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# PID Parameters. Intergral Time?

## 4 posts in this topic

Hi,

Can someone clarify this sentence for me from the instruction help of the PIDAT.

Note 1 When the unit is designated as 1, the range is from 1 to 8,191 times the period. When the unit is designated as 9, the range is from 0.1 to 819.1 s. When 9 is designated, set the integral and derivative times to within a range of 1 to 8,191 times the sampling period.

It relates to the integral constant value or C2 in the PIDAT blocks.

I would like to know what I would have to move into D2 to have an integral time of 0.2 seconds

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This is describing the time units for the Integral and derivative times.  It refers to  the second digit, when monitored in HEX (from right, bits 4-7) of C+6. 00x0 <- the x is the digit to which this is referring.  See below:

If you have a 9 in that digit when monitored in HEX, then your Integral and Derivative times (see below) will be straight values of time in 100ms units.  So, if the second digit of C+6 is a 9 and the value in C+2 is 255, then your integral time is 25.5 seconds (255 x 100ms = 25500ms = 25.5s).

If on the other hand, the value of the second digit of C+6 is a 1, then the setting in C+2 and C+3 is multiplied by the sampling period to determine your Integral and Derivative times.  For example:  if the second digit of C+6 is set to 1, the sampling period (see below) is set to 0.05s and the integral value (C+2) is set to 50, then the integral time is 50 x 0.05s = 2.5 seconds.

I assume that this option exists so that you can have your integral and derivative times automatically change if the sampling period is changed.

The last bit of the note informs you of the maximum value when the second digit of C+6 is set to 9:

" When 9 is designated, set the integral and derivative times to within a range of 1 to 8,191 times the sampling period. "

This means that the maximum time amount that you can set for the integral and derivative times when the second digit is set to 9 (entering derivative and integral values as 100ms units) is 8191 x the sampling period that is set.  So, the second digit of C+6 is set to 9, and the sampling period (C+4) is set to 0.25s, then the max value that can be set in C+2 and C+3 is 8191 * 0.25s = 2047s.

I agree that this is a bit confusing, but hopefully this explanation helps.

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On 7/5/2017 at 1:45 PM, PenneyInstruments said:

I would like to know what I would have to move into D2 to have an integral time of 0.2 seconds

I see that I did not answer this question above for you.

If you set the second digit of D6 (bits 4-7) to 9 and then set D2 to 2, you will have an integral time of 0.2s.

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Awesome info Michael. Thanks a million. Needless to say, I miss using Sysmac Studio right about now! Thanks again

Colin