The "D" in PID refers to a DERIVATIVE function that is necessary when the
process model includes rapid "Disturbances". In this case, if the oven
enclosure could be opened during a process cycle the PV would drop
suddenly - this is a typical disturbance. Derivative provides an
appropriately large adjustment to the result, or output, which causes the
system to recover quickly, with less hysteresis.
If your process will not regularly be disturbed you will only need a
Proportional and Integral control system. If the process is expected to be a
slow one - minutes of heating to reach the SV, and there is considerable
room for PV hysteresis, say 10% or more of the overall deltaT, you can get
away with Proportional control only.

As a side note: When you get to the point of writing your algorithm, you may
consider representing the Proportional value as a "Band" instead of Gain.
Most EE's & programmers tend to think in terms that make it easier to use a
"Gain" value, but in the real world, tuning such a system is made about 10X
easier when you can plot and visualize your system on a simple 2D plot -
which a PB, or Proportional Band allows you to do easily. In other words,
more work up front = easier system set-up later.

Chris
>
> I don't know your experience with control...
> First you have to take the model of the system.
> Termal system generaly are very slow,
> a first order system should be enoght.
> It will result in a second order system you
> consider the resistence Tranfer Function.
> In this case I think that just a PI controler is
> necessary. The diferencial controler wont
> contribute too much and will insert noise
> problems.
>
> The PI algorithm is something like this:
>
> Loop any 10ms (SampleRate)
>     Error = Temperature_Reference - Temperature_Read
>     ProportionalValue = kc * Error
>     IntegralValue = LastIntegralValue + ( ki * (Error + LastError) *
> (SampleRate / 2))
>     Output = ProportionalValue + IntegralValue
>     SaveLastValues()
> EndLoop
> kc is the proportional gain and
> ki the integral gain
> The problem, that is not too much problem
> is to calculate this.
>
> I hope bein usefull
>
> Augusto Marasca de Conto
> augusto@automacao.eng.br
>
>
>
> > He told us what he really needs:
> >
> > > In fact I need chocolate to reach 44 o C as quickly
> > > as i can .... The point is not to pass this temperture border.
> >
> > This is a control-systems problem.  I'm not sure how I'd proceed
> > but this sounds like a job for a PID algorithm.
> >
> > Barry

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