what is "I" in pid ?

Farid

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Jul 2004
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UAE
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HI, I have been reading all the pid related posts on this site and had an Idea what it is, but still have a question..now why do I have to set for example 30 minuts for the integral ("I")?? What will happen in 30 minutes of time ??


I understand how proportional works...it scale the input signal eg. from an analog input(this is the PV), then compare it with the setpoint
this is the SP, and act accordingly, here I am a little confused..why do I need the integral term ??


any help that will enlighten me I will appriciate alot.

ps: I'd read all the post related to pid so please don't suggest to use the search button on the top of this page loooooooool :D


🍺 thanks in advance
 
thank you Mike.. ;)


I'll check this one out... but still need some pointers if anyone else can explain in an easy way (away from math as I don't like math at all, actually I hate it).




thanx. :D
 
Basically elevmike is right when he says if you don't like math you're in trouble, although lots of experience (say 20 - 30 years) is also pretty good.

The basic problem is that depending on your other parameters, it can happen that for a relatively small offset the output may not change sufficiently to cause a response in the loop. By integrating the error over a period of time you cause the output to react as though the error was increasing with time so eventually causing a reaction in the loop.

Unfortunately, to know how much integral term to use you either need the maths, or a very good gut feeling for the process. As with all the parameters if you're not sure, start out VERY gently, too much integral term can cause chaos!
 
Whenever I try to explain PID loops to the maintenance guys, I tell them this.

P= how far off am I
I= how long have I been here
D= how fast am I going


For some reason that seems to work best with them.


Ken
 
hi all,

prince Mike,


I hate math , though I know basic math, but it seems to be frustrating every time, I think about tuning PID, its goes like mt ,kc bla bla ...well why not using a simple terms?? or is it just about keeping humble persons like me out of the track ?? :(


RMA ,

I have alot of experience , as I have done many tasks and I am in the feild now for 15 years..I used to work on hydraulic presses and cut-to-length machines and I have made lots of systems runing effeciently since long time now.


thank you alot Ken,

I am lookin for this kind of answers, somthin'readable to me..
what will happen when the "I" time elapse, say 30 minuts?? is the loop will stop? or is it gonna increase the output more rapidly? what is it ? or it will start to popen the output at its maximun value gradualy till it reaches 30 mnts?? I'm kinda lost out here :(

I have been reading Ron beaufort's threads , very intersting, especialy the buick thing :p and the story of his wife's car. :D

I am not an engineer, I have graduated from high school, And worked in a factory for roll formming machines and , I tell you, one can learn alot in such factories...hydraulics,pneumatics,etc..

what I need is an answer to my question : what will happen when the integral time elapse??


thanx to all for baring with me.
 
I like the thumbnail description you gave of PID, Ken. I'm going to steal it. :)

Anyway, for an understanding, you have to know what the PID is doing.

A PID is a controller for some process. When the process is in balance (output is doing what the input wants it to do) there is no error, and everything stays at the present level.

Then, something happens. Perhaps the operator increases the input, which we call a reference, for example. Now, the controller output is too low, and we have to increase it, but by how much? This is where the PID comes in.

If the operator changed the input very quickly, then the D term becomes the dominant term. We change our controller output quickly, because the input (actually the error) changed quickly. The D change occurs because the RATE of the change of the error is high. "Change the output by an amount proportional to the rate of change," says the D term.

Now, we're probably very close to the right output, but still off by a bit. At this point, the P becomes dominant. The P term is like a multiplier. Using 6 as the P gain: "I'm 2 away (error), so I'll change my control by 12... Now, I'm 1.5 away, so I'll change it by 9... Now, I have an error of 1.0, so I'll change it by 6."

As you can see, eventually the error becomes so small, that no real change is made in the controller. "Change the output by an amount proportional to the error itself," says the P term.

This is where the I term comes in. The I just keeps track of changes made so far, and says, "and keep making them." The P only wanted to change by 0.6 when the error was 0.1, but the I term says, "Hey, you changed by 12 just a little while ago, so I'm going to make you keep changing by a bit more," and *adds* its adjustment (it's gain times the "average" error) to the P adjustment. "Change the output by an amount proportional to the summation of errors I've had so far," says the I term.

As one last note, the I term averages the error over a set period of time, so errors that occurred before this are ignored.

I hope this helps.

Don
 
One other confusion factor

Proportional v/s Gain

Intergral v/s Resets

The terms are interchangeable but not the same. Any good PID reference should explain the difference. Also note that resets can be resets/min or min/reset and even use seconds in some systems. Keep this in mind while tuning and you should be ok.
 
I've always been a little confused by integral times myself, mostly because the involve an implied basis and I'm not sure what that basis is. But I'll take a shot here. Even if my basis is wrong the concept is the same.

An 'integral time' or 'reset time' (same thing) of 30 minutes will cause a one unit change in output given a 1 unit error in 30 minutes. As an example, consider a system with a setpoint of 10 and an actual value of 9. This produces an error of 1 (10-9). Assume the error is constant. An integral time of 30 minutes would cause the control output to ramp up slowly by one additional output unit over 30 minutes. So if your output at the start of the 30 minutes was 100, all other things being constant, your output after 30 minutes would be 101. And this is a continuous process so after 15 minutes your output would be 100.5. Also, as a continuous process, there is no beginning or end to the 30 minute period. Look at it relatively. Over ANY arbitrary 30 minute period the output would increase by one unit.

Now, if the error were 2 units, the output would increase by 2 output unite in 30 minutes. Also, if the integral time were 15 minutes with an error of one unit the output would change by one unit in 15 minutes. And this increase (or decrease if the error is negative) will continue until the output reaches it's positive (or negative) limit.

I hope this helps.
Keith

Edit: I really do type slow!!!
 
I use an explanation along the lines of Ken's.

Proportional looks at the present state of the control loop

Integral looks backward

Derivative looks forward

Technically the integral term is there to eliminate "droop". This is the inherrent tendency for a "pure" proportional loop to have, at steady state conditions, an offset or permanent error between the setpoint and the process variable.

Integral isn't really a time delay, which is how intuitively (and incorrectly) we tend to interpret the integral tuning term "resets per minute". It actually sums over time the difference between the setpoint and the process variable and adds a little push to the controller output to eventually eliminate the offset or droop. The tuning "resets per minute" defines how hard to make each each little push of the output. The loop is actually executing continuously in the mathematics.

The tuning term's name is a historical artifact of the old names for a three mode controller - Proportional, Reset, and Rate. For the old analog or pneumatic single mode Proportional Only controllers the operators would manually get rid of droop by tweaking the setpoint adjustment until the droop disapeared - in effect "resetting" the setpoint.

And before you ask, Derivative adjusts the controller output based on the Rate of change of the process variable in order to minimize overshoot. Mathematically it calculates the derivative, which is the slope of the process variable - the rate of change over time of the process variable. The tuning name, derivative rate per minutes or seconds, actually defines the relative effect of the rate on the controller output. Again the mathematics is usually continuously executed in the controller.

Confused yet? Join the club! Difficulty in explaining the tuning concepts to operators is one of the many reasons I aboandoned PID control years ago for floating control.
 
Last edited:
Originally posted by Tom Jenkins:

The tuning term's name is a historical artifact of the old names for a three mode controller - Proportional, Reset, and Rate. For the old analog or pneumatic single mode Proportional Only controllers the operators would manually get rid of droop by tweaking the setpoint adjustment until the droop disapeared - in effect "resetting" the setpoint.

Thanks, Tom. I was always curiuos about where that term came from.

Keith
 
Some references will also use different terms.

Gain=proportional
reset=integral
rate=derivitive.

oops should have read Tom's post sorry bout the repeat
 
Last edited:

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