HOMEBREW Digest #5912 Fri 09 March 2012

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  PID ("A. J. deLange")

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---------------------------------------------------------------------- Date: Fri, 09 Mar 2012 01:23:25 -0500 From: "A. J. deLange" <ajdel at cox.net> Subject: PID Devonna is confusing the 'proportional band' of PID with proportional output. These are separate things. The PID algorithm determines how much output is required based on where the PV is in the proportional band (P) i.e. the error, the integral of the error (I) and its differential (D). The 'tuneset' consists of the three gains applied to each of these before they are summed to form the output. How the output is controlled is a separate matter. In a 4-20 ma loop full output is requested by a current of 20 ma, 0 output by a current of 4. In a 0 - 5V output, 0 volts corresponds to a demand for 0 output and 5 volts to a demand for full output. In an on/off output half power is signaled by logic 1 for half the output cycle time, 70% by logic one for 70% of the cycle time and so on. This gives infinite control just as the 4-20 loop and 0-5 volt outputs do and it can, thus, do it for the full PID algorithm. I use full PID with on/off output (to open and close a steam valve on a 30 second cycle) and recordings of temperature vs time show the PV eventually settling out right on the set point value (0 error) which would not happen if I were not active. What I am doing is, of course, PWM. It's just that the PRI is 1/30 Hz. As the bandwidth of the loop (determined by the thermal mass of the HLT and its rate of heat loss) is less than 1/30 Hz the recordings are smooth. A.J. Return to table of contents
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