PID convert proportional band degrees to percent

Transistor · Oct 19, 2013

I'm trying to improve tuning on Watlow 'Series PD' PID controller and am using some tuning guides to help. Application is simple temperature control of a heatsealing station using electric heater elements.

My tuning guides want either proportional term gain or proportional band percent. The Watlow controller only uses proportional band in degrees.

So, if I'm using SP = 330°F, P = 50[FONT=&quot]°F and the Watlow manual gives allowable operating ranges 'Type J: -210 to 1,200°C or -346 to 2,192°F' then how do I calculate the gain or P term in %?

(I am trying to improve PID control when the thermal load changes between run and standby.)

Many thanks.
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Transistor · Oct 19, 2013

I think I figured out how to measure it:

Turn off I and D so we just have a P (proportional) controller.
Set the P setting.
Adjust SP to normal operating point.
Allow loop to stabilise.
Record the PV (actual value) and MV (the manipulated value or output).
Kp (gain) = MV / (SP - PV)

e.g.

P = 50°F.
SP = 330°F
PV = 296°F -> e (error) = 34°F.
MV (output power) = 69.5%.
Kp = 69.5% / 34°F = 2%/°F.

I checked it at a couple of other settings and this seems correct. It seems that for the Watlow controller 100°F = 100% proportional band. I don't know what would happen if I were to use °C, etc.

Most references seem to consider gain as unitless. Clearly from my calculations they do have units (%/°F). Anyone?

danw · Oct 19, 2013

Usually, those stand-alone PID controllers use the span of the input to define 100% proportional band.

Supposing you're on a K type thermocouple on the wide range, -454 to 1832 degC. That's a span of 2286 deg C (1832 + 454).

100% PB = 2286 deg.
10% PB = 228.6 deg.
1% PB = 22.86 deg.

Gain = 100/PB

Gain of 1 = PB 100 (100/100)
Gain of 10 = PB 10 (100/10)
Gain of 100 = PB =1 (100/1)

danw · Oct 19, 2013

Proportional only control splits the proportional band at the setpoint, 50% above, 50% below.

So a 34 deg error at a SP of 330 deg which produces an output of 69.5% would look like this:

Although the proportional band in this example is 174 deg 'wide', going from 243 deg to 417 deg, one can't say what the proportional band setting is because 174 deg is a percentage of what? 10% of 1740 deg? 12 % of 1450 deg? 29% of 600 deg?

One has to establish what the proportional band is based on.

Peter Nachtwey · Oct 19, 2013

Transistor said:
LIST]
[*]P = 50°F.
[*]SP = 330°F
[*]PV = 296°F -> e (error) = 34°F.
[*]MV (output power) = 69.5%.
[*]Kp = 69.5% / 34°F = 2%/°F.
[/LIST]
I checked it at a couple of other settings and this seems correct. It seems that for the Watlow controller 100°F = 100% proportional band. I don't know what would happen if I were to use °C, etc.

Most references seem to consider gain as unitless. Clearly from my calculations they do have units (%/°F). Anyone?

Yes, 2% output/℉ however you could have simply used the proportional band data where Δ50℉=100% output. You can change the offset of the proportional band. It could be at 300℉-350℉ or 290℉-340℉.
It looks like your range is 280℉-330℉ so you don't get any output unless the temperature drops below 330 and there is an error. If you moved the PB to 300℉-350℉ you would get 69.5% output at 316℉.

If the proportional band can be moved up and down as a function of error you would have the same effect as an integrator.

Transistor · Oct 19, 2013

Thank you, DanW.

Usually, those stand-alone PID controllers use the span of the input to define 100% proportional band.
Supposing you're on a K type thermocouple on the wide range, -454 to 1832 degC. That's a span of 2286 deg C (1832 + 454).

That's what I expected but it's too arbitrary and definitely isn't the case with Watlow.

Proportional only control splits the proportional band at the setpoint, 50% above, 50% below.

That's how I always understood it until I started to go through the maths. For a P controller the output is given by

Code:

Output = Kp x e

where Kp = gain and e = error between PV and SP. But when PV = SP this gives output = 0. That's why, with this simple control, the temperature always droops. If PV exceeds SP the output goes negative which implies a request for cooling.

Peter, I'll have to think about your comments a little longer.

... you could have simply used the proportional band data where Δ50℉=100% output.

I've never seen a controller offer an offset on the PB. Anything I've worked on has used the integral function to correct. None of the PID equations show an offset function either.

danw · Oct 20, 2013

I found the PD series manual and it shows the PB with 0% output at the SP:

The PB band "shift" Peter refers to is called manual reset by Honeywell.

I couldn't find any means of doing so in the Watlow PD series manual; there is only automatic PI action using the minutes/repeat setting.

Calistodwt · Oct 20, 2013

The manual reset is in fact a bias so the output of a proportional only controller would be (Kp * e) + Bias. Because of the inherent problem of offset with a proportional only controller this allows you to do an initial adjustment such that you can eliminate any offset between SP and PV at the normal set point. However with any process load change an offset will probably occur again. This offset could be eliminated again by adjusting the manual reset value (bias) of the controller. If this occurs frequently then you will get pretty tired of being asked to adjust it again and again etc. Best option would be to install a P + I controller where output = (Kp * e) + integral action (replaces manual reset). The old pneumatic controllers were always aligned in the workshop with 50% bias as standard. Offset in the field could be removed where some controllers had a manual reset dial or else the controller adjustment could be tweaked to eliminate offset.

PID convert proportional band degrees to percent

Transistor

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