Talk:Negative resistance

Opening paragraphs

The opening paragraphs of this article seem a little confusing to me. It says that as voltage is increased the current decreases, but that seems a little backward. How do you increase voltage when resistance is falling? That seems like trying to increase the pressure of a waterfall by spraying a garden hose into it. As I understand it (and please correct me if I'm wrong) as current is increased, both the voltage and resistance decrease.

When most people talk about voltage, they're talking about open-circuit voltage (OCV), which is when voltage is at its highest. When the circuit is closed, there is a voltage drop as amperage increases. Resistance is what limits this voltage drop and, per Ohm's law, the accompanying increase in current. In a normal resistor, heat is generated which actually increases its resistance. A hydraulic analogy is found in a simple water faucet. When the faucet is closed, pressure in the line is highest. When fully open, the pressure is lowest and flow is highest. Opening the faucet a little bit increases its resistance to flow, so the pressure drop is smaller and so is the flow. (Keep in mind that "open" and "closed" mean exactly the opposite with electricity as they do in hydraulics.)

For negative resistance, when the circuit is closed, the voltage drops accordingly, and the amperage increases. The difference is that as the plasma heats up, the resistance begins dropping along with the voltage, thus the current runs away like a freight train. (A hydraulic analogy is that as soon as you crack the faucet, it blows off and water flows just as fast as the pipe will allow, and pressure drops to almost nothing.) Zaereth (talk) 20:07, 30 March 2017 (UTC)

This is a pretty complicated subject, and hydraulic analogies are not going to get you very far. I don't think there is any effect in hydraulics which is analogous to negative resistance (a decrease in flow rate through a hydraulic system with an increase in the pressure across it?). I'll try to answer your questions

• "When the circuit is closed, there is a voltage drop as amperage increases." This is caused by source resistance, a positive resistance. Any source of power has internal resistance resulting in a drop below the open circuit voltage when current is drawn.
• "As I understand it (and please correct me if I'm wrong) as current is increased, both the voltage and resistance decrease." You're right. Keep in mind there are two kinds of resistance; the "absolute" or "static" resistance, V/I which is what you are talking about, and the differential resistance ΔV/ΔI. Anything can happen to the differential resistance; it can increase, stay the same, or decrease.

I-V curve of gas discharge. Note that the vertical current scale is logarithmic

The fluorescent tube is not the best example to introduce negative resistance, because due to hysteresis the current through it doesn't change continuously as a function of voltage, but in "jumps". I just added the picture because it is the most common device with negative resistance readers will be familiar with.

A gas discharge tube like a fluorescent light has a type of negative resistance called "current controlled" or "S type" (see Negative resistance#Types). The graph (right) shows its current-voltage curve; the voltage across the tube at any given current. The negative resistance region is between D and G. When a voltage source (the power line) is connected across it that is greater than its breakdown voltage D (about 700 V on this graph), this causes an effect called hysteresis. The current becomes unstable and increases exponentially, only limited by the external circuit. This is caused by a "chain reaction" of ionization in the tube. You could say the tube's absolute resistance "drops", but keep in mind the tube doesn't have a "resistance" at all currents because the current changes in jumps. In this regime the tube cannot operate stably at points on the curve where the differential resistance is negative (sloping down to the right, such as between E and F or I and J)(see Negative resistance#Stability conditions), the current can only be stable at points of positive differential resistance (A, B, C, D, G, H, J, K).

If the external circuit has sufficient resistance or other current-limiting device (such as a ballast) the current can stabilize at a positive resistance point higher on the curve (around G and H). At this point the current will be higher but the voltage across the tube will be lower, reduced by the voltage drop across the current limiter, so the tube's absolute resistance is lower. The tube will have a glow discharge and emit light. This is what happens in a working fluorescent light. If the external circuit has insufficient resistance, the current will continue increasing to the regions J and K where the discharge becomes an electric arc which releases a great deal of heat, causing failure of the tube. --Chetvorno^TALK 10:29, 31 March 2017 (UTC)

Ok, I see what you mean. I was under the impression that a fluorescent lamp operates in the arc regime, but could be wrong. None of the glow discharge lamps I've built ever needed a ballast to control the current; they could be connected directly to the transformer or line voltage, like a neon sign. The ballast seems to be necessary only when I crossed into the arc regime, where the current runs away on its own. (The exception of course is a flashtube.) Once this happens, no increase in voltage is going to extinguish the arc (which is what I was getting from the opening paragraphs.) Zaereth (talk) 17:17, 31 March 2017 (UTC)

You make neon lights? Cool. I know the term "arc light" is used for some gas discharge tubes, but I'm pretty sure fluorescent lamps and neon lights operate in the "glow discharge" regime. I don't know exactly how the current is limited in a neon tube light. I believe I read somewhere that the neon sign transformer is designed with large leakage inductance so it acts as a ballast, limiting the current so it doesn't run away? --Chetvorno^TALK 18:37, 31 March 2017 (UTC)

A photo of of one of my few surviving argon-flashtubes. For this one I used Viton O-rings to seal the electrodes initially, and then an epoxy to permaneantly seal them after bombardment.

I've made some simple ones in my younger days. Mostly I was trying to figure out how to make flashtubes that wouldn't explode or lose pressure over time. (Made well over 400, of which only four survive to this day. It really wasn't until I began talking to experts, like Harold Edgerton or Don Klipstein, and even had one owner of a company give me a bunch that he was going to throw away, that I finally succeeded in making my first operational laser.) I've got a nice, big neon-sign transformer that's perfect for bombarding (removing oxygen that adheres to the glass) small flashtubes and arc lamps. The transformers needed to bombard a full-size neon sign are monstrous. (Typically about the size of a 55 gallon drum. Way out of my price range.)

Nearly every book I've seen on fluorescents seem to call the discharge an arc. Of course, that doesn't mean they are always correct, so now I'm curious to dig deeper. Neon signs definitely operate as glow discharges, though. The small neons (like you often see in an extension cord to indicate it's plugged in) are connected directly to the line. Zaereth (talk) 19:06, 31 March 2017 (UTC)

I talked to a friend of mine who knows more about this than I, and he turned me on to this source, which gives wonderfully detailed explanations of the various discharges. (Cleared up a few of my misconceptions.) Fluorescent tubes are definitely low-pressure arc lamps, because they require the thermionic emission of electrons from the cathode to support their operation. The exception to this are cold-cathode fluorescents (often used in computer backlighting), which are are basically fluorescent tubes operated like a neon light. Neon signs also fall under the category of cold-cathode lamps. When a fluorescent is perpetually lit, the main process of wear becomes adsorption of the mercury onto the glass, which eventually reduces its pressure enough to move it into the glow-discharge regime, causing it to dim and eventually form visible Faraday cones that move from the ends toward the center. (This may also bee seen during start-up, especially in 8 foot T12s, before the tube heats up and increases in pressure as the mercury vaporizes.) Zaereth (talk) 00:28, 4 April 2017 (UTC)

Say Zaereth, I'm sorry, when I wrote the above comments I forgot that you have a great deal of experience with this stuff in building lasers (guess it was a senior moment). Most of what I wrote is probably not news to you. You met Harold Edgerton? Wow.

Yes, that gas discharge page by Calvert is absolutely great, I ran across it before, that's where I learned what little I know about the subject. It would be good to put some of that stuff into the Glow discharge article, which in my opinion needs a rewrite. BTW, editor Glrx has the most expertise in this area of anyone I know, and he would be the person to ask questions of. --Chetvorno^TALK 01:55, 4 April 2017 (UTC)

Hmm. This comment is going to take some time.

I'm always in an extended senior moment.

The negative resistance issue is just that at some point on a transfer curve the dV/dI slope becomes negative.

One issue is whether current or voltage (or neither) is controllable. The WP article intro uses voltage as the independent controllable variable, but that is probably the less common one. As the voltage increases across a positive resistance, we expect the current to montonically increase. For a (differential) negative resistance device, increasing the voltage a little bit might cause the current to decrease from its previous value. For each voltage value, we can plot a stable current.

Sometimes, the current is controllable: for each current, there is a specific output voltage. Somewhere in that transfer curve the voltage falls as the current increases. The classic example is a tunnel diode.

If the other variable is driven, then strange, counter-intuitive, things happen. If I increase the voltage across a tunnel diode, the current will slowly increase. At some point, I hit the negative resistance point and the current takes a discontinuous jump to a much higher value. Increasing the voltage when the resistance is falling becomes harder to do.

Now we get into disputed territory.

Many people use a glow discharge lamp as an example of a negative resistance. Chetvorno supplies the plot above, but the plot has problems. It shows a continuous transfer function on the hysteresis path, but most references do not show that. Instead they show a breakdown transition or an unstable region. I think a breakdown is the better explanation because the Townsend discharge to glow discharge involves a reorganization of the charge within the breakdown. During the Townsend discharge, the electric field is evenly distributed. At breakdown, the world changes: most of the field is at the cathode fall and there is a small field in the rest of the tube (positive column). It takes some time for that field to develop. The plot of the discharge tube is a steady state transfer function rather than an instantaneous IV characteristic (like a tunnel diode). The hysteresis is there because once the charge distribution is set up for a glow, it will keep that distribution until the current falls much below the current needed to start the glow in the first place.

In many cases, one can plot a steady-state transfer function with an apparent steady negative resistance. IIRC, Laser tubes are an example: the more DC current run through them, the lower the voltage drop across the tube. Zaerth points out that the gas is heating up, and that means there an additional state change rather than a negative resistance. Consider a negative temperature coefficient thermistor; if I increase the voltage across it, it self heats and becomes a lower resistance. We don't consider that a negative resistance because a steady state DC transfer function could show a negative resistance characteristic.

Similarly, some arc discharges have a thermal explanation. Joule heating at the cathode increases the temperature enough that thermionic emission takes over. The steady state transfer curve shows a negative resistance, but there's an underlying state change.

Yes, Calvert is an excellent source that explains a lot of this material; I stumbled across it a couple years back. User:Glrx/sandbox#Glow character.

BTW, the small neon lamps are not connected directly to the line; there's a limiting resistor. A friend and I were looking a neon night light when he suddenly realized the night light took more power when it was off than when it was on. A photoconductive cell was used to short out the lamp when it was too bright.

Chetvorno, Zaereth knows much more about gas discharges than I do. Years ago, I asked a friend about gas breakdown, and he gave me a fabulous introduction. Most of what I've picked up since then is from some ancient text books and a few research missions to the library. Much of that effort has raised more questions in my mind than they've answered. I've been meaning to read Penning's book, but I haven't gotten around to it. I've also been meaning to measure the dynamic characteristics of an NE2 and see if shielding it can increase turn on time.

I had a lot of fun talking to Doc Edgerton. I got an invite to be on the boat where he was trying out the side scan sonar just before going on his Nessie quest. The monster search was a good excuse to try out some interesting equipment. Driving a piezoelectric sonar transducer got me to understand inductors and impedance matching much better.

PS, say hi to Steve.

Glrx (talk) 04:17, 4 April 2017 (UTC)

Thanks, I'm flattered. The math is my weakness so I have to really try hard to follow along. (Some of the best books on lasers are written in nearly 80% math.) My strength and weakness is that I have to try to understand things from a mechanical perspective. I really envy your ability to read the complicated math equations. I did learn quite a lot from both yours and Chetvorno's explanations. I had always assumed the glow discharges were stable, in the sense that they wouldn't runaway, but it turns out I was mistaken. Checking some books, Chet is correct. The high reactance of the neon transformers provides the current limiting needed, which makes them perfect for He-Ne lasers or small CO2 lasers. Otherwise there needs to be a current-limiting resistor or ballast. However high-powered CO2s require a ballast, or else you need a current limiting resistor about the size of a 1500 watt space-heater. These lasers can't perform as arcs because the temperature increase tends to depopulate the lower energy states, reducing output to nil. Zaereth (talk) 07:40, 4 April 2017 (UTC)

The hydraulic analogy is called a non-Newtonian fluid. Normally for friction, you expect that increased force leads to increased velocity. In non-Newtonian fluids, increased force gives decreased velocity. Most common example is cornstarch in water. Gah4 (talk) 21:38, 23 November 2020 (UTC)

Differential resistance is only relevant to time-varying currents.

Seems to me that it is also relevant to static stable equilibrium, so not time-varying after equilibrium is reached. Once they are on, discharge lamps run at constant current. That changes when you turn them off. Gah4 (talk) 21:41, 23 November 2020 (UTC)

That's an interesting interpretation. But actually, discharge lamps are not at stable equilibrium and do not run at constant current - without a ballast. The ballast adds sufficient resistance or impedance (AC resistance) that the series combination of lamp and ballast has positive differential resistance and thus is stable (see image at top of article for source) A circuit with net negative differential resistance, like a lamp without a ballast, cannot operate stably at a constant current; it will oscillate or (in the case of the lamp) undergo runaway and burn out. --Chetvorno^TALK 00:34, 24 November 2020 (UTC)

Yes I meant with ballast. Though that is mostly because we like constant voltage power sources. It constant current power sources were more common, then that wouldn't be a problem. In either case, you choose the point on the IV curve where you want to operate, and, with appropriate power supply, operate at that point. Gah4 (talk) 01:47, 24 November 2020 (UTC)

Well, here's the deal as I see it, and maybe I'm wrong, but for constant current you need positive resistance in the circuit of some kind. For most lamps, they run on an AC signal, so the current is never constant. In fact, it's cycling on and off at twice the frequency, or 120 cycles per second. The current in this case is limited by the ballast, which uses inductance to produce a positive resistance to counteract the lamp's desire to runaway. The lamp may be operating within that unstable current regime, but the circuit as a whole must exhibit a positive resistance/impedance to limit the current. And this won't work with a truly stable, DC current, because inductors only limit AC or pulsed signals. Consider then a DC arc lamp. For this you need to replace the ballast with some actual resistors or space heater in the circuit, because a ballast won't work with DC, but the overall resistance in the circuit remains positive as long as the current is stable. Zaereth (talk) 03:04, 24 November 2020 (UTC)

Ok, I had to dig through some old emails and letters. Where I live, we don't have a lot of great libraries, so when I was learning about this stuff, I tended to contact experts and manufacturers and even inventors like Edgerton and Townes directly, and ask them a million questions. Negative resistance was one of those things I just couldn't wrap my head around. The name makes it sound like it's actually pulling the electricity rather than resisting it, but that's not the case.

This explanation, although it's OR, I thought I would share if it helps. This comes from Don Klipstein. The resistance in the gas is always positive, in the sense that it is always resisting the flow of electricity. "Negative resistance" means simply that the resistance behaves the opposite of a normal resistor when it encounters a change in voltage/current. The gas always resists the flow of current, or else no light would be emitted (ie: xenon is such an efficient gas because its resistance is so much higher than others), but when current increases, both the voltage and the resistance decrease.

So, when the current is held stable, the resistance in the gas is positive. For example, in a DC glow discharge, the current is kept low, and the resistance and voltage are high. In a DC arc lamp, the current is higher and both resistance and voltage are lower. But the resistance is always positive, in that is always resists the flow. It is called negative only when it is behaving abnormally to a change in voltage/current. I hope that helps. Zaereth (talk) 04:24, 24 November 2020 (UTC)

First, WP:OR is allowed in talk pages for discussions, just not for the actual article page.

Yes AC and current limiting inductance or capacitance (the latter is usual for electronic ballasts), but as above, that is because we like constant voltage power sources. Hot and cold cathode tubes will run fine on DC, though the glow tends toward one end. Short arc lamps are commonly run on DC, though. Small neon lamps also run fine on DC, and it isn't so unusual. Only one electrode glows, though. (Also for Nixie tubes, there the numeral glows, and not the mesh anode.) But otherwise, the purpose of the ballast is to convert a constant voltage power source to a (more or less) constant current power source. Gah4 (talk) 10:02, 24 November 2020 (UTC)

That makes sense, although that's not necessarily the case with every type of lamp. To use an example of a lamp I'm most familiar with, a flashtube operates without any type of ballast. When the lamp is triggered, it goes through a brief stage of negative resistance. It experiences and huge drop in voltage and a huge spike in current, but then, depending upon the impedance of the gas, it reaches a point of equilibrium where the voltage will be equal to some percentage of the square root of current. At this point, the resistance is no longer negative anymore, but positive. Of course, that level of current can't be maintained for more than a few hundred microseconds, maybe a couple of milliseconds at the outside edge, before it starts vaporizing the electrodes, but there is a point for most of the flash where the resistance is positive. Unfortunately, you don't reach that point until the voltage is extremely low compared to the level of current, and at that point it's not negative resistance anymore. Zaereth (talk) 01:13, 24 November 2020 (UTC)

Current voltage curve of a gas discharge

Most of the above is covered in the article (which I wrote most of) with sources. DC resistance is called static resistance, and all passive devices have positive static resistance. Negative resistance devices have negative differential resistance. The sections "Stability conditions" and "Operating regions" give conditions for stable and unstable operation. Discharge tubes have a type of negative resistance called "current controlled" (CCNR), which is stable when driven by a source resistance larger in magnitude than the negative resistance. The source resistance can be a resistor, or a current source to save energy. Other negative resistances, such as tunnel diodes, are a type called "voltage controlled" (VCNR) which are stable when driven from a positive source resistance of smaller magnitude than the negative resistance, such as a voltage source. See the "Types" section. Negative resistances can be biased to be stable, unstable (oscillate), or bistable, see the graphs in the "Operating regions" section.

As can be seen from the graph at right, discharge tubes actually have two regions of negative resistance, one beginning at breakdown voltage covering the "normal" glow discharge region from D to G, and the other beginning at the onset of arcing I. As Zaereth indicated, the resistance in the arc region eventually becomes positive, at point J. The operating point is determined by where the load line crosses the curve, which depends on the source resistance. Fluorescent and neon lamps have ballast resistances which establish an operating point in the glow region, around G or H. As Zaereth said, flashlamps are driven with lower source resistance so the operating point is beyond J, in the positive resistance arc region. --Chetvorno^TALK 12:07, 24 November 2020 (UTC)

Yes. I think I am only arguing against voltagism, the bias toward thinking about problems in terms of voltage sources. I do note that your graph has current as the independent variable. If driven by a constant current source, there is one choice for voltage, and so operating power. Driven by voltage sources, not only is there a need to limit current, but also to start the current flow. Many lamps won't start with an appropriate voltage and current limiting device. Many different ways have evolved to start lamps, to allow them to reach the operating point. I used to know the story about the original Edison lighting system, with a generator design that was especially good at keeping the voltage constant. I believe that there are also generator designs for constant current, though not so often used. Gah4 (talk) 17:20, 24 November 2020 (UTC)

Current/voltage curve of gas discharge, with current as a function of voltage

I agree with you about "voltagism". Our common sources of electricity have low source resistance and can be best modeled as voltage sources, but it is just as valid to model them as current sources. Gas discharge tubes are current controlled devices, and the current/voltage graph of a gas discharge is traditionally shown with the current as the independent variable, as in the above graph. I drew one with the axes switched, at left, that I thought might be more understandable to people used to thinking of voltage as the independent variable, which emphasizes the discharge tube's multivalued current function. --Chetvorno^TALK 18:03, 24 November 2020 (UTC)

@Zaereth: You have a lot of experience with gas discharges; does the curve in these graphs look right to you? Specifically the point (I). These were drawn from a 1975 graph in a text that I thought was reliable, but since then I've seen a number of gas discharge IV curves, and most show the transition to arcing at (I) occurring at a lower voltage, lower than the breakdown voltage (D). --Chetvorno^TALK 18:12, 24 November 2020 (UTC)

As far as I know, that particular graph comes from the 1940 paper by Druyvesteyn and Penning, called The Mechanism of Electrical Discharges in Gases of Low Pressure. If you have an account, you can view it here, but like most scientific journals you have to pay per view. I've never seen the original graph, but books I've read that reproduce the original all have the arc-transition voltage being higher than the breakdown voltage, just like in your chart. Now this has been reproduced thousands of times, and a lot of the variations I've seen are not too big on specifics as much as just showing a rough curve for demo purposes, and quite a few of those rough drawings do show the arc voltage as being lower, but the books that draw theirs from the original all have the transition voltage being higher than the breakdown voltage.

And I do agree that current is the controllable variable here. Once the current starts to runaway you can't just turn up the voltage to reel it back in. You could call a negative-resistance device one that the voltage is controlled by the current rather than current controlled by voltage. Zaereth (talk) 23:19, 24 November 2020 (UTC)

What I would say about this chart is that it's a rather static description of a very dynamic process (as graphs always are). Keep in mind this is a measure of a discharge between two flat plates parallel to each other in very close proximity, like a capacitor separated by an air gap. When I factor in the other dynamics of pressure and such, I am reminded of the works of Loeb and Meeks, who defined a spark as an unstable, transient state between a less stable and a more stable state. Nearly all discharges begin with a spark. This starts in the Townsend regime, and is only visible in a cloud chamber. This then leads to the formation of streamers, which are visible with a corona type discharge. These may start from each electrode, or may actually begin at any point between them and work their way toward each electrode. Once the streamers bridge the electrodes, the discharge begins. The glow regime is a more stable state than the spark was, and current can easily be limited within this range, but as it crosses into the arc regime it sparks again (although not what we commonly picture as a spark, as Loeb and Meeks point out, it fits their definition) and current enters the more-stable arc regime. That's what I see when I look at this graph. Zaereth (talk) 00:59, 25 November 2020 (UTC)

Hey Chetvorno. In studying this further, the original experiment by Druyvesteyn and Penning used flat electrode plates of 10 square centimeters, separate by a distance of 50 cm, in an atmosphere of neon gas at 1 torr of pressure. The only thing I would recommend changing is the label "Current" to "Current density (A/cm²)", because that's how the books that reproduce the original graph label it (for example, see; Non-Equilibrium Air Plasmas at Atmospheric Pressure, page 29), and it simply makes more sense to describe it in those terms. Otherwise I'd say it is correct. I hope that helps, and Happy Thanksgiving! Zaereth (talk) 19:59, 25 November 2020 (UTC)

I suppose current density sounds good, but I am not sure how it scales in all cases. In some regions, it isn't a uniform current density between the electrodes, and also is affected by heating of the gas due to currents, and then convection as it warms. As well as I know, neon signs and fluorescent lamps run about 0.1 atm, though newer lamps might run higher. The smaller lamp tube allows higher current density, with T5HO being pretty amazing in light output for a small area. Gah4 (talk) 20:21, 25 November 2020 (UTC)

That's exactly the point. 1 amp across 1 square meter is not going to generate enough heat to arc. Not even enough to glow. However, 1 amp across 1 square millimeter is going to be plenty hot for arcing. Of course, pressure plays a huge role, as does distance and diameter, but most of the cold-fill pressure depends upon the desired pressure at operating temp, and that is determined by the tube length and diameter, and the gas type. You want the arc to fill the tube, but not overfill it, if you know what I mean. The neon tube is a good example. When you build a neon tube, you run a current through the gas, much higher than the normal operating current. Then you vacuum the gas out, and as you do the discharge expands, until, by the time you reach just a few torr, the bulk of the plasma is located at the surface of the glass, bombarding the glass with high speed electrons and ions. That pressure is no good for operating current, so the tube is refilled with neon until the discharge just fills the tube at normal operating current. But it's the density of the current at any given spot that determines just how hot the gas will get. Zaereth (talk) 21:53, 25 November 2020 (UTC)

I think it doesn't scale, though. 1 amp/square mm sounds high, but say 1 amp/square cm. Maybe not so bad. But I don't think the same current density, 10,000 amps over a square meter, will work the same way. Gah4 (talk) 22:08, 25 November 2020 (UTC)

Even without getting the $40 paper, this one explains the density-electrode spacing scaling, but also mentions the filamentary nature of the discharge. Gah4 (talk) 22:14, 25 November 2020 (UTC)