直流电动机闭环控制系统毕业论文中英文资料外文翻译文献
直流电动机中英文对照外文翻译文献
中英文对照外文翻译文献(文档含英文原文和中文翻译)外文文献：DC Motor CalculationsOverviewNow that we have a good understanding of dc generators, we can begin our study of dc motors. Direct-current motors transform electrical energy into mechanical energy. They drive devices such as hoists, fans, pumps, calendars, punch-presses, and cars. These devices may have a definite torque-speed characteristic (such as a pump or fan) or a highly variable one (such as a hoist or automobile). The torque-speed characteristic of the motor must be adapted to the type of the load it has to drive, and this requirement has given rise to three basic types of motors: 1.Shunt motors 2. Series motors 3. Compound motors Direct-current motors are seldom used in ordinary industrial applications because all electric utility systems furnish alternating current. However, for special applications such as in steel mills, mines, and electric trains, it is sometimes advantageous to transform the alternating current into direct current in order to use dc motors. The reason is that the torque-speed characteristics of dc motors can be varied over a wide range while retaining high efficiency. Today, this general statement can be challenged because the availability of sophisticated electronic drives has made it possible to use alternating current motors for variable speed applications. Nevertheless, there are millions of dc motors still in service and thousands more are being produced every year.Counter-electromotive force (cemf)Direct-current motors are built the same way as generators are; consequently, a dc machine can operate either as a motor or as a generator. To illustrate, consider a dc generator in which the armature, initially at rest, is connected to a dc source E s by means of a switch (Fig. 5.1). The armature has a resistance R, and the magnetic field is created by a set of permanent magnets.As soon as the switch is closed, a large current flows in the armature because its resistance is very low. The individual armature conductors are immediately subjected to a force because they are immersed in the magnetic field created by the permanent magnets. These forces add upto produce a powerful torque, causing the armature to rotate.Figure 5.1 Starting a dc motor across the line.On the other hand, as soon as the armature begins to turn, a second phenomenon takes place: the generator effect. We know that a voltage E o is induced in the armature conductors as soon as they cut a magnetic field (Fig. 5.2). This is always true, no matter what causes the rotation. The value and polarity of the induced voltage are the same as those obtained when the machine operates as a generator. The induced voltage E o is therefore proportional to the speed of rotation n of the motor and to the flux F per pole, as previously given by Eq. 5.1:E o = Zn F/60 (5.1)As in the case of a generator, Z is a constant that depends upon the number of turns on the armature and the type of winding. For lap windings Z is equal to the number of armature conductors.In the case of a motor, the induced voltage E o is called counter-electromotive force (cemf) because its polarity always acts against the source voltage E s. It acts against the voltage in the sense that the net voltage acting in the series circuit of Fig. 5.2 is equal to (E s - Eo) volts and not (E s + E o) volts.Figure 5.2 Counter-electromotive force (cemf) in a dc motor.Acceleration of the motorThe net voltage acting in the armature circuit in Fig. 5.2 is (E s- E o) volts. The resulting armature current /is limited only by the armature resistance R, and soI = (E s- E o)IR (5.2)When the motor is at rest, the induced voltage E o= 0, and so the starting current isI = E s/RThe starting current may be 20 to 30 times greater than the nominal full-load current of the motor. In practice, this would cause the fuses to blow or the circuit-breakers to trip. However, if they are absent, the large forces acting on the armature conductors produce a powerful starting torque and a consequent rapid acceleration of the armature.As the speed increases, the counter-emf E o increases, with the result that the value of (E s—E o)diminishes. It follows from Eq. 5.1 that the armature current / drops progressively as the speed increases.Although the armature current decreases, the motor continues to accelerate until it reaches a definite, maximum speed. At no-load this speed produces a counter-emf E o slightly less than the source voltage E s. In effect, if E o were equal to E s the net voltage (E s—E o) would become zero and so, too, would the current /. The driving forces would cease to act on the armature conductors, and the mechanical drag imposed by the fan and the bearings would immediately cause the motor to slow down. As the speed decreases the net voltage (E s—E o) increases and so does the current /. The speed will cease to fall as soon as the torque developed by the armature current is equal to the load torque. Thus, when a motor runs at no-load, the counter-emf must be slightly less than E s so as to enable a small current to flow, sufficient to produce the required torque.Mechanical power and torqueThe power and torque of a dc motor are two of its most important properties. We now derive two simple equations that enable us to calculate them.1. According to Eq. 5.1 the cemf induced in a lap-wound armature is given byE o = Zn F/60Referring to Fig. 5.2, the electrical power P a supplied to the armature is equal to the supply voltage E s multiplied by the armature current I:P a = E s I (5.3)However, E s is equal to the sum of E o plus the IR drop in the armature:E s = E o + IR (5.4)It follows thatP a= E s I= (E o + IR)I=E o I + I2R (5.5)The I2R term represents heat dissipated in the armature, but the very important term E o I is the electrical power that is converted into mechanical power. The mechanical power of the motor is therefore exactly equal to the product of the cemf multiplied by the armature currentP = E o I (5.6)whereP = mechanical power developed by the motor [W]E o= induced voltage in the armature (cemf) [V]I = total current supplied to the armature [A]2. Turning our attention to torque T, we know that the mechanical power P is given by the expressionP = nT/9.55 (5.7)where n is the speed of rotation.Combining Eqs. 5.7,5.1, and 5.6, we obtainnT/9.55 = E o I= ZnFI/60and soT =Z F I/6.28The torque developed by a lap-wound motor is therefore given by the expressionT =Z F I/6.28 (5.8)whereT = torque [N×m]Z = number of conductors on the armatureF = effective flux per pole [Wb]*/ = armature current [A]6.28 = constant, to take care of units[exact value = 2p]Eq. 5.8shows that we can raise the torque of a motor either by raising the armature current or by raising the flux created by the poles.Speed of rotationWhen a dc motor drives a load between no-load and full-load, the IR drop due to armature resistance is always small compared to the supply voltage E s. This means that the counter-emf E s is very nearly equal to E s.On the other hand, we have already seen that Eo may be expressed by the equationE o = Zn F/60Replacing E o by E s we obtainE s = Zn F/60That is,wheren = speed of rotation [r/min]E s = armature voltage [V]Z = total number of armature conductorsThis important equation shows that the speed of the motor is directly proportional to the armature supply voltage and inversely proportional to the flux per pole. We will now study how this equation is applied.Armature speed controlAccording to Eq. 5.8, if the flux per pole F is kept constant (permanent magnet field or field with fixed excitation), the speed depends only upon the armature voltage E s. By raising or lowering E s the motor speed will rise and fall in proportion.In practice, we can vary E s by connecting the motor armature M to a separately excited variable-voltage dc generator G . The field excitation of the motor is kept constant, but the generator excitation I x can be varied from zero to maximum and even reversed. The generator output voltage E s can therefore be varied from zero to maximum, with either positive or negative polarity. Consequently, the motor speed can be varied from zero to maximum in either direction. Note that the generator is driven by an ac motor connected to a 3-phase line. This method of speed control, known as the Ward-Leonard system, is found in steel mills, high-rise elevators, mines, and paper mills.In modem installations the generator is often replaced by a high-power electronic converter that changes the ac power of the electrical utility to dc, by electronic means.What happens to the dc power received by generator G? When G receives electric power, it operates as a motor, driving its own ac motor as an asynchronous generator!* As a result, ac power is fed back into the line that normally feeds the ac motor. The fact that power can be recovered this way makes the Ward-Leonard system very efficient, and constitutes another of its advantages.Rheostat Speed ControlAnother way to control the speed of a dc motor is to place a rheostat in series with the armature . The current in the rheostat produces a voltage drop which subtracts from the fixed source voltage E s, yielding a smaller supply voltage across the armature. This method enables us to reduce the speed below its nominal speed. It is only recommended for small motors because a lot of power and heat is wasted in the rheostat, and the overall efficiency is low. Furthermore, thespeed regulation is poor, even for a fixed setting of the rheostat. In effect, the IR drop across the rheostat increases as the armature current increases. This produces a substantial drop in speed with increasing mechanical load.中文译文：直流电动机的计算概述现在，我们对直流发电机有一个很好的了解，我们可以开始对直流电动机的研究了。
毕业设计外文资料翻译汇编
毕业设计（论文）外文资料翻译附件1 ：外文资料翻译译文双闭环直流调速系统的说明一、系统分析与综合1. 系统分析（1）在转速、电流双闭环调速系统中，若要改变电动机的转速，应调节什么参数？改变转速调节器的放大倍数Kn行不行？改变电力电子变换器的放大系数Ks 行不行？改变转速反馈系数行不行？若要改变电动机的堵转电流，应调节系统中的什么参数？答：若要改变电动机的转速，改变转速调节器的放大倍数Kn 和电力电子变换器的放大系数Ks 都不行，稳定时n=Un=Un*,所以只有改变给定值Un*和反馈系数才行。
若要改变电动机的堵转电流，同样只须改变给定值Uim*和反馈系数，因为，稳定时，Uim* = Idm, 从式中可得出。
（2）转速、电流双闭环调速系统稳态运行时，两个调节器的输入偏差电压和输出电压各是多少？答：转速、电流双闭环调速系统稳态运行时，两个调节器的输入偏差电压均是零，由式子n=Un=Un*，n=n0 ; Uim* = Idm, Idm=Idl 。
（3）在转速、电流双闭环调速系统中，两个调节器均采用PI 调节器。
当系统带额定负载运行时，转速反馈线突然断线，系统重新进入稳态后，电流调节器的输入偏差电压Ui 是否为零？为什么？答：当系统带额定负载运行时，转速反馈线突然断线，则Un=0，Un =Un*-Un=Un*, 使Ui 迅速达到Uim ，Ui 0 ，速度n 上升，当系统重新进入稳态后，即Id=Idl ，那么，Ui = Uim*- Idl 0，Ui 也不再变化，转速n也不再变化，但，此时的转速n 比反馈线断线时的转速要大。
（4）为什么用积分控制的调速系统是无静差的？答：在积分调节器的调速系统中，能实现无静差，是由于积分调节器控制特点，即积分的记忆和积累作用。
（5）双环调速系统（PI），负载变化，Idl>Idm, 问双环调速系统ACR和ASR 怎么调节，结果如何？答：当负载变化时，Idl>Idm, 转速迅速下降，电流Id 很快增加到Idm，而达限幅值，速度ASR迅速饱和，ACR一直在限流状态下，形成堵转现象，长时间运行会损坏系统。
直流调速系统设计与仿真外文翻译
Double Loop DC Speed Control System Description工学部专业自动化班级学号姓名指导教师负责教师沈阳航空航天大学北方科技学院2010年6月Double Loop DC Speed Control System DescriptionⅠ.system analysis and synthesis1．Analysis（1）In the speed and current dual closed-loop speed control system, in order to change the motor speed, what parameters should be regulating? Change speed regulator Kn magnification work? Power electronic converter to change the magnification factor Ks work? Change the speed of the feedback coefficient of work? To change the motor's stall current system should adjust the parameters of what?A: To change the motor speed, change speed regulator Kn magnification and power electronic converters will not work magnification factor Ks, stable when n = Un = Un *, so the only change in the value of a given coefficient of Un * and feedback before. To change the motor's stall current, only need to change the same value given Uim * and feedback coefficient, because the stability, Uim * = Idm, can be drawn from the type(2) Speed, the current double closed-loop speed control system when the steady-state operation, the two regulator input voltage and output voltage deviation is the number?A: The speed and current dual closed-loop speed control system when the steady-state operation, the two regulators are the input bias voltage is zero, by the formula n = Un = Un *, n = n; Uim * = Idm, Idm = Idl.(3) In the speed and current dual closed-loop speed control system, the two regulators are PI regulator. When the system is running with rated load, the speed feedback line suddenly disconnected, the system re-enter the steady-state, the current regulator is the input bias voltage to zero? Why?A: When the system is running with rated load, the speed feedback line suddenly disconnected, then Un = 0, = Un *- Un = Un *, so that Ui to reach Uim, 0, rate of increase in n, when the system after re-entering the steady-state , that is, Id = Idl, then, = Uim *- Idl 0, are no longer changes, changes in rotational speed n is no longer, but at this time than the rotational speed n at the time of the feedback line speed to break big.(4) Why is the speed with integral control system is not static poor?A: Speed regulator integral system, to achieve non-static error is due to the characteristics of integral control regulator, that is, the accumulation of points and the role of memory.(5) Double-loop speed control system (PI), load changes, Idl> Idm, asked bicyclol speed control system ASR and ACR how-conditioning, the result?A: When the load changes, Idl> Idm, speed decreased rapidly, the current Id soon to Idm, and of limited amplitude, rapid rate of ASR saturation, ACR has been limiting conditions, to form a blocking phenomenon, long-running will damage the system.2. SystemSpeed regulator and current regulator in the Double Loop DC Motor Control System can be summarized as follows:(1). The role of speed regulatorSpeed regulator is a speed control system of the dominant regulator, which allows speed n will soon change with a given voltage Un * changes in steady-state speed error can be reduced, if the PI regulator can achieve the non-static error.1) The effect of load changes in the role of anti-disturbance.2) The output amplitude of the decision limit the maximum allowable motor current.(2) The role of current regulator1) As a regulator of the inner ring, outer ring at the speed of the adjustment process, it makes the current closely followed the given voltage Ui * (that is, the outer ring modulator output) changes.2) Fluctuations in voltage from the role of disturbance rejection in time.3) The speed of the dynamic process of ensuring that the maximum allowable motor current, thereby speeding up the dynamic process.4) When the motor overload or stall when the armature current limit of the maximum, automatic protection from the role of acceleration. Once the fault disappears, the system automatically return to normal. Yesterday, the role of the reliable operation is very important.Ⅱdouble-loop speed control system common faults analysis1.Introduction of a system(1). Double-loop speed control system components in Figure 1.1 Current loop: from the current regulator LT, trigger CF (input transformation for the CSR), silicon-controlled rectifier bridge, motor armature and current loop transform LB component. the speed of outer ring: the speed regulator from ST, current loop, such as link inertia, motor and load moment of inertia and the speed of transformationcomponents SB. in the double-loop speed control system, the speed of the decision loop of the running characteristics of the whole system and stability, and play a leading role, and to change the current ring plays the role of the internal structure of the system is dependent, but since it is as a whole to participate in the closed-loop speed to the speed of a direct impact on the work of the closed-loop, it must first good debugging current loop, and then testing the speed of outer ring, so that the whole system has good dynamic performance.(2). Double-loop speed control system of the typical working condition1) Start (or the speed):ST in the start-up process has been saturated, so that the speed of this loop in the equivalent open-loop state. System only in the constant current loop under the regulation to ensure the motor at a constant current of the maximum allowable under the start-up.Figure 1 double-loop speed control system structure2) slow down (or stop):ST at this time to reach the output amplitude of the reverse limit. Main circuit current by the bridge is reduced to zero after the inverter. LT and CSR output will soon reach the maximum reverse. CF pulse output to reach βmin, current loop for open-loop. motor torque under deceleration until the motor speed close to the given new value, current loop and speed loop one after another into the closed-loop work, motor in the new value of a given run .3) Grid voltage fluctuation: This motor because of the larger moment of inertia, which caused the first change in armature current, ST output also did not change the effect of current loop, LT rapidly changing the output so that angle α be adjusted quickly, so the impact on the speed .4）Small changes in load: in the operation, load changes, will cause the motor speed deviation from the given value. speed up the recovery process and the aforementioned speed (or deceleration) is similar to the process.2. Common Fault Analysis and Processing(1) The normal supply voltage, thyristor rectifier output waveform arrhythmia caused by this phenomenon is due to trigger sawtooth slope caused inconsistency. Sawtooth slope adjust potentiometer, the output waveform uniformity could be achieved. in the adjustment process to strike a balance between Qi, this point should be paid attention to the actual debugging.(2) DC Motor Analysis of mechanical properties of soft thyristor DC motor system, when the current intermittent mechanical properties when the first no-load speed is characterized by high ideals, and the second is characterized by mechanical properties of soft . The so-called mechanical features soft, that is, small changes in load will cause great changes in speed. oscilloscope to observe when using the bridge rectifier output waveform, one may find that missing relative. at this time need to check whether the trigger has pulse output, fast whether the fuse melting, whether the breakdown or thyristor circuit, synchronous transformer is damaged, whether the lack of power. to identify the problems, can be resolved.(3) The speed of the speed of instability caused by many factors of instability:1) Electric guns are not firmly fixed or with the host of different axis .2) The parameters of the speed regulator inappropriate. Respond to the dynamic parameters to adjust (change the ratio of integral parameters ) .3) Of a speed control system there are(or bad)4) May be caused as a result of interference. should be found to interfere with the reasons for taking anti-jamming measures.(4) A little to the set rated motor speed is higher than that should first check whether it is normal for the external control system, such as outside the normal control system, it may be given points, speed regulator, current regulator, such as caused by link failure. Should be cut off the main circuit power supply, only the control system to the electricity, not a given in the case, testing each of the key points (such as the current regulator, voltage regulator, etc. of the potential. and then given together with the former to one by one after each of the key points to check the potential changes, you can find out the fault lies.(5) Bridge rectifier output voltage is not high stressed1) the speed regulator and current regulator limiter too small, should be liberalizedin accordance with appropriate amplitude threshold .2) than the speed feedback signal, in that they can reduce the rate of appropriate feedback signals.(6) To the timing system still in the absence of low-speed operation (that is, the phenomenon of emergence of reptiles) This is because the system of "zero drift" caused by. When the input signal is zero, the output voltage by the input amplification stage of the offset potentiometer decisions can be offset by adjusting the potential allows α = 90 °, at this time to zero output voltage rectification system, the electrical will not crawl.(7) With a given system can not runShould first check whether it is normal for the external control system, such as outside the normal control system, it may be given points, speed regulator, current regulator, such as caused by link failure. Shall be cut off main circuit power supply, only the control system to the electricity, not Add the given circumstances, the key points of each test (such as the current regulator, voltage regulator, etc.) of the potential. and then combined with a given, after the previous one by one to each of the key points to check the potential changes, that is, where to find fault.(8) Lack of control accuracy in the distributor for a given run-time external control often requires precision sufficient parking in order to work properly. If poor precision parking, you can adjust the speed of the appropriate regulator of the PI link, generally by reducing the the ratio of the integral part of efforts to get satisfactory results.(9) Reversible system oscillation1) open-loop system in the state (the main circuit disconnect) the oscillation can be changed at this time given the previous inspection to the key points of the potential changes. If a given unchanged, but the potential is still a point of change, here is the crux of .2) system in the state when the closed-loop oscillation, in which case in order to ensure the safety of the electrical load should be replaced by the general resistance of the load, if there is no suitable resistance box which can be used in place of the two electric sub-series. inspection methods and similar open-loop, focusing on the link to check is: given points, level detection, operation control, such as the speed regulator. oscillations are often caused as a result of operational amplifiers, such as damage to electroniccomponents , system parameters caused by improper, according to the specific circumstances, properly addressed.双闭环直流调速系统的说明一、系统分析与综合1.系统分析（1）在转速、电流双闭环调速系统中，若要改变电动机的转速，应调节什么参数？改变转速调节器的放大倍数Kn行不行？改变电力电子变换器的放大系数Ks 行不行？改变转速反馈系数∂行不行？若要改变电动机的堵转电流，应调节系统中的什么参数？答：若要改变电动机的转速，改变转速调节器的放大倍数Kn和电力电子变换器的放大系数Ks都不行，稳定时∂n=Un=Un*,所以只有改变给定值Un*和反馈系数∂才行。
《基于simulink的双闭环直流调速系统的设计与仿真》毕业设计的英文翻译
《基于simulink的双闭环直流调速系统的设计与仿真》毕业设计的英⽂翻译《基于simulink的双闭环直流调速系统的设计与仿真》毕业设计的英⽂翻译基于Simulink的双闭环直流电机调速系统的参数优化与仿真摘要在控制系统中当动态性能的要求很⾼，并且单回路控制系统不能满⾜的要求时，我们实施了多环控制和在线对内环和外环进⾏参数优化。
本⽂以双回路直流电动机调速控制系统为例，采⽤仿真优化⽅法设计两个PI调节器的参数，使系统的动态和稳态指标达到设计要求。
关键字：参数优化直流电动机双闭环系统仿真正⽂：I简介在经典控制理论中，通常对控制电路中的每个物理量设⽴⼀个调节器，当有多个物理参数要被控制时，就需要设置多个调节器来控制这些参数，这样的系统被称为多回路控制系统。
双闭环直流电动机调速系统是⼀个典型的多环控制系统。
根据⽂献[1]我们很清楚的了解到速负反馈控制的单闭环直流调速系统⽤PI调节器可以保证系统的稳定性。
然⽽当控制系统对动态性能要求很⾼，例如快速制动、突加负载时转速降落⼩等要求，单闭环调速系统将难以满⾜需求。
经典控制理论解决这个问题的唯⼀的⽅法是实现电流负反馈控制，在电流控制环中设置⼀个调节器，特别是⽤于调节电流量的调节器。
控制系统中就建⽴了转速、电流两个调节器。
这样的系统被称为直流电动机的电流和转速控制系统（这样的系统也被称为双闭环直流电动机调速系统）。
II 系统模型为了充分发挥转速和电流负反馈在系统中的控制作⽤，以及它们不会相互抑制影响系统的性能，我们在系统中设置了两个调节器调节速度和电流，实现它们之间的联接。
也就是说，我们把速度调节器的输出作为电流调节器的输⼊，电流调节器的输出控制整流电路的触发装置。
从整个闭环反馈控制回路的结构来看，电流调节环在转速调节环之内，称之为内环；转速调节环在外，称之为外环。
这就形成了⼀个双闭环直流调速系统。
为了获得良好的静、动态性能，双闭环直流调速系统的两个调节器⼀般设计为PI 调节器。
直流电动机双闭环调速系统地研究外文翻译
直流电动机双闭环调速系统的研究外文翻译1．概述直流电动机具有良好的起、制动性能，宜于在大范围内实现平滑调速，并且直流调速系统在理论和实践上都比较成熟，是研究其它调速系统的基础。
而用 MATLAB 软件对直流调速系统进行虚拟环境下的仿真研究，不仅使用方便，也大大降低了研究成本。
本文叙述了直流电动机的基本原理和调速原理，介绍了直流电动机开环和双闭环调速系统的组成及静、动态特性，并且根据直流电动机的基本方程建设立了调速系统的数学模型，给出了动态结构框图，用工程设计方法设计了直流电动机双闭环调速系统。
最后，用 MATLAB 仿真软件搭建了仿真模型，对调速系统进行了仿真研究。
通过对直流电动机双闭环调速系统动态特性的研究与仿真，可以清楚地看到，直流电动机双闭环调速系统具有较好的动态性能，可以在给定调速范围内，实现无静差平滑调速，这为直流电动机调速系统的硬件实验提供了理论依据。
2．双闭环直流调节系统的研究背景直流调速是现代电力拖动自动控制系统中发展较早的技术。
在 20 世纪 60 年代，随着晶闸管的出现，现代电力电子和控制理论、计算机的结合促进了电力传动控制技术研究和应用的繁荣。
晶闸管 -直流电动机调速系统为现代工业提供了高效、高性能的动力。
尽管目前交流调速的迅速发展，交流调速技术越趋成熟，以及交流电动机的经济性和易维护性，使交流调速广泛受到用户的欢迎。
但是直流电动机调速系统以其优良的调速性能仍有广阔的市场，并且建立在反馈控制理论基础上的直流调速原理也是交流调速控制的基础。
现在的直流和交流调速装置都是数字化的，使用的芯片和软件各有特点，但基本控制原理有其共性。
对于那些在实际调试过程中存在很大风险或实验费用昂贵的系统，一般不允许对设计好的系统直接进行实验。
然而没有经过实验研究是不能将设计好的系统直接放到生产实际中去的。
因此就必须对其进行模拟实验研究。
当然有些情况下可以构造一套物理装置进行实验，但这种方法十分费时而且费用又高，而且在有的情况下物理模拟几乎是不可能的。
外文翻译---双闭环直流调速系统的说明
外文翻译---双闭环直流调速系统的说明of motor used.2.SynthesisThe double loop DC speed control system consists of two closed loops: a speed loop and a current loop。
The speed loop is___ the motor speed。
while the current loop is ___ the motor current。
The system can be adjusted by changing the value of the feedback coefficient and the given value of the speed and current。
It is ___ factor of the power electronic ___.Ⅱ。
System components and their nsThe double loop DC speed control system is composed of several components。
including the motor。
the power electronic converter。
the speed sensor。
the current sensor。
the speed regulator。
and the current regulator。
Each component has a specific n in the system.The motor is the main component of the system and is ___ the DC voltage into an AC voltage to power the motor。
The speed sensor and current sensor are used to measure the motor speed and current。
电力系统自动化毕业论文中英文资料外文翻译
毕业设计(论文)外文资料翻译专业名称：电力系统自动化英文资料：INDUCTION MOTOR STARTING METHODSAbstract -Many methods can be used to start large AC induction motors. Choices such as full voltage, reduced voltage either by autotransformer or Wyes - Delta, a soft starter, or usage of an adjustable speed drive can all have potential advantages and trade offs. Reduced voltage starting can lower the starting torque and help prevent damage to the load. Additionally, power factor correction capacitors can be used to reduce the current, but care must be taken to size them properly. Usage of the wrong capacitors can lead to significant damage. Choosing the proper starting method for a motor will include an analysis of the power system as well as the starting load to ensure that the motor is designed to deliver the needed performance while minimizing its cost. This paper will examine the most common starting methods and their recommended applications.I. INTRODUCTIONThere are several general methods of starting induction motors: full voltage, reduced voltage, wyes-delta, and part winding types. The reduced voltage type can include solid state starters, adjustable frequency drives, and autotransformers. These, along with the full voltage, or across the line starting, give the purchaser a large variety of automotives when it comes to specifying the motor to be used in a given application. Each method has its own benefits, as well as performance trade offs. Proper selection will involve a thorough investigation of any power system constraints, the load to be accelerated and the overall cost of the equipment.In order for the load to be accelerated, the motor must generate greater torque than the load requirement. In general there are three points of interest on the motor's speed-torque curve. The first is locked-rotor torque (LRT) which is the minimum torque which the motor will develop at rest for all angular positions of the rotor. The second is pull-up torque (PUT) which is defined as the minimum torque developed by the motor during the period of acceleration from rest to the speed at which breakdown torque occurs. The last is the breakdown torque (BDT) which is defined as the maximum torque which the motor will develop. If any of these points are below the required load curve, then the motor will not start.The time it takes for the motor to accelerate the load is dependent on the inertia of the load and the margin between the torque of the motor and the load curve, sometimes called accelerating torque. In general, the longer the time it takes for the motor to accelerate the load, the more heat that will be generated in the rotor bars, shorting ring and the stator winding. This heat leads to additional stresses in these parts and can have an impaction motor life.II. FULL VOLTAGEThe full voltage starting method, also known as across the line starting, is the easiest method to employ, has the lowest equipment costs, and is the most reliable. This method utilizes a control to close a contactor and apply full line voltage to the motor terminals. This method will allow the motor to generate its highest starting torque and provide the shortest acceleration times.This method also puts the highest strain on the power system due to the high starting currents that can be typically six to seven times the normal full load current of the motor. If the motor is on a weak power system, the sudden high power draw can cause a temporary voltage drop, not only at the motor terminals, but the entire power bus feeding the starting motor. This voltage drop will cause a drop in the starting torque of the motor, and a drop in the torque of any other motor running on the power bus. The torque developed by an induction motor varies roughly as the square of the applied voltage. Therefore, depending on the amount of voltage drop, motors running on this weak power bus could stall. In addition, many control systems monitor under voltage conditions, a second potential problem that could take a running motor offline during a full voltage start. Besides electrical variation of the power bus, a potential physical disadvantage of an across the line starting is the sudden loading seen by the driven equipment. This shock loading due to transient torques which can exceed 600% of the locked rotor torque can increase the wear on the equipment, or even cause a catastrophic failure if the load can not handle the torques generated by the motor during staring.A. Capacitors and StartingInduction motors typically have very low power factor during starting and as a result have very large reactive power draw. See Fig. 2. This effect on the system can be reduced by adding capacitors to the motor during starting.The large reactive currents required by the motor lag the applied voltage by 90 electrical degrees. This reactive power doesn't create any measurable output, but is rather the energy required for the motor to function. The product of the applied system voltage and this reactive power component can be measured in V ARS (volt-ampere reactive). The capacitors act to supply a current that leads the applied voltage by 90 electrical degrees. The leading currents supplied by the capacitors cancel the laggingcurrent demanded by the motor, reducing the amount of reactive power required to be drawn from the power system.To avoid over voltage and motor damage, great care should be used to make sure that the capacitors are removed as the motor reaches rated speed, or in the event of a loss of power so that the motor will not go into a generator mode with the magnetizing currents provided from the capacitors. This will be expanded on in the next section and in the appendix.B. Power Factor CorrectionCapacitors can also be left permanently connected to raise the full load power factor. When used in this manner they are called power factor correction capacitors. The capacitors should never be sized larger than the magnetizing current of the motor unless they can be disconnected from the motor in the event of a power loss.The addition of capacitors will change the effective open circuit time constant of the motor. The time constant indicates the time required for remaining voltage in the motor to decay to 36.8% of rated voltage after the loss of power. This is typically one to three seconds without capacitors.With capacitors connected to the leads of the motor, the capacitors can continue to supply magnetizing current after the power to the motor has been disconnected. This is indicated by a longer time constant for the system. If the motor is driving a high inertia load, the motor can change over to generator action with the magnetizingCurrent from the capacitors and the shaft driven by the load. This can result in the voltage at the motor terminals actually rising to nearly 50% of rated voltage in some cases. If the power is reconnected before this voltage decays severe transients can be created which can cause significant switching currents and torques that can severely damage the motor and the driven equipment. An example of this phenomenon is outlined in the appendix.Ⅲ. REDUCED VOLTAGEEach of the reduced voltage methods are intended to reduce the impact of motor starting current on the power system by controlling the voltage that the motor sees atthe terminals. It is very important to know the characteristics of the load to be started when considering any form of reduced voltage starting. The motor manufacturer will need to have the speed torque curve and the inertia of the driven equipment when they validate their design. The curve can be built from an initial, or break away torque, as few as four other data points through the speed range, and the full speed torque for the starting condition. A centrifugal or square curve can be assumed in many cases, but there are some applications where this would be problematic. An example would be screw compressors which have a much higher torque requirement at lower speeds than the more common centrifugal or fan load. See Fig. 3. By understanding the details of the load to be started the manufacturer can make sure that the motor will be able to generate sufficient torque to start the load, with the starting method that is chosen.A. AutotransformerThe motor leads are connected to the lower voltage side of the transformer. The most common taps that are used are 80%, 65%, and 50%. At 50% voltage the current on the primary is 25% of the full voltage locked rotor amps. The motor is started with this reduced voltage, and then after a pre-set condition is reached the connection is switched to line voltage. This condition could be a preset time, current level, bus volts, or motor speed. The change over can be done in either a closed circuit transition, or an open circuit transition method. In the open circuit method the connection to the voltage is severed as it is changed from the reduced voltage to the line level. Care should be used to make sure that there will not be problems from transients due to the switching. This potential problem can be eliminated by using the closed circuit transition. With the closed circuit method there is a continuousVoltage applied to the motor. Another benefit with the autotransformer starting is in possible lower vibration and noise levels during starting.Since the torque generated by the motor will vary as the square of the applied voltage, great care should be taken to make sure that there will be sufficient accelerating torque available from the motor. A speed torque curve for the driven equipment along with the inertia should be used to verify the design of the motor. A good rule of thumb is to have a minimum of 10% of the rated full load torque of the motor as a margin at all points of the curve.Additionally, the acceleration time should be evaluated to make sure that the motor has sufficient thermal capacity to handle the heat generated due to the longeracceleration time.B. Solid State or Soft StartingThese devices utilize silicon controlled rectifiers or Scars. By controlling the firing angle of the SCR the voltage that the device produces can be controlled during the starting of the motor by limiting the flow of power for only part of the duration of the sine wave.The most widely used type of soft starter is the current limiting type. A current limit of 175% to 500% of full load current is programmed in to the device. It then will ramp up the voltage applied to the motor until it reaches the limit value, and will then hold that current as the motor accelerates.Tachometers can be used with solid state starters to control acceleration time. Voltage output is adjusted as required by the starter controller to provide a constant rate of acceleration.The same precautions in regards to starting torque should be followed for the soft starters as with the other reduced voltage starting methods. Another problem due to the firing angle of the SCR is that the motor could experience harmonic oscillating torques. Depending on the driven equipment, this could lead to exciting the natural frequency of the system.C. Adjustable Frequency DrivesThis type of device gives the greatest overall control and flexibility in starting induction motors giving the most torque for an amount of current. It is also the most costly.The drive varies not only the voltage level, but also the frequency, to allow the motor to operate on a constant volt per hertz level. This allows the motor to generate full load torque throughout a large speed range, up to 10:1. During starting, 150% of rated current is typical.This allows a significant reduction in the power required to start a load and reduces the heat generated in the motor, all of which add up to greater efficiency. Usage of the AFD also can allow a smaller motor to be applied due to the significant increase of torque available lower in the speed range. The motor should still be sizedlarger than the required horsepower of the load to be driven. The AFD allows a great degree of control in the acceleration of the load that is not as readily available with the other types of reduced voltage starting methods.The greatest drawback of the AFD is in the cost relative to the other methods. Drives are the most costly to employ and may also require specific motor designs to be used. Based on the output signal of the drive, filtered or unfiltered, the motor could require additional construction features. These construction features include insulated bearings, shaft grounding brushes, and insulated couplings due to potential shaft current from common mode voltage. Without these features, shaft currents, which circulate through the shaft to the bearing, through the motor frame and back, create arcing in the bearings that lead to premature bearing failure, this potential for arcing needs to be considered when applying a motor/drive package in a hazardous environment, Division2/Zone2.An additional construction feature of a motor used on an AFD may require is an upgraded insulation system on the motor windings. An unfiltered output signal from a drive can create harmonic voltage spikes in the motor, stressing the insulation of the motor windings.It is important to note that the features described pertain to motors which will be started and run on an AFD. If the drive is only used for starting the motor, these features may not be necessary. Consult with the motor manufacturer for application specific requirements.D. Primary Resistor or Reactor StartingThis method uses either a series resistor or reactor bank to be placed in the circuit with the motor. Resistor starting is more frequently used for smaller motors.When the motor is started, the resistor bank limits the flow of inrush current and provides for a voltage drop at the motor terminals. The resistors can be selected to provide voltage reductions up to 50%. As the motor comes up to speed, it develops a counter EMF (electro-magnetic field) that opposes the voltage applied to the motor. This further limits the inrush currents. As the inrush current diminishes, so does t>e voltage drop across the resistor bank allowing the torque generated by the motor to increase. At a predetermined time a device will short across the resistors and open the starting contactor effectively removing the resistor bank from the circuit. This provides for a closed transition and eliminates the concerns due to switchingtransients.Reactors will tend to oppose any sudden changes in current and therefore act to limit the current during starting. They will remain shorted after starting and provide a closed transition to line voltage.E .Star delta StartingThis approach started with the induction motor, the structure of each phase of the terminal are placed in the motor terminal box. This allows the motor star connection in the initial startup, and then re-connected into a triangle run. The initial start time when the voltage is reduced to the original star connection, the starting current and starting torque by 2 / 3. Depending on the application, the motor switch to the triangle in the rotational speed of between 50% and the maximum speed. Must be noted that the same problems, including the previously mentioned switch method, if the open circuit method, the transition may be a transient problem. This method is often used in less than 600V motor, the rated voltage 2.3kV and higher are not suitable for star delta motor start method.Ⅴ. INCREMENT TYPEThe first starting types that we have discussed have deal with the way the energy is applied to the motor. The next type deals with different ways the motor can be physically changed to deal with starting issues.Part WindingWith this method the stator of the motor is designed in such a way that it is made up of two separate windings. The most common method is known as the half winding method. As the name suggests, the stator is made up of two identical balanced windings. A special starter is configured so that full voltage can be applied to one half of the winding, and then after a short delay, to the second half. This method can reduce the starting current by 50 to 60%, but also the starting torque. One drawback to this method is that the motor heating on the first step of the operation is greater than that normally encountered on across-the-line start. Therefore the elapsed time on the first step of the part winding start should be minimized. This method also increases the magnetic noise of the motor during the first step.IV .ConclusionThere are many ways asynchronous motor starting, according to the constraints of power systems, equipment costs, load the boot device to select the best method. From the device point of view, was the first full-pressure launch the cheapest way, but it may increase the cost efficiency in the use of, or the power supply system in the region can not meet their needs. Effective way to alleviate the buck starts the power supply system, but at the expense of the cost of starting torque.These methods may also lead to increased motor sizes have led to produce the required load torque. Inverter can be eliminated by the above two shortcomings, but requires an additional increase in equipment costs. Understand the limitations of the application, and drives the starting torque and speed, allowing you for your application to determine the best overall configuration.英文资料翻译：异步电动机起动的方法摘要：大容量的交流异步电动机有多种启动方法。
双闭环直流调速系统外文翻译
对直流电机的速度闭环控制系统的设计钟国梁机械与汽车工程学院华南理工大学中国，广州5 1 0640电子邮件：机械与汽车工程学院华南理工大学中国，广州5 1 064 0江梁中电子邮件：该研究是由广州市科技攻关项目赞助(No. 2004A)。
(赞助信息) 摘要木文介绍了直流电机的速度控制原理，阐述了速度控制P I C16F 8 7 7单片机作为主控元件，利用捕捉模块的特点，比较模块和在PIC16F87 7单片机模数转换模块将触发电路，并给出了程序流程图。
系统具有许多优点，包括简单的结构，与主电路同步，稳定的移相和足够的移相范围，1 0 000步控制的角度，对电动机的无级平滑控制，陡脉冲前沿，足够的振幅值，设定脉冲宽度，良好的稳定性和抗干扰性，并且成本低廉，这速度控制具有很好的实用价值,系统可以容易地实现。
关键词：单片机，直流电机的速度控制，控制电路，PI控制算法lo简介电力电子技术的迅速发展使直流电机的转速控制逐步从模拟转向数字，目前，广泛采用晶闸管直流调速传动系统设备（如可控硅晶闸管，SCR）在电力拖动控制系统对电机供电已经取代了笨重的F -D发电机电动机系统，尤其是单片机技术的应用使速度控制直流电机技术进入一个新阶段。
在直流调速系统中，有许多各种各样的控制电路.单片机具有高性能，体积小，速度快，优点很多，价格便宜和可靠的稳定性,广泛的应用和强劲的流通，它可以提高控制能力和履行的要求实时控制（快速反应）。
控制电路采用模拟或数字电路可以实现单片机。
在本文中，我们将介绍一种基于单片机PIC16F8 7 7单片机的直流电机速度控制系统的分类。
2.直流电机的调速原理在图1中，电枢电压匕，电枢电流人，电枢回路总电阻心，电机常数C。
和励磁磁通①，根据KVL方程，电机的转速为叶U厂展60a7；伙）=7；伙一1）+ G胆伙）一a x e{k一1） = 7；伙一1） + 0・84e伙）一0・63亡伙一1）4 =切（1+壬）a严KpTf = T + T dIa图i直流电机调速原理图这里，卩是极对数，w是圈数，为电机电枢支路数，电机常数是5 工,这意味着当电机被证实，该值是固定的.但是当U a-I a R a , 60G只有绕组电阻心，所以人心非常小。