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
数字集成电路(中文)第六章
第六章习题1.使用互补CMOS电路实现逻辑表达式X A B C D E F G=++++。
当反相器的NMOS W/L=2, PMOS (()())W/L=6时输出电阻相同,根据这个确定该网络中各个器件尺寸。
哪一种输入模式将会有最差和最好的上拉和下拉电阻?2.考虑下图,a.下面的CMOS晶体管网络实现的是什么逻辑功能?反相器的NMOS W/L=4, PMOS W/L=8时输出电阻相同,根据这个确定该网络中各个器件尺寸。
b.当输入是什么模式时t和p L H t最差。
最初的输入模pH L式是什么,必须采用哪一种输入才能取得最大传输延时?考虑在内部节点中的电容的影响。
3.CMOS组合逻辑a.下图中的两个电路A和B是否实现同一个逻辑函数?如果是的话,是什么逻辑;如果不是的话,给出两个电路的布尔表达式。
b.这两个电路的输出电阻是否总是相同?分析解释。
c.这两个电路的上升下降时间是否总是相同?分析解释。
4. 使用DCVSL实现F ABC ACD=+。
假设A,B,=+和F A BC ACDC,D和他们的反作为输入是允许的。
要求使用最少的晶体管。
5.一个复杂逻辑门电路如下图所示。
a.写出输出F和G的布尔表达式。
并说明这个电路实现的是什么功能。
b.这个电路属于哪一类电路。
6.分析下图所示电路实现什么功能。
7.使用NMOS传输管逻辑实现F ABC ABC ABC ABC=+++。
设计一个DCVSL门实现同样的功能。
假设A,B,C和他们的反都可以实现。
数字集成电路习题答案
W / L 1 根据VGS和VDS确定其处于线性、饱和还是截止状态,并求 I D
的值。
解: (1)nm os:
VGT VGS VT 0 2.5 0.43 2.07 VDS
nm os 处于饱和区 ,Vmin VGT 2.07
2 W V ' I D kn ( )(VGT Vmin min )(1 VDS ) L 2 2 2 . 07 115 (2.072 )(1 0.06 2.5) 2 283.3A
VT0(V)
NMOS PMOS 0.43 -0.4
(V0.5)
0.4 -0.4
VDSAT(V)
0.63 -1
k’(A/V2)
115×10-6 -30×10-6
(V-1)
0.06 -0.1
1.假设设计一个通用0.25m CMOS工艺的反相器,其中PMOS晶体管的 最小尺寸为(W=0.75m,L=0.25m,即W/L=0.75/0.25) , NMOS晶体管 的最小尺寸为(W=0.375m,L=0.25m,即W/L=0.375/0.25) 求出g,VIL,VIH,NML,NMH
( R1 R2 R5 )C5
DCLK 3 R1C1 R1C2 ( R1 R3 )C3 R1C4 R1C5
(b)
DCLK 1 R1C1 ( R1 R2 )C2 R1C3 ( R1 R2 R4 )C4 ( R1 R2 )C5
9 RC
DCLK 2 R1C1 ( R1 R2 )C2 R1C3 ( R1 R2 )C4
( R1 R2 R5 )C5 9 RC
DCLK 3 R1C1 R1C2 ( R1 R3 )C3 R1C4 R1C5
【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--电路、系统与设计(第二版)课后练习题第六.
数字集成电路--电路、系统与设计(第⼆版)课后练习题第六.Digital Integrated Circuits - 2nd Ed 11 DESIGN PROJECT Design, lay out, and simulate a CMOS four-input XOR gate in the standard 0.25 micron CMOS process. You can choose any logic circuit style, and you are free to choose how many stages of logic to use: you could use one large logic gate or a combination of smaller logic gates. The supply voltage is set at 2.5 V! Your circuit must drive an external 20 fF load in addition to whatever internal parasitics are present in your circuit. The primary design objective is to minimize the propagation delay of the worst-case transition for your circuit. The secondary objective is to minimize the area of the layout. At the very worst, your design must have a propagation delay of no more than 0.5 ns and occupy an area of no more than 500 square microns, but the faster and smaller your circuit, the better. Be aware that, when using dynamic logic, the precharge time should be made part of the delay. The design will be graded on themagnitude of A × tp2, the product of the area of your design and the square of the delay for the worst-case transition.。
模拟cmos集成电路设计课后题
模拟cmos集成电路设计课后题CMOS(Complementary Metal-Oxide-Semiconductor)集成电路设计是现代电子技术的关键领域之一。
该领域涉及到各种基本电路以及整个系统的设计与优化。
本文将模拟一篇CMOS集成电路设计的课后题,其中包括对基本电路的设计以及系统级优化的考察。
第一部分:基本电路设计(2000字左右)1. 设计一个2输入与门的CMOS电路。
给出电路图,并写出相应的布尔表达式。
2. 为了减小功耗并提高响应速度,经常需要将电路设计为动态逻辑电路。
请设计一个动态逻辑的非门电路,给出电路图,并写出相应的时钟脉冲控制信号。
第二部分:CMOS集成电路设计(2000字左右)3. 设计一个3输入与门的CMOS电路,并对其功耗进行优化。
4. 设计一个4位二进制全加器的CMOS电路,并考虑功耗和面积的优化。
第三部分:系统级优化(2000字左右)5. 将两个2输入与门和一个2输入或门组合成一个3输入与门。
请给出详细的设计流程和最终的电路图。
6. 设计一个8位互补码加法器的CMOS电路,并考虑功耗、面积和延迟的优化。
第一部分:基本电路设计1. 设计一个2输入与门的CMOS电路。
给出电路图,并写出相应的布尔表达式。
CMOS与门的基本电路由PMOS管和NMOS管组成。
在输入A和B分别接入与门电路的两个输入端,而输出则连接到NMOS管和PMOS管接口的并联电路的输出端。
当A和B同时为高电平时,输出才为高电平。
其布尔表达式可以写为:Z = A * B。
2. 为了减小功耗并提高响应速度,经常需要将电路设计为动态逻辑电路。
请设计一个动态逻辑的非门电路,给出电路图,并写出相应的时钟脉冲控制信号。
动态非门电路的设计可以采用PMOS管串联的结构。
当输入S 为高电平时,NMOS管导通,输出结果为0;当输入S为低电平时,PMOS管导通,输出结果为1。
其时钟脉冲控制信号可以表示为:NAND(A, A)。
数字集成电路分析与设计 第六章答案
CHAPTER 6P6.1. The on-resistance of a unit-sized NMOS device.LINEAR | SATURATIONOn-resistance of a unit-sized NMOS device051015202500.20.40.60.811.2V DSR D SThe average on-resistance is approximately 15kΩ. The expression for the average resistance value between DD V and 2DDV .()()()()()()()()222,,22,2223344V DD DSV DS DD DDDD DDV DDDS DD DS D satD satV V V V V I V I I ON DD ON ON DD GS T CN N DDD sat sat ox GS T R V R R V V VE L V I Wv C V V +++===-+==-P6.2. Since the signal must go around the ring twice for one oscillation, the period is :()()()()()()()()()()()()()()331531517301012.51021100.32727.5103100.3173ps TOT PLH PHL P LOAD N LOAD P N W EQP EQN g eff P N P N t N t t N R C R C N R R C W L L N R R C C W W W W --=+=+=+⎛⎫=+++ ⎪⎝⎭⎛⎫=⨯+⨯+ ⎪⎝⎭=⨯⨯=115.77GHz 173TOTf t ps=== Independent of inverter size.P6.3. SPICE.P6.4. The self-capacitance in these cases are the capacitances that will make the transition from0 to DD V or vice versa.a. In this case, all the internal nodes will be charged so the self-capacitance is :()2233313SELF eff eff C C W W W W W C W =++++=b. In this case, all the internal nodes but the one above the bottom NMOS transistor will be charged:()223310SELF eff eff C C W W W W C W =+++=c. If we assume a worst-case scenario, this node will be charged up to DD V from 0.()2233313SELF eff eff C C W W W W W C W =++++=d. The node above the bottom-most NMOS transistor has already been discharged.()223310SELF eff eff C C W W W W C W =+++=P6.5. SPICEP6.6. For optimum sizing given four inverters.()()()()()()()()()()()()44332214111111120012005.8911200203.895.891203.8934.645.89134.64 5.895.8915.8915.8945.890.525.5OUT NPE LE FO SE LE C C SE LE C C SE LE C C SE LE C C SE D LE FO P SE P =⨯=====⨯===⨯===⨯===⨯====⨯+=+=+=∏∑∑For the number of devices for optimum delay:log log log log log log12005.11log log 4N N SE SE PE SE PE N SE PEPE N SE =======Setting 5N =gives:()()()()()()()()5544332215114.1211200290.634.121290.6370.394.12170.3917.054.12117.05 4.124.1214.1214.1244.120.518.5OUT N SE LE C C SE LE C C SE LE C C SE LE C C SE LE C C SE D LE FO P SE P ===⨯===⨯===⨯===⨯===⨯====⨯+=+=+=∑∑P6.7. Solution for NAND3For the first NAND3, LE=5W/3W=5/3. For the second NAND3, the delay is not the same asthe basic inverter. So use the more general formula:310/25/33nand W R LE WR⨯== Same as the first case.a.For equal rise and fall time, we double the sizes of the transistors which leads to:313LE==b.For the pseudo-NMOS, we must first calculate the currents, which are different forpull-up and pull-down in the case of a pseudo-NMOS.For the case of the pull-up, only the PMOS is charging the output, for equal delays,we double the size of the PMOS and NMOS to obtain:23LE=P6.9.a.53 LE=b.53 LE=c.82,33 R FLE LE==d.4,23R FLE LE==P6.10.a.813RLE st gate =b.523FLE nd gate =()()()()()()()()()()()()()()45343433221411451110002222336.8711000145.656.87145.6535.366.8735.36 6.876.871 6.8716.8746.870.51 1.50.5OUT NN N PE LE FO SE LE C C SE LE C C SE LE C C SE LE C C SE D SE P SE P ⎛⎫⎛⎫=⨯== ⎪⎪⎝⎭⎝⎭===⨯===⨯===⨯===⨯====+=+=++++∏∑∑31=P6.12.()()()()()()()()()()()()()()()6345434332211546410001777833311.5510001173.2111.55173.2112511.5525411.5511.55111.551111.55OUT N N PE LE FO BE SE LE C BE C SE LE C BE C SE LE C BE C SE LE C BE C SE D SE P ⎛⎫⎛⎫⎛⎫=⨯⨯== ⎪⎪⎪⎝⎭⎝⎭⎝⎭===⨯⨯===⨯⨯===⨯⨯===⨯⨯====+∏()()41411.550.51 1.5251.2N SE P =+=++++=∑∑()()()()()()()()()()()635735445712(2)(4)800066730333314.6800011095.814.610951175.114.64512(4175.1500)533533OUT PE LE FO BE SE LE C BE C SE LE C BE C SE PE LE FO BE SE ⎛⎫⎛⎫⎛⎫=⨯⨯== ⎪⎪⎪⎝⎭⎝⎭⎝⎭===⨯⨯===⨯⨯===⎛⎫⎛⎫=⨯⨯=⨯+= ⎪⎪⎝⎭⎝⎭==∏∏()()()()()()()()()()()5343322151117.4712001114.317.54114.32317.517.5117.51117.53(17.5)214.60.51 1.5 2.25288.9NN N LE C BE C SE LE C BE C SE LE C BE C SE D SE P SE P =⨯⨯===⎛⎫⎪⨯⨯⎝⎭===⨯⨯====+=+=++++++=∑∑To minimize the delay, a estimate of the number of needed stages can be performed :log log 6637049.610log log 4SE PE N SE =∴===≈ The additional stages can be implemented as inverters attached at the input.P6.14. Consider the following situations :C LV inC LV outOutput high-to-low Output low-to-highIn the first case, the output is making a transition from high to low. The next inverter (not shown) has the PMOS in the cutoff region and the NMOS in the linear region. In these regions, the input capacitance of the next gate can be computed as follows:PMOS: C GP =C g x 2W x (1/2) NMOS: C GN =C g x WFor the output low-to-high transition, we have the PMOS linear and the NMOS cutoff: PMOS: C GP =C g x 2W NMOS: C GN =C g x W (1/2)Clearly, the second case has a larger total capacitance and hence a larger effective C g .P6.15. For this problem we examine ramp inputs as compared to step inputs. In both cases below,the transistors being driven enter the linear region and experience larger gate capacitances than the step input case. Therefore, C g is always larger for ramp inputs.C LC LV outpositive-going input ramp negative-going input rampV V DDP6.16. The FO4 delay for 0.18um is approximately 75ps. For 0.13um it is 55ps. Therefore, theconstant for the equation is roughly 420ps/um.。
数字集成电路(中文)第六章
第六章习题1.使用互补CMOS电路实现逻辑表达式X A B C D E F G=++++。
当反相器的NMOS W/L=2, PMOS (()())W/L=6时输出电阻相同,根据这个确定该网络中各个器件尺寸。
哪一种输入模式将会有最差和最好的上拉和下拉电阻?2.考虑下图,a.下面的CMOS晶体管网络实现的是什么逻辑功能?反相器的NMOS W/L=4, PMOS W/L=8时输出电阻相同,根据这个确定该网络中各个器件尺寸。
b.当输入是什么模式时t和p L H t最差。
最初的输入模pH L式是什么,必须采用哪一种输入才能取得最大传输延时?考虑在内部节点中的电容的影响。
3.CMOS组合逻辑a.下图中的两个电路A和B是否实现同一个逻辑函数?如果是的话,是什么逻辑;如果不是的话,给出两个电路的布尔表达式。
b.这两个电路的输出电阻是否总是相同?分析解释。
c.这两个电路的上升下降时间是否总是相同?分析解释。
4. 使用DCVSL实现F ABC ACD=+。
假设A,B,=+和F A BC ACDC,D和他们的反作为输入是允许的。
要求使用最少的晶体管。
5.一个复杂逻辑门电路如下图所示。
a.写出输出F和G的布尔表达式。
并说明这个电路实现的是什么功能。
b.这个电路属于哪一类电路。
6.分析下图所示电路实现什么功能。
7.使用NMOS传输管逻辑实现F ABC ABC ABC ABC=+++。
设计一个DCVSL门实现同样的功能。
假设A,B,C和他们的反都可以实现。
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。
数字电路与系统分析第六章习题答案复习进程
数字电路与系统分析第六章习题答案解:1)分析电路结构:该电路是由七个与非门及一个JKFF组成,且CP下降沿触发,属于米勒电路,输入信号X1,X2,输出信号Z。
2)求触发器激励函数:J=X1X2,K=⎺X1⎺X2触发器次态方程:Q n+1=X1X2⎺Q n+⎺X1⎺X2Q n=X1X2⎺Q n+(X1+X2)Q n电路输出方程:Z= ⎺X1⎺X2Q n+X1⎺X2⎺Q n+⎺X1X2⎺Q n+X1X2Q n3)状态转移表:输入X1X2S(t)Q nN(t)Q n+1输出Z0 0 0 0 0 10 11 0 1 0 1 1 1 1 01111111111114)逻辑功能:实现串行二进制加法运算。
X1X2为串行输入的被加数和加数,Q n 为低位来的进位,Q n+1表示向高位的进位。
且电路每来一个CP,实现一次加法运算,Z为本位和,Q在本时钟周期表示向高位的进位,在下一个时钟周期表示从低位来的进位。
例如X1=110110,X2=110100,则运算如下表所示:LSB MSB6.2 试作出101序列检测器的状态图,该同步电路由一根输入线X ,一根输出线Z ,对应与输入序列的101的最后一个“1”,输出Z =1。
其余情况下输出为“0”。
(1) 101序列可以重叠,例如:X :010101101 Z :000101001 (2) 101序列不可以重叠,如:X :010******* Z :0001000010 解:1)S 0:起始状态,或收到101序列后重新检测。
S 1:收到序列“1”。
S 2:连续收到序列“10”。
0/01/0X/Z0/011…100…S 2S 1S 01/00/01/12)0/01/0X/Z0/011…100…S 2S 1S 01/00/01/1解:(1)列隐含表:A B CDC B ×A B CD C B ×AD BC ××(a)(b)进行关联比较得到 所有的等价类为:AD ,BC 。
《数字电子技术(第二版)习题册》答案
《数字电⼦技术(第⼆版)习题册》答案数字电⼦技术(第⼆版)》习题册部分参考答案课题⼀认识数字电路任务⼀认识数制与数制转换⼀、填空题1.时间数值1 02.1 8 153.1 128 2554.75.96.16⼆、选择题1.C 2.B 3.C 4.B 5.C 6.A 7.D三、判断题1.V2. V3. X4.X5.V6.X7.V 8.V 9.X四、问答题1.答:数字电路中的信号为⾼电平或低电平两种状态,它正好与⼆进制的 1 和0 相对应,因此,采⽤⼆进制更加⽅便和实⽤。
2.答:⼗六进制具有数据读写⽅便,与⼆进制相互转换简单,较直观地表⽰位状态等优点。
五、计算题1. (1)7 (2)15 (3)31 (4)2132. (1)[1010]2 (2)[1 0000]23)[100 0000 0000]2 (4)[100 0000 0110]23. (1)[650]8 (4)[3153]84. (1)[010 111]2 (2)[001 101 110]23)[010 000 000]2 (4)[001 110 101 101]25. (1)0FH (2)1FH3)36H (4)0AE63H6. (1)0001 0110 B (2)0010 1010 1110 B3)1011 1000 1111 1100B (4)0011 1111 1101 0101B 任务⼆学习⼆进制数算术运算⼀、填空题1.加减乘除2. 0+0=0 0+1=1 1+0=1 1+1=103.0-0=0 1-0=1 1-1=0 10-1=14.0X0=0 0X1=0 1X0=0 1X1=15.1 06.最⾼正负原码7.字节 8.半字节 9.字⼆、选择题1.C 2.B 3.C 4.B 5.B 6.B 7.C3.x4. V四、问答题1.答:将⼆进制数 001 1移位⾄ 0110,是向左移动⼀位,应做乘2运算。
2.答:将⼆进制数 1010 0000 移位⾄ 0001 0100,是向右移动三位,应做除以 8运算。
《数字电路与系统设计》第6章习题答案概要
e a er a410101…X /Z0/01/0X /Z11…100…6.3对下列原始状态表进行化简: (a解:1列隐含表: 2进行关联比较3列最小化状态表为:a/1b/0bb/0a/0aX =1X =0N (t/Z (tS(t(b解:2进行关联比较: 3e:e r a94行' 试分析题图6.6电路,画出状态转移图并说明有无自启动性。
解:激励方程:略状态方程:略状态转移表:状态转移图该电路具有自启动性。
6.7 图P6.7为同步加/减可逆二进制计数器,试分析该电路,作出X=0和X=1时的状态转移表。
解:题6.7的状态转移表X Q 4nQ 3nQ 2nQ 1nQ 4n +1Q 3n +1Q 2n +1Q 1n +1Z 0 0 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 0 0 0 0 1 1 00 1 0 1 1 0 0 1 0 1 1 1 0 1 0 0 0 1 0 1 0 1 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 01 1 1 0 1 1 0 0 0 0 1 1 0 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 1 0 00 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 11 0 1 0 0 1 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 1 1 1 0 0 0 0 1 1 0 0 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 1 0 1 1 0 1 1 1 1 0 0 0 1 1 1 0 0 1 1 0 1 0 1 1 1 0 1 1 1 1 0 0 1 1 1 1 0 1 1 1 1 1 11 1116.8分析图6.8电路,画出其全状态转移图并说明能否自启动。
集成电路原理第六章习题解答
rds 2 1 g m3 g m1 rds1 rds 3 g m3 rds 3 g mbs 3 rds 3 rds 3 1 g m 2 rds 2
rds 3
rds 2 1 g m3 g m1 rds1 rds 3 g m3 1 rds 3 1 g m 2 rds 2
g m rds1 rds 3 设g m1 g m 2 g m3 g m
g m1 g m3 rds1 rds 3 g m2
3、已知如下图所示运放,其中MOS管器件参数:K-=25A/V2, n K-=10A/V2,n=p==0.04V-1, p 则: (1)、根据图中所示电路结构判断宜选N阱还是P阱工艺实 现集成?简述理由。 (2)、若VDD=5V,VSS=-5V,求运放的静态功耗Pdiss 和小信 号电压增益AV(忽略输出级产生的负载)。
此电路的输出等效电阻
rout g m3 g m1 rds1 1 rds 2 vout rds 2 g mbs 3 rds 2 rds 3 1 i0 1 g m 2 rds 2 1 g m 2 rds 2 1 g m 2 rds 2
第六章 习题
1、MOS型和双极型模拟集成电路相比各有何特点? 2、画出威尔逊电流镜的电路图、小信号等效电路,并求出其 输出电阻r0=? 解: 由Wilson电流镜电路可得其小信号等效电路:
v1 ir g m1 v2 rds1 v1 g m1 rds1 v2
因此应考虑将nmos差分对做在p阱中利用阱的电位浮动技术将其bs短接确保vbs运放的静态功耗是指从vddss的直流通路所消耗的功率此电路共有3条直流通路即分别通过m20240a静态功耗为ss20120401800w18mw此两级放大器的增益应为差分放大级和共源放大级增益的乘dsdsdsds10960257440106004600440044004将以上各值代入增益表达式中可求出该电路小信号增益
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)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)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.
(完整word版)《数字逻辑》(第二版)习题答案-第六章
习 题 六1 分析图1所示脉冲异步时序逻辑电路。
(1) 作出状态表和状态图; (2) 说明电路功能。
图1解答(1)该电路是一个Mealy 型脉冲异步时序逻辑电路。
其输出函数和激励函数表达式为211221212Q D x C Q D x Q CQ x Q Z =====(2)电路的状态表如表1所示,状态图如图2所示。
现 态 Q 2 Q 1次态/输出ZX=10 0 0 1 1 0 1 1 01/0 11/0 10/0 00/1图2(3) 由状态图可知,该电路是一个三进制计数器。
电路中有一个多余状态10,且存在“挂起”现象。
2 分析图3所示脉冲异步时序逻辑电路。
(1) 作出状态表和时间图; (2) 说明电路逻辑功能。
图3解答○1 该电路是一个Moore 型脉冲异步时序逻辑电路,其输出即电路状 态。
激励函数表达式为 1321123132233Q C C CP;C 1;K K K 1J ; Q J ; Q Q J =========○2 电路状态表如表2所示,时间图如图4所示。
表2图4○3 由状态表和时间图可知,该电路是一个模6计数器。
3 分析图5所示脉冲异步时序逻辑电路。
(1) 作出状态表和状态图; (2) 说明电路逻辑功能。
图5时 钟CP 现 态 Q 3 Q 2 Q 1 次 态 Q 3(n+1)Q 2(n+1)Q 1(n+1)11111111000 001 010 011 100 101 110 111 001 010 011 100 101 000 111 000解答○1 该电路是一个Moore 型脉冲异步时序逻辑电路,其输出函数和激励函数表达式为322111132212122212x y x R ; x S y x y x x R ; y y x S y y Z +==++===○2该电路的状态表如表3所示,状态图如图6所示。
表3现态 y 2y 1次态y 2(n+1)y 1(n+1)输出 Zx 1 x 2 x 3 0001 11 1001 01 01 0100 11 00 0000 00 10 000 0 0 1图6○3 该电路是一个“x 1—x 2—x 3”序列检测器。
《数字电路-分析与设计》第六章习题及解答(部分) 北京理工大学出版社
第六章习题6-1 略。
6-2 此时相当于触发器在前级Q的上沿翻转,所以是减法计数器。
6-3 异步可逆计数器。
UP/DOWN=0时,加法计数;UP/DOWN=1时,减法计数6-4该电路为异步置位法任意模计数器,置位状态为4(M-1),所以该计数器的模M=5计数器;时序图略。
有效状态循环:0、1、2、3、7、0;4,6为过渡状态,其次态为7;5的次态为6。
由于由状态011(M-2)到100(M-1)时,Q1、Q0由1变0,所以这两个Q端上会出现毛刺。
6-5 用4位T’触发器;因用复位法,故用状态1010清0(R=Q3Q1)。
有效状态循环为0~9;10、11、14、15均为过渡状态,其次态均为0;12的次态是13,13的次态是14;毛刺出现在Q1上。
6-6用4位T’触发器;因用置位法,故用状态1001置位(S=Q3Q1Q0)。
有效状态循环为0~8,15;9、13为过渡状态,其次态均为15;10→11→12→13→15,14→15;无毛刺。
6-7用4位T’触发器;因用置位法,故用状态1101置位(S=Q3Q2Q1Q0)。
有效状态循环为0~12,15;13为过渡状态,其次态为15;14→15;毛刺出现在Q0上。
6-8~6-11 略。
6-12 电路由T触发器组成;CP i=CP,T0=1,T1=Q0,T2=Q1Q0,所以它是同步二进制减法计数器。
时序图略。
6-13 该题未要求是同步还是异步计数器,可以两种都做,也可以只做一种。
异步:先将JK转换为T’,即令J=K=1;同步:先将JK转换为T,即令J=K=T;M=8,需要3个触发器;异步可逆、同步可逆,参见图6.39;6-14 异步计数器电路简单,速度慢;同步计数器则相反。
6-15可能产生毛刺,也可能不产生。
如果产生毛刺,则它(们)出现在由M-2到M-1时由1变0的Q A、Q D端上和/或由M-2到M-1时由0变1的Q B、Q C端上。
6-16 先分别将‘290接为8421和5421计数器,再分别用M=7(Q D Q C Q B Q A =0111)8421和(Q A Q D Q C Q B=1010)5421复位即可,应特别注意高低位的顺序。
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Chapter 6 Problem Set
Chapter 6 PROBLEMS
1. [E, None, 4.2] Implement the equation X = ((A + B) (C + D + E) + F) G using complementary CMOS. Size the devices so that the output resistance is the same as that of an inverter with an NMOS W/L = 2 and PMOS W/L = 6. Which input pattern(s) would give the worst and best equivalent pull-up or pull-down resistance? Implement the following expression in a full static CMOS logic fashion using no more than 10 transistors: Y = (A ⋅ B) + (A ⋅ C ⋅ E) + (D ⋅ E) + (D ⋅ C ⋅ B) 3. Consider the circuit of Figure 6.1.
2
VDD E 6 A A 6 B 6 C 6 D 6 E F A B C D 4 4 4 4 E 1 A B C D 4 4 4 4 E 1 6 F 6 B 6 C 6 D
Chapter 6 Problem Set
VDD 6
Circuit A
Circuit B
Figure 6.2 Two static CMOS gates.
5.
6.
[E, None, 4.2] The transistors in the circuits of the preceding problem have been sized to give an output resistance of 13 k Ω for the worst-case input pattern. This output resistance can vary, however, if other patterns are applied. a. What input patterns (A–E) give the lowest output resistance when the output is low? What is the value of that resistance? b. What input patterns (A–E) give the lowest output resistance when the output is high? What is the value of that resistance? [E, None, 4.2] What is the logic function of circuits A and B in Figure 6.3? Which one is a dual network and which one is not? Is the nondual network still a valid static logic gate? Explain. List any advantages of one configuration over the other.
VDD B
2.
D
C
A Y
B
A
C
பைடு நூலகம்
D
Figure 6.1 CMOS combinational logic gate.
4.
a. What is the logic function implemented by the CMOS transistor network? Size the NMOS and PMOS devices so that the output resistance is the same as that of an inverter with an NMOS W/L = 4 and PMOS W/L = 8. b. What are the input patterns that give the worst case tpHL and tpLH. State clearly what are the initial input patterns and which input(s) has to make a transition in order to achieve this maximum propagation delay. Consider the effect of the capacitances at the internal nodes. c. Verify part (b) with SPICE. Assume all transistors have minimum gate length (0.25µm). d. If P(A=1)=0.5, P(B=1)=0.2, P(C=1)=0.3 and P(D=1)=1, determine the power dissipation in the logic gate. Assume VDD=2.5V, Cout=30fF and fclk=250MHz. [M, None, 4.2] CMOS Logic a. Do the following two circuits (Figure 6.2) implement the same logic function? If yes, what is that logic function? If no, give Boolean expressions for both circuits. b. Will these two circuits’ output resistances always be equal to each other? c. Will these two circuits’ rise and fall times always be equal to each other? Why or why not?
VDD A B B A B A F B A Circuit A A B B A VDD B A F B A Circuit B Figure 6.3 Two logic functions.
7.
8.
[E, None, 4.2] Compute the following for the pseudo-NMOS inverter shown in Figure 6.4: a. VOL and VOH b. NML and NMH c. The power dissipation: (1) for Vin low, and (2) for Vin high d. For an output load of 1 pF, calculate tpLH, tpHL, and tp. Are the rising and falling delays equal? Why or why not? [M, SPICE, 4.2] Consider the circuit of Figure 6.5.
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.
A Y Y Y Y 50% of VDD A Y Y 0 0 1 1 1 0
Figure 6.7 Differential Buffer.
12.
What is the function of the circuit in Figure 6.8?
VDD (W/L) = 0.6 F A B C D (W/L) = 1.5 Figure 6.5 Pseudo-NMOS gate.
9. 10.
[M, None, 4.2] Implement F = ABC + ACD (and F) in DCVSL. Assume A, B, C, D, and their complements are available as inputs. Use the minimum number of transistors. [E, Layout, 4.2] A complex logic gate is shown in Figure 6.6. a. Write the Boolean equations for outputs F and G. What function does this circuit implement? b. What logic family does this circuit belong to? c. Assuming W/L = 0.5u/0.25u for all nmos transistors and W/L = 2u/0.25u for the pmos transistors, produce a layout of the gate using Magic. Your layout should conform to the following datapath style: (1) Inputs should enter the layout from the left in polysilicon; (2) The outputs should exit the layout at the right in polysilicon (since the outputs would probably be driving transistor gate inputs of the next cell to the right); (3) Power and ground lines should run vertically in metal 1. d. Extract and netlist the layout. Load both outputs (F,G) with a 30fF capacitance and simulate the circuit. Does the gate function properly? If not, explain why and resize the transistors so that it does. Change the sizes (and areas and perimeters) in the HSPICE netlist.