Ziegler-Nichols 'tuing'

I should first say that by tuning, I only mean guessing at starting values for a control loop. (Thermal PI in this case... thread started here.)

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So I've got a ~60 W power source, a big hunk of copper, ~1 lb? and a weak thermal link. I can post a pic, if that would help.

So I if I oscillate this at ~15% maximum power, and then again at 50% max power is there some obvious reason that the "ultimate" gain should change a lot? (these are two different temperatures and the gain of the diode temp sensor goes up a little. a few percent.. not a factor of two, which is about what I see.

George H.

Reply to
George Herold
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I would like to see a picture, but it's as much for my own curiosity as anything else.

How are you limiting the maximum power? Is the power applied to the heater equal to gain * (max power) -- this would explain things pretty quickly. How are you driving the heater? I'm pretty sure from prior discussions that it's a PWM, meaning that the power should be linear to the command.

Is the oscillation staying nicely sinusoidal, or is it (or the power command to the heater) distorted? I'm pretty sure that the Z-N method is predicated on a fairly sinusoidal oscillation (although keep in mind that it's an ad-hoc method -- if you read the original paper they're quite up front about the fact that they applied it to a whole bunch of plants and came up with their tuning rules empirically).

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Tim Wescott 
Wescott Design Services 
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Tim Wescott

Is the temperature the same in each case?

--sp

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Best regards,  
Spehro Pefhany
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Spehro Pefhany

Hi Tim, (as always thanks for the pid help) I think I figured it out in bed this morning. The error signal (voltage) is changed to current and sent through heater resistors and control FET. (with constant supply voltage)

At the low end most of the heat is generated in the resistors (which is also the point at which I measure temperature.) At the mid point the FET is doing ~1/2 the lifting (dissipating 1/2 the power.) So I guess I'm not surprised that the gain is different. (It's only overall, resistors and fet that the power is proportional to the error signal.) At first blush I would have guess that I'd need more gain to get the thing to oscillate at the mid power point. But I'm going to have to think harder about why it oscillates.. and maybe I'll "get it".

I also want to try oscillating it up at the ~85% power point. That should be somewhat similar to ~15%.

Yeah with a weaker thermal link. (I've now got four thermal links... pics below) It was hard to get a "nice" oscillation. And too easy to have the thing just banging into the rails, with attendant distortions.

Right, I'm mostly just thinking about writing the manual for students to use and I thus need (at least hand wavy) explanations of why it does what it does.

Here are some pics.

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This is just a mock-up. the real thing is a vacuum can type thing that lives in a cryostat. I've gotta do testing there too.

George H.

Reply to
George Herold

No! at 15% temperature is ~320K (50C) and 380K (110C) at half power.

The gain of the diode (delta V/ delta T) is a little bigger at 380K

2.8mV/K at 320K, 2.86mV/K at 380K

George H.

Reply to
George Herold

Don't be too sure. The total heating power is Vsupply * current, the only thing that changes is the share sourced by the FET vs. the resistor. Now, if the path from the resistor to the load is different than the path from FET to load they may have different dynamics, and possibly even different gains, if the losses to ambient are different.

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Tim Wescott 
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Tim Wescott

How is heat escaping back to ambient? I know that radiant heat is proportional to T^4, and conducted is proportional to T -- how about convected?

(380/320)^4 is close to 2, and heat back to ambient would be part of the system's overall gain.

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Tim Wescott 
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Tim Wescott

OK the above (by me) is horse hockey.

data at various set point. Tset = ~350 K, (I=0.85A) Gu = 250, Pu = 2 sec. "good tune" G=80, T = 2 sec.

Tset = ~335 K, (I=0.5A), Gu = 400, Pu = 1.8 sec "good tune" G=150, T= 2 sec.

Tset = ~315 K, (I = 0.15A) Gu =~800, Pu = 1.7 sec. "good tune" G = 250, T = 1.5 sec.

where, I is the heater current (voltage = 60 V) Gu is the ultimate gain, Pu the oscillation period at Gu. And "good tune" just me eyeballing the step response.

Except for small non-linearities in the temp sensor, the gain sets how many watts I put into the heater for a certain change in temperature of the sensor. I expected Gu would be mostly constant at different power levels.

I get the feeling I'm missing something simple. (which is typical.)

According to Zeigler and Nichols (on my SEB post there's a dropbox link to their article) here,

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The gain relates to the ratio between two times. The (thermal) RC time of my hunk of stuff, divided by the lag. (with some scaling factors...) The lag (as measured by the oscillation period) changes a little, but not enough to account for factor of ~four change. I'd also think the heat capacity, and thermal resistance of the copper and weak link wouldn't change that much with temperature... at least in this small range.

Confused, George H.

Reply to
George Herold

One thing I've done to help with that sort of problem is to put one thermistor next to each actuator (usually dual TECs in my case) and wire them in series.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Phil Hobbs

I've got a little cardboard box that I put over it to reduce air disturbances, that should also cut down on radiation losses.

At the moment I've got for 3/8"x 1/2" hex brass standoff that tie the plate to a big aluminum bread board. The bread board is my "thermal bath".

For the earlier data I had only one (longer) stand off.

George h.

Reply to
George Herold

does it not depend on how you define GAIN?

at 50% another 50% doubles the power.

at 15%, you have to change only another 15% to double the power.

So if gain is defined by on time, then it would be non-linear.

m
Reply to
makolber

I tried to talk him into a dual loop when he started in on this. Didn't succeed.

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Tim Wescott 
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Tim Wescott

Duh... A plot of my data (delta T vs Power) (assuming room temp is ~295) shows it's not at all linear. I'm guessing convection loses. (Between hot plate and bread board heat sink.)

George H.

Reply to
George Herold

Saving dropbox link/

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Sorry for having to beat this to death... but if I don't kill it now... OK so far all my ideas are wrong... I've added more data to my dropbox link. In the voice of hunchback Igor, "dada gooood!"

1.) the temperature is linear with the power.
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(the temp sensor is un-calibrated ~+/- 3 K.) I had to put a temp sensor on my "thermal bath", (1/2 inch Al bread board (2'x3', but the gizmo is in one corner.) it's not that great, changes from ~30 to 50 C. 2.) the ultimate gain still changes a lot with power (or delta temperature.)
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Oh the low power data point is a guess... Gu>1k which was the max... Sorta Q=4-5 at 1k, put big error bars on the first data point.

I did the step response at various powers, and it's the same, lag doesn't change (much), but RC time constant is a lot faster. (I figure Tim or Phil (or both) has already told me the answer to this, but I wasn't in a listening mind... )

The step response data is in the dropbox link if you care.

So I'm thinking of this as a linear hunk of stuff, heater, sensor on one end and weak link (to a weak "bath") on the other. At low power I've got to raise whole the hunk by delta T, and set up this thermal gradient. I've gotta keep making a bigger gradient, (as I go to higher power power/ temps.) Until the gradient across the hunk of stuff (TeCu in this case) is equal to that across the weak link?

Is that close to right?

George H.

Reply to
George Herold

On a sunny day (Tue, 2 May 2017 16:57:25 -0700 (PDT)) it happened George Herold wrote in :

Not sure what you are doing, but are you driving the heating transistor as current source? Seems similar to my very small hotplate:

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Been working now OK for nearly 4 years...

I dunno what it is with me and control loops. I did read Tim's paper long ago, and yea. And a whole lot of math .. I guess my neural net configuration changed to what it is now when I worked with old Ampex quadruplex - and then the new Ampex servo systems (in the seventies) the new ones digtal, and since then it all works first time, even in drones, but do not ask me WHY, I am but a neural net.

So I obviously cannot help you. ?

Ignore this posting is better. ;-)

Reply to
Jan Panteltje

Hi Jan, I've had a lot of threads here about my heater, (everyone else is sick of it...) The heater is driven from ~60V DC supply (63 Vmax.) There are three(3) 22 ohm resistors (to-200 packs.) in series. One of them has a temp sensor (to-220 pac tip32 diode connected.) on top of it. The pass element fet (started as isolated to-220, now a beefy to-247.) is also on the hot plate. The fet feeds back to a 1 ohm resistor, that's the sense resistor in a V-I converter. with an opamp driving the whole thing. (Vin max = ~1V.. 1 amp)

(sense resistor and opamp are not on the hot plate.)

I'm trying to figure out the PI tuning parameters, (I've got this time delay plus RC time constant model. (C is heat capacity and R is thermal resistance of a weak link.) Well, Phil H. turned me on to it, and Z-N had the same model, way back when, so it can't be all bad. (there's a dropbox link with the Z-N paper, but you can find it online too.)

Anyway, it seems like the thermal resistance changes with the heat flow.... at low power (heat flow) there's a lot more 'resistance'. but with heat flowing there is less.. but with a limit.

It's not what I expected, so I'm confused.

George H.

Reply to
George Herold

You need to compute the Raleigh number for your heat dissipation circuit.

If it is less than 500, you don't have to worry about convection. If it's more than 100,000 you will have turbulent convection.

The point about convection is that the thermal difference drives the convection, and the thermal resistance decreases as the convecting air moves faster.

Once you get into turbulent convection, the heat only has to diffuse through the boundary layer at the surface of your radiator (which gets thinner as the air moves faster).

If memory serves you need quite a lot of space to convect in to let the Rayleight number hit 100,000.

--
Bill Sloman, Sydney
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bill.sloman

On a sunny day (Wed, 3 May 2017 16:39:49 -0700 (PDT)) it happened George Herold wrote in :

OK, but I understand your sensor is on the heater (resistor) and not on the hotplate? Don't you think it is easier to sense temperature at the place where you want it controlled? Now you introduce a whole lot of variables, for example air flow, moisture, etc?

If you want your room temperature at some level, then you do not put the thermostat in the basement next to the kettle or whatever you have?

Reply to
Jan Panteltje

Hi Bill, I stuck foam insulation around it. (so no, or only a little convection.) The temperature difference is linear with power. ie. To first order the thermal resistance is linear.

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But if I look at the open loop step response, it responds much slower for a lower power step. (say I let it equilibrate at 10 watts and then step up to 15W.) versus if I let it equilibrate at 40 watts and then step to 45 watts. That doesn't make sense in a simple RC model of my system. And I think* that even if I add more RC's to my model... to try and account for thermal resistance of my hot plate, I still will have a linear system. The time constants I measure change by factors of 4 or more!

George H.

*I'm playing around with LTspice models now.... sometimes that's easier than writing (and solving) equations.
Reply to
George Herold

Right. But the heater is right on the hot plate. I can move the sensor away ~1cm on the hot plate, and now instead of the hunk of copper right under the heater being at my set point temp, the piece of copper 1cm away is. Which is fine, one point is as good as another. But what this movement does is cause lag (time delay) from my heater to sensor. And that means I have a longer integrating time in my control loop and less proportional gain.

Hmm that does give me an idea... what if I move the sensor away, and look at this low and high power behavior.. maybe that will change something. And give me a clue.

George H.

Reply to
George Herold

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