negative resistor

if you look at the voltage current relationship at the input side of a voltage regulator, you'll note that it acts like a negative resistor. as the voltage source increases, the current decreases. my question is, could you get a super conducting (zero resistance) element by putting a resistor in series with the negative resistor?

Reply to
STUARTe
Loading thread data ...

Only a switcher, not a linear regulator.

You can get as close to zero resistance as tolerances allow, but it's not a superconductor, it's just a synthesized zero resistance between two nodes, and it needs a power supply to keep working.

You can do some fun experiments with a 2-terminal negative resistor. All the normal circuit equations work, but with unusual results, like voltage dividers with gain and stuff.

John

Reply to
John Larkin

it's a negative DIFFERENTIAL resistor.

The voltage and current are still both positive so the actual resistance is positive...

the CHANGES behave as if the positve resistance value has a region of negative slope..

Mark

Reply to
Mark

Yes, but you can easily build a 2-terminal true negative resistor with a opamp, a few resistors, and a couple of batteries. And you can put it in series (or parallel) with a regular resistor and tune the sum to as close to zero ohms as you have patience for.

John

Reply to
John Larkin

Not in any useful way.

But one way of getting constant speed out of DC motor over a range of loads is to feed it from a circuit which is designed to look like a negative resistance that exactly compensates the widing resistance of the motor coils.

IIRR Philips got a patent on this many years ago (which has long ago expired). I built one back in 1993 and it worked a treat - admittedly, at low revs, where stick-slip was a problem, only the average speed of the motor came out right, but the circuit did the job that was asked of it.

--
Bill Sloman, Nijmegen
Reply to
bill.sloman

The same sort of thing can be used to get a constant torque out of a motor. As the motor speeds up, its voltage rises and the circuit increases the current to make up for the motor's losses.

--
--
kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

For a while one of the useful amplifying elements at microwaves was the tunnel diode. One segment of the tunnel diode's V-I curve has a pronounced negative slope, so you can use it for building amplifiers and oscillators.

Better microwave transistors pretty much killed off the tunnel diode, but the Gunn diode (also a negative resistance device in it's own peculiar way) is still a useful critter at 10GHz, where transistors tend to be too slow.

--

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

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

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

About as much fun as using a sub-harmonic generator where the plate is at zero volts, the power input is applied to the plate, and the power output is taken from the grid at some negative (mumble) dB gain.

Reply to
Robert Baer

only if you could match them exactly, thermal noise may scvrew things up.

--

Bye.
   Jasen
Reply to
jasen

In principle, all you need for constant torque a constant current drive

- a much easier circuit to design. In fact, because progressively more of the current being fed into the motor goes to charge the winding capacitances as the motor speed increases, you would need to make the current increase with the applied voltage, which does imply a measure of negative resistance. I've never tried to do this in practice, so I don't know whether this is ever worth doing.

--
Bill Sloman, Nijmegen
Reply to
bill.sloman

You already have that situation at the voltage regulator output. The voltage regulator's "negative resistance" is compensating for the positive resistance of the source.

Note this isnt a useful superconductor-- it won't do superconductor-like things, like hold a permanent current flow, expel magnetic fields, dissipate zero power, etc....

Reply to
Ancient_Hacker

If you want to play around with 4kV - a MO magnetron also has a portion of negative slope (so I'm told).

Reply to
ian field

For power at microwave, IMPATTs surpassed GUNNs and transistors.

The Phoenix Missile for example, used phase locked GUNNs for the LO and TO and then a three-stage "amplifier" (really injected-locked oscillators) using IMPATTs.

The first stage used a single IMPATT; the second, three IMPATTS; the third had sixteen IMPATTs in a cavity combiner. Twenty-five+ years ago when we were developing this, getting sixteen nearly matched diodes was a real challenge.

Reply to
Wes Stewart

[....]

That would only lead to a constant torque if the losses were constant. They are nowhere near constant.

There are core losses too to consider.

--
--
kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

Why would you want to do that and throw away a perfectly good battery charger? Juz' put that thar' negitiv resisdr in parallel with eny battry and recharge it for free, right? Imagination can really explode off simple-minded word reasoning devoid of the slightest underlying reality. Let's have fun with words, let's see how ridiclous we can get while remaining dumb as hell. Let's be USENET insects.

Reply to
Fred Bloggs

Fred, You're a good start ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson

IMPATTs are really noisy, though, even compared with Gunns. The phase noise of an IMPATT oscillator is only slightly better than what you'd get by frequency multiplying a 555.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

Which won't matter much at low rpm either. I'd go with the constatn current source, and see how high an rpm I could get before the torque began to fall way. The test gear might be interesting.

--
Bill Sloman, Nijmegen
Reply to
bill.sloman

OPA544 .-----+-----. | | | .-. /^\\ .-. R2| | /+ -\\ | |2.2 | | '-----' | | '-' | | '-' ___ | | | | -|___|---+----' '----+---o | R1 | /+\\ .-. ( )Vref | | \\-/ Rload| | '-' | Ri=-R1 x 2.2/R2 | === === GND GND (created by AACircuit v1.28 beta 10/06/04

formatting link

--
ciao Ban
Apricale, Italy
Reply to
Ban

It depends a lot on what you call "fall away". The way you do the test is to rig the motor so that its body is supported but free to rotate. You prevent the rotation with a torque wrench. The shaft of the motor can be connected to another motor of a similar type as a load.

This is all fairly easy to do so have at it if you are interested. IIRC the losses rise as about the square of the RPM so yes it doesn't matter a lot at low RPMs but does matter a lot at higher speeds.

I did the experiment with a DC motor for a winch. We wanted to avoid having to sense the cable tension and yet to be able to limit it. The user had a knob that adjusted the upper limit on current and hence the limit on torque. We added a bit of the speed setting to the current setting to make it more closely follow torque and hence cable tension.

--
--
kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.