Lecture5-CMOS_2
basicCMOSanalogicdesignLecture5CMOScurrentsinkandcurrentmirror
Lecture 5 CMOS
current sink and current mirror
Micro-electronics Department EIS Soochow University
CMOS Analog IC Design
index
1.current sink / source
CMOS Analog IC Design
Simple current sink(cont.)
3. Gate-source voltage component:
It is important to note that the gate-source voltage consists of two parts as illustrated below:
This is sometimes called “degeneration”.
For an example:
Micro-electronics Department EIS Soochow University
CMOS Analog IC Design
Simple current sink(cont.)
W W2 1
/ /
L1 L2
ID2
Micro-electronics Department
EIS Soochow University
CMOS Analog IC Design
Cascode current sink(cont.)
2. Gate-Source Matching Principle(cont.) :
VGS = VT0 + VON = Part to enhance the channel +N = VDS(sat) = VGS - VT0
CMOS_Analog_Lec5-Single-Stage Amplifier
Spring 10
CMOS Analog IC Design
5
Single Stage MOS Amplifiers
Spring 10
CMOS Analog IC Design
6
Common Source Topology
• Load types
– Passive: R, L, and C – Active: diode, R, and current source
• Source degeneration improves linearity and output resistance at the cost of gain and voltage headroom.
Spring 10
CMOS Analog IC Design
28
Source Follower
Spring 10
CMOS Analog IC Design
19
Triode Region Load
• Consumes less voltage headroom than a diode connected MOS.
Spring 10
CMOS Analog IC Design
20
Common Source with Source Degeneration
Spring 10
CMOS Analog IC Design
4
Functions of an Amplifier
• An essential block in analog circuits • Increase the signal magnitude
– To drive a load – To overcome the noise of a subsequent stage – To provide logical levels to a digital circuit – To play a critical role in feedback circuit
lecture05
EE 338L CMOS Analog Integrated Circuit DesignLecture 5, Single-Stage Amplifiers (2)Calculations of Small Signal Input and Output ImpedancesHow to calculate input and output impedances (or admittances) of an amplifier? In the following sections, we assume that the amplifier is a voltage amplifier, whose input and output are both voltages. But we can easily extend the principles to any other types of amplifiers, such as current amplifiers (input and output are both currents), transimpedance amplifiers (input: current, output: voltage), and transconductance amplifiers (input: voltage, output: current).1. Input impedanceMethod A:i) Apply tst v at the input (* seenote below), draw the smallsignal diagram.ii) Calculate )(tst tst v f i =.iii) The input impedance isgiven by tsttst in i vz =, and the inputadmittance is tsttst in v iy =.Method B:i) Apply tst i at the input, draw thesmall signal diagram. ii) Calculate )(tst tst i f v =.iii) The input impedance isgiven by tsttst in i vz =, and the inputadmittance is tsttst in v iy =.* Note: If the amplifier requires an output termination, we should terminate the output accordingly. The load condition may affect the input impedance.iv tst2. Output impedanceMethod A:i) Set 0=in v , or if the input is a signal current, set 0=in i (** see note below).ii) Apply tst v at the output, draw the small signal diagram. iii) Calculate )(tst tst v f i =.iv) The output impedance is given by tsttst out i vz =, and the output admittanceis tsttst out v iy =.Method B:i) Set 0=in v , or if the input is a current, set 0=in i (** see note below).ii) Apply tst i at the output, draw the small signal diagram.iii) Calculate )(tst tst i f v =.iv) The output impedance is given by tsttst out i vz =, and the output admittanceis tsttst out v iy =** Note: If the amplifier requires some input termination, we should terminate the input accordingly. The input termination may affect the output impedance.Example : Calculate the output impedance of the following circuit, assuming both M1 and M2 work in saturation region. The small signal parameters of M1 and M2 are shown in the following table.Transconductance Bulk transconductance Drain-sourceconductanceM1 g m1 g mb1 g ds1 M2 g m2 g mb2 g ds2Note: As g mb is not 0, we take bulk (or body) effect into consideration. As g ds is not 0, we also consider channel length modulation effect.v tsttstSolution:(1) Set v in =0, and draw the small signal digram.(2) Apply i tst at the output. The small signal diagram is shown in Fig. 1.v intstFig. 1.Note that v gs1=0, and v bs1=0, thus the two voltage controlled current sources inFig. 1 are actually 0 (see Fig. 1). We redraw Fig. 1 as Fig. 2.tstFig. 2(3) For Fig. 2, according to KCL,tst i i i i i =++=2322211Thusv V1112ds tst ds s g i g i v ==Note that, v g2=v b2=0, thus22222222221)()(s m s m s g m gs m v g v g v v g v g i −=−=−== (3)22222222223)()(s mb s mb s b mb bs mb v g v g v v g v g i −=−=−== (4)and)()(222222222s tst ds s d ds ds ds v v g v v g v g i −=−== (5)Eq. (3)+Eq. (4)+Eq. (5), and combine with Eq. (1), we arrive at,tsts mb ds m tst ds s mb s tst ds s m i v g g g v g v g v v g v g i i i =++−=−−+−=++22222222222232221)()( (6)From Eq. (6) we can write,222221ds tsts ds mb m tst g i v g g g v +⎟⎟⎠⎞⎜⎜⎝⎛++=Substitute Eq. (2) into Eq. (7),tst ds ds ds ds mb m ds tst ds tst ds mb m tst i g g g g g g g i g i g g g v ⎟⎟⎠⎞⎜⎜⎝⎛+++=+⎟⎟⎠⎞⎜⎜⎝⎛++=21212221222111 (8) Thus,21212211ds ds ds ds mb m tst tst out g g g g g g i v r +++==(9a)Note that 111ds ds g r =, 221ds ds g r =, and m mb g g η=, Eq. (9a) can be re-written as, 21222122221212212122]1)1([]1)[()1()(ds ds ds m ds ds ds mb m ds ds ds ds m ds ds ds ds mb m tst tstout r r r g r r r g g r r r r g r r r r g g i v r +++=+++=+++=+++==ηη (9b)v V 122122212122)()(ds ds ds mb m ds ds ds ds mb m out r A r r g g r r r r g g r =+≈+++=where2222)(ds mb m r g g A +=Example : 1) Assuming 0=λ, and 0=γ, what is the input impedance of the amplifier? 2) if 0≠λ, and 0≠γ, please repeat 1). Note that V B is a DC bias voltage.Solution:1) As 0=λ, and 0=γ, we have 0=ds g and 0=mb g . Draw the small signal diagram, and apply v tst at the input.g =0iFig. 1From Fig. 1, we have,tst m tst m s g m gs m d tst v g v g v v g v g i i =−−=−−=−=−=)0()(Thus,V DD v inm tst tst in g i v r 1==2) As 0≠λ, and 0≠γ, we have 0≠ds g and 0≠mb g .Draw the small signal diagram as Fig. 2, and apply v tst at the input. According to KCL,)(232221i i i i tst ++−=(1)gs =-g m v tsti 23=g mb v bs =-g i 22=g ds v dsFig. 2From Fig. 2, we can write tst m tst m s g m gs m v g v g v v g v g i −=−=−==)0()(21 (2))()(22tst out ds s d ds ds ds v v g v v g v g i −=−== (3) tst mb tst mb s b mb bs mb v g v g v v g v g i −=−=−==)0()(23 (4)D tst out R i v =(5)Substitute Eqs. (2)-(5) into Eq. (1), we havetst D ds tst ds mb m out ds tst ds mb m tst i R g v g g g v g v g g g i −++=−++=)()( (6a) Simplify Eq. (6) as tst ds mb m tst D ds v g g g i R g )()1(++=+(6b)Thusdsmb m D ds tst tstin g g g R g i v r +++==1Source Followers (Common Drain Amplifiers)(b)Input terminal: gate; output terminal: source.Note that, not all process technologies allow the source and the bulk of NMOS transistors to be connected together. The above schematics only show the concept.Large signal behavior Input/Output Voltage(V)Vout(d)VinVout(b)Input Voltage(V)(V DD=3.3V, W/L=8(6/0.9), R S=5K, I B=120u)When V in<V T, M1 is off, and Vout is 0.When V in>V T, M1 turns on in saturation. As V in increases further, V out follows Vin with a difference of V GS.When V in increases to a certain voltage (exceeding V DD), M1 enters triode region, the output voltage flattens out and clips at V DD.Small signal analysisWe only perform small signal analysis for the schematic above. The small signal diagram is shown in the right side.Note that, 2322211i i i i ++=(1)and SoutR v i =1 (2) )()(21out in m s g m gs m v v g v v g v g i −=−== (3) out ds out ds ds ds v g v g v g i −=−==)0(22(4)out mb out mb s b mb bs mb v g v g v v g v g i −=−=−==)0()(23 (5) Substitute Eqs. (2)-(5) into Eq. (1), and after some simplification, we obtain, out ds mb m in m Soutv g g g v g R v )(++−= (6)ThusSds mb m minout v R g g g g v v A 1+++==(7)Example . For the amplifier shown below (left side), assuming, 0≠λ, and 0≠γ.1) What is the small signal voltage gain inout v v vA =?2) What is the small signal output resistance out r ?outsSmall signal diagramFig. 1, Original schematic and small signal diagram for calculating small signalvoltage gain1) What is the small signal voltage gain inoutv v v A =? Solution:The small signal diagram is drawn in the right side in Fig. 1 above. According to KCL, we have 0232221=++i i i (1a) where )()(21out in m s g m gs m v v g v v g v g i −=−== (1b) out ds out ds s d ds ds ds v g v g v v g v g i −=−=−==)0()(22 (1c)out mb out mb s b mb bs mb v g v g v v g v g i −=−=−==)0()(23 (1d)Combine Eqs. (1a)-(1d), we have dsmb m min out g g g g v v ++= (2)2) What is the small signal output resistance out r ? Solution:Following the steps to calculate output resistance. (i) Set v in = 0.(ii) Apply v tst at the output, and draw the small signal diagram as shown below.Fig. 2, Small signal diagram for calculating output resistance(iii) According to KCL, we have0232221=+++tst i i i i (3) where tst m tst m s g m gs m v g v g v v g v g i −=−=−==)0()(21 (4a) tst ds out ds s d ds ds ds v g v g v v g v g i −=−=−==)0()(22 (4b)tst mb out mb s b mb bs mb v g v g v v g v g i −=−=−==)0()(23 (4c)Substitute Eqs. (4a)-(4c) into Eq. (3)tst ds mb m tst v g g g i i i i )()(232221++=++−= (5)(iii) Thus dsmb m tst tst out g g g i v r ++==1Example . For the amplifier shown below, assuming, 0≠λ, and 0≠γ.1) What is the small signal voltage gain inout v v vA =,2) What is the small signal output resistance out r ?V outFig. 1, Original schematic and small signal diagram for calculating small signalvoltage gain1) What is the small signal voltage gain inoutv v v A =? Solution:The small signal diagram is drawn in the right side in Fig. 1 above. Since there is no body effect , according to KCL, we have 02221=+i i (1a)where )()(21out in m s g m gs m v v g v v g v g i −=−== (1b)out ds out ds s d ds ds ds v g v g v v g v g i −=−=−==)0()(22 (1c)Combine Eqs. (1a)-(1c), we have dsm min out g g g v v += (2)2) What is the small signal output resistance out r ? Solution:Following the steps to calculate output resistance. (i) Set v in = 0.(ii) Apply v tst at the output, and draw the small signal diagram as shown below.i 21=g m v i 22=g v =-g v Fig. 2, Small signal diagram for calculating output resistance(iii) According to KCL, we have02221=++tst i i i (3) where tst m tst m s g m gs m v g v g v v g v g i −=−=−==)()(021 (4a)tst ds out ds s d ds ds ds v g v g v v g v g i −=−=−==)()(022(4b) Substitute Eqs. (4a)-(4b) into Eq. (3)tst ds m tst v g g i i i )()(+=+−=2221 (5)(iii) Thus dsm tst tst out g g i v r +==1Common Gate AmplifiersV DDv inFig. 1, Common-gate stage with direct coupling at inputIn a common-gate amplifier, the input signal is applied to the source terminal, as is shown in Fig.1. It senses the input at the source and generate the output at the drain. The gate is connected to a dc voltage to establish proper operating conditions. Note that the bias current of M1 flows through the input signal source.Large signal behaviorFig. 2 Large signal behaviorWhen V in >V B -V T , M1 is off, and Vout is V DD .As V in decreases, so does Vout, M1 is in saturation until()T B D T in B ox n DD V V R V V V LW C V −=−−−2121)(µ (1)After that, M1 is driven into the triode region.Small signal analysisDraw the small-signal diagram below when the transistor works in saturation, assuming 0=λ and 0≠γ,i 22=g mb v bs =-g i 21=g m v gs =-g m v inFig. 3 Small signal diagram to calculate the voltage gainWe can have,in D mb m D in mb in m out v R g g R v g v g v ⋅⋅+=⋅−−−=)()( (1)Thus,)()(η+⋅⋅=⋅+==1D m D mb m inout v R g R g g v vA (2)Example : What is the input impedance of the common gate amplifier discussed above (assuming 0=λ, and 0≠γ)?Solution : To obtain the impedance seen at the source, we use the following equivalent circuit:i 21=g m v gs =-g m v tsti 22=g mb v bs =-gFig. 4 Small signal diagram to calculate the input impedanceAccording to KCL, we have,)(2221i i i tst +−=(1)From Fig. 1, we have tst m tst m s g m gs m v g v g v v g v g i −=−=−==)()(021 (2)tst mb tst mb s b mb bs mb v g v g v v g v g i −=−=−==)()(022 (3)Substitute Eqs. (2)-(3) into Eq. (1), we havetst mb m tst mb m tst v g g v g g i )()(+=+= (4a) Simplify Eq. (4a), we have tst mb m tst v g g i )(+=(4b)Thus mbm tst tst in g g i v r +==1Example: What is the voltage gain of the following amplifier (assuming 0=λ, and 0=γ)?VFig. 1, Original schematic and small signal diagram for calculating small signalvoltage gainSolution:According to KCL, we have, 1121i i =(1)From Fig. 1, we haves m gs m v g v g i 22221−== (2))(in s m sg m v v g v g i −==11122 (3)Substitute Eqs. (2)-(3) into Eq. (1), we have in m m m s v g g g v 211+=(4)ThusD m m m m in D s m in out v R g g gg v R v g v v A 21212+===(5)。
黑色反光玻璃5英寸速度域摄像头指南说明书
Hiwatch series HWP-T5225I-A(D) IR Turbo 5-Inch Speed Dome is able to capture high quality images in poor light environment. The black anti-reflective glass increases the luminousness which helps IR distance reach up to 150 m.The embedded CMOS chip makes WDR, and real-time 1920 × 1080 resolution possible. With the help of the 25× optical zoom, and IR cut filter, the camera offers more details over an expansive area.•1/2.8" HD progressive scan CMOS•1920 × 1080 resolution•25× optical zoom•120 dB true WDR (Wide Dynamic Range)•Up to 150 m IR distance•3D intelligent positioning•Switchable TVI/AHD/CVI/CVBS video outputSpecification CameraModel HWP-T5225I-A(D)Image Sensor1/2.8"progressive scan CMOSMax. Image Resolution1920 × 1080Frame Rate 50Hz: 25fps @(1920 × 1080) 60Hz: 30fps @(1920 × 1080)Min. Illumination Color: 0.005 Lux @(F1.6, AGC ON) B/W: 0.001 Lux @(F1.6, AGC ON) 0 lux with IRDigital Zoom16×White Balance Auto/Hauto/Manual/ATW/Indoor/Outdoor AGC Auto/ManualDNR3D DNRWDR≥ 120 dBShutter Time PAL: 1/1 s to 1/10,000 s NTSC: 1/1 s to 1/10,000 sDay & Night IR cut filterPrivacy Mask8 programmable privacy masks LensFocus Mode Auto/Semiauto/ManualFocal Length 4.8 mm to 120 mm, 25 × Optical Aperture Range F 1.6 to F 3.5Horizontal Field of View 57.6° to 2.5° (wide to tele)Min. Working Distance10 mm to 1500 mm (wide to tele) Zoom Speed Approx. 3.2 s (optical, wide to tele) Pan and TiltPan Range360° endlessPan Speed Pan manual speed: 0.1° to 120°/s Pan preset speed: 120°/sTilt Range-15° to 90° (auto flip)Tilt Speed Tilt manual speed: 0.1° to 80°/s Tilt preset speed: 80°/sProportional Zoom Rotation speed can be adjusted automatically according to zoom multiplesPresets256Patrol 10 patrols, up to 32 presets per patrolPattern 5 patterns, with the total recording time no less than 10 minutesPower-off Memory SupportPark Action Preset/Patrol/Pattern/Pan Scan/Tilt Scan/Panorama Scan/Day Mode/Night Mode/None PTZ Position Display ON/OFFPreset Freezing SupportScheduled Task Preset/Patrol/Pattern/Pan Scan/Tilt Scan/Panorama Scan/Day Mode/Night Mode/Zero Calibration/NoneInfraredIR Distance Up to 150 mIR Intensity Automatically adjusted depending on the zoom ratioInput/OutputVideo Output Switchable TVI/AHD/CVI/CVBS video output, (NTSC or PAL composite, BNC)RS-485 Interface Half-duplex modeSelf-adaptive HIKVISION, Pelco-P, Pelco-D protocolUTC function UTC protocol (or HIKVISION-C protocol in previous DVR) GeneralMenu Language EnglishPower 24 VACMax.30 W (IR: 10 W)Working Temperature-30° C to 65° C (-22° F to 149° F) Working Humidity90% or lessProtection Level IP66 standard (outdoor dome)TVS 4,000 V lightning protection, surge protection and voltage transient protectionMounting Various mounting modes optional DimensionØ 213.4 mm × 345 mm (Ø 8. 4" × 13.8") Weight (approx.) 3.3 Kg (7.28 Ib.)Order Model HWP-T5225I-A(D) Dimension345mm(13.6")35mm(12.")AccessoryDS-1602ZJ-ConnerConner MountDS-1602ZJ-PolePole MountDS-1602ZJ-BoxBox MountDS-1619ZJSwan-neck MountDS-1662ZJPendant mountDS-1661ZJPendant mountDS-1663ZJCelling mountDS-1602ZJWall Mount305101090624。
CMOS_scaling
/technology/mooreslaw/index.htm
MOSFET fundamentals
Device schematic of MOSFET S G D
Cross-section SEM image
B
Gate oxide: SiO2 (1~2nm) Channel length ~ 40nm
Moore’s Law on Device scaling ……
Alienware
Commodore 64
Source: P. Asbeck, UCSD
• Shrinking of transistors to nanoscale dimensions has enabled Shrinking of transistors to nanoscale dimensions has enabled computers to become ~10,000 times more powerful over past 25 years
ΔVT can be minimized by Reducing Tox Reducing rj Increasing NA (however, trade off is degraded m, μ)
MOSFET vertical dimension should be scaled along with horizontal dimensions!
At high VDS, Saturation region
Modified: Lundstrom, Purdue,
MOSFET operation: I-V (iii)
VS = 0 VG > VT VD
Courtesy: Lundstrom, Purdue, nanoHUB com Lundstrom Purdue
实训2 CMOS设置及应用
实训2 CMOS设置及应用 (1)2.1 实训准备 (1)2.2 实训操作 (3)2.2.1 实验一常用项设置 (3)2.2.2 实验二高级项设置 (7)习题 (10)实训2 CMOS设置及应用2.1 实训准备1、实训目标（1）了解CMOS设置主界面上各功能设置菜单项的含义，并熟悉常用项所在位置；（2）熟练掌握CMOS设置中常用项的设置方法，并能够合理进行配置。
2、实训环境一台具备Phoenix－Award BIOS程序的微型计算机3、实训要点CMOS参数设置是微机装配调试过程中必须掌握的一个重要步骤，该步骤完成的好坏将直接影响到微机的整体性能。
准确设置系统时间，对于正确检测硬盘信息，合理配置开机引导优先顺序，快速加载稳定系统参数设置，有效优化系统参数等CMOS参数设置中常用项设置方法的熟练掌握，将有助于合理调配微机的硬件资源，使其各部件发挥最大效能。
操作之前，应明确各项值的基本含义，能够熟练运用各控制键进出各界面并对各类参数进行设置。
练习过程中，应重点对常用项进行操作，非常用项可加以了解，严禁对不熟悉、易造成硬件故障的项值进行操作。
4、知识准备1）BIOS与CMOS简介BIOS和CMOS是相互关联又完全不同的两个概念。
BIOS（Basic Input/Output System）即基本输入输出系统，它是被固化在主板ROM芯片中的一组系统程序，包括微机系统最重要的BIOS基本输入输出中断服务程序，系统信息设置、开机上电自检程序和系统启动自举程序等内容。
BIOS为计算机提供了最底层、最直接的硬件控制。
CMOS（Complementary Metal Oxide Semiconductor）即互补金属氧化物半导体存储器，它是一种大规模应用于集成电路芯片制造的原料，这里特指集成在微机主板上的一块可读写的RAM芯片。
这一RAM芯片也称为CMOS芯片，它主要保存当前系统的硬件配置信息和用户对某些参数的设置。
basicCMOSanalogicdesignLecture2CMOStechnology
tox <1000 angstrom tox >1000 angstrom
dry oxidation wet oxidation
Micro-electronics Department EIS Soochow University
3) The temperature for oxidation is commonly 700—1000oC; 4） tox determines the reliability and current
CMOS Analog IC Design
Classification of Si-technology
Note: iCMOS technology from ADI
Micro-electronics Department EIS Soochow University
CMOS Analog IC Design
second bake
etching
getting rid of the resist
Micro-electronics Department EIS Soochow University
CMOS Analog IC Design
major CMOS process steps (N-well example)
Presonus R-Series R65 V2和R80 V2激活AMT工作室扬声器快速启动指南说明
®®18011 Grand Bay Ct. • Baton Rouge,Louisiana 70809 USA• 1-225-216-7887R-Series R65 V2 and R80 V2Active AMT Studio Monitors Quick Start Guide R-Series R65 V2 and R80 V2Active AMT Studio MonitorsQuick Start GuideVisit us at to get a PDF manual and view instructional videos for this product.Visit us at to register your product for warranty and technical support.Visítenos en para conseguir un manual en PDF y ver vídeos ilustrativos para este producto.Visítenos en para registrar su producto para obtener la garantía y el soporte técnico.Auf unserer Webseite finden Sie ein Handbuch im PDF-Format und Video-Tutorials zu diesem Produkt.Registrieren Sie Ihr Produkt auf unserer Webseite , um auf den Garantiestatus und den technischen Support zuzugreifen.Visitez notre site pour obtenir un manuel au format PDF et visionner des vidéos de formation sur ce produit.Visitez notre site pour enregistrer votre produit et ainsi bénéficier de la garantie et de l’assistance technique.访问我们的网站 以获取PDF手册并观看该产品的教学视频访问我们的网站 以注册您的产品保修与技术支持。
CMOS图像传感器培训
4
曝光方式
目前sensor多采用滚动曝光方式，即逐行曝光逐 行读出，每行复位到读出的时间间隔即曝光时 间。曝光控制寄存器中数值代表曝光多少行： 曝光时间=曝光行数*行长。注意：当曝光时间大 于1帧时，会插入vblank来获得较长的曝光时间。
为了便于调试和计算，行长通常用多少个clk表 示，如sp2508行长为1158，指的是1158个 DAC_CLK周期，用时间表示：行长 =1158*T_dac_clk。请注意不同sensor可能用不同 的clk来表示，参数给出时会注明。 帧长=有效数据行+vblank行（单位：行） VSYNC定义一帧的有效数据输出时间 HSYNC定义一行的有效数据输出时间
SP2508 Digital colorbar 使能寄存器
SP2508 MIPI colorbar 使能寄存器
MIPI colorbar
Digital YUV colorbar
17
尺寸和窗口
• Sensor通常都支持比自身全尺寸小的各种格式尺寸输出
如sp2508全尺寸输出为1616*1216，就支持720p(1288*728), 900p(1288*908), 1600HD+(1608*908), 2*2 binning(808*608)等等输出尺寸，通过寄存器控制选择。
FSI (Front Side Illumination) BSI (Back Side Illumination) Color filter Micro-lens
• Position（定义颜色位置） • 行 row & 列 column
!
通常长方向为行，短方向为列，在调试中一定要正确区分，因为行列噪声或行列线 出现问题的原因是完全不一致的！
计算机知识-基本CMOS设置
更改CMOS设置
更改CMOS设置的原因
如果主板上的电池没电了，那CMOS原来记忆 的设置就全丢失了，电脑就可能无法启动或者 运行不正常，这时必须要全部重新设置；
想改变A盘、C盘的启动顺序； 想设置或更改开机密码； 加一个或换一个硬盘，或者改变软驱设置等； 想调节一下高级参数的设置，好让电脑能运行
得更好； 安装其它硬件设备时，可能有些设置需要改变。
如何进入CMOS设置
当开机后屏幕上常显示如下信息，马上 敲一下“Delete”键，就进到了CMOS设 置的主菜单。(有些电脑是按 Ctrl+Alt+Esc三键、有些是按F10键，具 体要看屏幕上的提示)
CMOS设置的界面
不同的电脑可能有不同界面，但常见的 也就是AWARD、AMI、Phoenix等几种。 CMOS界面形式虽然不同，但功能基本 一样，所要设置的项目也差不多。我们 刚才看到是AWARD的CMOS设置画面， 是最常见的一种。
基本CMOS设置
什么是CMOS？
CMOS(CMOS RAM或CMOS SRAM)， 叫做“互补金属氧化物半导体存储器”， 属于内存的一种，它需要很少的电源来 维持所存储系统设置或配置的信息。
CMOS存储的信息
CMOS记录计算机的日期、时间、硬盘 参数、软驱情况及其它的高级参数。平 常人们说的BIOS设置或CMOS设置指的 就是这方面的内容。电脑每次启动时都 要先读取里面的信息。
特别注意，一旦设置了密码，就要牢牢记住。如果你 给电脑设置了开机密码，又把它忘了，你就无法使用 电脑工作了，这会很耽误事情。
第五步：设置密码（6）
要是真的忘记了密码，无法启动电脑了，但是只要我 们拆开电脑主机，然后进行CMOS放电，就可以让电 脑将密码忘掉，但这要找对电脑硬件非常熟悉的人而 且要参照主板说明书才可以办到，而且，CMOS在忘 掉密码的同时，把所有设定好的值也都忘掉了，必须 重新全部设置。最好把密码记牢了！
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
#
!!"#"
#
4$/$-#')*+90"+".$/&:')784
$%% ()" ()+ ()()" ()+ ()/%/12345)67 &'()"*()+*,()-. /0-
/0-48)94/%-48:;49<86 65=>?4);@A5:BC /0-48)94/%-4D<)?@>5)C48:;4?5EF6;E;)@8:74
""
+,
#-.
!"##$%&'()*+,
120
B,�$C)D"
!'
!!"#"
!'
4$/.@/&@')"002
E&(%$+&<&+,# ;&FFG0&3< -3G(&<'
/00
E&(%$0&#&.&;";$ ;&FFG0&3<
/00
&'' () $% (*
120 120
!"#
+,
#-.
+,
#-.
"!
!!"#"
"!
)784'>&*9"&$-"2
&<664:8>6J@5J:8>64CA>)=Z4T>=T4)5>C;4E8:=>)C ! I5=>?46;V;6C4)5@49;F;)9;)@4<F5)4@T;4:;68@>V;4 9;V>?;4C>U;CZ :8@>56;CC ! P6A87C484F8@T4@54$99 5:4M)9 >)4C@;8974C@8@;Z4 65A45<@F<@4>EF;98)?; ! !W@:;E;674T>=T4>)F<@4:;C>C@8)?;Z4);8:674U;:54 C@;897JC@8@;4>)F<@4?<::;)@ ! -549>:;?@4F8@T4C@;8974C@8@;4S;@A;;)4F5A;:4 8)94=:5<)9Z4)54C@8@>?4F5A;:49>CC>F8@>5) ! /:5F8=8@>5)49;6874D<)?@>5)45D465894 ?8F8?>@8)?;48)94:;C>C@8)?;45D4@:8)C>C@5:C
314*1"#5),(,*0'(,
!"
!!"#"
!"
)"00'A"2-6.
! 3@8)98:94H;66C
" M;);:864F<:F5C;465=>? " H8)4S;4C7)@T;C>U;9 " 38E;4T;>=T@*4V8:7>)=4A>9@T
! %8@8F8@T
H;66C
" &5:4:;=<68:*4C@:<?@<:;949;C>=)C4'8:>@TE;@>?. " ()?6<9;C4C5E;4A>:>)=4>)4@T;4?;66 " &>W;94T;>=T@48)94A>9@T
9*:;%#+*-(%32':*#-P% .<9% 904%.<= 9,*%9#%.// 904%>?/@%,*7$*3#/+*:-<% ?+%-@2@*4%A+B('%4+#-,3A@*+#P =6*,*%0*+*,%*2/7#7%9%4/,*3#%$9#6%A*#8**0%.// 904% .CC '>?/(%/0%7#*94-;7#9#*%1.4*<% D+3A2'2EF(%'*-(%2#0%!2FF%@*3(-G '"04*,%9$$,.$,/9#*%7/Q/05%3.04/#/.07(%
$%
!"#$%&'(%"&') /"0%1 !%21.%(
$% !"#
!"#$%&'(%"&') /"0%1 !%21.%(
!"#
+('(,
!"#$%&'(%"&')* !"#$"#%&%!'"#(
!!"#"
+,-.,&(%') !"#$"#%&%!'"#$%&'()*+,-%"#(
" "
4$/$-#')784')-&#%-$
) E ) ? M M%&%E% />%)%)BF%?%&%)%N%?
E
?
M
M%&%E%/>%)%!G%?%&%)?
OC!D%=,907/7#.,7%$977%9%H7#,.05I%L%A"#%9%H8*9KI%J
'
!!"#"
'
<?&"2?*0@'A&*92
/0$%%
3
$%% $%% G4! $%% HI $M3
!+
)*+90"C')784'D/$"
R P H % 20K4N4%4O4P4Q4'R4O4H. P % R H
!!
!!"#"
!!
)*.2$&%#$-.6'/')*+90"C'D/$"
+'' & ( ) ' * & ' * & +:; ( ) +:< +:= +:> ' ( 3'4*5.))67"8&*&,(8"29 3$4*?,2%@%&0*(A,*5.))6.5*&,(8"29 A%,2'21A%1'))B*$B*%7,&(%CB%&0 D.$6&,(D ) ' * & ) * +''
"(
!!"#"
d d
+
"(
O/.MR.')*.2-@"&/$-*.2
P R P R H % HY H+ H" H % HI
%>C@:>S<@;94XH4E59;6 '!6E5:;49;687.
@FaI N4G]^_4X;[)'H"O+H+OYHYO#HI. /:5F8=8@>5)49;68749;@;:>5:8@;C4 :8F>96748C484D<)?@>5)45D4D8)J>)4\ [<89:8@>?8667 >)4@T;4A5:C@4?8C;]
!
a>=T4@5465A4@:8)C>@>5)
" S5@T4>)F<@C4=54T>=T
\ 9;6874>C4G]^_4+X) HI
"$
!!"#"
"$
A"0/:'A"9".@".#"'*.'R.9%$'>/$$"&.2
' &#$ &
PNRN"!G PN"4!G*4RN" PN"*4RN"!G
5(1'6$/*6* 7*66"+( PNRNG!" PN"*4RNG!" PN4G!"* RN" PNRN"!G PN"*4RN"!G
"&
!!"#"
<&/.2-2$*&'4-S-.6
XF + P X) + + R X) P H>)@ X) P R XF + HI # R # XF P X) R " HI H>)@ XF
"
"'
!!"#"
"'
<&/.2-2$*&'4-S-.6'/')*+90"C' )784'D/$"
R P # H % # 20K4N4%4O4P4Q4'R4O4H. P % " R +H +
!#
!!"#"
!#
4$/.@/&@')"00'!/:*%$'7"$?*@*0*6:'H IJKL2