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Section 1 Introduction 第一节介绍The modern society depends on the electricity supply more heavily than ever before.现代社会比以往任何时候对电力供应的依赖更多。
It can not be imagined what the world should be if the electricity supply were interrupted all over the world. 如果中断了世界各地的电力供应，无法想像世界会变成什么样子Electric power systems (or electric energy systems), providing electricity to the modern society, have become indispensable components of the industrial world. 电力系统（或电力能源系统），提供电力到现代社会，已成为产业界的不可缺少的组成部分。
The first complete electric power system (comprising a generator, cable, fuse, meter, and loads) was built by Thomas Edison –the historic Pearl Street Station in New York City which began operation in September 1882. 托马斯爱迪生建立了世界上第一个完整的电力系统（包括发电机，电缆，熔断器，计量，并加载）它就是位于纽约市具有历史意义的珍珠街的发电厂始于1882年9月运作。
This was a DC system consisting of a steam-engine-driven DC generator supplying power to 59 customers within an area roughly 1.5 km in radius. The load, which consisted entirely of incandescent lamps, was supplied at 110 V through an underground cable system. 这是一个直流系统，由一个蒸汽发动机驱动的直流发电机其供电面积约1.5公里至59范围内的客户。
第一章⚫Section1习题答案一.Choose the best answer into the blank1.B2.D3.C4.A5.B二.Answer the following questions according to the text1.No. The current need not be a constant-valued function because charge can vary with2.Time.2.The current increases when the time rate of charges is greater.3.The uab=-1V can be interpreted in two ways:①point b is 1 V higher than point a;②the Potential at point a with respect to point b is -1V.4.w=∫pdt5.Because by the passive sign convention,current enters through the positive polarity ofThe voltage,p=ui>0 implies that the element is absorbing power and p=ui<0 impliesThat the element is releasing or supplying power.⚫Section2习题答案一.Choose the best answer into the blank1.B2.A3.B4.C5.B二.Answer the following questions according to the text1.The difference between an independent source and a dependent source is: the source2.Quantity of a dependent source is controlled by another voltage or current,but the source Quantity of an independent source maintains a specified value.3.An ideal independent source is an active element that provides a specified voltage or4.Current that is completely independent of other circuit variables.3.No.The current through an independent voltage source can be calculated by the4.External circuit.4.A voltage-controlled voltage source(VCVS),A current-controlled voltage source (CCVS),A voltage-controlled current source (VCCS), A current-controlled current source (CCCS)5.No,it isn’t.三.Translate the following into Chinese（译文）在随后内容中提及的所有简单电路元件，根据通过它的电流和其两端电压之间的关系进行分类。
晶体管和电子管在大多数电器和电子设备，晶体管几乎完全取代电子管。
晶体管作为电子管执行相同的功能。
但是，它们也有几个重要的优点。
大公较小，从而使更紧凑的产品成为可能。
晶体管也比电子管更坚固耐用。
它通常会提供更好的性能，在一段较长的时间。
最重要的是，晶体管通常需要少得多的电流和电压下正常工作。
这样可以节省能源。
例如，12V汽车收音机使用管吸引约2.5A。
一个类似的晶体管汽车收音机提请只有一小部分的安培。
低功耗晶体管电路的需求尽可能小，重量轻，随身便携产品的工作很长一段时间，小，低小的电池。
各种各样的晶体管最常见的两种类型的晶体管是NPN型晶体管和PNP晶体管。
它们通常被称为双极型晶体管，因为他们的操作取决于被布置为二极管连接在一个“背背”的方式这两种材料的移动。
这样的安排形成三个区域的发射极，基极和集电极。
这些地区被确定由符号E，B，和C。
的一晶体管的区域接合引线或标签，它连接在晶体管电路。
晶体管封装在金属外壳经常有第四铅被称为盾铅的。
将此导线安装在壳体内部，并连接到电路中的一个公共点。
金属外壳的屏蔽层附近晶体管表格的静电和磁场。
符号解释: 有一个方便的方式来记住的符号是否代表了一个结晶体管NPN 或PNP型。
注意代表发射器的箭头指向什么方向。
如果箭头指向相差形成的基，它可以被认为是“不指向N”，因此，该符号代表一个NPN晶体管。
如果箭头指向底座，它可以被认为是的“指向N”。
因此，这个符号代表的P-N-P晶体管。
鉴定: 大多数晶体管标识由一些字母代码，例如2N，然后通过一系列的数字，例如，2N104，2N337，2N556。
其它晶体管都确定了一系列的数字或数字和字母，例如40050，40404，和4D20的组合。
晶体上手册: 设备是否是NPN或PNP型的晶体管的识别码不表示。
晶体管手册或规格表中发现这样的技术数据。
这些手册也给各种不同的电路中使用的晶体管的信息。
晶体管外形图提供了详细的信息，它们的大小，形状和连接。
Semiconductor switches are very important and crucial components in power electronic systems.these switches are meant to be the substitutions of the mechanical switches,but they are severely limited by the properties of the semiconductor materials and process of manufacturing. 在电力电子系统，中半导体开关是非常重要和关键部件。
半导体开关将要替换机械开关,但半导体材料的性质和生产过程严重限制了他们。
Switching losses开关损耗Power losses in the power eletronic converters are comprised of the Switching losses and parasitic losses. 电力电子转换器的功率损耗分为开关损耗和寄生损耗the parasitic losses account for the losses due to the winding resistances of the inductors and transformers,the dielectric losses of capacitors,the eddy and the hysteresis losses. 寄生损失的绕组电感器、变压器的阻力、介电损耗的电容器,涡流和磁滞损耗the switching losses are significant and can be managed. 这个开关损耗是非常重要的,可以被处理。
they can be further divided into three components:(a)the on-state losses,(b)the off-state losses and the losses in the transition states. 他们可以分为三个部分: 通态损耗，断态损耗和转换过程中产生的损耗。
电气自动化专业英语第三单元专业英语第三单元3 Analog Electronics3.1 INTRODUCTION3.1.1 The Contrast between Analog and Digital ElectronicsWe have already explored how transistors and diodes are used as switching devices to process information which is represented in digital form. Digital electronics uses transistors as electrically controlled switches: transistors are either saturated or cut off. The active region is used only in transition from one state to the other.By contrast, analog electronics depends on the active region of tran sistors and other types of amplifiers. The Greek roots of “analog” mean “in due ratio”, signifying in this usage that information is encoded into an electrical signal which is proportional to the quantity being represented.713宿舍In Fig.3.1 our information is some sort of music, originating physically in the excitation and resonance’s of a musical instrument. The radiated sound consists in the ordered movement of air molecules and is best understood ad acoustic waves. These produce motion in the diaphragm of a microphone, which in turn produces an electrical signal. The variation in the electrical signal are a proportional representation of the sound waves. The electrical signal is amplifiedelectronically, with an increase in signal power occurring at the expense of the input AC power to the amplifier. The amplifier output drives a recording head and produces a wavy groove on a disk. If the entire system is good, every acoustic variation of the air will be recorded on the disk and, when the record is playedback through a similar system and the signal reradiated ad sound energy be a loudspeaker, the resulting sound should faithfully reproduce the original music.Electronic systems based on analog principles form an important class of electronic devices. Radio and TV broadcasting are common examples of analog systems, as are many electrical instruments used in monitoring deflection(strain gages, for example), motion (tachometers), and temperature (thermocouples).Many electrical instruments-voltmeters, ohmmeters, ammeters, and oscilloscopes-utilize analog techniques, at least in part.Analog computers existed before digital computers were developed. In an analog computer, the unknowns in a differential equation are modeled with electrical signals. Such signals are integrated, scaled, and summed electrically to yield solutions with modes effort compared with analytical or numerical techniques.3.1.2 The Contents Of This ChapterAnalog techniques employ the frequency-domain viewpoint extensively. We begin by expanding our concept of the frequency domain to include periodic, nonperiodic, and random signals. We will see that most analog signals and processes can be represented in the frequency domain. We shall introduce the concept of a spectrum, that is, the representation of a signal as the simultaneous existence of many frequencies. Bandwidth (the width of a spectrum) in the frequency domain will be related to information rate in the time domain.714宿舍This expanded concept of the frequency domain also helps us distinguish the effects of linear and nonlinear analog devices. Linear circuits are shown to be capable of“filtering” out unwanted frequency components. By contrast, new frequencies can be created by nonlinear devices such as diodes and transistors. This property allows us to shift analog signals in the frequency domain through AM and FM modulation techniques, which are widely used in public and private communication systems. As an example we shall describe the operation of an AM radio.Next we study the concept of feedback, a technique by which gain in analog systems is exchanged for other desirable qualities such as audio amplifiers or TV receivers would at best offer poor performance. Understanding of the benefits of feedback provides the foundation for appreciating the many uses of operational amplifiers in analog electronics.Operational amplifiers (op amps, for short) provide basic building blocks for analog circuits in the same way that NOR and NAND gates are basic building blocks for digital circuits. We will present some of the more common applications of op amps, concluding with their use in analog computers.3.3.2 OPERATIONAL-AMPLIFIER CIRCUITS3.2.1 Introduction(1) The Importance of OP Amps. An operational amplifier isa high-gain electronic amplifier which is controlled by negative feedback to accomplish many functions or “operations” in analog circuits. Such amplifiers were developed originally to accomplish operations such as integration and summation in analog computers for the solving of differential equations. Applications of op amps have increased until, at the present time, most analog electronic circuits are based on op amp techniques. If, for example, you required an amplifier with of 10, convenience, reliability, and cost considerations would dictate the use of an opamp. Thus op amp from the basic building blocks of analog circuits much as NAND and NOR gates provide the basic building blocks of digital circuits.(2) An OP-Amp Model Typical Properties. The typical op amp is a sophisticated transistor amplifier utilizing a dozen or more transistors,several diodes, and many resistors. Such amplifiers are mass produced on semiconductor chips and sell for less than ＄1 each. These parts are reliable, rugged, and approach the ideal in their electronic properties.Fig.3.2 shows the symbol and the basic properties of op amp. The two input voltages, u＋and u－, are subtracted and amplified with a large voltage gain, A, typically 105～106. The input resistance, Ri, is large, 100KΩ～100MΩ. The output resistance, Ro, is small, 10～100Ω. The amplifier is often supplied with DC power from positive (+Ucc)and negative(﹣Ucc) power supplies. For this case, the output voltage lies between the power supply voltages, ﹣Ucc﹤Uo﹤+Ucc. Sometimes one power connection is grounded (i.e., “﹣Ucc”=0). In this case the output lies in the range, 0﹤Uo﹤+Ucc. The power connections are seldom drawn in circuit diagrams; it is assumed that one connects the op amp to the appropriate power source. Thus the op amp approximates an ideal voltage amplifier, having high input resistance, low output resistance, and high gain.The high gain is converted to other useful features through the use of strong negative feedback.All the benefits of negative feedback are utilized by op-amp circuits. To those listed earlier in this chapter, we would for op-amp circuits add three more: low expanse, ease of design, and simple construction.(3)The Contents of This Section. We begin by analyzing twocommonop-amp applications, the inverting and uninverting amplifiers. We derive the gain of these amplifiers by a method that may be applied simple and effectively to any op-amp circuit. We then discuss active filters, which are op amp amplifiers with capacitors added to shape their frequency response. We then deal briefly with analog computers and conclude by discussing some nonlinear application of op-amp.3.2.2 Op-amp Amplifiers712宿舍(1) The Inverting Amplifier. The inverting amplifier, show in Fig.3.3, use an op-amp plus two resistors. The positive (+)input to the op-amp is grounded (zero signal); the negative (﹣)input is)and to the feedback signal from the connected to the input signal (via R1output (via R). One potential source of confusion in the followingFdiscussion is that we must speak of two amplifiers simultaneously. The op amp is an amplifier which forms the amplifying element in a feedback amplifier which contains the op amp plus associated resistors. To lessen confusion, we shall reserve the term “amplifier”to apply only to the overall, feedback amplifier. The op-amp will never be call ed an amplifier; it will be called the op-amp. For example, if we refer to the input current to the amplifier, we are referring to the current through Ri, not the current into the op-amp.We could solve for the gain of the inverting amplifier in Fig.3.3 either by solving the basic circuit laws (KCL and KVL) or byattempting to divide the circuit into main amplifier and feedback system blocks. We shall, however, present another approach based on the assumption that the op-amp gain is very high, effectively infinite. In the following, we shall give a general assumption, which may be applied to any op-amp circuit; then we will apply this assumption specifically to the present circuit. As a result, we will establish and input resistance of the inverting amplifier.We assume that the output is well behaved and does not try to go to infinity. Thus we assume that the negative feedback stabilizes the amplifier such that moderate input voltages produce moderate output voltages. If the power supplies are +10 and﹣10V, for example, the output would have to lie between these limits.Therefore, the input voltage to the op-amp is very small, essentially zero, because it is the output voltage divided by the large voltage gain of the op-ampu+﹣u_≈0?u+≈u_For example, if ∣Uo∣﹤10V and A=105, then ∣u+﹣u_∣﹤10\105=1+and u_ are equal with 100μV or less,for any op-amp circuit. For the inverting amplifier in Fig.3.3, u+is ground;therefore, u_≈0. Consequently, the current at the input to the amplifierwould bei 1= 1_R u Ui - ≈1R Ui (3.1) Because u +≈u_ and Ri is large, the current into the + and – op-ampinputs will be very small, essentially zero∣i +∣=∣i -∣=||RiU U +--≈0 (3.2) For example, for Ri =100KΩ, |i_|﹤104-/105＝109-A.For the inverting amplifier, Eq. (3.2) implies that the current at the input, i i , flows through R f , as shown in Fig.3.4. This allows us to compute the output voltage. The voltage across R F would be i i R F and, because one end of R F is connected to u_≈0 Uo=-i i R F =-1R U i R F Thus the voltage gain would beA u =Ui Uo =1R R F - (3.3) The minus sign in the gain expression means that the output will be inverted relative to the input: a positive signal at the input: a positive signal at the input will produce a negative signal at the output, Eq. (3.3) shows the gain to depend o the ratio R F to R 1. This would imply that onlythe ratio and not the individual values of R F to R 1 matter. This would betrue if the input resistance to the amplifier were unimportant, but the input resistance to an amplifier is often critical. The input resistance to the inverting amplifier would follow from Eq. (3.1);R i =i i i U ≈R 1 (3.4)For a voltage amplifier, the input resistance is an importantfactor, for if R i were too low the signal source (of U i ) could be loaded down by R i . Thus in a design, R 1 must be sufficiently high to avoid his loadingproblem. Once R 1 is fixed, R F may be selected to achieve the requiredgain. Thus the values of individual resistors become important because they affect the input resistance to the amplifier.Let us design an inverting amplifier to have a gain of ﹣8. The input signal is to come from a voltage source having an output resistance of 100Ω. To reduce loading, the input resistor, R 1, must be much larger than100Ω. For a 5﹪loading reduction, we would set R 1=2000Ω. To achievea gain of -8(actually 95﹪of -8, considering loading ), we require that R F =8×2000=16KΩ.Feedback effects dominate the characteristics of the amplifier. When an input voltage is applied, the value of u_ will increase. This will cause U 0 to increase rapidly in the negative direction . This negative voltagewill increase to the value where the effect of U 0 on the –input via R F cancels the effect of U i through R 1. Put another way, the output willadjust itself to withdraw through R F any current that U i injects through R 1, since the input current to the op-amp is extremely small. In this waythe output depends only R F and R 1.711宿舍 The Noninverting Amplifier. For thenoninverting amplifier show in Fig.3.5 the input is connected to the +input. The feedback from the output connects still to the– op amp input, as required for negative feedback. T o determine the gain, we apply the assumptions outlined above.①Because u +≈u_, it follows thatu_ ≈U i (3.5)②Because i ≈0, R F and R 1 carry the same current. Hence U0 is related to u_ through a voltage-divider relationshipu i =U 0 FR R R +11(3.6) Combining Eqs. (3.5) and (3.6), we establish the gain to beU i =U 0F R R R +11=A u =+(1+1R R F ) (3.7) The + sign before the gain expression emphasize that the output of the amplifier has the same polarity as the input: a positive input signal produces a positive output signal. Again we see that the ratio of R F and R 1 determines the gain of the amplifier.When a voltage is applied to the amplifier, the output voltage increase rapidly and will continue to rise until the voltage across R 1 reaches theinput voltage. Thus little input current will flow into the amplifier, and the gain depends only on R F and R 1. The input resistance to the noninvertingamplifier will be very high because the input current to the amplifier is also the input current to the op-amp, i +, which must be extremely small.Input resistance values exceeding 1 000 MΩ are easily achieved with this circuit. This feature of high input resistance is an important virtue of the noninverting amplifier.3.2.3 Active Filters(1)What Are Active Filters? An active filter combines amplification with filtering. The RC filters we investigated earlierare called passive filters because they provide only filtering. An active filter uses an op-amp to furnish gain but has capacitors added to the input and feedback circuits to shape the filter characteristics.We derived earlier the gain characteristics of an inverting amplifier in the time domain. In Fig.3.6 we show the frequency-do-main version. We may easily translate the earlier derivation into the frequency domainU i ?U i (ω) U 0?U 0（ω）A u =﹣1R R F ?F u （ω）=﹣)()(1ωωZ Z F The filter function, F u （ω）, is thus the ratio of the two impedances,and in general with give gain as well as filtering. We could have written the minus sign a s 180°, for in the frequency domain the inversion is equivalent to a phase shift of 180°.(2) Low-pass Filter. Placing a capacitor in parallel with R F (seeFig.3.7) will at high frequencies tend to lower Z and hence the gain of the amplifier; consequently, this capacitor an inverting amplifier into a low-pass filter with gain. We may writeF Z (ω)=R F ∣∣F C j ω1=F F C j R ω+)/1(1=FF F C R j R ω+1(3.8) Thus the gain would be)/(11111c u F F F u j A C R j R R F ωωω+=+-=(3.9) Where 1/R R Au F -=, the gain without the capacitor, andF C C R R /1=ωwould be the cutoff frequency. The gain of the amplifier isapproximately constant until the frequency exceeds C ω, after which thegain decreases with increasing ω. The Bode plot of this filter function is shown in Fig.3.8 for the case where R F =10K ωΩ, R1=1KΩ, and C F =1μF.(3) High-pass filter. The high-pass filter show in Fig.3.9 usesa capacitor in series with R 1 to reduce the gain at low frequencies. Thedetails of the analysis will be left to a problem. The gain of this filter isu c c F u A j j R R F =+-=)/(1)/()(1ωωωωω)/(1)/(c c j j ωωωω+ Where 1/R R Au F -= is the gain without the capacitor and 11/1C R c =ω is the cutoff frequency, below which the amplifier gain is reduced. The Bode plot of this filter characteristic is show in Fig.3.10.(4) Other Active Filter. By using more advanced techniques, one can simulate RLC narrowband filters and, by using additional op-amps, many sophisticated filter characteristics can be achieved. Discussion of such applications lies beyond the scope of this text, but there exist many handbooks showing circuits and giving design information about active filters.3. 2. 4 Analog ComputerOften a differential equation is Fig.3.10 solved by integration. The integration may be accomplished by analytical methods or by numerical methods on a digital computer. Integration may also be performed electronically with an op-amp circuit. Indeed, op-amps were developed initially for electronic integration of differential equations.⑴ An Integ rator . The op-amp circuit in Fig.3.11 uses negative feedback through a capacitor to perform integration.We have charge the capacitor in the feedback path to an initial value of U 1, and then removed this prebias(预偏置)voltage at t=0. Let usexamine the initial state of the circuit before investigatingwhat will happen after the switch is opened. Since +u is approximately zero, sowill be _u , and hence the output voltage is fixed at ﹣1U . The inputcurrent to amplifier, R U i /, will flow through the 1U voltage will remainat ﹣1U until the switch is opened.After the switch is opened at t=0, the input current will flow through the capacitor and hence the U C will be,0,0)()0()(dt RC t U U t Uc ti ?+= Thus the output voltage of the circuit is0)(1)()(,,010≥--=-=?t dt t U RC U t U t U ti c （3.10）Except for the minus sign, the output is the integral of U i scaled by1/RC, which may be made equal to any value we wish by proper choiceof R and C.⑵ Scaling and Summing . We need two other circuits to solve simple differential equations by analog computer methods. Scaling refers to multiplication by a constant, such as 12KU U ±=Where K is a constant. This is the equation of an amplifier, and hence we would use the inverting amplifier in Fig.3.5 for the – sign or the noninverting amplifier in Fig.3.5 for the + sign.A summer produces the weighted sum of two or more signals.Fig.3.12 shows a summer with two inputs. We may understand the operation of the circuit by applying the same reasoning we used earlier to understand the inverting amplifier. Since 0≈-u , the sum of the currents through 1R and 2R is22111R U R U i +=(3.11) The output voltage will adjust itself to draw this current through RF, and hence the output voltage will be)(221110R R U R R U R i U F F F ?+?-=-= The output will thus be sum of 1U and 2U , weighted by the gainfactors, 1/F R R and 2F R R , respectively. If the inversion produced by thesummer is unwanted, the summer can followed by an inverted, a scalier with a gain of -1. Clearly, we could add other inputs in parallel withR R and 21. In the example to follow, we shall sum three signals to solve a second order differential equation.(3) Solving a DE. Let us design an analog computer circuit tosolve the differential equation t u dt du dtu d 10cos 65222=++ t>0 U(0)=﹣2 and at dtdu 3+= t=0 （3.12） Moving everything except the highest-order derivative to the right side yields t u dt du dtu d 10cos 32222+--=（3.13）女生宿舍The circuit which solves Eq. (3.12) is shown in Fig.3.13. The circuit consists of two integrators to integrate the left side of Eq. (3.13), a summer to represent the right side, and two inverts to correct the signs. The noninverting inputs are grounded, and the inputs and feedback are connected to the inverting input of the op-amps. Hence we have shown only the inverting inputs. With 22/dt u d the input to the integrators, the output of the first integrator will be-du/dt [with the battery giving the initial condition of 3V , as in Eq. (3.13)], and hence the output of the second integrator will be +u (withan initial condition of -2 V ). This output is fed into the summer, along with du/dt after inversion, and the driving function cos10 t, which must also be inverted to cancel the inversion in the summer. The input resistors connecting the three signals into the summer produce the weighting factors in Eq. (3.13), and hence the output of the summer represents the right side of Eq. ( 3.13 ). Wetherefor e connect that output to our “input” of 22/dtd to satisfy Eq.u(3.12 ). To observe the solution to Eq. (3.12 ), we merely open the switches at t=0.Clearly, these techniques can be applied to higher-order equations. Sophisticated use of analog computer requires a variety of refinements. Often, the equations being solved are scaled in time (time is sped up or slowed down on the computer) to accommodate realistic resistor and capacitor values. Also, voltage and current values can be scaled to bring the unknowns within the allowable range of the computer. In the next section we show how nonlinear operations can be introduced to solve nonlinear differential equations by analog methods.3. 2. 5 Nonlinear Applications of Op-ampsOp-amps can be combined with nonlinear circuit elements such as diodes and transistors to produce a variety of useful circuits. Below we discuss a few such applications. Many more circuits are detailed in standard handbooks and manufacturers’ application literature for their products.An Improved Half-Wave Rectifier. The op-amp in Fig 3.14 drives a half-wave rectifier. When the input voltage is negativethe output of the op-amp will be OFF; hence the output will be zero. When the output is positive the diode will turn ON and the output will be identical to the input, because the circuit will perform as a non-inverting amplifier shownin Fig.3.5 with R F=0. Use of the op-amp effectively reduces the diode turn-on voltage. If the input voltage is greater than 0.7/A, where A is the voltage gain of the op-amp, the output voltage exceed 0.7V and turn on the diode. Hence the turn-on voltage is effectively reduced from 0.7~0.7/A.This circuit would not be used in a power supply circuit; rather, it would be used in a detector or other circuit processing small signals, where the turn-on voltage of the diode would be a problem.。
电气工程及其自动化专业英语Section I basic electric circuitChapter 1 Introduction to electric circuitsNew Words and Expressions1. electrical circuit n. 电路2. voltage n. 电压，伏特3. curre nt n. 电流，通用的，流通的，现在的4. curre nt flow n. 电流5. resistor n. 电阻，电阻器6. battery n. 电池7. load n. 负载，负荷8. performa nee n. 性能9. circuit diagram n. 电路图10. idealized model n. 理想模型Introduction*A simple circuit and its components.idealized model of the circuit*Model can be cha nged if n ecessary.*summarizeIn elementary physics classes you undoubtedly have been introduced to the fun dame ntal con cepts of electricity and how real comp onen ts can be put together to form an electrical circuit. A very simple circuit, for example, might consist of a battery, some wire, a switch, and an incandescent light bulb as shown in Fig.1-1. The battery supplies the en ergy required to force electro ns around the loop, heati ng the filame nt of the bulb and caus ing the bulb to radiate a lot of heat and some light.Energy is transferred from a source, the battery, to a load, the bulb———You probably already know that the voltage of the battery and the electrical resista nee of the bulb have something to do with the amount of curre nt that will flowin the circuit. From your own practical experienee you also know that no current will flow until the switch is closed. That is, for a circuit to do anything, the loop has to be completed so that electro ns can flow from the battery to the bulb and the n back aga in to the battery. And fin ally, you probably realize that it doesn t much matter, whether there is on e foot or two feet of wire connecting the battery to the bulb, but that it probably would matter if there is a mile of wire between it and the bulb.Also shown in Fig. 1-1 is a model made up of idealized components. The batteryis modeled as an ideal source that puts out a constant voltage, VB, no matter what amount of curre nt, i, is draw n. The wires are con sidered to be perfect con ductors that offer no resista nee to curre nt flow. The switch is assumed to be ope n or closed. There is no arcing of curre nt across the gap whe n the switch is ope ned, nor is there any bounce to the switch as it makes con tact on closure. The light bulb is modeled as a simple resistor, R, that never changes its value, no matter how hot it becomes or how much curre nt is flow ing through it.Fig. 1-1 (a) A simple circuit(b) An idealized represe ntati on of thecircuitFor most purposes, the idealized model shown in Fig. 1-1b is an adequate represe ntati on of the circuit; that is, our prediction of the current that will flow through the bulb whenever the switch is closed will be sufficiently accurate that we can consider the problem solved. There may be times, however, when the model is in adequate. The battery voltage, for example, may drop as more and more curre nt is drawn, or as the battery ages. --------------------------------- T he light bulb' s resistance may change as it heats up, and the filame nt may have a bit of inductance and capacitance associated with it as well as resistance so that when the switch is closed, the current may not jump in sta ntan eously from zero to some fin al, steady state value. The wires may beundersized, and some of the power delivered by the battery may be lost in the wires before it reaches the load. These subtle effects may or may not be important, depending on what we are trying to find out and how accurately we must be able to predict the performa nee of the circuit. If we decide they are importa nt, we can always cha nge the model as n ecessary and then proceed with the an alysis. The point here is simple. The comb in ati ons of resistors, capacitors, in ductors, voltage sources, curre nt sources, and so forth, that you see in a circuit diagram are merely models of real comp onents that comprise a real circuit, and a certa in amount of judgme nt is required to decide how complicated the model must be before sufficie ntly accurate results can be obta in ed. For our purposes, we will be using very simple models in general, leav ing many of the complicati ons to more adva need textbooks.Chapter 2Definitions of key electrical quantitiesNew Words and Expressionscharge n. vt.电荷；充电nu cleus n.原子核(pl.); nuclear adj.n egative n.否定，负数，底片adj.否定的，消极的，负的，阴性的positive adj.［数］正的adj.［电］阳的in gen eral 通常,大体上，一般而言，总的说来algebraic adj.代数的，关于代数学的soluti on to the circuit problem n.关于电路问题的解法the un its of power n.功率的单位direct curre nt (dc) n 直流电alter nat ing curre nt(ac) n.交流电sinu soidally adv.正弦地tran sistor n.晶体管Part 1 Charge and CurrentAn atom con sists of a positively charged nu cleus surro un ded by a swarm of n egativelycharged electr ons. The charge associated with one electr on has bee n found to be 1.602 x 10- 19 coulombs; or, stated the other way around, one coulomb can be defined as the charge on 6.242 x 1018 electro ns. While most of the electr ons associated with an atom are tightly bound to the nu cleus, good con ductors, like copper, have free electrons that are sufficie ntly dista nt from their nu clei that their attract ion to any particular n ucleus is easily overcome. These con ducti on electr ons are free to wan der from atom to atom, and their moveme nt con stitutes an electric curre nt.In a wire, when one coulomb ' s worth of charge passes a given spot in one second, the current is defined to be one ampere (abbreviated A), named after the nineteenth-century physicist Andr ' e Marie Amp'ere. That is, curre nt i is the net rate of flow of charge q past a point, or through an area:i=d q/d t (1.1)In general, charges can be negative or positive. For example, in a neon light, positive ions move in one direct ion and n egative electr ons move in the other. Each con tributes to curre nt, and the total curre nt is their sum. By conven ti on, the direct ion of curre nt flow is take n to be the direct ion that positive charges would move, whether or not positive charges happen to be in the picture. Thus, in a wire, electrons moving to the right constitute a current that flows to the left, as shown in Fig.1-2.(〉)dq--- / =—dtFig. 1-2 By conven tio n, n egative charges movi ng in one direct ion con stitute a positive curre ntflow in the opposite direct ionW/hen charge flows at a steady rate in one direction only, the current is said to be direct current, or 血A battery, for example, supplies direct curre nt. When charge flows back and forth sinusoidally, it is said to be alternating current, or ac. In the United States the ac electricity delivered by tes of ac and dc are show n in Fig.1-3.Time ―(a)Fig. 1-3 (a) Steady-state direct curre nt (de) (b) Alter nat ing curre nt(ac)Part 2 Kirchhoff' s Current LawTwo of the most fun dame ntal properties of circuits were established experime ntally a cen tury and a half ago by a Germa n professor, Gustav Robert Kirchhoff (1824 - 1887). The first property, known as Kirchhoff ' s current law (abbreviated KCL), states that at every instant of time the sum of the curre nts flow ing into any node of a circuit must equal the sum of the curre nts leavi ng the no de, where a node is any spot where two or more wires are join ed. This is a very simple, but powerful con cept. It is in tuitively obvious once you assert that curre nt is the flow of charge, and that charge is con servative—n either being created nor destroyed as it en ters a no de. Uni ess charge somehow builds up at a no de, which it does not, the n the rate at which charge en ters a node must equal the rate at which charge leaves the no de.There are several alter native ways to state Kirchhoff ' s curre nt law. The most com monly used stateme nt says that the sum of the curre nts flow ing into a node is zero as show n in Fig. 1-4a, in which case some of those curre nts must have n egative values while some have positive values. Equally valid would be the stateme nt that the sum of the curre nts leav ing a node must be zero as show n in Fig. 1-4b(aga in some of these curre nts n eed to have positive values and some n egative). Fin ally, we could say that the sum of the curre nts en teri ng a node equals the sum of the curre nts leav ing a node (Fig. 1-4c). These are all equivale nt as long as we un dersta nd what is meant about the directi on of curre nt flow whe n we in dicate it with an arrow on a circuit diagram. Curre nt that actually flows in the directi on show n by the arrow is give n a positive sig n. Curre nts that actuallyflow in the opposite direct ion have n egative values.(a) The sum of the curre nts into a node equals zero(b) The sum of the curre nts leav ing the node is zero(c) The sum of the curre nts en teri ng a node equals the sum of the curre nts leavi ng the node Note that you can draw curre nt arrows in any directio n that you want — that much is arbitrary — but once havi ng draw n the arrows, you must the n write Kirchhoff ' s curre nt law in a manner that is con siste nt with your arrows, as has bee n done in Fig.1-4. The algebraic soluti on to the circuit problem will automatically determ ine whether or not your arbitrarily determ ined direct ions for curre nts were correct.Example 1.1 Using Kirchhoff ' s Current LawA node of a circuit is shown with current direction arrows chosen arbitrarily. Havingpicked those directi on s, i1 = - 5 A, i2 = 3 A, and i3 = - 1 A. Write an expressi on for Kirchhoff ' s current law and solve for i4.Solution. By Kirchhoff ' s current law,i1 + i2 = i3 + i4 so thatThat is, i4is actually 1 A flowi ng into the no de. Note that i2, i3, and i4 are all en teri ng the no de, and i1 is the only curre nt that is leavi ng the no de.Part 3 Kirchhoff ' s Voltage LawElectr ons won ' t flow through a circuit uni ess they are give n some en ergy to help send them on their way. That “ push ” is measured in volts, where voltage is defi ned to be the amount nodenodenode1 + i4 i4 = - 1 AFig. 1-4 lllustrating various ways that Kirchhoff ' s current law can be statedof en ergy (w, joules) give n to a un it of charge,v=dw/dq A 12-V battery therefore gives 12 joules of en ergy to each coulomb of charge that it stores. Note that the charge does not actually have to move for voltage to have meaning. Voltage describes the potential for charge to do work.While curre nts are measured through a circuit comp onent, voltages are measured across componen ts. Thus, for example, it is correct to say that curre nt through a battery is 10 A, while the voltage across that battery is 12 V. Other ways to describe the voltage across a comp onent in clude whether the voltage rises across the comp onent or drops. Thus, for example, for the simple circuit in Fig. 1-1, there is a voltage rise across the battery and voltage drop across the light bulb. Voltages are always measured with respect to someth ing. That is, the voltage of the positive terminal of the battery is“ so many volts ” with respect to the negative terminal; or, the voltage at a point in a circuit is some amount with respect to some other poin t. In Fig. 1-5, curre nt through a resistor results in a voltage drop from point A to point B of VAB volts. V A and VB arethe voltages at each end of the resistor, measured with respect to some other point.The reference point for voltages in a circuit is usually desig nated with a ground symbol. While many circuits are actually groun ded — that is, there is a path for curre nt to flow directly into the earth —some are not (such as the battery, wires, switch, and bulb in a flashlight). When a ground symbol is show n on a circuit diagram, you should con sider it to be merely a reference point at which thevoltage is defi ned to be zero. Fig.1-6 points out how cha nging the node labeled as ground cha nges the voltages at each node in the circuit, but does not cha nge the voltage drop across each comp onent.(1-2)Fig. 1-5 The voltage drop from point A to point B is V AB, where VAB = VA - VBThe sec ond of Kirchhoff ' s fun dame ntal laws states that the sum of the voltages around any loop of a circuit at any instant is zero. This is known as Kirchhoff ' s voltage law (KVL). Just as was the case for Kirchhoff ' s curre nt law, there are alter native, but equivale nt, ways of stat ing KVL. We can, for example, say that the sum of the voltage rises in any loop equals the sum of the voltagedrops around the loop. Thus in Fig. 1-6, there is a voltage rise of 12 V across the battery and avoltage drop of 3 V across R1 and a drop of 9 V across R2. ------------- Notice that it doesn' t matterwhich node was labeled ground for this to be true. Just as was the case with Kirchhoff ' s current law, we must be careful about labeli ng and in terpret ing the sig ns of voltages in a circuit diagram in order to write the proper vers ion of KVL. A plus (+) sig n on a circuit comp onent in dicates a reference direct ion un der the assumpti on that the pote ntial at that end of the comp onent is higher than the voltage at the other end. Aga in, as long as we are con siste nt in writi ng Kirchhoff ' s voltage law, the algebraic soluti on for the circuit will automatically take care of sig ns.Part 5 Summary of Principal Electrical QuantitiesThe key electrical qua ntities already in troduced and the releva nt relati on ships betwee n these quantities are summarized in Table 1-1.Since electrical quantities vary over such a large range of magnitudes, you will often find yourself work ing with very small qua ntities or very large qua ntities. For example, the voltage created by your TV antenna may be measured in millionths of a volt (microvolts, 卩V), while the power gen erated by a large power stati on may be measured in billi ons of watts, or gigawatts (GW). To describe quantities that may take on such extreme values, it is useful to have a system of prefixes that accompany the units. The most commonly used prefixes in electrical engineering are give n in Table 1-2.Part 6 Ideal Voltage Source and Ideal Current SourceElectric circuits are made up of a relatively small nu mber of differe nt kinds of circuiteleme nts, or comp onen ts, which can be in terc onn ected in an extraord in arily large nu mber of ways.At this point in our discussion, we will concentrate on idealized characteristics of these circuit eleme nts, realiz ing that real comp onents resemble, but do not exactly duplicate, the characteristics that we describe here.An ideal voltage source is one that provides a give n, known voltage vs, no matter what sort ofload it is conn ected to. That is, regardless of the curre nt draw n from the ideal voltage source, it will always provide the same voltage. Note that an ideal voltage source does not have to deliver a con sta nt voltage; for example, it may produce a sinu soidally vary ing voltage —the key is that voltage is not a fun ctio n of the amount of curre nt draw n. A symbol for an ideal voltage source is show n in Fig. 1-7.A special case of an ideal voltage source is an ideal battery that provides a con sta nt dc output, as show n in Fig. 1-8. A real battery approximates the ideal source; but as curre nt in creases, the output drops somewhat. To acco unt for that drop, quite ofte n the model used for a real battery is an ideal voltage source in series with the internal resista nee of the battery.An ideal curre nt source produces a give n amount of curre nt is no matter what load it sees. As show n in Fig. 1-9, a commo nly used symbol for such a device is circle with an arrow in dicati ng the directi on of curre nt flow. While a battery is a good approximati on to an ideal voltage source, there is nothing quite so familiar that approximates an ideal curre nt source. Some tran sistor circuits come close to this ideal and are ofte n modeled with idealized curre nt sources.Section II The electric power systemChapter 1 Brief Introduction to The Electric Power SystemNew Words and ExpressionsMinimum a 最小prime mover n 原动机gen erator n 发电机load n 负载furn ace n 炉膛boiler n 锅炉fissi on able n 可裂变的fissi on able material 核燃料Part 1 Minimum Power systemelevatio n n 高度，海拔internal combusti on engine 内燃机 steam-drive n turbi ne 汽轮机hydraulic turbi ne 水轮机convert v 变换，转换 shaft n 传动轴，轴 torquen 力矩servomecha nism n 伺服机构* Elements of a minimum electric power system *Types of energy source *Types of prime mover *Types of electrical load*Functions of the control systemA minimum electric power system is shown in Fig.1-1, the system consists of an energy source, a prime mover, a generator, and a load.The en ergy source may be coal, gas, or oil burned in a furnace to heat water and gen erate steam in a boiler; it may be fissi on able material which, in a nu clear reactor, will heat water to produce steam; it may be water in a pond at an elevatio n above the gen erat ing stati on; or it may be oil or gas burned in an internal combusti on engine.The prime mover may be a steam-driven turbine, a hydraulic turbine or water wheel, or aninternal combustion engine. Each one of these prime movers has the ability to convert energy in the form of heat, falling water, or fuel into rotation of a shaft, which in turn will drive theEnergy source Prime nioverGenerator Lx>adContjolFig* 1-1 The tninfnmm electric power systemgen erator.The electrical load on the gen erator may be lights, motors, heaters, or other devices, alone or in comb in ati on. Probably the load will vary from mi nute to min ute as differe nt dema nds occur. The control system functions (are ) to keep the speed of the machines substantially constant and the voltage within prescribed limits, even though the load may cha nge. To meet these load con diti on s, it is n ecessary for fuel in put to cha nge, for the prime mover in put to vary, and for the torque on the shaft from the prime mover to cha nge in order that the gen erator may be kept at con sta nt speed. In additi on, the field curre nt to the gen erator must be adjusted to maintain con sta nt output voltage. The con trol system may in clude a man stati oned in the power pla nt who watches a set of meters on the gen erator output term in als and makes the n ecessary adjustme nts manu ally .In a moder n stati on, the con trol system is a servomecha nism that sen ses gen erator-output con diti ons and automatically makes the n ecessary cha nges in en ergy in put and field curre nt to hold the electrical output with in certa in specificati ons.Part 2 More Complicated Systems*Foreword*Cases of power system with out circuit breaker *Power system with circuit breakerNew Words and Expressions1. associated2. circuit3. circuit breaker4. dee nergize5. dee nergized6. outage n7. diagram8. switch out of9. switch offIn most situati ons the load is not directly conn ected to the gen erator term in als. More com monlya 联接的 n 电路n 断路器 vt 切断，断电 adj 不带电的停电 n 简图退出来，断开 v 切断，关闭the load is some distanee from the generator, requiring a power line connecting them. It is desirable to keep the electric power supply at the load with in specificati ons. However, the con trols are near the generator, which may be in another building, perhaps several miles away.If the dista nce from the gen erator to the load is con siderable, it may be desirable to in stall transformers at the generator and at the load end, and to transmit the power over a high-voltage line (Fig.1-2). For the same power, the higher-voltage line carries less current, has lower losses for the same wire size, and provides more stable voltage., TransformerTransformerPrime 〔Mover Generator f C High-voltage line—Fig- 1-2 A generator connected through transformers anda high-voltage line to a distant loadIn some cases an overhead line may be un acceptable. In stead it may be adva ntageous to use an un dergro und cable. With the power systems talked above, the power supply to the load must be in terrupted if, for any reas on, any comp onent of the system must be moved from service for maintenance or repair.Additi onal system load may require more power tha n the gen erator can supply. Ano ther gen erator with its associated tran sformers and high-voltage line might be added.It can be shown that there are some advantages in making ties between the generators (1) and at the end of the high-voltage lines (2 and 3), as shown in Fig.1-3. This system will operate satisfactorily as long as no trouble develops or no equipment needs to be taken out of service.Kig. 1-3 A system with para)lei operation or the generators t of the transformers andof the transmission lintsThe above system may be vastly improved by the in troducti on of circuit breakers, which may be ope ned and closed as n eeded. Circuit breakers added to the system, Fig.1-4, permit selected piece of equipme nt to switch out of service without disturb ing the rema in der of system. With this arran geme nt any eleme nt of the system may be dee nergized for maintenance or repair by operati on of circuit breakers. Of course, if any piece of equipme nt is take n out of service, the n the total load must be carried by the remaining equipment. Attention must be given to avoid overloads duri ng such circumsta nces. If possible, outages of equipme nt are scheduled at times when load requireme nts are below no rmal.Low-voltageo=^GeneratorsFig.1-5 shows a system in which three gen erators and three loads are tied together by threeFig* 1-4 A system with necessary circuit breakerstran smissi on lin es. No circuit breakers are show n in this diagram, although many would berequired in such a system.Fis- 1-S Three generators supplying threeloads over hlgh-voltnge trAnsmlsston linesChapter 2 Faults on Power SystemNew Words and Expressions1. fault2. in terfere neen 干扰，防碍6. feed (fed)给。
