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1、Double Loop DC Speed Control System Description工 学 部 专 业班 级学 号姓 名指导教师负责教师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.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K
39、s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=n; Uim* =Idm, Idm=Idl。（3）在转速、电流
40、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*,使Ui迅速达到Uim ，0 ，速度 n 上升，当系统重新进入稳态后，即Id=Idl ，那么，= Uim*-Idl0，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，
41、IdlIdm,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，IdlIdm,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。2.系统综合转速调节器和电流调节器在双闭环直流调速系统中的作用可归纳如下：（1）.转速调节器的作用转速调节器是调速系统的主导调节器，它使转速n很快随给定电压变化Un*变化，稳态时可减小转速误差，如果采用PI调节器，则可实现无静差。1) 对负载变化起抗扰作用。 2) 其输出限幅值决定电动机允许的最大电流。（2）电流调节器的作用1）作为内环的调节器，在转速外环的调节过程中，
42、它的作用使电流紧紧跟随其给定电压Ui*（即外环调节器的输出量）变化。2）对电网电压的波动起及时抗扰的作用。3）在转速动态过程中，保证获得电动机允许的最大电流，从而加快动态过程。4）当电动机过载甚至堵转时，限制电枢电流的最大值，起加速的自动保护作用。一旦故障消失，系统立即自动恢复正常。这个作用对昨天的可靠运行来说是十分重要的。二、双闭环调速系统常见故障分析1系统工作原理简介（1）.双闭环调速系统的构成如图1.1电流内环:由电流调节器、触发器(其输入变换为)、可控硅整流桥、电动机电枢回路及电流变换组成.速度外环:由速度调节器、电流环等惯性环节、电动机及负载的转动惯量及速度变换组成.在双闭环调速系统
43、中,速度外环决定整个系统的运行特征及稳定工作状态,起主导作用,而电流内环起着改变系统内部结构的作用,是从属的,但由于它将作为一个整体参与到速度闭环中去,直接影响速度闭环的工作,因此必须要先调试好电流内环,再调试速度外环,这样才能使整个系统有较好的动态性能。（2）.双闭环调速系统的典型工作状态1）起动(或升速):在起动过程中一直是饱和,这样相当于使速度环处于开环状态.系统只在电流环的恒值调节作用之下,保证电动机在恒定的最大允许电流下起动。图1.1 双闭环调速系统结构图2）减速(或停车):此时的输出迅速达到反向限幅值.主回路电流经本桥逆变后很 降到零.和输出很快达到反向最大值.输出脉冲迅速达到,电
44、流环也为开环.电动机在阻力矩作用下减速,直到电动机转速降至接近新的给定值,电流环和速度环相继投入闭环工作,电机在新的给定值下运行.3)电网电压波动:此时由于电动机的转动惯量较大,因而最先引起电枢电流的变化,的输出还没有变化,在电流环的作用下,的输出迅速变化,使角迅速得到调整,因此对转速影响很小.4)负载变化:在运行中,负载发生变化时,会造成电动机转速偏离给定值.转速的恢复过程与前述升速(或减速)的过程相似。2.常见故障的分析与处理（1） 电源电压正常,可控硅整流桥输出波形不齐造成这种现象的原因是由于触发器锯齿波斜率不一致而引起的.适当调节锯齿波斜率电位器,便可达到输出波形整齐.在调整过程中要兼
45、顾齐,这一点在实际调试中应注意。（2）直流电动机机械特性变软分析晶闸管直流电动机系统,当电流断续时其机械特性的第一个特点是理想空载转速高,第二个特点是机械特性软2.所谓机械特性变软,就是负载很小的变化就会引起转速很大的变化.这时用示波器观察整流桥的输出波形,便可发现缺相.此时需要检查各触发器是否都有脉冲输出、快熔是否熔断、晶闸管是否击穿或断路、同步变压器是否损坏、电源是否缺相.找出问题所在,加以解决即可。（3）速度不稳造成速度不稳的因素较多:1)测速电机固定不牢固或与主机不同轴.2)速度调节器的参数不合适.应对动态参数进行调整(改变比例、积分参数).3)调速系统的某个环节有虚焊(或接触不良).
46、4)可能由于干扰引起.应查出干扰原因,采取抗干扰措施。（4）稍加给定,电机转速即超过额定值应首先检查外部控制系统是否正常,如外控系统正常,则可能是给定积分、速度调节器、电流调节器等环节出现故障造成.应切断主回路电源,只将控制系统给电,在不加给定的情况下,测试每个关键点(如电流调节器、电压调节器等)的电位.然后再加上给定,从前至后逐个检查每个关键点的电位变化情况,即可找出故障所在。（5）整流桥输出电压调不高1)速度调节器、电流调节器限幅偏小,应根据情况适当放开限幅值.2)速度反馈信号过强,可适当减小速度反馈信号。（6）系统在无给定时仍低速运转(即出现爬行现象)这是由于系统出现“零点漂移”造成的.
47、当输入信号为零时,输出电压是由输入放大环节中的偏移电位器决定的,可以通过调节偏移电位器使=90,此时整流系统输出电压为零,电机将不再爬行。（7）加上给定系统仍不能运行应首先检查外部控制系统是否正常,如外控系统正常,则可能是给定积分、速度调节器、电流调节器等环节出现故障造成.应切断主回路电源,只将控制系统给电,在不加给定的情况下,测试每个关键点(如电流调节器、电压调节器等)的电位.然后再加上给定,从前至后逐个检查每个关键点的电位变化情况,即可找出故障所在。（8） 控制精度不够在以分配器为给定的外部控制运行时,往往要求有足够的停车精度,才能正常工作.如果停车精度差,可以适当调整速度调节器的环节,一般通过减小比例部分,加大积分部分即可得到满意的效果。（9）可逆系统出现振荡现象1)系统在开环状态时(主回路断开)有振荡现象,此时可通过改变给定,从前至后检查各关键点电位的变化情况.如果给定不变,而某点电位仍在变化,此处即为症结所在.2)系统在闭环状态时出现振荡现象,这种情况下为了保证安全,一般应将电机负载改为电阻负载,如果没有合适的电阻箱,可用两个电炉子串联代替.检查方法和开环情况相似,着重检查的环节是:给定积分、电平检测、运转控制、速度调节器等.造成振荡的原因往往是由于运算放大器等电子元件损坏(或虚焊)、系统参数不当等原因引起,可根据具体情况,妥善解决。
