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Time clock overflow

Created by: auto_r at: 1/27/2010 6:20 AM (6 Replies)

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1/27/2010 6:20 AM	Rate (0)
auto_r Experienced Member Joined: 8/27/2008 Last visit: 10/14/2022 Posts: 52 Rating: (1)	Dear All, In order to read PLC clock, we use SFC64 "TIME_TCK". the output format is TIME and when the clock reachthe maximum ( 2147483647ms) the clock is reset and it count again. How can i detect this overflow? Regards, Rami

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1/27/2010 7:30 AM	Rate (0)
fritz Diamond Expert Joined: 10/7/2005 Last visit: 6/9/2023 Posts: 2969 Rating: (1025)	auto_r How can i detect this overflow? Hello auto_r it depends on what you do with the system time. Let's say you use SFC64 to for example to count or measure a certain time period. In this case you would saved a "start timetick" value in a variable andcalculate the elapsedtime by subtracting the saved"start time" from the current SFC64 "timetick" value. Should you now have an "overflow" of SFC64 (which means SFC64 timetick value has reached 2147483647 msand starts again "ticking" from0 ms), you're subtraction will give you a negative result. This can be easily deteted by the "A <0" STL command (in case you use STL) ORby using the "<0" box from the "Status bits" Elements (in case you use LAD or FBD) ORby simply programming an DINT comparator to check if the subtraction result is negative (i.e. <0). The rectification then comprises of adding L#2147483647 to the subtraction result to make up for the occured SFC64 "overflow" of the timetick value. In case you don't use a subtraction of a SFC64 "start timetick" valuefrom it's current value you could of course always implement one purely for the "overlow" detection I hope this helps
	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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1/27/2010 8:26 AM	Rate (0)
auto_r Experienced Member Joined: 8/27/2008 Last visit: 10/14/2022 Posts: 52 Rating: (1)	Dear Fritz, thank you for your reply. In fact, i am using SFC64 for calculating runtime by initializing a T0 for first scan and then T1 for tracking the time clock. I have seen the FAQ regarding this method, and as you have mention when T1<T0 that mean the clock has overflown.but this isnot always the case: Suppose that initially T0=5ms and T1= 10ms (after overflowing), you wil not have T1<T0. how then you will detect the overflowing? Another case is that T0=0ms exactly (by concidence) you will never have T1<T0. therefore the method T1<T0 is risky. Please advice. regards, Rami

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1/27/2010 9:06 AM	Rate (0)
fritz Diamond Expert Joined: 10/7/2005 Last visit: 6/9/2023 Posts: 2969 Rating: (1025)	auto_r In fact, i am using SFC64 for calculating runtime by initializing a T0 for first scan and then T1 for tracking the time clock. Another case is that T0=0ms exactly (by concidence) you will never have T1<T0. therefore the method T1<T0 is risky. Hello again auto_r you are correct, that's why the method that I described is "(T1-T0)<0" to detect the "overflow" auto_r Suppose that initially T0=5ms and T1= 10ms (after overflowing), you wil not have T1<T0. how then you will detect the overflowing? You will of course "miss" the overflow event in this case (even with "(T1-T0)<0"), but this doesn't matter sincethe resulting "measured" time difference between the two "checkpoints" would give you acorrect result of 5 ms in this case(i.e. why worry that a timetick "overflow" took place). I hope this helps
	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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1/27/2010 9:57 AM	Rate (0)
auto_r Experienced Member Joined: 8/27/2008 Last visit: 10/14/2022 Posts: 52 Rating: (1)	I have got your point. But in fact if i miss one overflow i will not catch it until the next 24D20H... has been elapsed. because if overflow occur and the first T1after oveflow > T0, all the T1 afterthen will be >T0. therefore if my process is running for more than 24D20H... it will reset again. My concern is to count the number of overflow in order to used to count the days and hours in separate INT. I am thinking about an idea: instead of comparing T1 and T0, i will compare T1a and T1b where: T1a: time clock at odd scan T1b:time clock at even scan at odd scan if T1a<T1b or at even scan if T1b<T1a then there is an overflow ,and overflow count will be incremented. odd scan and even scan can be detected by creating a flip-flop bit. I willprepare the code and send it for your interpretation. thank you regards, Rami

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1/27/2010 2:20 PM	Rate (0)
auto_r Experienced Member Joined: 8/27/2008 Last visit: 10/14/2022 Posts: 52 Rating: (1)	Here by the code: //Calculating pump run time A #P_run_i // P_run_i: pump start JNB a01 R #P_stop_scan1_set CALL "TIME_TCK" RET_VAL:=#Time_clock A #P_scan_flip // even scan JNB a010 R #P_scan_flip L #Time_clock T #P_start_time_flip1 L #P_start_time_flip0 <D JNB a011 L #P_runtime_clk_ov_c //overflow counter L 1 +I T #P_runtime_clk_ov_c JU a011 a010: S #P_scan_flip //odd scan L #Time_clock T #P_start_time_flip0 L #P_start_time_flip1 <D JNB a011 L #P_runtime_clk_ov_c //overflow counter L 1 +I T #P_runtime_clk_ov_c a011: A #P_start_scan1_set //first scan only (T0) JCB a012 L #Time_clock T #P_stop_time S #P_start_scan1_set a012: L #Time_clock T #P_start_time L #P_start_time L #P_stop_time -D T #P_delta_time //in case delta time <0 L T#0MS <D JNB a013 TAK L T#24D20H31M23S647MS +D T #P_delta_time a013: L #P_delta_time T #P_runtime_o JU a02 a01: NOP 0 I have already simulated and the runtime is working. But i didin't check the overflow otherwise i should wait more than 24 days. Please advice regards, Rami

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1/27/2010 4:09 PM	Rate (0)
uel123 Gold Member Joined: 1/29/2006 Last visit: 6/8/2023 Posts: 438 Rating: (63)	Hey, auto_r! A "serious" discussion about measuring of working time was held here: /tf/WW/en/Posts/29422 Have fun! uel123

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