外文翻譯雙閉環(huán)直流調(diào)速系統(tǒng)的說明

1、double loop dc speed control system description工 學(xué) 部 專 業(yè)班 級學(xué) 號姓 名指導(dǎo)教師負(fù)責(zé)教師2010年6月double loop dc speed control system description.system analysis and synthesis1analysis （1）in the speed and current dual closed-loop speed control system, in order to change the motor speed, what parameters should be regu

2、lating? 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 motors stall current system should adjust the parameters of what?a: to change the motor speed, change speed

3、 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 motors stall current, only need to change the same value given uim * and fee

4、dback 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 spee

5、d 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 wi

6、th 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 t

7、o 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

8、) 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

9、 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 phenom

10、enon, long-running will damage the system.2. system speed regulator and current regulator in the double loop dc motor control system can be summarized as follows: (1). the role of speed regulator speed regulator is a speed control system of the dominant regulator, which allows speed n will soon chan

11、ge 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 cur

12、rent regulator 1) 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 sp

13、eed 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 auto

14、matically return to normal. yesterday, the role of the reliable operation is very important.double-loop speed control system common faults analysis 1.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

15、 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 transformation components sb. in the

16、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

17、 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 condition 1)

18、 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-

19、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 min, current loop f

20、or 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 cau

21、sed 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 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 giv

22、en 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 inc

23、onsistency. 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,

24、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

25、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. t

26、o 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 (ch

27、ange 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 wheth

28、er 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

29、, 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 hi

30、gh stressed 1) 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 l

31、ow-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 = 90 , at

32、 this time to zero output voltage rectification system, the electrical will not crawl. (7) with a given system can not run 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, currentregulator, su

33、ch 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 prev

34、ious 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 adju

35、st 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 oscillation 1) open-loop system in the state (the main circuit disconnect) the oscillation can be changed at this time give

36、n 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 replace

37、d 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. oscill

38、ations are often caused as a result of operational amplifiers, such as damage to electronic components , system parameters caused by improper, according to the specific circumstances, properly addressed.雙閉環(huán)直流調(diào)速系統(tǒng)的說明一、系統(tǒng)分析與綜合1.系統(tǒng)分析（1）在轉(zhuǎn)速、電流雙閉環(huán)調(diào)速系統(tǒng)中，若要改變電動(dòng)機(jī)的轉(zhuǎn)速，應(yīng)調(diào)節(jié)什么參數(shù)？改變轉(zhuǎn)速調(diào)節(jié)器的放大倍數(shù)kn行不行？改變電力電子變換器的放大系數(shù)k

39、s行不行？改變轉(zhuǎn)速反饋系數(shù)行不行？若要改變電動(dòng)機(jī)的堵轉(zhuǎn)電流，應(yīng)調(diào)節(jié)系統(tǒng)中的什么參數(shù)？答：若要改變電動(dòng)機(jī)的轉(zhuǎn)速，改變轉(zhuǎn)速調(diào)節(jié)器的放大倍數(shù)kn和電力電子變換器的放大系數(shù)ks都不行，穩(wěn)定時(shí)n=un=un*,所以只有改變給定值un*和反饋系數(shù)才行。若要改變電動(dòng)機(jī)的堵轉(zhuǎn)電流，同樣只須改變給定值uim*和反饋系數(shù)，因?yàn)?，穩(wěn)定時(shí)，uim* =idm,從式中可得出。（2）轉(zhuǎn)速、電流雙閉環(huán)調(diào)速系統(tǒng)穩(wěn)態(tài)運(yùn)行時(shí)，兩個(gè)調(diào)節(jié)器的輸入偏差電壓和輸出電壓各是多少？答：轉(zhuǎn)速、電流雙閉環(huán)調(diào)速系統(tǒng)穩(wěn)態(tài)運(yùn)行時(shí)，兩個(gè)調(diào)節(jié)器的輸入偏差電壓均是零，由式子n=un=un*，n=n; uim* =idm, idm=idl。（3）在轉(zhuǎn)速、電流

40、雙閉環(huán)調(diào)速系統(tǒng)中，兩個(gè)調(diào)節(jié)器均采用pi調(diào)節(jié)器。當(dāng)系統(tǒng)帶額定負(fù)載運(yùn)行時(shí)，轉(zhuǎn)速反饋線突然斷線，系統(tǒng)重新進(jìn)入穩(wěn)態(tài)后，電流調(diào)節(jié)器的輸入偏差電壓是否為零？為什么？答：當(dāng)系統(tǒng)帶額定負(fù)載運(yùn)行時(shí)，轉(zhuǎn)速反饋線突然斷線，則un=0，=un*-un=un*,使ui迅速達(dá)到uim ，0 ，速度 n 上升，當(dāng)系統(tǒng)重新進(jìn)入穩(wěn)態(tài)后，即id=idl ，那么，= uim*-idl0，也不再變化，轉(zhuǎn)速n也不再變化，但，此時(shí)的轉(zhuǎn)速n比反饋線斷線時(shí)的轉(zhuǎn)速要大。（4）為什么用積分控制的調(diào)速系統(tǒng)是無靜差的？答：在積分調(diào)節(jié)器的調(diào)速系統(tǒng)中，能實(shí)現(xiàn)無靜差，是由于積分調(diào)節(jié)器控制特點(diǎn)，即積分的記憶和積累作用。（5）雙環(huán)調(diào)速系統(tǒng)（pi），負(fù)載變化，

41、idlidm,問雙環(huán)調(diào)速系統(tǒng)acr和asr怎么調(diào)節(jié)，結(jié)果如何？答：當(dāng)負(fù)載變化時(shí)，idlidm,轉(zhuǎn)速迅速下降，電流id 很快增加到idm，而達(dá)限幅值，速度asr迅速飽和，acr一直在限流狀態(tài)下，形成堵轉(zhuǎn)現(xiàn)象，長時(shí)間運(yùn)行會(huì)損壞系統(tǒng)。2.系統(tǒng)綜合轉(zhuǎn)速調(diào)節(jié)器和電流調(diào)節(jié)器在雙閉環(huán)直流調(diào)速系統(tǒng)中的作用可歸納如下：（1）.轉(zhuǎn)速調(diào)節(jié)器的作用轉(zhuǎn)速調(diào)節(jié)器是調(diào)速系統(tǒng)的主導(dǎo)調(diào)節(jié)器，它使轉(zhuǎn)速n很快隨給定電壓變化un*變化，穩(wěn)態(tài)時(shí)可減小轉(zhuǎn)速誤差，如果采用pi調(diào)節(jié)器，則可實(shí)現(xiàn)無靜差。1) 對負(fù)載變化起抗擾作用。 2) 其輸出限幅值決定電動(dòng)機(jī)允許的最大電流。（2）電流調(diào)節(jié)器的作用1）作為內(nèi)環(huán)的調(diào)節(jié)器，在轉(zhuǎn)速外環(huán)的調(diào)節(jié)過程中，

42、它的作用使電流緊緊跟隨其給定電壓ui*（即外環(huán)調(diào)節(jié)器的輸出量）變化。2）對電網(wǎng)電壓的波動(dòng)起及時(shí)抗擾的作用。3）在轉(zhuǎn)速動(dòng)態(tài)過程中，保證獲得電動(dòng)機(jī)允許的最大電流，從而加快動(dòng)態(tài)過程。4）當(dāng)電動(dòng)機(jī)過載甚至堵轉(zhuǎn)時(shí)，限制電樞電流的最大值，起加速的自動(dòng)保護(hù)作用。一旦故障消失，系統(tǒng)立即自動(dòng)恢復(fù)正常。這個(gè)作用對昨天的可靠運(yùn)行來說是十分重要的。二、雙閉環(huán)調(diào)速系統(tǒng)常見故障分析1系統(tǒng)工作原理簡介（1）.雙閉環(huán)調(diào)速系統(tǒng)的構(gòu)成如圖1.1電流內(nèi)環(huán):由電流調(diào)節(jié)器、觸發(fā)器(其輸入變換為)、可控硅整流橋、電動(dòng)機(jī)電樞回路及電流變換組成.速度外環(huán):由速度調(diào)節(jié)器、電流環(huán)等慣性環(huán)節(jié)、電動(dòng)機(jī)及負(fù)載的轉(zhuǎn)動(dòng)慣量及速度變換組成.在雙閉環(huán)調(diào)速系統(tǒng)

43、中,速度外環(huán)決定整個(gè)系統(tǒng)的運(yùn)行特征及穩(wěn)定工作狀態(tài),起主導(dǎo)作用,而電流內(nèi)環(huán)起著改變系統(tǒng)內(nèi)部結(jié)構(gòu)的作用,是從屬的,但由于它將作為一個(gè)整體參與到速度閉環(huán)中去,直接影響速度閉環(huán)的工作,因此必須要先調(diào)試好電流內(nèi)環(huán),再調(diào)試速度外環(huán),這樣才能使整個(gè)系統(tǒng)有較好的動(dòng)態(tài)性能。（2）.雙閉環(huán)調(diào)速系統(tǒng)的典型工作狀態(tài)1）起動(dòng)(或升速):在起動(dòng)過程中一直是飽和,這樣相當(dāng)于使速度環(huán)處于開環(huán)狀態(tài).系統(tǒng)只在電流環(huán)的恒值調(diào)節(jié)作用之下,保證電動(dòng)機(jī)在恒定的最大允許電流下起動(dòng)。圖1.1 雙閉環(huán)調(diào)速系統(tǒng)結(jié)構(gòu)圖2）減速(或停車):此時(shí)的輸出迅速達(dá)到反向限幅值.主回路電流經(jīng)本橋逆變后很 降到零.和輸出很快達(dá)到反向最大值.輸出脈沖迅速達(dá)到,電流環(huán)也為開環(huán).電動(dòng)機(jī)在阻力矩作用下減速,直到電動(dòng)機(jī)轉(zhuǎn)速降至接近新的給定值,電流環(huán)和速度