讲解：通过授课阐述数字电路基本概念及设计方法。
2.设计实验：设计组合逻辑电路及时序逻辑电路，通过仿真验证电路功能。
3.小组讨论：分组讨论如何对数字电路进行优化改进，提高电路的性能。
第二部分：深亚微米工艺1.理论讲解：介绍深亚微米工艺及其应用场景和特点。
2.设计实验：设计晶体管和CMOS电路，以及反演放大电路，通过仿真验证电路功能。
3.个人实验：学生个人选定一个应用场景，设计并实现相应的电路。
教学评估与考核教学评估主要分为两种类型：课上表现和实验成绩。
课上表现1.出席率，考虑缺课情况。
2.参与度，考虑学生在教学讨论中的发言次数和内容质量。
3.作业提交，考虑作业的完成情况。
实验成绩1.完成实验的数量和质量。
2.实验报告的书写和质量。
参考资料1.R. Jacob Baker, Harry W. Li, and David E. Boyce.CMOS: Circuit Design, Layout, and Simulation, ThirdEdition. John Wiley & Sons, Inc., Hoboken, NJ. 2010.2.Sung-Mo (Steve) Kang and Yusuf Leblebici. CMOSDigital Integrated Circuits: Analysis and Design.McGraw-Hill, Boston, MA. 2003.3.Jun Xu. Deep Submicron CMOS Circuit Design. JohnWiley & Sons, Inc., Hoboken, NJ. 2002.结论通过本教学设计，学生将会掌握数字电路的基本理论和设计方法，熟悉深亚微米工艺，了解其应用场景及特点。
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问答：Point out design objects in the figure such as :design, cell, reference, port, pin, net, then write a command to set 5 to net ADesign: topReference: ADD DFFCell: U1 U2Port: A B clk sumPin: A B D QNet: A B SINSet_load 5 [get_nets A]why do we not choose to operate all our digital circuits at these low supply voltages? 答：1）不加区分地降低电源电压虽然对减少能耗能正面影响，但它绝对会使门的延时加大 2）一旦电源电压和本征电压（阈值电压）变得可比拟，DC特性对器件参数（如晶体管阈值）的变化就变得越来越敏感3）降低电源电压意味着减少信号摆幅。
虽然这通常可以帮助减少系统的内部噪声（如串扰引起的噪声），但它也使设计对并不减少的外部噪声源更加敏感）问道题：1.CMOS静态电路中，上拉网络为什么用PMOS,下拉网络为什么用NMOS管2.什么是亚阈值电流，当减少VT时，V GS =0时的亚阈值电流是增加还是减少？3.什么是速度饱和效应4.CMOS电压越低，功耗就越少？是不是数字电路电源电压越低越好，为什么？5.如何减少门的传输延迟？ P2036.CMOS电路中有哪些类型的功耗？7.什么是衬垫偏置效应。
8.gate-to-channel capacitance C GC,包括哪些部分VirSim有哪几类窗口3-6. Given the data in Table 0.1 for a short channel NMOS transistor withV DSAT = 0.6 V and k′=100 µA/V2, calculate V T0, γ, λ, 2|φf|, and W / L:解答:对于短沟道器件：2'min min [()](1)2DGS T DS V WI k V V V V L λ=--+ min min[(),,]GS T DS DSAT V V V V V =-在选择公式的时候，首先要确定工作区域，表格中的所有V DS 均大于V DSAT ，所以不可能工作在线性区域。
如果工作在饱和区域则： V T 应该满足 : V GS -V T <V DSAT 2-V T <0.6 1.4<V T这是不可能的，所以可以假设所有的数据都是工作在速度饱和区域 所以：2'[()](1)2DSAT D GS t DSAT DS V WI k V V V V L λ=--+由 1&22'00.6[(2.5)0.6](1 1.8)18122D W I k Vt L λ=--+=2'00.6[(2)0.6](1 1.8)12972D W I k Vt L λ=--+= 20200.6(2.5)0.61812212970.6(2)0.62Vt Vt --=--0.44VtV=(01.4Vt V<) 所以 1,2,3是在速度饱和区由 2&312971 1.813611 2.5λλ+=+10.08V λ-=由 2&4 1297/1146=[（2-Vt0）x0.6-o.62/2]/[(2-Vt)x0.6-0.62/2] Vt=0.587V由 2 &5 Vt=0.691V这两个值都满足 Vt<1.4, 所以表中的数据都是工作的速度饱和状态0(22)SB f fVt Vt V γφφ=++-由4 &5 和 00.44Vt V=可以计算出20.6f Vφ= 和1/20.3V γ=2'1297[()]2DSAT DGS t DSAT V WI A k V V V L μ==-- 得到 W/L=1.53-7 Giv en Table 0.2 ,the goal is to derive the important device parametersfromthese data points. As the measured transistor is processed in a deep-submciron technology, the‘unified model ’ holds. From the material constants, we also could determine that the saturation voltage V DSAT equals -1V. You may also assume that -2ΦF = -0.6V.NOTE: The parameter values on Table 3.3 do NOT hold for this problem. a. Is the measured transistor a PMOS or an NMOS device? Explain your answer.b. Determine the value of VT0.c. Determine γ.d. Determine λ.e. Given the obtained answers, determine for each of the measurements the operation region of the transistor (choose from cutoff, resistive, saturated, and velocity saturated ). Annotateyour finding in the right-most column of the above.解答:a) 这是 PMOS 器件 b) 2'min min[()](1)2D GS T DS V WI kV V V V Lλ=--+比较各表中(),,GS T DS DSAT V V V V -的值知道1，4为工作在速度饱和状态由 1&42'[()](1)2DSAT D GS t DSAT DS V WI k V V V V L λ=--+2'01[( 2.5)(1)](1*( 2.5))84.3752D W I k Vt L λ=----+-=-2'01[( 2.0)(1)](1*( 2.5))56.252D W I k Vt L λ=----+-=-20201( 2.5)(1)84.375256.251( 2.0)(1)2Vt Vt -----=-----Vt 0=0.5Vc) 由 1&5和上面求出的Vt0的值: 1，5工作在速度饱和区域 则：（-84.375）/(-72.0)=[（-2.5-Vt0）*(-1)-12/2]/[(-2.5-Vt)*(-1)-12/2]求出Vt ，代入下面公式：0Vt Vt γ=+-求出：γ=0.538V 1/2d)由 1&6，因为1，6均工作在速度饱和区域:84.3751( 2.5)80.6251( 1.5)λλ-+-=-+-λ=0.05V -1e)1-vel. Sat, 2-cutoff, 3-saturation , 4-5-6 vel. Sat, 7-linear3-8 An NMOS device is plugged into the test configuration shown below in Figure0.4. The input V in =2V. The current source draws a constant current of 50 µA. R is a variableresistor that can assume values between 10k Ω and 30 k Ω. Transistor M1 experiences short channel effects and has following transistor parameters: k ’ = 110*10-6 V/A2, V T = 0.4 ,and V DSAT = 0.6V. The transistor has a W/L = 2.5µ/0.25µ. For simplicity body effect andchannel length modulation can be neglected. i.e λ=0, γ=0. .a. When R =10k Ω find the operation region, V D and V S .b. When R= 30k Ω again determine the operation region V D , V Sc. For the case of R = 10k Ω, would V S increase or decrease if λ ≠ 0. Explain qualitatively解答:1)当 R=10k, V D =V DD -IRV D =2.5-50x10-6x104=2.5-0.5=2V假设器件工作在饱和区 ( 需要以后验证)则：2'()50D GS t WI K V V A Lμ=-= GS t V V -=0.3V 所以 V GS =0.3+0.4=0.7VV S =2-0.7=1.3VVmin=min(V GS -Vt, V DSAT , V DS )=min(0.3,0.6,0.7)=V GS -Vt 所以是饱和区V D =2VV S =1.3V saturation operationb) V D =2.5-30x103x50x10-6=2.5-1.5=1V assume linear op:2'()]502DS DGS t DS V WI K V V V A L μ=--=26(1)1101010(20.4)(1)]502S S S V V V Aμ--⨯⨯----= 0.93S V V=Min(V GS -V T ,V DS ,V DSAT )=min((1-0.93-0.4).0.07,0)=V DS SO linearc) increas e , R = 10k Ω2'()(1)D GS t DS WI K V V V Lλ=-+ R 变化，则V D 必须变化以保持电流稳定，(1)DS V λ+试图增加电流，而为了恒定电流值，V GS 必须减小，即V S 必须增加 1、（10）P137Assume an inverter in the generic 0.25 mm CMOS technology designed with a PMOS/NMOS ratio of 3.4 and with the NMOS transistor minimum size (W = 0.375 mm, L = 0.25 mm, W /L =1.5). V M = 1.25 V, please compute V IL , V IH , N ML , N MH . the processparametersispresentedintable1()(/2)(1)DM n DSATnin Tn DSATn n out IV k VV V V V λ=--+由此可以得到 V IL , V IH , NM L , NM H :因为V IH =V M -V M /g , V IL =V M +(V DD -V M )/g NM H =V DD -V IH , NM L =V IL V IL =1.2V, V IH =1.3V, NM L =NM H =1.25.3、For the inverter of Figure 1 and an output load of 3 pF ，at Vout=2.5V, I DVsat =0.439mA, at Vout=1.25V, I D vsat=0.41mAfig 1a. Calculate t plh , t phl , and t p .b. Are the rising and falling delays equal? Why or why not? 解答：t pLH =0.69R L C L = 155 nsec.对于 tp HL :首先计算 R on for V out at 2.5V and 1.25V. 因为 Vout=2.5V, I DVsat =0.439mA 所以 Ron= 5695 当 Vout=1.25V, IDvsat=0.41m 所以Ron= 3049.这样， Vout=2.5Vand Vout=1.25V 之间的平均电阻 Raverage=4.372k t pLH =0.69RaverageC L =9.05nsec. t p =av{t pLH , t pHL }=82.0nsecb. Are the rising and falling delays equal? Why or why not? Solutiont pLH >> t pHL 因为 R L =75k 远大于有效线性电阻 effective linearized on-resistance of M1.5-5 The next figure shows two implementations of MOS inverters. The first inverter uses onlyNMOS transistors. Calculate V OH , V OL , V M for each case. 有的参数参考表1解答：电路 A.V OH: 当 M1关掉， M2 的阈值是:当下面条件满足的时候，M2将关闭:所以 V OUT=V OH=1.765VV OL: 假设V IN=V DD=2.5V.我们期望 V OUT为低, 因此我们可以假设M2工作在速度饱和区，而M1工作在线性区域.因为 I D1= I D2 , 所以 V OUT=V OL=0.263V, 假设成立V M: 当V M=V IN=V OUT.假设两晶体管均工作在速度饱和区域, 我们得到下面两个方程:设 I D1=I D2, 得到 V M=1.269V电路 B.当 V IN=0V, NMOS 关掉，PMOS 打开，并把V OUT拉到VDD, so V OH=2.5. 同样, 当 V IN=2.5V, the PMOS关掉，NMOS 把 V OUT拉到地, 所以V OL=0V.为了计算 V M： V M=V IN=V OUT.假设两晶体管均工作在速度饱和区域，可以得到下面两组方程.设 I D3+ I D2 =0 ,可以得到r V M = 1.095V.所以假设两晶体管均工作在速度饱和区域是正确的.5-7Consider the circuit in Figure 5.5. Device M1 is a standard NMOS device. Device M2 has allthe same properties as M1, except that its device threshold voltage is negative and。