Exercise for a power supply design

Negative output impedance, easy. And as JW suggests, you could compensate for ambient temperature effect on wire resistance. For extra credit, simulate the dynamics of wire self-heating and compensate that, too.

I've built several buck switchers (out of discrete parts, not SimpleSwitcher IC's or such) that had mildly negative output resistances; never bothered to find out why.

John

Frank, If you look at commercial power supplies made by outfits like Lambda, this is the way they do it.

Tam

I read in sci.electronics.design that Frank Bemelman wrote (in ) about 'Exercise for a power supply design', on Wed, 24 Aug 2005:

If the cable resistance varies only with temperature, it is a known and can be allowed for in the design.

I don't understand any of that. It doesn't seem to me to be a route to a solution.

You can include a resistance equal to the cable resistance somewhere in your power supply circuit (with a PTC to deal with temperature). You can also sense the load current. So surely it isn't all that difficult to devise a closed-loop voltage controller that uses as a reference a fixed

12 V plus the voltage produced by a current proportional to the load current flowing through the cable-simulating resistance? Just make sure the overall loop gain is less than 1!

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Assume this a very very long cable with a resistance of perhaps 10, 20 or 50 ohms. At the other end is a device that needs 12V. Current comsumption may vary between 0 and 1A. Just some figures for the sake of the discussion.

The powersupply, not yet designed of course, has an adjustment potentiometer. This potentiometer is set to the same resistance value as the long wire to the device. Ideally, with auto-setup, by shorting the wires at the other end. Hmm, this may require a small PIC. The PIC could disconnect the load, and check if a capacitor is at the other end. If the voltage drops immediately, there is a dead short on the other end, if not, the device is still connected.

This miraculous power supply now regulates it's output voltage, so that the device get's the 12V that it wants, regardless the current it consumes.

All we need is a brilliant designer who makes the schematic.

(No, I don't need one, honestly)

--
Thanks, Frank.
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Why can't you use a 4-terminal regulator with high-impedance sense inputs? You know, where you run a seperate pair back from the load to the sense inputs. Because the sense input is high-impedance, there is negligible voltage drop in the sense wires. Or are 4 wires not allowed?

Many DC motor speed controls work like this- there is a pot which is used to adjust the negative resistance of the supply so that it mostly cancels out the armature resistance. As the output current increases, so does the output voltage.

Best regards, Spehro Pefhany

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"Frank Bemelman" a écrit dans le message de news:430cd2c5$0$576$ snipped-for-privacy@dreader31.news.xs4all.nl...

Uh... Why all those complications?

Either have:

- point of load regulation

- kelvin sense.

Plus those open loop negative output impedance supplies always give me some uncomfortable feelings.

--
Thanks,
Fred.

[...]

Sounds like the 'boosters' they used in tramway generating stations to cope with the voltage drop on long feeders, they were in use from about

1900 onwards.

Of course, in those days they were set up manually.

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hmm, why not a 1A current source, lm317 etc and a shunt reg at the distal end. might need an opamp, something high spec like a 741, or am I missing something

martin

On Wed, 24 Aug 2005 23:05:21 +0200, "Fred Bartoli" wroth:

You're giving a sensible answer to a nonsense question. That is NOT what SED is all about.

Otherwise known as "the willies".

Jim

[....]

How much ripple/noise can there be on the 12V? I assume we can put a modestly large capacitor out there, can we?

I've got a circuit that needs only the two wires and no pot. The only problem is there's a bit of AC on the load end.

--
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"Fred Bartoli" schreef in bericht news:430ce115$0$28869$ snipped-for-privacy@news.free.fr...

de

some

Remote sensing or a regulator at the other end is the simplest. But I was thinking about 'power over ethernet' situations where cable lengths can be very long and causing serious voltage drops. Or situations where people run long wires and not knowing what they are doing, and not want to be bothered with 4 wire cables.

I envisioned this smart wallwart that somehow figures out the cable resistance and compensates for it, without asking (too many) questions. The highly appreciated NoDrop(tm) wallwart.

But it's all hypothetical ;)

--
Thanks, Frank.
(remove \'q\' and \'.invalid\' when replying by email)

I did that once, but the motor got all herky-jerky as the positive feedback approached the ideal value. I think I should have seriously rolled off the time constant of the feedback path, to make the motor's mechanical output impedance zero only at low frequencies, but softer transiently.

I did have a big inertial load, a human body on a tiltable table, with some belts and gears and stuff to complicate the dynamics. Must have been some resonances here and there.

John

ISTM a buck-boost regulator at the far end is a better choice than anything at the near end. Dump 48V down the line, perhaps even ACC, and let the load figure it out.

--
  Keith

"John Larkin" a écrit dans le message de news: snipped-for-privacy@4ax.com...

You can't perfectly compensate for the system losses: any perturbation will give you permanent oscillations as your system is always of order > 2.

The human body internal losses probably provided some welcomed damping.

--
Thanks,
Fred.

The old Boschert open frame supplies introduced a negative resistance element in series with the main output, in order to reduce cross-regulation to the auxiliaries.

Any regulator looks like a negative rsistance, to the source.

I get my exercise on a bicycle, or chasing the damn dog.

RL

On Fri, 26 Aug 2005 07:03:51 GMT, legg wroth:

Most linear regulators look like a constant current sink to the source, not a negative resistance at all.

Jim

Ah yes. Linears. I remember those.....

RL

Sensing the voltage drop in one wire should be sufficient to fiddle the regulation with. The assumption is that both power conductors are identical. If one is a system ground plane, you can't make this assumption. A ground reference that doesn't carry distribution current might serve as a rough sensing 'point' to the remote load.

A wall wart can have whatever cabling it needs - the end user is seldom expected, or permitted, to fiddle with it.

You might examine industrial long-wire or down-hole configurations.

Just what everyone needs - another revision to a com standard. Get your new (backwards incompatible) toys out now.... before the next revision!

RL