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How to use FB250？

Created by: Ystick at: 6/25/2009 5:23 AM (3 Replies)

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6/25/2009 5:23 AM	Rate (0)
Ystick Posts: 8 Rating: (0)	JU:FB250 NAME：RLG：AE BG:KF+128 KNKT:0,4 ORG:KF+1000 UGR:KF+0 EINZ:F12.0 XA:FW10 FB:F12.1 BU:F12.2 TBIT:F12.3 This is an example from the S5-115U manual. I know what is mean and what the parameters present,but I have another problem when I use it . Well, the current is measured and FB250 is used like this(source code wrote by other people) : JU FB 250 Name :RLG:AE BG : KF +192 KNKT : KY 0,5 OGR : KF +2048 UGR : KF -2048 EINZ : F 13.0 XA : DW 0 FB : F 13.1 BU : F 13.2 TBIT : F 13.3 You know, the current can't be negative , but why they use bipolar (KNKT second parameter is 5) ? Also,if the current limit is 0 ~ 97 A , what is mean by the 2048 ?
Ystick Posts: 8 Rating: (0)	Last edited by: Ystick at: 6/25/2009 5:26 AM
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6/25/2009 5:53 AM	Rate (0)
fritz Diamond Expert Joined: 10/7/2005 Last visit: 6/8/2023 Posts: 2968 Rating: (1025)	Dear Ystick Ipresume you have the 115U system manual (you are listing an example from it) and am not too sure what the "real"problem is. All that your logic does is to takePW 192 (the first AI (channel 0) from the AI card with start address 192) and "scales" it to an "engineering range" of -2048-->+2048. This of course just so happens to be the range/resolution of the RAW AI value which means yourlogics call to FB250 really only removes the Activity, Fault and Overflow bits from the AI. Wether or not this type ofscaling is useful is of course another thing. Maybe the person who wrote this just wanted to ge the AI"scaled" withfull resolutionof theAI card and does the "real" scaling to engineering unitson the SCADA. I hope this helps cheers fritz
	Cheers fritz The advice and opinion given in this thread is that of the author and does not necessarily reflect the views of I S Systems Pty Limited. To find out how I S Systems can help you with your automation needs please visit www.issystems.com.au.
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6/26/2009 4:06 AM	Rate (0)
Ystick Posts: 8 Rating: (0)	Thanks sir, fritz Another question.For the same example mentioned before, JU FB 250 Name :RLG:AE BG : KF +192 KNKT : KY 0,5 OGR : KF +2048 UGR : KF -2048 EINZ : F 13.0 XA : DW 0 FB : F 13.1 BU : F 13.2 TBIT : F 13.3 Assume that the scope of current is 0 ~ 97A , can I say that the actual current value is ( 97 / 2048 ) * XA ?
Ystick Posts: 8 Rating: (0)	Last edited by: Ystick at: 6/26/2009 4:15 AM
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6/26/2009 8:33 AM	Rate (0)
fritz Diamond Expert Joined: 10/7/2005 Last visit: 6/8/2023 Posts: 2968 Rating: (1025)	Hello again Ystick Yes you can assuming that your current measurement device delivers a unipolar output that equates to themeasured 0 to 97 Amps. For example, lets say therange module in your AI card is of type +/-10V itwill work if your instrument delivers 0to +10V for a measured value of0 to 97 Amps. It will NOT work if your instrument is configured to deliver -10 to +10V for a measured value of 0 to 97 Amps (I highly doubt this will be the case and mention it here only for completness). Alterantively, you could simply change the call of FB250 or add another FB250 call to the logic which would look like this: JU FB 250 Name :RLG:AE BG : KF +192 KNKT : KY 0,5 OGR : KF +97 UGR : KF -97 EINZ : F x.y XA : DW x FB : F x.y BU : F x.y TBIT : F x.y I hope this helps cheers fritz
	Cheers fritz The advice and opinion given in this thread is that of the author and does not necessarily reflect the views of I S Systems Pty Limited. To find out how I S Systems can help you with your automation needs please visit www.issystems.com.au.
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