AB冗余系统内存使用详解

1、RockwellAutomationPage 1of 5Un dersta nd Con trolLogix Redu ndancy Memory UsageCreation Date:6-30-04DISCLAIMERBecause of the variety of uses for this information, the user of and those responsible for applying this in formati on must satisfy themselves as to the acceptability of each applicati on an

2、d use of the program. In no eve nt will Alle n-Bradley Compa ny be resp on sible or liable for in direct or con seque ntial damages result ing from the use or applicati on of this in formati on.The illustrations, charts and examples shown in this application note are intended solely to illustrate th

3、e principles of programmable controllers and some of the methods used to apply them. Particularly because of the many variables and requirements associated with any particular installation, Allen- Bradley Company cannot assume responsibility or liability for actual use based upon the illustrations u

4、sed and applicati ons.Alle n-Bradley Compa ny assumes no pate nt liability with respect to use of i nformatio n, circuits, equipme nt, or software described in this text.Reproduction of the contents of the application note, in whole or in part, without the express written consent of the Allen-Bradle

5、y Company is prohibited.Docume nt PurposeThis document is not a manual or training material, but an Application Note, which could be useful in helpi ng a Rockwell Automati on customer with un dersta nding Con trolLogix Redu ndancy memory usage.Intended AudienceThis docume nt is to be used by Rockwel

6、l Automati on employees/customers support ing and selli ng Con trolLogix Redu ndan cy.V1.0RockwellAutomationPage 3of 5Con cept of Applicati on NoteIn a Con trolLogix con troller (L55 and L6X) there are two separate areas of memory, I/O memory and data and logic memory. When Redundancy is enabled, ce

7、rtain areas of memory will approximately double in usage; those are I/O memory and data memory. We will discuss why these areas of memory double.First let us look at an example using an L55M13 to illustrate:Non-Redu ndant System:I/O memory:Total 208 KbytesUsed 75 KbytesData and Logic memory:Total 1.

8、5 MegUsed 250 Kbytes data and 250 Kbytes logicNow when Redundancy is enabled the amount of I/O memory and data memory will double.Redu ndant System:I/O memory:Total 208 KbytesUsed 125 KbytesData and Logic memory:Total 1.5 MegUsed 500 Kbytes data and 250 Kbytes logicAno ther example of the I/O and da

9、ta memory doubli ng can also be show n by using V13 software and the offline memory estimati on tool.First we will look at the memory estimation of a blank non-redundant controller. That is no I/O, data, or logic. This is only an estimati on.Estimated 170 MemoryTotal:Free:H Used：嘴 Max Used:Largest B

10、lock. Free:229J76 bytes211744 bvtes17J632 bytes17J632 bytes211744 bytesEstimated Data and Logic Memory1,605,632 bytes1.59QJ92 bytes24,740 bytes24740 bytesLargest Block Free:1,530.892 bytesBaseli neThe figure above will give us a baseline from which to work. There is some memory used even if the con

11、troller program is empty.Now we will add some I/O to the program. The following I/O was added.台 B i/c conf kgura EtonEl- T l 1 17CNGPty> CNBL t r 2 0 17tj6-CNDR/D RtMOTCL 0 I l756-IB3Z/ti 1*11 用2 17S6 IB33/B til 1 °l F3 1756-IB32/B M12 J +Cl 130 |> 1756 IB32/B MW °l53 7产11旷2泪拥l石PJ f8

12、 176-IB32/B M17“0 C9 1756 -I M16=- 1DJ 756-JE32 /B Mt P白 J 3T0 17ES-CwBF(/t FEMOTEB“0 11 1756-OB3Z DlJ LH 1756-OB32 Dll3 釦 17S6-OB92 D12 C+ I75i-OB3Z D13El 15 17S6-OB32 D14 fl 6 17S& OD32 DIG 3 M 1756OB3Z D16 3 8 1756*OB32 D17 0 pg 1750-OD32 Die 9 El Fl 75t-OR3ZrJ19Redu ndancy not en abled:Estim

13、ated I/O MemoryEstimated Data and Logic MemoryTotd：Zl Free: 口 Used:< Max UsedLargest Bkick Free：229.376 bytes203,488 bytes 19UBB8 bytes19J88 bytes209*88 bytesTotal:Zl Free:Zl Used:A Max Used1,605,632 bytes 15S1J64 bytes 44468 bytes44,468 bytesLargest Block Free：1.561J64 切teaV1.0RockwellAutomation

14、Page #of 5V1.0RockwellAutomationPage #of 5Figure 1Redu ndancy en abled:Estimated I/O MemorvT otal：Free：Used:嚼 Max Used:229J7G bytes 207,568 byjtes 21,800 by 怕 g21.609 bvtesEstimated Dak and Logic Memory1JG05S32 bytes1,551,136 btes 54,496 bytes54X96 by怕*就 Block Free：207.568 by怕占 Larged Block Free：L55

15、L136 bytesV1.0RockwellAutomationPage #of 5V1.0RockwellAutomationPage #of 5Figure 2V1.0RockwellAutomationPage 4of 5By subtracting the Max Used bytes under Estimated I/O Memory from figures Baseline and 1 you can see that 2256 bytes were added to memory. Then by subtract ing the Max Used bytes un der

16、Estimated I/O Memory from figures Baseline and 2 you can see that 4176 bytes were added to memory, approximately double the size.Now we will now add a 10,000 element DINT array to the program.Redu ndancy not en abled:Eshrnatad 11/0 Memarped Dwb and Logic MemoryT oUl: 二| Free； _ Used四 Maw UsedL旳est B

17、IogK Free-229,376 bytes209,4S0 bjXes8SB bites19,888 b秋钢却也4讯byte?Totd:I I Free；1 Used;Figure 3Max Used;1,605,632 tjXes 1,521,0946?S4,54S bytes84,5JS tyJes1 1,084 bytesV1.0RockwellAutomationPage #of 5V1.0RockwellAutomationPage #of 5Redu ndancy en abled:Estimated IAJ MemoryTotd:Free:Used:V Max Used:Lar

18、gest Block Free：229,376 bytes207.568 bytes21,809 bytes21 ,S08 btes307.568 bjHesEstimated Date and Logic M亡mmyH Total：Free: Used.Max Used:Largest Block Free:1,605,632 bes1X71 JEB 应 es134.604 b畑134.604 bytes1J71.O20 如V1.0RockwellAutomationPage #of 5V1.0RockwellAutomationPage 5of 5Figure 4By subtractin

19、g the Max Used bytes under Estimated Data and Logic Memory from figures 1 and 3 you can see that 39880 bytes were added to memory, approximately the size of the 10,000 element DINT array. Then by subtracting the Max Used bytes under Estimated Data and Logic Memory from figures 2 and 4 you can see th

20、at 80108 bytes were added to memory, approximately double the size of the 10,000 element dint array.The two examples above confirm that I/O and data memory are approximately doubled whe n redu ndancy is en abled. The questi on that comes out of all this is why is I/O and Data memory usage double whe

21、 n redu ndancy is en abled? Each secti on of memory will be addressed separately.Why is I/O memory doubled? Simply to have bumpless outputs for the highest priority user task. Another way to say this is it prevents a momentary reversal of outputs immediately after a switchover for outputs of the hig

22、h priority user task.The basic objective of the output han dli ng approach in the redu ndant Logix Con troller is to en sure that no data is sent to the output modules from the primary Logix con troller without the sec on dary Logix con troller hav ing the same values in its output image. This is ac

23、complished via an output bufferi ng mecha ni sm.There are two copies of the output image:1. Program Output Image2. Output Tran smit ImageThe Program Output Image (POI) is the copy of the outputs that are directly accessed by the user program. All in structi ons, both output (write) in structi ons an

24、d in put (read) in structi ons which refere nee output data values, will use the POI.The sec ond buffer, the Output Tran smit Image (OTI), is the copy of the outputs that are actually sent to the output modules. If there is a qualified sec on dary con troller in the system, the n the cha nges in the

25、 primary con troller POs are first sent to the sec on dary con troller. The OTI in the primary is not updated with the POIs'data un til the sec on dary con troller ack no wledges that it has received the dataveand moved the data to its own OTI. If there is no sec on dary con troller prese nt the

26、 data will be copied from the POI to the OTI at the end of every program.So because of the creati on of the POI we double the amou nt of I/O memory.But you may ask youonly talked about “ outputs ” what about“ Ifputpufsa re the only I/O that need to be bufferedthen my I/O memory should only in crease by the amou nt used for outputs. Yes that is true, but for in creased performa nce it is easier and faster to make an exact copy of I/O memory. The