電動(dòng)機(jī)無級(jí)調(diào)速畢業(yè)設(shè)計(jì)--外文翻譯資料

1、<p>　　Ⅰ.system analysis and synthesis</p><p>　　1．Analysis </p><p>　?。?）In the speed and current dual closed-loop speed control system, in order to change the motor speed, what parameters s

2、hould 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 st

3、all current system should adjust the parameters of what?</p><p>　　A: To change the motor speed, change speed regulator Kn magnification and power electronic converters will not work magnification factor Ks,

4、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 coeffici

5、ent, because the stability, Uim * = Idm, can be drawn from the type</p><p>　　(2) Speed, the current double closed-loop speed control system when the steady-state operation, the two regulator input voltage an

6、d 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

7、 *, 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</p><p>　　2. System Speed regulator and current regulato

8、r 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 change wit

9、h 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 ampl

10、itude of the decis</p><p>　　Ⅱdouble-loop speed control system common faults analysis 1.Introduction of a system (1). Double-loop speed control system components in Figure 1. Current loop: from the current

11、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

12、, such as link inertia, motor and load moment of inertia and the speed of transformation components S</p><p>　　Figure 1 double-loop speed control system structure</p><p>　　2) slow down (or stop)

13、: 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 r

14、each 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 o

15、f a given run .</p><p>　　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 curren

16、t loop, LT rapidly changing the output so that angle α be adjusted quickly, so the impact on the speed .</p><p>　　4）Small changes in load: in the operation, load changes, will cause the motor speed deviation

17、 from the given value. speed up the recovery process and the aforementioned speed (or deceleration) is similar to the process.</p><p>　　2. Common Fault Analysis and Processing (1) The normal supply voltage,

18、 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 t

19、he 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 cu</p>

20、<p>　　1) Electric guns are not firmly fixed or with the host of different axis .</p><p>　　2) The parameters of the speed regulator inappropriate. Respond to the dynamic parameters to adjust (change the

21、 ratio of integral parameters ) .</p><p>　　3) Of a speed control system there are(or bad) </p><p>　　4) May be caused as a result of interference. should be found to interfere with the reasons fo

22、r 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

23、 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 give</p><p><b>　　一、

24、系統(tǒng)分析與綜合</b></p><p><b>　　1.系統(tǒng)分析</b></p><p>　　（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)Ks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)？</p><p&g

25、t;　　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)椋€(wěn)定時(shí)，Uim* =Idm,從式中可得出。</p><p>　?。?）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？</p><

26、p>　　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。</p><p>　　（3）在轉(zhuǎn)速、電流雙閉環(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？</p><p>　　答：當(dāng)系統(tǒng)帶額定負(fù)載

27、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。</p><p>　?。?）為什么用積分控制的調(diào)速系統(tǒng)是無靜差的？</p><p>　　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

28、分調(diào)節(jié)器控制特點(diǎn)，即積分的記憶和積累作用。</p><p>　?。?）雙環(huán)調(diào)速系統(tǒng)（PI），負(fù)載變化，Idl>Idm,問雙環(huán)調(diào)速系統(tǒng)ACR和ASR怎么調(diào)節(jié)，結(jié)果如何？</p><p>　　答：當(dāng)負(fù)載變化時(shí)，Idl>Idm,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(zhǎng)時(shí)間運(yùn)行會(huì)損壞系統(tǒng)。</p>

29、<p><b>　　2.系統(tǒng)綜合</b></p><p>　　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：</p><p>　?。?）.轉(zhuǎn)速調(diào)節(jié)器的作用</p><p>　　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。

30、 </p><p>　　1) 對(duì)負(fù)載變化起抗擾作用。</p><p>　　2) 其輸出限幅值決定電動(dòng)機(jī)允許的最大電流。</p><p>　?。?）電流調(diào)節(jié)器的作用</p><p>　　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，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。</p>&

31、lt;p>　　2）對(duì)電網(wǎng)電壓的波動(dòng)起及時(shí)抗擾的作用。</p><p>　　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。</p><p>　　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(duì)昨天的可靠運(yùn)行來說是十分重要的。</p><p>　　二、雙閉環(huán)調(diào)速系統(tǒng)常見故

32、障分析</p><p><b>　　1系統(tǒng)工作原理簡(jiǎn)介</b></p><p>　?。?）.雙閉環(huán)調(diào)速系統(tǒng)的構(gòu)成如圖1.電流內(nèi)環(huán):由電流調(diào)節(jié)器ＬＴ、觸發(fā)器ＣＦ(其輸入變換為ＣＳＲ)、可控硅整流橋、電動(dòng)機(jī)電樞回路及電流變換ＬＢ組成.速度外環(huán):由速度調(diào)節(jié)器ＳＴ、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ＳＢ組成.在雙閉環(huán)調(diào)速系統(tǒng)中,速度外環(huán)決定整個(gè)系統(tǒng)的運(yùn)行特征及

33、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。</p><p>　?。?）.雙閉環(huán)調(diào)速系統(tǒng)的典型工作狀態(tài)</p><p>　　1）起動(dòng)(或升速):</p><p>　　在起動(dòng)過程中ＳＴ一直是飽和

34、,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)。</p><p>　　圖1 雙閉環(huán)調(diào)速系統(tǒng)結(jié)構(gòu)圖</p><p>　　2）減速(或停車):</p><p>　　此時(shí)ＳＴ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.ＬＴ和ＣＳＲ輸出很快達(dá)到反向最大值.ＣＦ輸出脈沖迅速達(dá)到βｍｉｎ,電流環(huán)也為開環(huán).

35、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)和速度環(huán)相繼投入閉環(huán)工作,電機(jī)在新的給定值下運(yùn)行.3)電網(wǎng)電壓波動(dòng):此時(shí)由于電動(dòng)機(jī)的轉(zhuǎn)動(dòng)慣量較大,因而最先引起電樞電流的變化,ＳＴ的輸出還沒有變化,在電流環(huán)的作用下,ＬＴ的輸出迅速變化,使α角迅速得到調(diào)整,因此對(duì)轉(zhuǎn)速影響很小.4)負(fù)載變化:在運(yùn)行中,負(fù)載發(fā)生變化時(shí),會(huì)造成電動(dòng)機(jī)轉(zhuǎn)速偏離給定值.轉(zhuǎn)速的恢復(fù)過程與前述升速(或減速)的過程相似。</p><p

36、>　　2.常見故障的分析與處理</p><p>　　（1） 電源電壓正常,可控硅整流橋輸出波形不齊造成這種現(xiàn)象的原因是由于觸發(fā)器鋸齒波斜率不一致而引起的.適當(dāng)調(diào)節(jié)鋸齒波斜率電位器,便可達(dá)到輸出波形整齊.在調(diào)整過程中要兼顧齊,這一點(diǎn)在實(shí)際調(diào)試中應(yīng)注意。</p><p>　?。?） 直流電動(dòng)機(jī)機(jī)械特性變軟分析晶閘管直流電動(dòng)機(jī)系統(tǒng),當(dāng)電流斷續(xù)時(shí)其機(jī)械特性的第一個(gè)特點(diǎn)是理想空載轉(zhuǎn)速高,第二個(gè)

37、特點(diǎn)是機(jī)械特性軟〔2〕.所謂機(jī)械特性變軟,就是負(fù)載很小的變化就會(huì)引起轉(zhuǎn)速很大的變化.這時(shí)用示波器觀察整流橋的輸出波形,便可發(fā)現(xiàn)缺相.此時(shí)需要檢查各觸發(fā)器是否都有脈沖輸出、快熔是否熔斷、晶閘管是否擊穿或斷路、同步變壓器是否損壞、電源是否缺相.找出問題所在,加以解決即可。</p><p>　?。?）速度不穩(wěn)造成速度不穩(wěn)的因素較多:</p><p>　　1)測(cè)速電機(jī)固定不牢固或與主機(jī)不同軸.&l

38、t;/p><p>　　2)速度調(diào)節(jié)器的參數(shù)不合適.應(yīng)對(duì)動(dòng)態(tài)參數(shù)進(jìn)行調(diào)整(改變比例、積分參數(shù)).</p><p>　　3)調(diào)速系統(tǒng)的某個(gè)環(huán)節(jié)有虛焊(或接觸不良).</p><p>　　4)可能由于干擾引起.應(yīng)查出干擾原因,采取抗干擾措施。</p><p>　?。?）稍加給定,電機(jī)轉(zhuǎn)速即超過額定值應(yīng)首先檢查外部控制系統(tǒng)是否正常,如外控系統(tǒng)正常,則可能

39、是給定積分、速度調(diào)節(jié)器、電流調(diào)節(jié)器等環(huán)節(jié)出現(xiàn)故障造成.應(yīng)切斷主回路電源,只將控制系統(tǒng)給電,在不加給定的情況下,測(cè)試每個(gè)關(guān)鍵點(diǎn)(如電流調(diào)節(jié)器、電壓調(diào)節(jié)器等)的電位.然后再加上給定,從前至后逐個(gè)檢查每個(gè)關(guān)鍵點(diǎn)的電位變化情況,即可找出故障所在。</p><p>　　（5）整流橋輸出電壓調(diào)不高1)速度調(diào)節(jié)器、電流調(diào)節(jié)器限幅偏小,應(yīng)根據(jù)情況適當(dāng)放開限幅值.2)速度反饋信號(hào)過強(qiáng),可適當(dāng)減小速度反饋信號(hào)。</p>

40、<p>　?。?）系統(tǒng)在無給定時(shí)仍低速運(yùn)轉(zhuǎn)(即出現(xiàn)爬行現(xiàn)象)這是由于系統(tǒng)出現(xiàn)“零點(diǎn)漂移”造成的.當(dāng)輸入信號(hào)為零時(shí),輸出電壓是由輸入放大環(huán)節(jié)中的偏移電位器決定的,可以通過調(diào)節(jié)偏移電位器使α=90°,此時(shí)整流系統(tǒng)輸出電壓為零,電機(jī)將不再爬行。</p><p>　　（7）加上給定系統(tǒng)仍不能運(yùn)行</p><p>　　應(yīng)首先檢查外部控制系統(tǒng)是否正常,如外控系統(tǒng)正常,則可能是給定

41、積分、速度調(diào)節(jié)器、電流調(diào)節(jié)器等環(huán)節(jié)出現(xiàn)故障造成.應(yīng)切斷主回路電源,只將控制系統(tǒng)給電,在不加給定的情況下,測(cè)試每個(gè)關(guān)鍵點(diǎn)(如電流調(diào)節(jié)器、電壓調(diào)節(jié)器等)的電位.然后再加上給定,從前至后逐個(gè)檢查每個(gè)關(guān)鍵點(diǎn)的電位變化情況,即可找出故障所在。</p><p>　　（8） 控制精度不夠在以分配器為給定的外部控制運(yùn)行時(shí),往往要求有足夠的停車精度,才能正常工作.如果停車精度差,可以適當(dāng)調(diào)整速度調(diào)節(jié)器的ＰＩ環(huán)節(jié),一般通過減小比例部

42、分,加大積分部分即可得到滿意的效果。</p><p>　　（9）可逆系統(tǒng)出現(xiàn)振蕩現(xiàn)象1)系統(tǒng)在開環(huán)狀態(tài)時(shí)(主回路斷開)有振蕩現(xiàn)象,此時(shí)可通過改變給定,從前至后檢查各關(guān)鍵點(diǎn)電位的變化情況.如果給定不變,而某點(diǎn)電位仍在變化,此處即為癥結(jié)所在.2)系統(tǒng)在閉環(huán)狀態(tài)時(shí)出現(xiàn)振蕩現(xiàn)象,這種情況下為了保證安全,一般應(yīng)將電機(jī)負(fù)載改為電阻負(fù)載,如果沒有合適的電阻箱,可用兩個(gè)電爐子串聯(lián)代替.檢查方法和開環(huán)情況相似,著重檢查的環(huán)節(jié)是:給