步進(jìn)電機(jī)和伺服電機(jī)的系統(tǒng)控制外文翻譯

1、<p>　　SELECTING THE MOTOR THAT SUITS YOUR APPLICATION</p><p>　　Motion control, in its widest sense, could relate to anything from a welding robot to the hydraulic system in a mobile crane. In the field

2、of Electronic Motion Control, we are primarily concerned with systems falling within a limited power range, typically up to about 10HP (7KW), and requiring precision in one or more aspects. This may involve accurate cont

3、rol of distance or speed, very often both and sometimes other parameters such as torque or acceleration rate. In the case of the two examples</p><p>　　Fig. 1 Elements of motion control system</p><

4、p>　　The motor，This may be a stepper motor (either rotary or linear), a DC brush motor or a brushless servo motor. The motor needs to be fitted with some kind of feedback device unless it is a stepper motor.</p>

5、<p>　　Fig. 2 shows a system complete with feedback to control motor speed. Such a system is known as a closed-loop velocity servo system.</p><p>　　Fig. 2 Typical closed loop (velocity) servo system<

6、/p><p>　　The drive，this is an electronic power amplifier that delivers the power to operate the motor in response to low-level control signals. In general, the drive will be specifically designed to operate wit

7、h a particular motor type – you can’t use a stepper drive to operate a DC brush motor, for instance.</p><p>　　Application Areas of Motor Types</p><p>　　Stepper Motors</p><p>　　Stepp

8、er Motor Benefits</p><p>　　Stepper motors have the following benefits:</p><p>　　? Low cost</p><p>　　? Ruggedness</p><p>　　? Simplicity in construction</p><p&

9、gt;　　? High reliability</p><p>　　? No maintenance</p><p>　　? Wide acceptance</p><p>　　? No tweaking to stabilize</p><p>　　? No feedback components are needed</p>

10、<p>　　? They work in just about any environment</p><p>　　? Inherently more failsafe than servo motors.</p><p>　　There is virtually no conceivable failure within the stepper drive module th

11、at could cause the motor to run away. Stepper motors are simple to drive and control in an open-loop configuration. They only require four leads. They provide excellent torque at low speeds, up to 5 times the continuous

12、torque of a brush motor of the same frame size or double the torque of the equivalent brushless motor. This often eliminates the need for a gearbox. A stepper-driven-system is inherently stiff, with known</p><

13、p>　　Stepper Motor Disadvantages</p><p>　　Stepper motors have the following disadvantages:</p><p>　　? Resonance effects and relatively long settling times</p><p>　　? Rough perform

14、ance at low speed unless a micro step drive is used</p><p>　　? Liability to undetected position loss as a result of operating open-loop</p><p>　　? They consume current regardless of load conditi

15、ons and therefore tend to run hot</p><p>　　? Losses at speed are relatively high and can cause excessive heating, and they are frequently noisy (especially at high speeds).</p><p>　　? They can e

16、xhibit lag-lead oscillation, which is difficult to damp. There is a limit to their available size, and positioning accuracy relies on the mechanics (e.g., ball screw accuracy). Many of these drawbacks can be overcome by

17、the use of a closed-loop control scheme. Note: The Comp motor Zeta Series minimizes or reduces many of these different stepper motor disadvantages. There are three main stepper motor types:</p><p>　　? Perman

18、ent Magnet (P.M.) Motors</p><p>　　? Variable Reluctance (V.R.) Motors</p><p>　　? Hybrid Motors</p><p>　　When the motor is driven in its full-step mode, energizing two windings or “p

19、hases” at a time (see Fig. 3), the torque available on each step will be the same (subject to very small variations in the motor and drive characteristics). In the half-step mode, we are alternately energizing two phases

20、 and then only one as shown in Fig. 4. Assuming the drive delivers the same winding current in each case, this will cause greater torque to be produced when there are two windings energized. In other wor</p><p

21、>　　Clearly, we would like to produce approximately equal torque on every step, and this torque should be at the level of the stronger step. We can achieve this by using a higher current level when there is only one wi

22、nding energized. This does not over dissipate the motor because the manufacturer’s current rating assumes two phases to be energized the current rating is based on the allowable case temperature). With only one phase ene

23、rgized, the same total power will be dissipated if the current is </p><p>　　Fig. 3 Full step current</p><p>　　Fig. 4 Half step current</p><p>　　Fig.5 Half step current, profiled<

24、/p><p>　　We have seen that energizing both phases with equal currents produces an intermediate step position half-way between the one-phase-one positions. If the two phase currents are unequal, the rotor positi

25、on will be shifted towards the stronger pole. This effect is utilized in the micro stepping drive, which subdivides the basic motor step by proportioning the current in the two windings. In this way, the step size is red

26、uced and the low-speed smoothness is dramatically improved. High-resolution mic</p><p>　　Fig. 6 Phase currents in micro step mode</p><p>　　Standard 200-Step Hybrid Motor</p><p>　　Th

27、e standard stepper motor operates in the same way as our simple model, but has a greater number of teeth on the rotor and stator, giving a smaller basic step size. The rotor is in two sections as before, but has 50 teeth

28、 on each section. The half-tooth displacement between the two sections is retained. The stator has 8 poles each with 5 teeth, making a total of 40 teeth (see Fig. 7).</p><p>　　Fig.7 200-step hybrid motor<

29、/p><p>　　If we imagine that a tooth is placed in each of the gaps between the stator poles, there would be a total of 48 teeth, two less than the number of rotor teeth. So if rotor and stator teeth are aligned

30、at 12 o’clock, they will also be aligned at 6 o’clock. At 3 o’clock and 9 o’clock the teeth will be misaligned. However, due to the displacement between the sets of rotor teeth, alignment will occur at 3 o’clock and 9 o’

31、clock at the other end of the rotor.</p><p>　　The windings are arranged in sets of four, and wound such that diametrically-opposite poles are the same. So referring to Fig. 7, the north poles at 12 and 6 o’c

32、lock attract the south-pole teeth at the front of the rotor; the south poles at 3 and 9 o’clock attract the north-pole teeth at the back. By switching current to the second set of coils, the stator field pattern rotates

33、through 45°. However, to align with this new field, the rotor only has to turn through 1.8°. This is equivalent to one </p><p>　　Note that there are as many detent positions as there are full steps

34、 per rev, normally 200. The detent positions correspond with rotor teeth being fully aligned with stator teeth. When power is applied to a stepper drive, it is usual for it to energize in the “zero phase” state in which

35、there is current in both sets of windings. The resulting rotor position does not correspond with a natural detent position, so an unloaded motor will always move by at least one half steps at power-on. Of course,</p&g

36、t;<p>　　Another point to remember is that for a given current pattern in the windings, there are as many stable positions as there are rotor teeth (50 for a 200-step motor). If a motor is de-synchronized, the resu

37、lting positional error will always be a whole number of rotor teeth or a multiple of 7.2°. A motor cannot “miss” individual steps – position errors of one or two steps must be due to noise, spurious step pulses or a

38、 controller fault.</p><p>　　Fig. 8 Digital servo drive</p><p>　　Digital Servo Drive Operation</p><p>　　Fig.8 shows the components of a digital drive for a servo motor. All the main

39、control functions are carried out by the microprocessor, which drives a D-to-A converter to produce an analog torque demand signal. From this point on, the drive is very much like an analog servo amplifier.</p>&l

40、t;p>　　Feedback information is derived from an encoder attached to the motor shaft. The encoder generates a pulse stream from which the processor can determine the distance traveled, and by calculating the pulse freque

41、ncy it is possible to measure velocity.</p><p>　　The digital drive performs the same operations as its analog counterpart, but does so by solving a series of equations. The microprocessor is programmed with

42、a mathematical model (or “algorithm”) of the equivalent analog system. This model predicts the behavior of the system. It also takes into account additional information like the output velocity, the rate of change of the

43、 input and the various tuning settings.</p><p>　　To solve all the equations takes a finite amount of time, even with a fast processor – this time is typically between 100ms and 2ms. During this time, the tor

44、que demand must remain constant at its previously-calculated value and there will be no response to a change at the input or output. This “update time” therefore becomes a critical factor in the performance of a digital

45、servo and in a high-performance system it must be kept to a minimum.</p><p>　　The tuning of a digital servo is performed either by pushbuttons or by sending numerical data from a computer or terminal. No pot

46、entiometer adjustments are involved. The tuning data is used to set various coefficients in the servo algorithm and hence determines the behavior of the system. Even if the tuning is carried out using pushbuttons, the fi

47、nal values can be uploaded to a terminal to allow easy repetition.</p><p>　　Some applications, the load inertia varies between wide limits – think of an arm robot that starts off unloaded and later carries a

48、 heavy load at full extension. The change in inertia may well be a factor of 20 or more, and such a change requires that the drive is re-tuned to maintain stable performance. This is simply achieved by sending the new tu

49、ning values at the appropriate point in the operating cycle.</p><p>　　步進(jìn)電機(jī)和伺服電機(jī)的系統(tǒng)控制</p><p>　　運(yùn)動控制，在其最廣泛的意義上說，可能與任何移動式起重機(jī)中焊接機(jī)器人液壓系統(tǒng)有關(guān)。在電子運(yùn)動控制領(lǐng)域，我們的主要關(guān)切系統(tǒng)范圍內(nèi)的有限功率的大小，通常高達(dá)約10hp（7千瓦），并要求在一個或多個方面有嚴(yán)格

50、精密。這可能涉及精確控制的距離或速度，但很多時候是雙方的，有時還涉及其它參數(shù)如轉(zhuǎn)矩或加速率。在以下所舉的兩個例子中，焊接機(jī)器人，需要精確的控制雙方的速度和距離;吊臂液壓系統(tǒng)采用驅(qū)動作為反饋系統(tǒng)，因此，它的準(zhǔn)確度會隨著操作者的技能的不同而不同。在嚴(yán)格意義上來說，這將不會被視為一項(xiàng)運(yùn)動控制系統(tǒng)。 我們的標(biāo)準(zhǔn)運(yùn)動控制系統(tǒng)由以下三個基本要素組成：</p><p>　　圖1運(yùn)動控制系統(tǒng)組成元件</p><

51、;p>　　電機(jī)，可能是一個步進(jìn)電機(jī)（要么旋轉(zhuǎn)或線性），也可能是直流無刷電機(jī)或無刷伺服馬達(dá)。電機(jī)必須配備一些種回饋裝置，除非它是一個步進(jìn)電機(jī)。</p><p>　　圖 2顯示了一個完善地反饋控制電機(jī)轉(zhuǎn)速的系統(tǒng)。這樣一個具有閉環(huán)控制系統(tǒng)的速度伺服系統(tǒng)。</p><p>　　圖2 典型的閉環(huán)（速度）伺服系統(tǒng)</p><p>　　驅(qū)動器是一個電子功率放大器，以提供電力

52、操作電動機(jī)來回應(yīng)低層次的控制信號。一般來說，驅(qū)動器將特別設(shè)計(jì)，其操作與特定電機(jī)類型相配合。例如，你不能用一個步進(jìn)驅(qū)動器來操作直流無刷電機(jī)。</p><p>　　不同電機(jī)適應(yīng)的不同領(lǐng)域</p><p><b>　　步進(jìn)電機(jī)：</b></p><p><b>　　步進(jìn)電機(jī)的好處。</b></p><p>

53、;　　（1）成本低廉（2）堅(jiān)固耐用（3）結(jié)構(gòu)簡單（4）高可靠性（5）無維修（6）適用廣泛（7）穩(wěn)定性很高（8）無需反饋元件（8）適應(yīng)多種工作環(huán)境（9）相對伺服電機(jī)更具有保險性。</p><p>　　因此，幾乎沒有任何可以想象的失敗使步進(jìn)驅(qū)動模塊出錯。步進(jìn)電機(jī)驅(qū)動簡單，并且驅(qū)動和控制在一個開放的閉環(huán)系統(tǒng)內(nèi)。他們只需要4個驅(qū)動器。低速時，驅(qū)動器提供良好的扭矩，是有刷電機(jī)同一幀大小5倍連續(xù)力矩，或相當(dāng)于無刷電機(jī)一倍扭矩

54、。這往往不再需要變速箱。步進(jìn)驅(qū)動系統(tǒng)遲緩，在限定的范圍內(nèi)，可以更好的減少動態(tài)位置誤差。</p><p><b>　　步進(jìn)電機(jī)弊端。</b></p><p>　　步進(jìn)電機(jī)有下列缺點(diǎn)：</p><p>　?。?）共振效應(yīng)和相對長的適應(yīng)性（2）在低速，表現(xiàn)粗糙，除非微驅(qū)動器來驅(qū)動（3）開環(huán)系統(tǒng)可能導(dǎo)致未被查覺的損失（4）由于過載，他們消耗過多電流。因

55、此傾向于過熱運(yùn)行。（5）虧損速度比較高，并可產(chǎn)生過多熱量因此，他們噪音很大（尤其是在高速下）。（6）他們的滯后現(xiàn)象導(dǎo)致振蕩，這是很難抑制的。對他們的可行性，這兒有一個限度，而他們的大小，定位精度主要依靠的是機(jī)器（例如，滾珠絲杠的精確度） 。許多這些缺點(diǎn)是可以克服的，通過使用一個閉環(huán)控制方案。</p><p>　　注：comp motor系列很多地減小或降低了這些不同的步進(jìn)電機(jī)不利之處。主要有3類步進(jìn)電機(jī)：（1）永

56、磁式步進(jìn)電機(jī) ，（2）可變磁阻式步進(jìn)電動機(jī)，（3）混合式步進(jìn)電機(jī)汽車。</p><p>　　當(dāng)電動機(jī)驅(qū)動，給兩個繞組通電時或"2相"通電的時候（見圖 3），扭矩可于每一個步將是相同（除極少數(shù)的變異和傳動特性）。在半步模式下，我們交替改變兩相電流，如圖4所示。假設(shè)該驅(qū)動器在每種情況下提供了相同的繞組電流，再通電時，這將導(dǎo)致更大的轉(zhuǎn)矩。換句話說，交替的步進(jìn)距將時強(qiáng)時若。對電動機(jī)表現(xiàn)來說，這并不代表

57、著一個重大的威懾。扭矩明顯受制于較弱的一步，低速平滑有一個顯著的改善。顯然，我們想在每一個步驟實(shí)現(xiàn)約相等扭矩對時，這扭矩應(yīng)該在水平較強(qiáng)的一步。們可以實(shí)現(xiàn)這個，當(dāng)只有一個繞組通電時，通過用高電流水平。這并不過度消耗電機(jī)，因?yàn)樵撾姍C(jī)的額定電流假定兩個階段被激活（目前的評級是基于許可的情況溫度） 。只有一相通電，如果目前是增加了40％的功率，同樣的總功率將會消散。利用這種更高的電流在一相中產(chǎn)生大致相等的扭矩，在交替的步進(jìn)距中。（見圖5 ）&l

58、t;/p><p><b>　　圖3半步電流</b></p><p><b>　　圖4 全步電流</b></p><p><b>　　圖5 側(cè)面全步電流</b></p><p>　　我們已經(jīng)看到，給兩相都通與相同的電流產(chǎn)生的一個中間的步進(jìn)，居于每一相的中間位置。如果兩相電流是不相等

59、的，轉(zhuǎn)子位置將轉(zhuǎn)向更強(qiáng)的一極。這種作用是利用細(xì)分驅(qū)動，其中細(xì)分的大小基于兩個繞組中的電流的大小。以這種方式，步長是減少了，而低速平滑度得到大幅度提高。高細(xì)分驅(qū)動電動機(jī)細(xì)分整步步進(jìn)到多達(dá)500個細(xì)分步，轉(zhuǎn)一圈可細(xì)分十萬步。在這種情況下，繞組中的電流極為相似的兩個正弦波有90°相移。（圖1.11）電機(jī)被驅(qū)動好像轉(zhuǎn)換成了交流同步電機(jī)。事實(shí)上，步進(jìn)電機(jī)可被驅(qū)動，從60赫茲美（50赫茲-歐洲）正弦波源頭起，包括電容器系列的一相。它將旋轉(zhuǎn)

60、72轉(zhuǎn)。</p><p>　　圖6步進(jìn)電機(jī)的相電流</p><p>　　標(biāo)準(zhǔn)200步混合電機(jī)</p><p>　　標(biāo)準(zhǔn)步進(jìn)電機(jī)運(yùn)行在同就如同我們的簡單模式，但有一個更大的數(shù)目齒數(shù)在轉(zhuǎn)子和定子中，從而有了一個較小的基本步長。轉(zhuǎn)子有2部分，但每部分有50個齒。該齒輪位于兩部分之間。定子每5個齒有8個極，完整的共有40個齒（見圖1.12）</p><p

61、>　　圖7 200步混合標(biāo)準(zhǔn)電機(jī)</p><p>　　如果我們想象一個齒，是擺在2個定子極點(diǎn)每一齒隙中，假設(shè)定子共有48個齒，少于轉(zhuǎn)子齒數(shù)兩個。因此，如果轉(zhuǎn)子和定子的齒排列一整圈，他們同樣也可以排列半圈。1/4和3/4圈也同樣可以排列。然而，由于轉(zhuǎn)子的齒排列位置，在另一端的轉(zhuǎn)子，排列將發(fā)生在1/4和3/4位置處。繞組4個一組，并對角線方向的極性相反。如圖7所示，北極在轉(zhuǎn)子前面的12點(diǎn)和6點(diǎn)位置，吸引著在在背

62、面3時和9時的南極。通過開關(guān)第二組線圈的電流，定子模式旋轉(zhuǎn)45。不過，要配合這個新的領(lǐng)域，轉(zhuǎn)子只轉(zhuǎn)過1.8°。相當(dāng)于轉(zhuǎn)子，這只轉(zhuǎn)過了四分之一的齒間距，每一次旋轉(zhuǎn)要200個全步。</p><p>　　注意到，每一次旋轉(zhuǎn)全部時這兒有很多定位點(diǎn)位置，通常是200個。該定位點(diǎn)的位置與轉(zhuǎn)子的齒能全面接軌定子齒時相對應(yīng)。的當(dāng)通電給步進(jìn)驅(qū)動器時，它通常是零狀態(tài)時最活躍，也就是兩套繞組都通電。因此產(chǎn)生的轉(zhuǎn)子位置并不符合

63、轉(zhuǎn)子自然定位點(diǎn)的位置。因此，空載時，一旦通電電機(jī)將至少步進(jìn)半步。當(dāng)然，如果系統(tǒng)關(guān)機(jī)，或在零相位位置，電機(jī)一旦通電將步進(jìn)一大步。</p><p>　　另一點(diǎn)要注意的是，對于一個給定電流的繞組，有很多穩(wěn)定的位置，正如轉(zhuǎn)子齒（200步進(jìn)電機(jī)有50個齒）。如果電機(jī)是同步電機(jī)，導(dǎo)致位置誤差將永遠(yuǎn)是一個整體倍轉(zhuǎn)子齒或能被7.2°整除。電機(jī)不能"細(xì)分"，如個別一個或兩個位置誤差，是由于噪聲，錯誤脈

64、沖或控制器故障造成的。</p><p><b>　　圖8數(shù)字伺服驅(qū)動</b></p><p>　　圖8顯示為伺服電機(jī)的數(shù)控驅(qū)動。所有的主控制功能是微處理器，驅(qū)動為DA模擬轉(zhuǎn)換器，以產(chǎn)生一個模擬扭矩需求信號。從這個角度上，這臺機(jī)器非常很像一個模擬伺服放大器。反饋的信息是來自隸屬該電機(jī)軸的一個編碼器。編碼器生成脈沖流可確定傳輸路程，并通過計(jì)算脈沖頻率，是可以測定轉(zhuǎn)速的。&

65、lt;/p><p>　　數(shù)碼驅(qū)動通過求解一系列的方程式，履行同樣類似的功能。微處理器是與數(shù)學(xué)模型（或“算法"）的等效的編程模擬系統(tǒng)。這模型預(yù)測系統(tǒng)的行為。它響應(yīng)一個給定輸入的信號并產(chǎn)生速度。它同樣也考慮到額外信息如輸出速度，速率轉(zhuǎn)變中的投入和各種調(diào)校設(shè)定。</p><p>　　解決所有方程需數(shù)額需有限的時間，即使是一個快速的處理器一次處理通常也是100ms和2ms之間。在此之間，在改

66、變輸入或輸出，先前的計(jì)算值將有沒有回應(yīng)時，扭矩要求必須保持恒定。因此更新時間成為數(shù)字伺服和一臺高性能系統(tǒng)關(guān)鍵的因素，它必須保持及時更新。</p><p>　　調(diào)試數(shù)字伺服電機(jī)可按鈕或從一個計(jì)算機(jī)或終端調(diào)試。電位器調(diào)整是涉及的。調(diào)試數(shù)據(jù)是設(shè)置在伺服算法的各種系數(shù)，因此，它決定了系統(tǒng)的性能。即使如果調(diào)諧進(jìn)行使用按鈕，終值也可以上傳到終端，讓其進(jìn)行簡單的重復(fù)。</p><p>　　在某些應(yīng)用中，