HOW to on compensating the feedback loop on a descrete linear regulator

National semi has an application note "A User's Guide to Compensating Low Dropout Regulators", also AN1148 -- "Linear Regulators: Theory of Operation and Compensation".

It's nice to have a network analyzer like an HP3577A and all the transformers but you can probably do reasonably well using the bode plotter in LTSpice.

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After years closing control loops I can assure you that given aggressive enough feedback _nothing_ is _inherently_ stable. The emitter follower by itself is pretty darn stable, however.

I assume that you're doing something like this (sans divider):

+V === | Vref |\\ | o-------|+\\ ___ |/ | >---|___|---| .--|-/ |>

| |/ | Vout '-------------------o--------o (created by AACircuit v1.28.6 beta 04/19/05

The emitter follower stage will have some non-zero resistance on the output, so hanging a cap on the thing will slow it down slightly. A low-ESR cap would make it worse (opposite of a PNP output stage) because it'll tend to introduce more pure delay with less of the lead that you'd get for free from a high series resistance on the cap.

If all these guesses are correct, you should be able to tame that beast with a nice little inner capacitor:

+V === | Vref |\\ | o-------|+\\ ___ |/ | >--o--|___|---| .--|-/ | |>

| |/ | | | | | | || | | o--||----' | | || ___ | Vout '-----------|___|-----o--------o

(created by AACircuit v1.28.6 beta 04/19/05

This will slow down the response of the thing a bit (but not too much, because of the emitter follower). The capacitor, however, will introduce a zero into the loop gain that should boost the phase around zero crossing enough to give you stability.

You can simulate this and test for stability using Bode plots. I think I would be more inclined to test for stability range by testing the circuit's output impedance. Do this by hanging a current source on Vout then doing a frequency sweep of Vout. If the phase of Vout goes negative at any frequency then the circuit has a negative real component of impedance, which means that there will be some reactance that you could hang on the thing to make the circuit unstable. Once you've found those points, look at the required reactance and ask yourself if you need to worry about it.

Come to think of it, now that I've written this -- have you checked the circuit operation under the real load? A nice high conductance in parallel to your capacitor would bring the total output impedance back into positive-real territory, and stabilize the circuit. If you're doing your testing with _just_ a capacitor on the output then perhaps you are asking too much from your simple circuit?

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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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What about the old classic regulator helper shown below?

+V === | o--- ----------. | \\ ^ | | --- | .-. | | | | | | | | .-. | '-' | | | | | | | | '-' ____ | | | | | | '----o--|7805|---o-------o |____| | | === GND (created by AACircuit v1.28.6 beta 04/19/05

You should be able to configure this to use very little current from the linear regulator, which would keep it's junction temperature down (unless you're running way up close to 125C).

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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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Oops -- that should be a PNP transistor, connected in common-emitter. The regulator provides the low impedance, the transistor provides a whopping current boost.

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Tim Wescott
Wescott Design Services
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Like I said, I can't use IC regulators because the ambient temperature is above their thermal shutoff at 160C. We do this all the time and have found parts that work (SIC and SOI) that hot and hotter but you have to go back to the old school and do things like discrete regulators.

I was looking for folks with old appnotes or know how on descrete regulators. The only guys doing it nowadays are the audio golden ears that claim the LM317 'is just way too noisy and slow for audio'. :)

thanks for what you have contributed so far.

If you have more, keep it comming.

It wasn't clear if you were above the thermal shutoff of the part without current or if you just couldn't pull any appreciable current. At 160C it's obviously the former.

It could be that your parts are no longer working as advertised because of the thermal shift -- have you characterized your op amps at these temperatures? Have you considered going a bit older school yet and building the regulators out of discretes? Depending on your accuracy and output impedance requirements you may end up getting something good enough that's smaller and uses less components.

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Tim Wescott
Wescott Design Services
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This is not a temperature problem. I'm just trying to get a design that I'm comfortable with understanding the control loop of so I won't get oscillations over a range of step loads and output capacitance that couldbe connected.

Believe me I looked at a dog simple zener follower but the digial guys had a cow about the voltage slowly sagging down from 5.1v to 4.8V during high load events (driving long capacitive lines). Just a droop no spikes on the line and they still don't like it.

Looks like I need a feedback loop.

I just did a set of 3 linear regulators today. simple as really, a feedback divider up the -ve pin, r-c feedback, driving npn pass transistor. initially use cap-only feedback, step value in spice to get best response. then insert resistor, repeat process. all the while bashing the ouput with 1% - 101% - 1% load steps.

If you are patient, you can also do that with a soldering iron.

Cheers Terry

And oven mitts, in this case.

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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google?  See http://cfaj.freeshell.org/google/