电气工程及其自动化专业英语第三章section_3-1
电气工程及自动化专业英语考试翻译课文Electric Power Systems 电力系统3.1
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。
电气工程及其自动化专业英语答案
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。
电气工程与自动化专业英语翻译(第三章)
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。
电气工程及其自动化专业英语第三章课文翻译
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3
System Design Methodology
The control system design should include appropriate feedback control loops to ensure stable and accurate system performance
Power electronic device selection
The selection of appropriate power electronic devices is critical for the performance and reliability of the power electronic system
Definition
Electrical automation has a wide range of application fields, which can achieve efficient, safe, reliable, and economical automation control, improve production efficiency, improve labor conditions, and provide strong technical support for the rapid development of modern industry.
Semiconductor technology
This involves the basic knowledge of semiconductors, including their properties and the operation of transformers and diodes
电气自动化专业英语第三单元
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第三章section_3-1
Section 1 Semiconductor Switches
give rise to power losses in the switching devices. We will examine these switching losses in two cases separately: the inductive and capacitive loads. Switching with Inductive Load The inductor is assumed to be large so that the current through it in steady state is nearly constant Io. Assume that initially the switch is off. The inductor current is +Io and freewheels through diode V1. When the switch is turned on, the current through the switch begins to build up linearly (an assumption) to +Io while the diode V1 is still on.
Poff usoff ir
(3-2)
Section 1 Semiconductor Switches
The usoff and ir are respectively the reverse bias voltage in the off-state and the reverse current through the switch. For example, the typical power diodes and the power transistors have high reverse voltages in hundreds to thousands of volts and microamps to milliamps through them in the off state. Transition-State Losses The practical switching devices have limited capabilities of rate of voltage transition and the rate of current steering. These nonabrupt transition rates
电气工程及其自动化专业英语
time-invariant 时不变的
self-(or mutual-)induction 自（互）感
displacement current 位移电流 conductance 电导
voltage drop 电压降 volt-ampere characteristics 伏安特性
metal-filament lamp 金属丝灯泡
seen, increase of current from zero to
I≈I1 causes the terminal voltage of the source to decrease linearly
V12=V=E－VS=E－RSI
Fig.1.3
In other words, the voltage drop VS across the source resistance rises in proportion to the current. This goes on until
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Section 1 Semiconductor Switches
give rise to power losses in the switching devices. We will examine these switching losses in two cases separately: the inductive and capacitive loads. Switching with Inductive Load The inductor is assumed to be large so that the current through it in steady state is nearly constant Io. Assume that initially the switch is off. The inductor current is +Io and freewheels through diode V1. When the switch is turned on, the current through the switch begins to build up linearly (an assumption) to +Io while the diode V1 is still on.
Section 1 Semiconductor Switches
over, the diode Vl begins to conduct and the voltage on the switch is clamped at Uo, and the current through the switch ramps linearly (again an assumption) down to zero. When the switch is closed, the current begins to build up linearly to IS while the diode V1 is still on. The voltage on the switch remains clamped at UO. After the switch current reaches IS, the diode turns off and the voltage on the switch begins to ramp down to zero.
Section 1 Semiconductor Switches
the current through the switch begins to decrease below Io , as the remaining current is now steered through the diode V1, which has now turned on. The current through the switch ramps down to zero ultimately. Switching waveforms with inductive load are shown in Fig.3-1.
Section 1 Semiconductor Switches
The on diode has zero voltage across it (an ideal diode), hence, the voltage on the switch is held constant at +Us. When the current buildup is over, the diode Vl ceases to conduct and the voltage on the switch ramps linearly (again an assumption) down to zero. When the switch is turned off, the voltage begins to build up linearly to +Us while the diode V1 is off. While the diode is off the current through the switch equals the inductor current, which is constant Io. After the switch voltage reaches zero,
Section 1 Semiconductor Switches
On-State Losses The electrical switches conduct heavy current and have nonzero voltage across the switch in the onstate. The on-state power losses are given by
Chapter 3
Power Electronic Technology
Section 1 Semiconductor Switches
TextБайду номын сангаас
New Words and Expressions Transition of part of speech
Exercises
End
Section 1 Semiconductor Switches
Fig.3-1 Switching waveforms with inductive load
Section 1 Semiconductor Switches
The switching losses are given by The switching power losses increase linearly with the switching frequency like in the resistive case but about six times more. The upper bound on the switching frequency is also about half.
Section 1 Semiconductor Switches
power losses are very significant. Off-State Losses The electrical switches withstand high voltages and have nonzero leakage current through the switch in the off-state. The off-state power losses are given by
Poff usoff ir
(3-2)
Section 1 Semiconductor Switches
The usoff and ir are respectively the reverse bias voltage in the off-state and the reverse current through the switch. For example, the typical power diodes and the power transistors have high reverse voltages in hundreds to thousands of volts and microamps to milliamps through them in the off state. Transition-State Losses The practical switching devices have limited capabilities of rate of voltage transition and the rate of current steering. These nonabrupt transition rates
f s max t on1 t on 2 1 t off 1 t off 2
(3-4)
1 Psw Us I o [ton1 ton 2 toff 1 toff 2 ] fs 2
(3-3)
Section 1 Semiconductor Switches
Switching with capacitive load The capacitor is assumed to be large so that the voltage through it in steady state is nearly constant Uo. Assume that initially the switch is on, hence, the current through the switch is IS. The capacitor voltage is Uo, the voltage across the switch is zero and the diode V1 is reverse biased. When the switch is turned off, the switch voltage begins to ramp up to + Uo while the diode V1 is still off. During this buildup, the current through the switch is held constant at IS . When the voltage buildup is
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 the process of manufacturing.