plc外文翻譯--根據(jù)需要選擇 plc

1、<p><b>　　附錄A：英文資料</b></p><p>　　Selecting a PLC for the application</p><p>　　Questions answered</p><p>　　1\ how do we understand and estimate the requirement for a P

2、LC?</p><p>　　2\ how do we select the I\O hardware?</p><p>　　3\ how are I\O circuits configured?</p><p>　　4\ how do we size the processor and memory for our application?</p>&

3、lt;p>　　5\ how do we select a supplier?</p><p>　　The choice of a PLC for a particular application can be bewildering. The range of suppliers is vast, many offering a number alternative product ranges, with

4、 any number of modules to perform choice must meet the job and customers' requirements, provide extra capacity to enable future modifications and provide an acceptable cost solution. </p><p>　　We have t

5、o make choices balancing the cost of extra, or more expensive hardware against the time required to program algorithms that allow us to use cheaper hardware to meet the system requirements, each case has to be considered

6、 on its merits. Beware of the common trap of underestimating the time taken to write such code!</p><p>　　9.1 estimating requirements</p><p>　　The starting point in determining any solution must

7、be to understand what is to be achieved. In an ideal world our customer (even if we are building a system for ourselves) will have produced a detailed specification of the requirements. If this is not the case, we must s

8、tart by preparing one.</p><p>　　9.1.1 System definition</p><p>　　In chapter 7 we discussed program design, breaking down the task into a number of simple understandable elements, each of which c

9、an be easily described. The same technique of functional decomposition is equally applicable to defining the whole system, both hardware and software, as it is defining the program alone. The most common mistake is to at

10、tempt to handle the entire system as one unit. When such an approach is made we will immediately select solutions for the parts of the system we ‘know’</p><p>　　A worked example can be found in the appendix,

11、 which show a typical decomposition of fairly complex application and sample I\O diagrams</p><p>　　9.2 choosing the correct I\O hardware</p><p>　　With an understanding of the entire system we ca

12、n start to estimate the PLC requirements. For each module the inputs and outputs can be categorized for type and speed of operation. Section 6.4 described the various types of input and output modules but here we will co

13、nsider the selection criteria.</p><p>　　By knowing the number of any type of I\O lines we need and the number of lines available on a given module, the final shopping list of modules and the size of the PLC

14、system are determined. In addition, burring in at least 20 percent extra capacity to allow for future modifications or to solve problem identified during commissioning.</p><p>　　9.2.1 Simple I\O timing consi

15、derations</p><p>　　For every element, we need to determine how fast the subsystem of input program and output must react to changing input condition. The speed of operation will be the sum of the input hardw

16、are delays plus the PLC scan time plus any output hardware delays. In the vast majority of cases a time delay of 100ms or greater is not significant. Typical instances in which this may not be the case are pulse counters

17、, or where a movement has to be stopped in mid-stroke. To determine the required response sp</p><p>　　Fig.9.1 Tank level control</p><p>　　In the second example (figure 9.2) we need to stop the c

18、ylinder mid-stroke to an accuracy of ±0.2mm .we know that is maximum speed is 100mm/s, so to achieve this we would need to be able to make a decision to set or clear the control valve every 2ms, i.e.the time taken f

19、or it to move 0.2mm/100mm/s.</p><p>　　Fig.9.2 Controlling pneumatic cylinder</p><p>　　To achieve this we would require scan speed that would be difficult to guarantee except in the smallest prog

20、rams using a fast PLC. An interrupting input is indicated .the switching speed involved will probably also cause a problem because of delays in the electronic and pneumatic hardware .This will require us to stop the move

21、ment a known distance before the target position .Figure 9.3 shows how these delays are introduced.</p><p>　　This, of course, assumes that the cylinder is moving at constant speed as it trips the switch and

22、that all the electronic and pneumatic delays are constant. If d.c. inputs and outputs are used, this is a reasonable assumption. When a.c. I/O lines are in use there is always an additional 10ms uncertainty, as described

23、 below. The more normal action would be to use a second switch to slow the cylinder down before it reaches its stop position and then use then use the final switch to stop it on stat</p><p>　　Input selection

24、</p><p>　　For each input we need to determine the following points:</p><p>　　1 Voltage level: most systems in the UK use 24V d.c. or 110V a.c. for inputs and/or outputs .24V is rapidly becoming

25、 more favored , particularly for input circuits using solid-state proximity switches .</p><p>　　2 Response speed: d.c. input modules typically have a response speed in the range 25ms. Inputs a slow filter

26、circuit (>10ms) has to be fitted within the input board to prevent an input which is ON appearing to go OFF every time the current reverses. This sets the response the response to 10-20ms (based on a mains frequency o

27、f 50 Hz).</p><p><b>　　Outputs </b></p><p>　　For each output we need to determine the following points:</p><p>　　1 Voltage level: the considerations are the same as for i

28、nput circuits.</p><p>　　2 The power that PLC outputs need to switch varies greatly. A lamp may only require 5W while a hydraulic solenoid can draw up to 400W.</p><p>　　3 Output resistance and el

29、ectrical noise can be issues in cases where low level signals are to be switched. For example, consider the case where a number of low level (<100 mV)analog voltages are to be switched into analog input (a multi-plexu

30、s).In such a case ,voltages drops or induced voltages across contacts designed to carry high current are not acceptable , and modules specified as using ‘signal’ or ‘mercury wetted’ relay are required.</p><p&g

31、t;　　4 The use of a.c. outputs can often be an advantage. In most cases the voltage is higher (commonly 110V a.c.),giving a fourfold reduction in current for any particular load, when compared to 24V outputs, the conseque

32、nt reduction in the wire size required giving a reduction in wiring costs. A second and often more important advantage is the reduction of electromagnetic interference (EMI). When a contact opens or closes the current wi

33、ll attempt to flow across the momentary gap between contacts. Th</p><p>　　9.2.2 Analog I/O modules </p><p>　　When we set out to select analog modules there is a need to understand a number of te

34、rms used to describe their performance; this allows us to match hardware performance to requirements. This section will describe those terms that are common to both input and output modules.</p><p><b>

35、　　Inputs</b></p><p>　　As discussed in section 6.5.1, to select an analog input module the following points need to be considered:</p><p>　　■Voltage level The maximum voltage of the input t

36、o be measured must be determined. A module with a maximum range just greater than this level would normally be the best alternative. If, however, there are a number of analog inputs to be measured at differing maximum vo

37、ltage, we must either use a module to cope with the highest of the voltages or use a lower voltage module and resistors as potential dividers to normalize all inputs to the same range. If any of the voltages to be measur

38、ed can swin</p><p>　　■Current input By using a 4-20mA current loop module all the problems of voltage level selection are avoided but at the cost of using process transmitter units to convert the measured va

39、riable into current signals. If the transducers have not been purchased at the time of control system specification, devices with built-in current output can be purchased at very little extra cost. The advantages are sim

40、pler wiring, better noise immunity (particularly over long distances) and avoidance of earth l</p><p>　　■Conversion speed There are two basic type of A/D converter. The first will perform a conversion ever

41、y 20ms (the period of the a.c. mains voltage), which gives us a good clean reading free from worriers of line frequency interference. The second will convert in 2-20us, giving the possibility of measuring transient data.

42、 When a module has more than one input to measure it will do a conversion of each in turn, reducing the data rate for each individual channel. The choice comes down to the conve</p><p><b>　　Outputs<

43、/b></p><p>　　The conversion speed of an analog output is generally <100us and rarely a problem. Once the resolution of the module is selected we have only to consider the following points:</p>&

44、lt;p>　　■Voltage level Most modules provide +-10V outputs which can be scaled with a potential divider to the required level. Univocal devices of 0-10V are which effectively doubles the resolution when using the same D

45、/A converter.</p><p>　　■Load resistance Voltage output modules are not designed to supply more than a few mill amperes. Typically the minimum load resistance is 300歐姆</p><p>　　■current output A

46、s discussed in section 6.5.1, it is often an advantage to use current loop output (4-10mA). Such modules are available for most PLCs.</p><p>　　附錄B：英文資料翻譯</p><p>　　根據(jù)需要選擇 PLC</p><p>

47、<b>　　疑問回答</b></p><p>　　1\ 我們?nèi)绾瘟私夂驼莆誔LC?</p><p>　　2\ 我們?nèi)绾芜x擇 I\O 設(shè)備 ?</p><p>　　3\ 如何配置 I\ O 線路 ?</p><p>　　4\ 根據(jù)需要如何選擇處理器和內(nèi)存型號(hào) ?</p><p>　　5\ 如何選

48、擇一個(gè)供應(yīng)者 ?</p><p>　　當(dāng)選擇一個(gè)有特殊功能的PLC時(shí)是令人困惑的。提供的產(chǎn)品種類是巨大的，多數(shù)通過提供產(chǎn)品型號(hào)代替選擇產(chǎn)品范圍，大多數(shù)單元模塊一定要根據(jù)工程的需要和用戶的需求來選擇，為了以后能夠有一個(gè)令人滿意的解決方案，可以提供一些額外的功能。</p><p>　　我們不得不在平衡額外支出和程序運(yùn)算所需相對(duì)時(shí)間內(nèi)的更昂貴的硬件中做出選擇，這一選擇使我們使用系統(tǒng)所需的更廉價(jià)的

49、硬件，每一個(gè)方案都不得不考慮其優(yōu)處，注意低估編寫代碼所耗隨時(shí)間而產(chǎn)生的普通陷阱。</p><p><b>　　9.1 預(yù)測(cè)需求</b></p><p>　　決定任何的解決方面的出發(fā)點(diǎn)是一定要了解要達(dá)到什么目的。較為理想的情況下是我們的客戶（甚至是我們自己假象的這種理想狀態(tài)）提供一個(gè)所需產(chǎn)品的詳細(xì)的規(guī)格說明，即便不存在這種情況，我們也一定要從一開始做出準(zhǔn)備。</p

50、><p><b>　　9.1.1系統(tǒng)定義</b></p><p>　　在第7章我們討論了計(jì)劃制定，一些簡(jiǎn)單可以理解的因素會(huì)損壞我們的任務(wù)，每個(gè)這樣的因素都是很容易描述的。當(dāng)單獨(dú)明確計(jì)劃時(shí)對(duì)硬件和軟件功能分解的技術(shù)與定義整個(gè)系統(tǒng)同樣可用的。</p><p>　　最普通的錯(cuò)誤是試圖把整個(gè)系統(tǒng)當(dāng)作一個(gè)組件來掌控。當(dāng)這樣的方法被我們掌握，我們將會(huì)‘知道‘問

51、題出在什么地方并迅速選擇解決方法來解決系統(tǒng)單元的問題，或者說我們會(huì)立刻‘知道‘問題該如何解決。這種轉(zhuǎn)移并分離設(shè)計(jì)與選擇設(shè)備的方法是解決真實(shí)問題所必須的，如果匆匆翻閱一些較為皮毛的解決方法的話會(huì)使我們與理想目的越來越遠(yuǎn)。</p><p>　　當(dāng)我們看到附錄中的一個(gè)例子時(shí)，我們會(huì)發(fā)現(xiàn)它復(fù)雜的運(yùn)用與I/O實(shí)例圖表分析的是十分公平而又典型的。</p><p>　　9.2 正確選擇I/O硬件<

52、/p><p>　　隨著對(duì)整個(gè)系統(tǒng)的了解，我們可以開始估算對(duì)PLC的需求。每個(gè)組件的輸入輸出都要通過類型和運(yùn)行速度進(jìn)行分類。第6.4節(jié)介紹了多種類型的輸入輸出的組件，但是我們這里需要考慮選擇的標(biāo)準(zhǔn)。</p><p>　　通過了解I/O線路類型的數(shù)量我們所需指定模塊的可用線路數(shù)量，可以決定最終的模塊和PLC系統(tǒng)大小的購貨單。此外，至少20％額外功能可以用于解決以后修復(fù)或用于解決任務(wù)內(nèi)的典型問題。&

53、lt;/p><p>　　9.2.1簡(jiǎn)單I/O時(shí)間安排考量</p><p>　　對(duì)于每個(gè)元素，我們必須確定子系統(tǒng)的輸入計(jì)劃和輸出反應(yīng)輸入條件的改變的響應(yīng)時(shí)間有多長(zhǎng)。運(yùn)行速度是輸入硬件延時(shí)、輸出硬件延時(shí)和PLC掃描時(shí)間的綜合。在絕大多數(shù)實(shí)例中一次延遲100ms或者更多都是不重要的。典型的實(shí)例可能不是脈動(dòng)的計(jì)數(shù)器或者運(yùn)行時(shí)不得不在中圖停止。為了確定必須的響應(yīng)速度，我們必須考慮每一個(gè)我們所定義的模塊的

54、輸入和輸出。各種不同的速率都會(huì)對(duì)控制結(jié)果產(chǎn)生影響，但由速率最慢的所決定。例如，圖9.1所示的一個(gè)簡(jiǎn)單的池中水平面控制響應(yīng)。如果我們已知液體的流速是10公升/s 并且我們需要維持容器中的液體在±1公升，我們需要讀取輸入的水平并且決定輸出水閥的開和關(guān)，最后確認(rèn)最慢需要每0.1s開關(guān)一次。這樣我們可以通過計(jì)算知道控制液體流入水箱的最低限度，也就是1公升/s 。</p><p>　　在第二個(gè)例子（圖9.2）中我

55、們需要讓活塞以±0.2mm的精確度停止。我們知道它的最大運(yùn)行速度是100mm/s，所以為了達(dá)到目的，我們必須每2ms控制閥門開關(guān)一次，也就是說0.2mm/100mm/s。</p><p>　　為了達(dá)到這個(gè)目的除非在最小的程序使用一個(gè)較快的PLC，否則我們所需要的掃描速度是很難保證的。一個(gè)輸入中斷是可以被暗示的。由于電氣開關(guān)和空氣開關(guān)等硬件的延時(shí)以及較為復(fù)雜的開關(guān)速度也會(huì)引起一些問題。這就需要我們知道在目

56、標(biāo)位置前停止移動(dòng)一個(gè)已知的距離。圖9.3只顯示了如何解釋這些延時(shí)。</p><p>　　也就是說，假設(shè)當(dāng)汽缸中活塞的運(yùn)動(dòng)速度固定時(shí)，那些開關(guān)的的過失和所有的電氣和空氣延時(shí)也是固定的。也就是說，當(dāng)輸入輸出被使用是，這種假設(shè)是較為合理的。或者說用下面的描述，當(dāng)I/O電纜處于使用狀態(tài)的時(shí)候總是會(huì)有10ms的時(shí)間是不確定的。更標(biāo)準(zhǔn)的運(yùn)動(dòng)是用第二個(gè)限位開關(guān)來控制汽缸活塞在到達(dá)指定位置前的減速，然后再用前端的限位開關(guān)控制其停

57、止在固定的位置。</p><p><b>　　輸入選擇</b></p><p>　　對(duì)于每個(gè)輸入裝置來說，我們必須確定一下幾點(diǎn)：</p><p>　　1、電壓標(biāo)準(zhǔn)：在大多數(shù)系統(tǒng)中使用直流24V或交流110V。對(duì)于輸入或輸出來說，特別是當(dāng)輸入線路使用電晶體管開關(guān)的時(shí)候，24V直流電壓是十分受大家所接受的。</p><p>

58、　　2、響應(yīng)速度：直流輸入模塊的響應(yīng)速度在25ms以內(nèi)是較為典型的。輸入一較慢的過濾線路（>10ms）時(shí),為了避免開顯示與關(guān)顯示每次都相反我們不得不使這個(gè)輸入在在合適的輸入范圍內(nèi).這個(gè)裝置的響應(yīng)時(shí)間在10-20ms(主頻率為50Hz)。</p><p><b>　　輸出選擇</b></p><p>　　對(duì)于每個(gè)輸出裝置來時(shí)我們必須確定以下幾點(diǎn)：</p>

59、;<p>　　1、電壓標(biāo)準(zhǔn)：所需要考慮的與輸入裝置相同。</p><p>　　2、PLC輸出端口的能量需要進(jìn)行極大的改變，若液壓螺線管可以升至400W則一個(gè)指示等只需5W。</p><p>　　3、當(dāng)開關(guān)量我較微小的信號(hào)輸出時(shí)電阻和電噪音會(huì)對(duì)其產(chǎn)生影響。舉例來說，輸出較為微弱（<100mV）的電信號(hào)會(huì)改變輸入量。有這樣的情況發(fā)生，比如電壓下降或者感應(yīng)電壓擊穿觸點(diǎn)從而高于

60、現(xiàn)行的傳輸是不能被接受的，指定的單元組件需要使用‘信號(hào)’或‘水銀變濕’來傳達(dá)信號(hào)。</p><p>　　4、使用交流輸出一般情況下是較為有利的。一般情況下電壓是比較高的（常為交流110V），對(duì)于一些特殊的負(fù)載需要提供一個(gè)是平時(shí)四倍的電壓，跟直流24V相比，可以通過減少電纜的尺寸來減少可以減少在電纜上的損耗。第二個(gè)比較重要的優(yōu)點(diǎn)是減小電磁干擾。當(dāng)接觸開關(guān)開或關(guān)時(shí)電流會(huì)試圖瞬間流過快關(guān)之間的間隙。原因是以很高的頻率產(chǎn)

61、生的無線點(diǎn)播所產(chǎn)生的電火花。這會(huì)在控制面板的控制下逐步轉(zhuǎn)好。PLC就是為了防止這種沖突和動(dòng)作失常而設(shè)計(jì)的。然而，沒有任何設(shè)備是絕對(duì)完善的，也就是說，交流電晶體管輸出的設(shè)計(jì)是為了使開關(guān)在開和關(guān)時(shí)產(chǎn)生盡可能小的電磁干擾。</p><p>　　9.2.2 特殊I/O模塊</p><p>　　當(dāng)我們決定選擇一個(gè)相似的I/O模塊時(shí)，必須了解一些限期內(nèi)這些產(chǎn)品的運(yùn)行情況，這些有助于我們根據(jù)需要比較硬件

62、的性能。本章節(jié)我們將介紹對(duì)于輸入輸出模塊都比較普通的要求。</p><p><b>　　輸入</b></p><p>　　如第6.5.1章節(jié)所討論的，擇輸入模塊需要從以下幾點(diǎn)進(jìn)行考慮：</p><p>　　■電壓標(biāo)準(zhǔn) 輸入的最高電壓是通過測(cè)量來確認(rèn)的。模塊的最大值的范圍大于這個(gè)值是較為正常的選擇也是最好的。然而，有很多輸入單元的測(cè)量是在不同的

63、最大電壓下進(jìn)行的，我們必須用一個(gè)擁有較高的電壓的模塊或者有一個(gè)較低的電壓的模塊串聯(lián)一個(gè)分壓電阻是所用輸出處于一個(gè)正常的標(biāo)準(zhǔn)。如果所有的電壓都是在最高或是在最低之間的話，我們需要選擇一個(gè)具有兩極的模塊。如圖9.4所示為分壓電路。</p><p>　　■電流輸入 通過使用一個(gè)4-20mA的電流環(huán)可以忽略根據(jù)電壓選擇標(biāo)準(zhǔn)的所有問題，但是需要考慮在使用時(shí)發(fā)送單位的過程中將其精確的轉(zhuǎn)化成可變的電流信號(hào)。如果轉(zhuǎn)化器沒有在掌握

64、系統(tǒng)規(guī)格時(shí)購買的話，在建的裝置會(huì)在電流輸出方面花費(fèi)一點(diǎn)額外的支出。還有的優(yōu)點(diǎn)是較為簡(jiǎn)單的接線，更好的噪聲免疫和忽略地域等問題。缺點(diǎn)是電流環(huán)系統(tǒng)響應(yīng)的較慢，而且只適用于改變頻率低于10Hz的信號(hào)。</p><p>　　■轉(zhuǎn)換速度 A/D轉(zhuǎn)換器有兩種較為基本的類型。第一種是每20ms執(zhí)行一次轉(zhuǎn)換（交流電壓的周期）這給我們提供的閱讀自由，不必?fù)?dān)心在通信線路上各種噪聲的頻率的干擾。第二種是在2-20us內(nèi)進(jìn)行轉(zhuǎn)換，使測(cè)量

65、瞬時(shí)數(shù)據(jù)成為可能。當(dāng)一個(gè)模塊不止一個(gè)輸入需要測(cè)量時(shí)，他會(huì)按順序轉(zhuǎn)換每一個(gè)輸入信號(hào)，并減小每個(gè)單獨(dú)通道的傳輸速度。選擇這種轉(zhuǎn)換器會(huì)降低轉(zhuǎn)換時(shí)間和規(guī)范PLC載入和讀取數(shù)據(jù)。當(dāng)高速傳輸時(shí)PLC只起到一個(gè)中轉(zhuǎn)的作用，載入數(shù)據(jù)時(shí)只是閱讀，在此之后進(jìn)行分析從而進(jìn)行記錄和顯示。如果需要一個(gè)更高的傳輸速度時(shí)，我們需要只讀取一個(gè)通道的信號(hào)來適應(yīng)整個(gè)系統(tǒng)的外部電壓表的顯示。如果這不是足夠快的話會(huì)通過外部電壓表來讀取一個(gè)達(dá)到2000轉(zhuǎn)/s是的真實(shí)時(shí)間并且將這

66、個(gè)時(shí)間裝入PLC用于連續(xù)通訊。</p><p><b>　　輸出</b></p><p>　　一個(gè)特殊功能輸出模塊的轉(zhuǎn)換速度是個(gè)問題，它一般低于100us。如果要選擇一個(gè)這樣的模塊我們只需要考慮以下幾點(diǎn)：</p><p>　　■電壓標(biāo)準(zhǔn) 大多數(shù)模塊根據(jù)提供輸出±10V的電壓進(jìn)行分類，這些可以提供所需的標(biāo)準(zhǔn)。較為簡(jiǎn)單的提供0-10V的裝