Need help understanding HP10544 osc oven circuit.

If the ujt is off, the loop gain is very high and the thing pretty much works in bang-bang mode. A very small change in opamp output will slam the heater full on or full off.

The Q1-Q2 differential pair compares the opamp output to the swatooth created by the ujt. So the amp output has to span a roughly 7 volt range to move the heater from full off to full on, which is effectively a much lower gain.

One less obvious advantage of pwm, as compared to a linear system, it that it makes heater power linear on error. A linear voltage or current drive into a heater is a square function.

You can model the thermal stuff as a group of resistors (thermal resistance) and capacitors (thermal masses). The approximate (within

5%) mapping is

1 ohm == 1 degc/watt

1 amp == 1 watt of heat

1 volt == 1 deg C

1 farad == 1 gram of aluminum

the catch being that the components tend to be distributed, not lumped, so nasty diffusion math applies.

John

Also R11 and Q4 seems to be drawn wrong.

Hi John - Thanks for the explanation. Seems obvious now. Another case of not seeing the forest through the trees.

One last thing, do you have a suggestion as to a literature reference or tutorial in regards to the thermal modeling?

Again - Thanks a bunch... Take care - Jim

Yeah. The PNP should point the other way. Weird.

Maybe Jim can check and see how it's actually built.

SRS sells some clones of the old HP ocxo boxes, with nice SC-cut rocks. Their stability and phase noise are impressive, but their thermal design is bizarre. They use TO-220 voltage regulators as heaters, and it looks to me like whoever designed the loop didn't really understand the dynamics, so they used a proportional-only loop with fairly low gain, then added feedforward compensation from an ambient temp sensor to improve temperature regulation.

John

I can't seem to retrieve this thread's references, but it sounds like an inverted bipolar device used for AGC.

Can someone repost the schematic?

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
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I actually meant that Jim Flanagan might check the actual oscillator so see how the heater driver PNP works.

John

[.....]

That is very strange. SRS seems to know what they are doing in the other stuff they do. I can even see an argument for using a 3 pin regulator as the heater. They give a huge current gain, are self protecting and don't have their own tempco.

The components needed for a PID controller wouldn't add much to the size of the circuit. These days op-amps with very low bias currents can be obtained easily so the capacitors don't have to be huge.

I caught that also. My unit didn't come with a controller board. It had been removed for some odd reason. That is why I built a replica and the subsequent initial posting concerning the thermal feedback question. I built my unit the way we believe it should be. All is well now, functionally.

John Larkin wrote: > On Sat, 26 May 2007 21:23:46 -0400, Jim Flanagan > wrote: >

Ummmm, looks like U1 is integrating the thermistor bridge error with a

0.001Hz time constant and ~40dB gain, so hard to visualize a "slam" on/off action in that circuit.

Ummm, since the Q4 current gain is so high, U1 is required to inject just a few uA into Q1 base for full-on of the Darlington. There is not going to be much of a "span" about this Q1 base voltage threshold.

The UJT will be oscillating at something around 4KHz, and its purpose is to chop the Q4 drive to minimize power dissipation, nothing exotic there in the way of control loop processing.

The funny labeling on the heater wires and the diode in Q4 lead me to suspect the heater drive may be AC...but then the color coding makes me think it should not be.

On May 29, 4:53 am, Fred Bloggs wrote: [....]

[....]

How do you get 0.001Hz?

51.1K * 2 uF -> 1.55Hz

5.6M * 2 uF -> 0.014Hz

The latter being the zero in the PI controller, I'd expect the gain cross over to be near that point.

[....]

It is the voltage on the base of Q1 that matters here. Q2 is fed with a ramp from C1.

Then maybe you should give it another try.

Roughly 7 volts.

Turning on the ujt cuts loop gain by a factor several hundred, and changes it from super-nonlinear to mostly linear. Do you think that might affect loop dynamics?

John

Looks like someone needs to review their basic arithmetic. In the simplest case of linear PWM, the change in output power per unit of error voltage drive into the UJT modulator remains the same over the span (7V) of modulator input voltage. This is not a gain reduction of several hundred. The circuit is a simple dominant pole regulator with temperature->voltage->power->temperature forming the loop states.

No, the sawtooth is not linear, but let's assume it's close enough.

The gain reduction is relative to when the ujt oscillator is stopped. In that case, the system becomes continuous/nonlinear with very high gain, and becomes unstable as the op noted.

John

The gain reduction is relative to a completely different circuit topology when the ujt is stopped, so the comparison is not meaningful. Also, I'm not sure it even makes sense to use the word "gain" in that context because the loop becomes a simple two-state in each of which the incremental gain is zero. The experiment may be used to infer the thermal time constant of the heater system and that's about it.