at nanovolts, do resistors still experience heat at high amperage? - Page 1
Forum section: electronics / beginners

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 09:16:04 am
content: so if i am at 1 nanovolt, and i have 10,000 amps. does the resistor heat up?

author: tooki
author_rank: Super Contributor
timestamp: July 27, 2023, 09:28:22 am
content: Well, 1 nV * 10 kA = 10 μW. So yes, it’ll heat up, technically speaking. Just not by much…Of course, that requires your resistor to have just 0.1 pico ohms of resistance, so basically it’d have to be a superconductor. Or really, really, really big.

author: larsdenmark
author_rank: Regular Contributor
timestamp: July 27, 2023, 10:20:27 am
content: The formula isP = I^2 * RwhereP is power (W)I is current (A)andR is resistance (Ohm)so it is easy to calculate the power generation for a known resistor.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 10:27:39 am
content: so when they say copper wire can take about 65 amps, is a little uninformative.because it can actually be giga-amps, if the voltage is low enough.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 27, 2023, 10:56:41 am
content: The copper wire also defines the resistance. So the voltage is very small only if the current is not very high or the material is a superconductor.So normally high current and low votlage don't go togehter. It takes extra effort to make a 1 n Ohm resistor (AFAIR I have seen one with superconducting connecting wires) and it would be hard to get much below this.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 11:01:23 am
content: Low voltage and high current dont usually happen, unless ur using a step down transformer.

author: Shonky
author_rank: Frequent Contributor
timestamp: July 27, 2023, 12:45:31 pm
content: Quote from: robowaffe on July 27, 2023, 10:27:39 amDo you understand V=IR? How do you get "giga amps" at a low voltage? You need to have an a very low resistance i.e. getting near super conductor...Rating copper at a certain current rating is based on practical temperature rise / acceptable voltage drop per unit length.Quote from: robowaffe on July 27, 2023, 11:01:23 amA transformer has nothing to do with it.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 12:48:00 pm
content: u can get 20 volts at 100 milliamps, and if u step it down to 2 nanovolts youll get 1 gigamp in the circuit, as long its a really pure conductor.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 12:54:59 pm
content: nah... it wasnt that good.https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWK0wHYwA4MDYCcAmBDfDXAZhOxAUmutoQFMBaMMAKABUR9V9xstMmSr58AFhAYQkmJAjM5uBPlwZUGceIRkw2MhpC5o4sJFzi8BhNj1gyyXE-YA3HnwFDIkvbT8zafH8-aAR2ACcQMm9PKJFYwLh2AHd3fkJRDzFZFLjM-mFRCSgItJ4bPLKQuA4gAI only could step down 100 into 1 milliohm, 2 millivolts and i only got 10 amps.heres version with led.https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWK0AsKAcYBs6CckB2SAZmPUixSxCJCQUjoFMBaMMAKABUQAmA3uCyNS1XrxQh0dZJAgsYkXAl64ClNbjC5NIXKjCk1KMAXTFKyHbg4A3PgKEjIk7I3cgUjXh-fQEHABOIMQuTiEW4d5wHADuEWKOonwSUEEJDoLJoZJ+cJzx-IK8CIlZYZDcmSFh2eiS0mCsjfCMLPomRgQmZhaSLFiwcPwI6igIxKa46KzsVjoc2FmRZJJFIfV8IAAmTABmAIYArgA2AC4sJ0zb4FB3igXVJWUpuRxAAso the step down transformer is 5 henries, but i only need 1 for the whole system.the step up transformer which gets the output, where i only need a low amount of them, only about 16 probably, is only 10 nanohenrys and i still get a fairly good pulse out of it, but i dont how much time this is in seconds.But I have to add one thing to it yet.

author: mikerj
author_rank: Super Contributor
timestamp: July 27, 2023, 01:01:32 pm
content: Quote from: robowaffe on July 27, 2023, 10:27:39 amThe voltage drop depends on the current and the length of the cable, so it makes far more sense to specify current. Having a power cable with a maximum rating of 10mV/meter is not very useful to anyone.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 01:08:52 pm
content: https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWEBmaCCc7nLADl8ggGwBMB6RIA7JCEgrQgKYC0YYAUACoglUngitbJRIkALCFx0osCCxiR0JMMnFUi7FWE3oIuWOPwUSkcenNUcszOg4A3Xv0HCzLqB-G1TH2jAQcAE4objrCyJRhHqaQHADuKBFOAiK8ElBBiaLOqchufrCQnAl8AiTEySGSsTylIOJGWXRkUuCF8B2dRaxU4B0gLFTQGqbImFRWVBRUkiwk0OhFCAjIJGokFjRg4qzsNpgcOilJCC11p9ICACZMAGYAhgCuADYALizPTFfgvrJF8ZVytkyukapU8pJckRpNJFEUfiwDGZjKQzBZ0FZqMNUWMMZNptUFgcSjloZUGrDMlDpKkvH4-v8ErS4E0KRkgAok so this sim is one step down transformer, to 1 milliohm conductor, to an isolation transformer at 1 millihenry, to a 100 nanohenry step up transformer to get the led output.Ive got a feeling that 1 millihenry transformers cant be 1 millohm on the winding, so this doesnt work yet....So I worked out im wrong about things, thankyou, Im wrong about the gigamps.(unless ur super conducting.)https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWEA2aAWB8CcBmSy0AOBBAdkJFMhCUxoFMBaMMAKABUQAmUr8ZajhzJuXNCAp0YkCI2lcEhHgS6RhWBGKRh0MksjBYwCHLxxYsUaBaysAbt179BkcWAFRPaaqs-UYCKwATiB4bh5CIu7+3HCsAO6hwo58kaLikMFJIjypyWF+0GyJudwIOU4FmZylaETZNFwUFNIy4PAgcrDIhKRYaKRoCj1YyKMQrXBgvASmXEakmGCQPFY2rO55IuU+TjvcIAAm9ABmAIYArgA2AC6MV-SH4H5WMgkpZRV8XGJQHB8FNLCZqvNqyHTeYzlQzGObmSzSHp9AZDcqEUbjNYWd6lYEfOotLJAnoNbwxGDFUlwUn1TIhNKaCgMjz+Io4vYeUqMv5AAhere this one is 100 milliohm windings.how long is this pulse duration in seconds??

author: Psi
author_rank: Super Contributor
timestamp: July 27, 2023, 01:10:59 pm
content: Quote from: robowaffe on July 27, 2023, 10:27:39 amWhen they say copper wire of X width can take about 65 amps they are not just giving you a wire thickness and number of amps. They are also giving you the resistance of the wire based on the fact it's copper and at a specific thickness.So its not uninformative. For the information giving it's accurate. Ideally they would also give you the max temp rise in the cable used to give that 65A rating, but meh, most of the time you can assume a typical temp rise. It does matter though, if the cable had silicone insulation it can handle more heat than PVC and so the same thickness wire can carry more amps without melting its insulation.If you have 65 amps flowing through the cable, and you know its resistance, then you know how much power/heat is being wasted in the cable. The voltage used in the circuit is irrelevant. Could be 100V or 100,000V. The only voltage that is relevant is the voltage across the length of the cable, but you don't need to know that because you can calculate the power lost in the cable just from the Resistance and the Amps. Power = R * I2

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 01:28:45 pm
content: I see, Psi. So the rest is computable off that.these. this is what I need now.https://www.zl2pd.com/Small_nH_Inductors.html

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 01:32:35 pm
content: https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWKsBMBOAHANjQFixmihtgOwa4imQhII0ICmAtGGAFAAqIKpK4YfgGYhWcKTEZaySBGYxIWHGDwpZuQiiFoxWaFjAV8fDJCGCU-GGhvsAbjz4DhkSmytQQuGms80YCOwATiBCrs6hohE+cOwA7pFivMJRlpSQwYmOKWJh6TIcCck8CElOeVBc2V4aWQjEIFJgLE3wMvAoCGC4EghCGKREAyo80PVg5GiQKLiCaGDaWCykyDZo7GBYObRYPk6lPiAAJowAZgCGAK4ANgAuzNeMR+B+BfHVnWX8aZXcxRUiXIERrgFovZh6AxGLAmMyCTYgeSwDpdHpYPpTRRmHpCVa2IrlYHFXC1DIhQGhYEU7z+N4JalwLIkqRkuoNCkHPywQrVTnFeos9gAZ2qE1yUTFnggF2uQsY7wpkopPwyBP4kuKKsyAp4kCkCBJupZyA4AGNaIa1PrdkaufA4BAENAtFt6KRSBNCOQwEgFLJ3ggiLbA99vJUEiGeGGuqH-AHBFGGAmrZUHDHgw0U7To68AhstjsYvqgyn+CcLjd7o9nhBaQoOEASo the blip speed here, unless falstad is confusing me, I think its about 1/5th of a second, through the transformers, because im going against shorting a 1 microfarad cap into an led, and I know that one, but I spose the sim could be misleading..

author: AnalogTodd
author_rank: Regular Contributor
timestamp: July 27, 2023, 02:11:05 pm
content: This is where you learn all about parasitic effects in electronics. A copper wire from point A to point B is not a pure short. It often gets treated as such because relative to the rest of the circuit the resistance is smaller than typical tolerances for devices and therefore can be ignored.This is why you can't get away with 30ga wire in the walls of your house for power. The resistance is so high relative to the length of the wire that the constant amperage of a load would create so much power dissipation in the wire that it heats to a point it can cause a fire or open up like a fuse. Hence the need for 14ga for a 15A circuit, 12ga for a 20A circuit, etc. in wiring codes.The second thing to think about is time in all of your equations. Not only do wires have resistance, but they have inductance as well. Again, the effect of this is often negligible in most circuits, but as you get to higher frequency signals (how fast does a short actually occur?) you now limit the current you can get through because of the V = L * dI / dt equation. The link you added shows that just a simple wire bent into a hairpin is 15nH of inductance, and the effect of the shape is actually going to be a lower inductance than a straight wire of the same length.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 02:18:27 pm
content: Yes. I'm planning on making a computer.And now if its going a gigahert, the whole thing is a transformer, not just the isolation transformers.If the "transformers" in my design end up looking just like little bits of wire, whats the rest of the wires there doing as well! its all inducting to itself!And the pcb its sitting on all turns into a capacitor at high enough frequency, So if an isolation transformer is discharging for example, the whole area around it is getting back-charged like a capacitor as it happens.So I dont know what this thing is going to shape into in the end, and how fast it can go, No idea yet, So once I've got the circuit down, I've got no idea how its going to perform in real life!!!But Im pretty sure my computer design idea is a good one, but who knows how good it actually is in the real, or what limits Im going to meet on the way to actually get it running!I dont know about these limits yet. I'm still drawing up the core, and I've got some things on the outside running it I have to do as well, But when it comes to running it for real, maybe I should be a bit more timid about going so quickly straight away and getting something slow working first. I dont know!!!

author: tooki
author_rank: Super Contributor
timestamp: July 27, 2023, 02:42:39 pm
content: Quote from: robowaffe on July 27, 2023, 10:27:39 amWell, a particular wire can be rated for 65 amps according to a particular standard.Quote from: robowaffe on July 27, 2023, 10:27:39 amNo.The practical current capability of a wire is ultimately a function of a) how hot you can tolerate the wire getting, and b) how much voltage drop you can tolerate in your application.Remember that the current passing through the wire, and the resistance of the wire itself,producethe voltagedropacross the wire. It's not the voltage you apply that matters for calculating with Ohm's law, it's the voltage drop.What I mean by that is this: if you are using a cable for a 230V power cord, for example, then the 230V doesn't matter. What matters is the resistance of the cable, plus the impedance of the load on the other end. Suppose that this load is drawing exactly 10A, and that you're using 10 meters of 1.5mm2cable, which has around 24mOhm/m resistance (total for both line and neutral conductors). Then the resistance in the cable is 10m*24mOhm/m = 240mOhm. From this you can calculate that the voltage drop is 240mOhm*10A = 2.4V, and that the power dissipated is 10A*10A*240mOhm = 24W.See how the voltage in question isn't 230V, but 2.4V? It's theresultof the material properties of the wire and the current drawn, not the other way around.24W of heat dissipating through 10m of cable laid out in the open is reasonable, since at room temperature that will have no trouble dissipating into the air. If the cable is coiled up or otherwise confined, the heat can't escape as readily and it could heat up more.The current ratings of cables and wires basically give you a rough idea of what current can be pushed through it without it overheating. (Different standards have different permissible temperature rises for different installation types and insulation materials.)The voltage ratings on wires don't have anything at all to do with the conductor size or thickness: they are the voltage rating of theinsulation. So for a 230V power cable, you need cable whose insulation can withstand 230V. (In practice, you'll use something a bit higher.)So as you can see, a huge amount of a wire or cable's voltage and current ratings have to do more with the insulation than with the copper inside.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 02:54:05 pm
content: Yes thankyou.The voltage does matter tho, you cant step down more than 100:1 without the resistance of the copper eventually dissipating the voltage in the secondary.voltages below 200millivolts cant even get through the copper.So extreme low voltages cant even exist in practical example.Thats what I didn't think of, when it came to the 20 gigamps... unless ur super conducting.

author: Shonky
author_rank: Frequent Contributor
timestamp: July 27, 2023, 03:28:51 pm
content: Quote from: robowaffe on July 27, 2023, 02:54:05 pmJust no. None of this is how it works.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 04:09:28 pm
content: Have you ever had 2 millivolts generating a high amount of current in copper wire? No, thats what I'm saying.

author: rstofer
author_rank: Super Contributor
timestamp: July 27, 2023, 05:28:56 pm
content: Ohm's Law is a law, not a suggestion like speed limits.V=I*RP=I2*RP=V2/R

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 05:47:05 pm
content: yeh but does ohms law explain voltage induced shorting? (as in yeh, i know but there is more interesting things for me to think about.)

author: themadhippy
author_rank: Super Contributor
timestamp: July 27, 2023, 05:54:53 pm
content: QuoteDefine low voltage,below 1000v AC/1500v DC is classed as some as low voltage.

author: golden_labels
author_rank: Super Contributor
timestamp: July 27, 2023, 06:08:12 pm
content: People. Doesn’t OP — with already two topics closed — remind you of one recently banned? Just a hint…

author: tggzzz
author_rank: Super Contributor
timestamp: July 27, 2023, 06:08:48 pm
content: Quote from: Shonky on July 27, 2023, 03:28:51 pmIt may be worth scanning this post and the reply fromanother of the OP's topics...https://www.eevblog.com/forum/beginners/analogue-risc-core-amazing-idea/msg4969408/#msg4969408https://www.eevblog.com/forum/beginners/analogue-risc-core-amazing-idea/msg4969414/#msg4969414I can't insert them as a direct quote, sincethat thread has been locked. The gist is....tggzzz « Reply #52 on: July 19, 2023, 04:36:25 am »Quote from: robowaffe on July 19, 2023, 01:21:43 amIsn't the output voltage of a transistor just whatever power you connect through the collector to the emitter - and u just open the switch with a little current?That's word salad.I'm beginning to wonder whether "robowaffe" => robot waffle or robot arms/weapon. I'm more than happy for it to bedemonstratedthat I'm wrong.robowaffe « Reply #53 on: July 19, 2023, 04:37:28 am »robot waffle will do. sounds like im going to ice cream parlour.

author: tggzzz
author_rank: Super Contributor
timestamp: July 27, 2023, 06:12:01 pm
content: Quote from: rstofer on July 27, 2023, 05:28:56 pmEr, Ohm's law is the codification of a simple relationship that can be valid in some materials under some conditions. It isn't a law that has to be obeyed. Under some conditions the voltage across a specimen is linearly proportional to the current flowing through it, the constant of proportionality being termed the resistance. Most specimens won't obey Ohm's law, e.g. a semiconductor junction, or an ionised gas, or a conductor that is overheating

author: MrAl
author_rank: Super Contributor
timestamp: July 27, 2023, 06:27:56 pm
content: Quote from: robowaffe on July 27, 2023, 09:16:04 amHi,When you get into high level values like 10000 amps the more physical aspect of the construction kicks in. That is, it is no longer just an electrical formula like Ohm's Law it becomes a deeper physics problem. That requires understanding how metal heats up over time as well as how surface area acts to cool the construction. This is because, under normal circumstances not in some advanced lab, the wire would have to be pretty thick and that means the surface area would be large.You also get into end-of-wire heat dissipation, where something connected to the ends of the wire acts to dissipate some of the heat, and that means that the physical length of the wire comes into play and can be quite significant in determining what actually happens in real life. Real life vs simpler theory can be very surprising sometimes.There are unusual situations like this where you really have to delve into the physics of the problem, and even that may not be enough you might have to look at examples where people did measurements and tabulated data and maybe graphed the results too. Another example is how electrical arcs form and maintain an arc. The physics is not very intuitive at all.What all this means is you have to specify the variables exactly and that includes the physical dimensions of anything and everything in the setup. If that is too much trouble, then you have to be satisfied with some basic theory like Ohms Law.

author: BillyO
author_rank: Super Contributor
timestamp: July 27, 2023, 06:29:01 pm
content: Quote from: robowaffe on July 27, 2023, 04:09:28 pmYou would need about 10 million amps to generate 2 millivolts across a 0.1mm length of 10mm diameter pure copper wire.Two things:1) If you could get 10 million amps, the copper would likely explode with such vigor you would blow yourself up as you would be dissipating 20,000 watts.2) Where the heck would you get 10 million amps?

author: MrAl
author_rank: Super Contributor
timestamp: July 27, 2023, 06:30:57 pm
content: And, when the wire gets that short there is a lot more to it than just the wire length and the amperes.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 27, 2023, 06:33:16 pm
content: Quote from: themadhippy on July 27, 2023, 05:54:53 pmLow voltage as in 200millivolts and 1 amp. it wasnt as much as i thought it was going to be out of failstad.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 27, 2023, 06:57:01 pm
content: If you have direct currentI(in amps) passing through a purely resistive thing with resistanceR(in ohms), it will generateR I²(watts) of heat. The voltage is irrelevant.To see huge currents without large voltage drops, you need tiny resistances, which do not typically exist in the human scale, specifically temperature and pressure. At truly low temperatures, or very high pressures, several metals and materials are superconductive. Niobium becomes superconductive below 9.26 Kelvin (-263,89°C or -443°F) at human-scale pressures down to zero pressure, for example. Even before becoming superconductive, many materials resistance does drop to close enough to zero to be considered having basically no resistance at all.That does not mean it does not happen in reality. For example, Jupiter's massive magnetosphere is induced by currents flowing in its outer core, composed of metallic hydrogen at stupendous pressures with almost zero resistance (some believe it might actually be superconductive). There there are almost certainly immense currents with low voltages, that cause some heating in its outer core.Plasma channels in space are another situation with very low resistances. Technically, they have negative resistance, because increasing current flow reduces the voltage drop. On Earth, plasma channels (thunder strikes) don't get into truly low resistances, but in space, they can contain immense currents for indefinite durations. The Jupiter-Io plasma channel (flux tube) current is estimated at 3 MA (three million amperes), but the voltage potential is also on the order of 400 kV; for a resistance of about 0.13Ω across a million kilometers or so as it arcs from Io's poles to Jupiter's poles. In any cubic meter there you see a significant current with an almost unmeasurably small voltage difference; so much so that it is easier to treat as a charged particle flow instead (which it physically is).Even with a significant z-pinch (Lorentz force due to the current flow squeezing the flux tube tighter/smaller) there is heating due to the current flow.I'm pretty sure you can pick a volume out of the Io flux tube, where the potential difference along the flux is about 1 nV, and the total current about 10 kA. You'd see that volume being heated by about 10 µW, too, if you were to exclude other radiation-induced effects acting upon the charged particle flux.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 27, 2023, 07:01:56 pm
content: A more Earthly example of relatively low voltages but high currents occurs in welding. Using a few hundred amps at a couple of dozen volts is common, of course depending exactly on what you are doing. In arc gouging, you might use 20V and several hundred amps ("as many as you got").

author: AnalogTodd
author_rank: Regular Contributor
timestamp: July 27, 2023, 08:22:34 pm
content: Quote from: Nominal Animal on July 27, 2023, 07:01:56 pmWhich is where you see guys using the transformers coming out of microwaves to create spot welders. You've got a massive core that normally can give you hundreds of watts of power that normally steps up from 110/230V to drive the magnetron at really high voltage. Take all of those turns of wire for high voltage and low current out on the secondary side and just put in a few turns of big cable and you instead step down to a few V at a lot of amps. Even with that, you can't neglect the resistance of the wire in there on the high current side.

author: tooki
author_rank: Super Contributor
timestamp: July 27, 2023, 11:27:46 pm
content: Quote from: tggzzz on July 27, 2023, 06:12:01 pmThis is one of those answers that could be viewed astechnicallycorrect (if absurdly pedantic), yet isfunctionallywrong because it suggests that Ohm’s law doesn’t always apply. But it does. It always applies, it’s just that in active components, the component’s own properties can vary in response to something else, giving the impression that Ohm’s law doesn’t apply.

author: tggzzz
author_rank: Super Contributor
timestamp: July 27, 2023, 11:44:56 pm
content: Quote from: tooki on July 27, 2023, 11:27:46 pmWhat is the resistance of a tunnel diode? (Hint: sometimes increasing the voltage increases the current, other times increasing the voltagedecreasesthe current)What is the resistance of a bog-standard silicon diode? (Hint: the dynamic small signal resistance depends on the "bias" current)What is the resistance of a zener diode? (Hint: give a single value for the resistance above and below the zener knee. Or fail to do so)For any of those, specify V1,I1,R. I'll then choose a different voltage V2, and you willfailto use Ohm's "law" to predict the current I2from I2=V2/R.If there is "something else" other than V and I, then please specify what that might be.Then go and look at wackypedia, which correctly states "Ohm's lawis an empirical relationwhich accurately describes the conductivity of the vast majority of electrically conductive materials over many orders of magnitude of current. However some materials do not obey Ohm's law; these are called non-ohmic".

author: Manul
author_rank: Super Contributor
timestamp: July 28, 2023, 01:11:53 am
content: Quote from: tggzzz on July 27, 2023, 11:44:56 pmIn my opinion, to say that Ohm's law does not apply for a diode is similar as saying that Ohms law did not apply for a resistor which just burned open. Ohm's law does not state that in any given circuit more voltage equals more current or implying linearity or anything of such nature. Ohm's law is a simple relation between voltage, current and resistance for a given moment (and resistance may be static, dynamic, non-linear, negative, whatever). If at a given moment diode voltage is 0.7V and current 1A, you can replace the diode with 0.7Ohm resistor and get same 1A current. That is what diode resistance was at that moment. Resistance being voltage dependent does not invalidate Ohm's law, because it is not even in the scope of the Ohm's law, it's a different matter.

author: alm
author_rank: Super Contributor
timestamp: July 28, 2023, 01:26:36 am
content: Quote from: tooki on July 27, 2023, 11:27:46 pmOhm's law was stated in hisoriginal 1827 work (page 178)(English translation, see page 416) as the current is the voltage between the two terminals of the circuit divided by the reduced length between those two terminals, where reduced length was something like the sum of each conductor divided by it's conductance per unit of length. So clearly he considered resistance a constant. This is still a common interpretation of Ohm's law.Under your interpretation Ohm's law is just V = f(I, t, T, Vbe, ...), which is not a very useful observation. What would constitute Ohmic and non-Ohmic resistors under your definition?

author: Manul
author_rank: Super Contributor
timestamp: July 28, 2023, 02:16:56 am
content: Well, I suppose I mostly use Ohm's law in a way which was not intended by the man himself back in the day. For me, the primary value is a unit itself, defining the voltage to current ratio, be it DC or AC (impedance). And dealing with what is "Ohmic" and "non-Ohmic" for me is a separate subject.

author: John B
author_rank: Frequent Contributor
timestamp: July 28, 2023, 02:46:10 am
content: Maybe Ohm's Relation, or Ohm's Suggestion is more appropriate?

author: tggzzz
author_rank: Super Contributor
timestamp: July 28, 2023, 08:42:31 am
content: Quote from: Manul on July 28, 2023, 01:11:53 amWrong in all respects, if V=IR.QuoteThat can only be true if V != IR. You seem to be proposing that V = f(I, temperature, humidity, orientation, phase of moon), but that is not auseful"law"!Let's be less theoretical. So what is thenumericalvalue of the resistance of this tunnel diode at V = 0.1, 0.4 and 0.25 volts?Next, use Ohm's Law and that constant value of resistance to predict the current at 0.05V and 0.5V. If V=IR, there must be a factor of 10 difference in the current. If not then Ohms Law does not hold. (Not surprising in a non-ohmic material !)

author: robowaffe
author_rank: Regular Contributor
timestamp: July 28, 2023, 09:11:50 am
content: as volts goes up, current goes up, then it goes down again, then it goes up again.

author: MrAl
author_rank: Super Contributor
timestamp: July 28, 2023, 09:37:25 am
content: Hello there,There have been MANY discussions about Ohm's Law which focus on whether or not it is a true Law of nature or not, and when it can be applied. There are different opinions but there is some science behind it that cannot be ignored. You have to realize though that some people will not care that much about how we classify it.First of all, if it is a Law it is an empirical law that was found through experimentation only and cannot be derived from other sources. This in itself does not really qualify it as a true Law.Second, the main focus of attention is on what the resistance is, and this gives us two views right off:1. V=I*R2. V=I*rThe difference is in #1 the variable R is capitalized and that means it is constant, ALWAYS.In #2 the resistance can vary, so it does not have to be constant. This means a more succinct definition would be written:3. V=I*r(x)where x can be time, temperature, etc., or both as in V=I*r(t,T) or just about anything else.Now we know that there are devices where the resistance changes with current or voltage, and one usual example of this is a silicon diode, where the resistance changes as the forward current changes, or we could look at it from the standpoint of the voltage:V=I*r(I) or V=I*r(V)You might notice one thing here though. If 'r' is dependent on I, then I is not allowed to be constant either, and if dependent on V, then V cannot be allowed to be constant either. That turns it into this:v=i*r(i) or v=i*r(v) or v=i*r(i,v)That means that if we allow 'r' to change, 'v' and 'i' must be allowed to change as well.So what's the big deal?The thing is, if we allow everything to change we have not really stated anything new about anything. If we don't know what the voltage is until we measure it and we don't know what the current is unless we measure it, then we don't know what the resistance is until we measure that either. What good would it be to have a "Law" that cannot be used to determine anything we don't already know.If we could call that a "Law", then we would have to call just about everything a Law. For example, the equation of a circle:r^2=x^2+y^2Is that a "Law"? I dont think so. If it was, we'd have to call it something like the "Law of the Circles" (ha ha). It's just the way to show the RELATIONSHIP between the dimensions of a circle in a most general way.This is the same thing we see with Ohm's Law. It is just a relationship not a true Law.In some forums it is more proper to call it "Ohm's Rule".Also note that we cannot use Ohm's Whatever to determine how the regular silicon diode works, we instead have to resort to another 'rule' that includes an exponential. Should we call that the "Law of Diodes" (ha ha). I don't think so.Originally it might have been called a Law either because it was thought to be a Law or just out of respect for all the work Ohm put into finding this relationship. After all, it does hold for pure resistances over a given temperature. Maybe it should be called, "Ohm's Approximation For Resistive Elements". I believe this is still very respectable to the man who found this relationship.

author: alm
author_rank: Super Contributor
timestamp: July 28, 2023, 10:42:51 am
content: Quote from: MrAl on July 28, 2023, 09:37:25 amThe original publication said that voltage is linearly related to current, and that their ratio is related to other measurable quantities in the system. It's indeed empirical just like Hooke's law that the force is proportional to the extension of the spring. Both are empirical and both are understood in modern physics to represent an ideal system (for example no temperature coefficient in the resistor) that real systems may or may not conform to.Physics has plenty of those laws. Anything that neglects quantum or relativistic effects, like Newtonian mechanics.

author: tggzzz
author_rank: Super Contributor
timestamp: July 28, 2023, 10:54:34 am
content: Quote from: alm on July 28, 2023, 10:42:51 amYup.It is worth remembering that this kicked off as a result of the statement...Quote from: tooki on July 27, 2023, 11:27:46 pm

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 28, 2023, 04:29:42 pm
content: Remember, in physics ,"law" ≡ "model". This is because physics does not define reality, it attempts to describe it.If you study physics long enough, you realize that the choice of word, "law", is actually fantastically apt: its connotations are thatwhen everyone behaves, this is how things happen; but also that not everyone always follows the law, which leads tobreakage. (In physics, this is alimitationof the model, and implies no moral wrong.) Especially in particle physics with many competing models, talking aboutbreakingthis or that is extremely common. Some of the terms, likebreaking parity(meaning for something symmetric to become asymmetric), have become widely used outside science, for example in Magic the Gathering card game.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 28, 2023, 05:10:43 pm
content: Apologies for this long post. In this, I try to explain how my earlier description of Io flux tube and how there you'll find very low voltages (potential differences in large volumes of the flux tube) with huge currents flowing through, and the above "law" ≡ "model", bind together. Feel free to skip!I was hoping that someone would point out that my usage ofresistanceto describe the process of how the charged particle flow within the flux tube heats up wasn't exactly correct, and that the "real" effect is actually more likePthermal= ΔVI, where ΔVis the voltage drop within the two cross sectional surfaces perpendicular to the particle flow, andIis the current in that volume.Then, I could have agreed, and pointed out that to describe that voltage drop, we could use another variable,R, because the rough magnitude of that voltage drop is determined by the rough location of the volume within the flux tube, but is otherwise pretty constant. It does vary in time, so much so that the radio emissions can easily be captured on Earth, but for an approximation of the magnitude of the effect –– i.e., no digits of precision, just the roughscaleof the effect, as in the approximate power of ten for the value –– it works quite well.And this isuseful. We don't actually care whether the current in that flux tube is two million or four million amps, but knowing that it is in the low single digit millions of amps, gives us the ability to estimate various effects to anuseful degree.Similarly, Ohms law is a very simple model that nevertheless can be applied in many situations. The above Io flux tube is one example: it is not a precise model there, but it gives you the correct magnitude and mental picture of what is happening.I particularly like the Scope part of the Wikipedia Ohm's law article:Quote from: WikipediaIt reflects how we're using Newton's laws of motion for atomic and molecular simulations, even though we are fully aware that they only apply in non-relativistic conditions (which such simulations basically always are, or the human constructing the simulation is doing something wrong).What difference is there between 'empirical law' and non-empirical 'law' in physics, then? Aren't all models empirical, when they model measurements?An empirical law, or an empirical model, is just a fit of some mathematical function that matches the observation, without any reasoning or basis on why exactly that form is used.A "proper" law in physics, also has a reasoning behind the specific mathematical equation or form used, with at most someconstantsin the equation chosen to fit the observations. (Then, those constants are said to be empirically defined. Only in very rare situations do you find a "law" that has no empirically determined constants at all, but typically, each such constant has been verified in many different ways and methods of measurement/experimentation.)In other words, if I have some measurementsyiwith my single control variable having corresponding valuesxi,if I fit any mathematical functionf(x) =yto the measurements, I get an empirical "law".If I have a reason to choose a specific form of a function, sayf(x) = C e-D(x-x0)², and only fit C, D, and x0to match my measurements, I get a "law", with an empirical fit to the constants C, D, and x0. Those constants can be (will be!) refined later by others, without proving my "law" incorrect. To prove my "law" incorrect, you only need to show that my reasons for choosing that specific form are incorrect, and show a better-suited one. If someone only points out a flaw in my reasoning, but has no better form to suggest, it is usually considered acomment; this is because until there is a better model to use, we still have to use the models we have, no matter how flawed we know them to be.

author: MrAl
author_rank: Super Contributor
timestamp: July 29, 2023, 07:25:02 am
content: Quote from: alm on July 28, 2023, 10:42:51 amHi,Most people take Ohm's "Law" to mean the proportionality between voltage and current in a conductor. That would mean that the device under examination would have to be a pure resistance.However, resistance is not the same as that described by Ohm's "Law". I think this is where the confusion often creeps in.We can have resistance that does not obey Ohm's Law. We sometimes use this term to describe something that produces heat with all of the energy applied to it, leaving no energy to accomplish anything else. That is often referred to as resistive heating.Resistive heating does not require a resistance that obeys Ohm's "Law". A diode produces heat when energized, but it does not obey Ohm's "Law". In fact, most of the energy in the diode produces heat. That makes it look resistive, but again it does not have to follow Ohm's idea of how a conductor behaves.Anything that has resistance will show a relationship like V=I*R, but if it does not follow that relationship over a wide range of currents and voltages, it could not be said to obey Ohm's "Law". If a diode has V=0.7 volts and I=1 amps, the resistance at that one operating point is 0.7 Ohms, but again that's just a resistance and it does not have anything to do with Ohm's "Law".We sometimes think in terms of Ohm's "Law" anyway though, because we sometimes KNOW the voltage or current at a given operating point, and that makes it seem like we can use Ohm's "Law" to work with it. That's just a superficial view though.Yes there are many "Laws", but they may not be true Laws. That's the whole point. What Laws are really Laws and what are not.A good example comes from control theory, where we often have what is referred to as a "Control Law". We call it a law, but it only applies in that ONE given application, because another application will have an entirely different "Law". It's just an equation that describes a relationship, that's all.So maybe we should just consider these 'laws' to be a subclass of all laws of physics and be happy with that.Here is a more comical look at some "Laws"...Law of the Circles: r^2=x^2+y^2Law of the Circumferences: C=pi*diameter (also known as the Law of pi, or just pi's Law)Law of the Slopes (in the time domain): dy/dt=a*t (also known as the Law of Rooftops)Law of the Laws: AnyResult=AnyFunction(AnySetOfParameters), also known as Law's Law.One thing this does accomplish though, it makes it easier to describe some relationships. If we did call the equation for a circle the Law of the Circles, we would not have to keep writing r^2=x^2+y^2 when we needed to talk about it. We could even abbreviate it LOTC or just LOC (Law of Circles) ha ha.The sad part I think is that the unit for resistance is the "Ohm", and I think that starts the confusion right off the bat. That might make us think that if it has units of Ohms then it must obey Ohm's "Law". To add to the confusion, we also call things that have resistance "Ohmic". It's just crazy how out of wack some things can get.

author: MrAl
author_rank: Super Contributor
timestamp: July 29, 2023, 07:41:52 am
content: Quote from: Nominal Animal on July 28, 2023, 05:10:43 pmHi,I don't think the "empirical" part of the discussion is that important. What is more important is how things work when it comes to voltage and current. That is, the relationship between voltage and current in a device. When we make measurements, we find that some devices can be described in a certain way, while other devices cannot be described in the same way. This particular way of describing this class of devices covers a wide variety of objects, while the other ones are outliers that need special formulations for each one.I think the way the "empirical" part works is there are some constants that are considered "very" empirical while others are not "as" empirical. This probably deserves a separate discussion.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 29, 2023, 01:11:47 pm
content: Quote from: MrAl on July 29, 2023, 07:41:52 amWell, Itriedto say that we always have at least empirically defined constants, like the speed of light in vacuumc.Then we have empirically or mathematically pulled-out-of-a-hat models and formulae, like choosing between e-|x|and 1/xforms just because one fits the measurements better, with the only rationale for choosing a particular function or form being "it fits the data better".While Ohm's law is the latter, it is also so simple it has no constants. You could even say that it embodies one of the very few mathematical ways of how three variables can directly (without any constants) relate to each other, whenever the situation is nice and regular.In materials physics and electomagnetics stuff, physicists tend to use theequivalentvector form, \$\vec{J} = \sigma \vec{E}\$, where \$\vec{E}\$ is the electric field at that point, \$\sigma\$ is the conductivity at that point, and \$\vec{J}\$ is the current density. Better physicists than myself would have immediately pointed out (to me earlier) that my approximation of V = I R in the Io flux tube is actuallly valid in electromagnetic field terms within any volume where the electric field and conductivity vary so little they can be considered a constant. Me, I had to sleep and wake up realizing that.

author: MrAl
author_rank: Super Contributor
timestamp: July 30, 2023, 10:43:58 am
content: Hi,That sounds like a counter argument in that Ohm's Law is one of the only true laws. That's an interesting viewpoint too.It is interesting that it has been taken further too into AC circuits using AC quantities and using impedance Z instead of resistance R.The bottom line I guess is that whatever we call it, it certainly is useful to know.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 30, 2023, 04:50:32 pm
content: Quote from: MrAl on July 30, 2023, 10:43:58 amI don't see how "true" applies here, unless you mean in the same sense as "basic". It is its very simplicity that makes it so useful; but also easy to break.When running simulations, or doing any physics at all, my own first thought after seeing some results is always"Does this make sense?", with the same connotations as "Am I breaking any 'laws' here?". By 'laws', I of course mean the assumptions and limitations of the models I use. With Ohm's law, that would be the purely resistive nature of the signal, and everything being so near to DC we don't need to worry about the conductors being transmission lines, nd so on.For example, my very first own atomic simulation a couple of decades ago was of copper, using a very simple pairwise Morse potential (and usually called 'Morse copper'). I had half of the simulation volume filled with regular face-centric cubic lattice of copper atoms, and the other half randomized (annealed in a separate simulation, representing liquid copper). Unfortunately, I miscalculated and put some of the atoms at the midline half too close, i.e. at half the typical distances. This lead to somefunkyeffects: in one run, I got a molten droplet inside a fully organized cubic lattice solid. In another, I got acoustic waves through the system with wavelengths many times larger than the simulated system (started by a shockwave emanating from that super-energetic boundary, the simulation having periodic boundaries, i.e. repeating infinitely in all directions). The only 'law' I broke was the typical interatomic distances, as putting them too close gives the atoms HUGE (utterly unrealistic) energies, with everything else being an effect of how those energies dissipate/equalize throughout the system.

author: tooki
author_rank: Super Contributor
timestamp: July 30, 2023, 06:18:01 pm
content: Quote from: alm on July 28, 2023, 01:26:36 amIt strikes me as interesting how differently the English and German Wikipedia articles handle the definition. The English article follows (and postulates) the very strict definition used by tgggz and others here. The German article includes basically what I said. The fact that I learned electronics formally in German may not be a coincidence.Both articles begin with a similar explanation, that in a conductorwith a constant resistance, the current is given by I=V/R, and how Ohm arrived at it.In the German article, it continues with the following statement:QuoteTranslated, it says:QuoteSo perhaps I should rephrase: Ohm’s law (or at minimum, the same I=V/R relationships described within it) applies to theinstantaneous valuesof I, V, and R, even with non-ohmic components. The R of a non-ohmic component cannot, however, be predicted by Ohm’s law, as it may change in response to another quantity.If Ohm’s Law strictly applies exclusively to conductors with constant resistance, then I guess I am wrong. But in practice, we commonly refer to the I=V/R relationship (and the equivalents R=V/I and V=IR) as “Ohm’s Law” even when evaluating circuits with non-ohmic components. It’s simply understood that Ohm’s law has no predictive value with non-ohmic components.Edit: slightly improved translation.

author: MrAl
author_rank: Super Contributor
timestamp: July 30, 2023, 07:28:41 pm
content: Quote from: Nominal Animal on July 30, 2023, 04:50:32 pmHello there,Maybe you can explain to me how you can "break" Ohm's Law, or Ohm's Rule.

author: MrAl
author_rank: Super Contributor
timestamp: July 30, 2023, 07:49:32 pm
content: Quote from: tooki on July 30, 2023, 06:18:01 pmHi there,Well see, Ohm's Law, or Ohm's Rule, only applies with constant resistance. If that was not so, it would not be stating anything. We can surely know that R=V/I when R is allowed to vary without the need for any rule of any kind. Here R is just resistance.I believe the confusion comes in because we use units of Ohms to measure resistances, and so when we hear Ohm's Law spoken, it sounds like it applies to every resistance simply because the units are also Ohms. However, units are different than laws or rules like that. Rules or laws show how things work over a wide range of variable conditions. That is, when we vary V or I we get something that holds the same relationship as with other V or I. If we didn't get some result that was the same, it wouldnt tell us anything more than just plain old R=V/I.The other source of confusion i believe is that too. R=V/I helps to describe Ohm's Law, so we think that R=V/I always has to involve Ohm's Law in some way.Another way to look at this is to examine the regular silicon diode. That's often used in these discussions because it is nonlinear with V and I, very much unlike a constant resistance, and so we can look at the contrast between the diode and the pure resistance.Now to see this contrast we can take one form of Ohm's Law to be R=V/I, and this describes something very specific. This does not describe a diode, only a pure resistance. Let's put Ohm's Law into another form: I=V/R and consider that.Next we can look at an ideal diode equation: I=Is*(e^(q*v/(n*k*T)))Now ask the question, how on earth can we use I=V/R to describe the ideal diode equation. That is, what use is I=V/R for use with a diode.The answer is that it is of no use when we want to describe the workings of a diode over a fair range of V and I. When we want to linearize a diode curve, we can use I=V/R, but only over a very limited range of V and I. That still doesn't mean that Ohm's Law holds for that situation though, because in reality, R still changes, we just pretend that it is the same over that small range of V and I.We also have to remember that it probably doesn't matter that much how we think of this "Law" or "Rule". As long as we know how to deal with the different devices as they work with V and I, I think we will all be just fine.

author: tggzzz
author_rank: Super Contributor
timestamp: July 30, 2023, 08:24:18 pm
content: Quote from: tooki on July 30, 2023, 06:18:01 pmA law without predictive value is completely and utterly useless and anti-scientific. Science isnothingwithout falsifiable predictions.I put 20mA through this tunnel diode; what is the voltage across its terminals? 40mV? 200mV? 500mV? Obviously it's resistance is 2ohms; for clarity that's 10ohms, and to be unambiguous it is 25ohms.

author: tooki
author_rank: Super Contributor
timestamp: July 30, 2023, 08:38:05 pm
content: Quote from: tggzzz on July 30, 2023, 08:24:18 pmI didn’t say it has no predictive value; I said it has no predictive value for specific things:Quote from: tooki on July 30, 2023, 06:18:01 pmInstead of robotically repeating what you’ve already said before, how about actually engaging with what I wrote?

author: tooki
author_rank: Super Contributor
timestamp: July 30, 2023, 08:46:43 pm
content: Quote from: MrAl on July 30, 2023, 07:49:32 pmBut I never said we could use Ohm’s law to describe a diode’s curve. What I said is that we can use Ohm’s law (or the relationship expressed therein) to calculate an instantaneous (=single-point) value. And we can. (tgggz just did so, in his snippy response to me above.)And while you are reiterating “it only applies with constant resistance”, what I am saying is thatat least colloquially, the term “Ohm’s law” is used to describe the I=V/R relationship even in situations where the resistance is not constant. E.g. saying that theinstantaneous(=single-point) resistance on a diode’s V/I curve is a particular number of ohms. Honest question: how else would you calculate the instantaneous resistance ifnotusing R=VI?!!

author: schmitt trigger
author_rank: Super Contributor
timestamp: July 30, 2023, 08:58:20 pm
content: Quote from: golden_labels on July 27, 2023, 06:08:12 pmIn plain English, the OP is a troll. The scourge of web forums.The best defense against trolling is simply to ignore them.

author: tggzzz
author_rank: Super Contributor
timestamp: July 30, 2023, 09:53:30 pm
content: Quote from: schmitt trigger on July 30, 2023, 08:58:20 pmBased on the OP's responses to his own posts, that is my conclusion too.I noted that in reply #24https://www.eevblog.com/forum/beginners/at-nanovolts-do-resistors-still-experience-heat-at-high-amperage/msg4983397/#msg4983397His subsequent posts/responses don't change that conclusion.

author: tggzzz
author_rank: Super Contributor
timestamp: July 30, 2023, 10:09:19 pm
content: Quote from: tooki on July 30, 2023, 08:46:43 pmAn "instantaneous" value of resistance would - if valid - make the whole precept of Ohms law valueless. Or maybe you prefer "preserving" Ohms law in the example I gave, by conceding that 2=10=25!Note, there is ausefulconcept of a diodes's resistance: the "small signal" resistance defined by theslopeof the VIcurveat a specified point. That's how you canvery usefullyconsider that the tunnel diode has a positive resistance at 20mA/40mV, anegativeresistance at +20mA/200mV, and a (lower) positive resistance at 20mA/500mV. The negative resistance is the key to the tunnel diode being useful as a threshold switching element and as a logic element.

author: tooki
author_rank: Super Contributor
timestamp: July 31, 2023, 12:05:40 am
content: Quote from: tggzzz on July 30, 2023, 10:09:19 pmYour “note” is literally the exact point I have been making the entire time: resistance can also be the resistance at a single point. “Instantaneous” means “at one point”. Resistance can always be regarded as the slope of the V/I relationship. In an ohmic material that slope is the same everywhere, in a non-ohmic material it is not. You’re literally restating what I have been saying all along, but when you say it it’s OK, and when I say it it’s wrong?!? I am genuinely confused as to why you are still arguing with me.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 31, 2023, 01:23:57 am
content: Quote from: MrAl on July 30, 2023, 07:28:41 pmPick any nonlinear semiconductor, like a diode, and try to apply it within a wide enough voltage range.Quote from: tggzzz on July 30, 2023, 10:09:19 pmYou and Tooki are having communications problem with the term 'instantaneous'.In this context, Tooki obviously referred to the concept of it being exactly correct at a specific point along the diode voltage-current relationship, making it a valid approximation in a voltage/current *range* surrounding that point –– say, the "useful" range in some circuit, depending on where that original point was along the curve –– which makes it a valid and useful tool. Exactly what tggzzz described as an approximation byslopeof the V-I curve.This is a common approximation technique in all of physics, and applies even to Newton's laws of motion, which technically only matches special relativity at zero velocity (where Lorentz contraction constant \$\gamma =1\$). However, Lorentz contraction is *approximately* \$\gamma = 1\$ for human-scale velocities, including velocities of atoms in typical materials humans will encounter, so Newton's laws of motion, being simple to apply, are used.Even in mathematics, such extensions are an useful tool for solving problems approximately. For example, \$\log n! \approx n \log n - n\$ (Stirling's approximation), which is "exact" only at infinity, and only really useful for large \$n\$: at \$n \gt 14\$, the relative error is less than 10%, at \$n \gt 90\$, the relative error is less than 1%, and so on. At \$n = e\$ the approximation is particularly bad, yielding zero, while the correct answer is approximately \$1.4495\$; for \$1 \lt n \lt e\$, the approximation is negative (while correct value is positive) and is therefore pretty much useless.Heck, both Newton's method (Newton-Raphson) and piecewise numerical integration using the trapezoidal rule, implement the very same idea in different contexts. As I see it, your (tggzzz-tooki) disagreement stems from language used, not from the concepts themselves.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 31, 2023, 05:17:54 am
content: Could you guys code a resistor sim? That's how far ohms law got you if you couldn't do it.

author: tggzzz
author_rank: Super Contributor
timestamp: July 31, 2023, 08:02:16 am
content: Quote from: tooki on July 31, 2023, 12:05:40 amThe small signal resistance is not R=V/I as you claimed (see myemphasisof your statement)The small signal resistance is the R=dV/dI = i.e. the slope of the VI curve at that point. That's R=deltaV/deltaI.

author: MrAl
author_rank: Super Contributor
timestamp: July 31, 2023, 09:34:43 am
content: Quote from: tooki on July 30, 2023, 08:46:43 pmHello again,It is going to take some time for you to understand these points entirely.Ohm's Law is not about a single point on a V,I curve, it's about a range of V and I.As i said before i think twice now, we don't need a "Law" for something that only occurs at one point. A Law is something that related variables over a range, not just at one point.What you are doing is forcing the relationship at a single point to look like Ohm's Law when it's not that at all, it's just a RESISTANCE period.I never said you wanted to use Ohm's Law for a diode, I said a diode is often used to contrast between what is not Ohm's Law and what is Ohm's Law. Apparently you did not understand that either.I can tell you firsthand that for some people it takes years to understand this completely because it is a very unusual situation. We have a lot of confusing things working together here that makes it harder than many other things to understand and so it is going to take some time. If you are patient though, you will see the way this works in the future as most people do.I can also tell you that many years ago when this was first told to me I did not understand it completely either. It took more study and understanding how laws work. I guess I could show some more examples at some point.Also, the point of whether or not it is a true law is really a different question so we should probably leave that for another time.You can also ask the question, again, if Ohm's Law works for a diode in any shape or form, then why do we need the diode equation. The diode equation explains the ENTIRE behavior of the diode why would we want to think about Ohm's Law.That's just one example though and that is used in these discussions, again, to show the contrast between what Ohm's Law is and what it is not.I also brought up the idea of the equation of a circle because a circle is a well-known 2d object:R^2=x^2+y^2If you look at this carefully, you will see that these 'variables' are interrelated. The idea is that if R is constant we get a circle. If R is not constant, we no longer have a circle:r^2=x^2+y^2Since r is lower case it indicates that it is not a constant (strictly speaking) and so if we draw it we would not see a circle. Of course we also have to know how 'r' varies too in order to draw anything at all. For example:r(t)^2=x^2+y^2and now we see that 'r' varies with time. This could make a spiral for example if r(t) varies in a linear fashion, but that's no longer a circle. So we can't call that second expression a circle anymore in the same way we can not call Ohm's Law an "Ohm's Law" for a resistance that varies.It might be a subtle point, and it may not even matter to some people. They might go on forever calling anything that involves R=V/I "Ohm's Law". If you really want to do that, that's up to you of course. I am just explaining why it should not be taken to be the same thing.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 31, 2023, 09:56:39 am
content: Quote from: robowaffe on July 31, 2023, 05:17:54 amSure I could. Even as a self-contained web page that works even locally, i.e. without a server. What's in it for me, though?

author: robowaffe
author_rank: Regular Contributor
timestamp: July 31, 2023, 10:20:40 am
content: Quote from: Nominal Animal on July 31, 2023, 09:56:39 amThat's well and good, I don't disbelieve in people for no reason, but my point was actually that ohms law really only handles 1 resistor, when it comes to a resistor network there is more involved, even just with resistors. and that's why just being proud of knowing ohms law is not the whole picture, and its silly IMCO.

author: tggzzz
author_rank: Super Contributor
timestamp: July 31, 2023, 10:31:42 am
content: Quote from: robowaffe on July 31, 2023, 10:20:40 am"IMCO" => in my clueless opinion, which seems valid.https://acronyms.thefreedictionary.com/IMCOOverall: not a bad troll, but I doubt Nominal Animal will fall for it.

author: robowaffe
author_rank: Regular Contributor
timestamp: July 31, 2023, 10:49:23 am
content: Quote from: tggzzz on July 31, 2023, 10:31:42 amHow was what I said not true.

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 31, 2023, 06:18:58 pm
content: Quote from: tggzzz on July 31, 2023, 10:31:42 amYep. It was too clear a nod towards Kirchhoff's circuit laws (that the sum of currents in each node is zero, and the sum of potential differences in each loop is zero), since that's all you need for simple (lumped model, at DC limit) circuit modeling. Add time steps and basic instantaneous models for nonlinears (diodes, capacitors, transmission lines, BJTs, FETs, JFETs, MOSFETs) and voltage and current sources, and you have built yourself a SPICE.

author: tggzzz
author_rank: Super Contributor
timestamp: July 31, 2023, 06:38:34 pm
content: Quote from: Nominal Animal on July 31, 2023, 06:18:58 pmHmm. That's not the first time the word "instantaneous" has appeared in this thread; I've not seen it in this context elsewhere. I wonder if that's due to an autocorrupt or gurgle translate error."Instantaneous" means at an instant in time, a concept which is obviously relevant to time domain simulation. Itdoesn'tmean "slope" or "small-signal" or "differential" in the vicinity of the bias/operating point.

author: EPAIII
author_rank: Super Contributor
timestamp: July 31, 2023, 07:11:36 pm
content: "Voltage induced shorting"? What are you talking about? Where did you get that?Ohm's law describes the relationship between Voltage, current and resistance. And it is about as close to an absolute law of nature as you are likely to encounter.I = E/RThe current is directly proportional to the Voltage and inversely proportional to the resistance in the circuit. Plain and simple and the units are defined in such a manner that no constant of proportionately is needed.It also applies on an instantaneous basis in a circuit where things are changing (an AC circuit). Strictly speaking you need to use calculus to apply it that way but there are shortcuts for common AC circuit configurations. Each situation, each circuit MUST be PROPERLY evaluated on an individual basis. You can't just pull a shortcut out of thin air and apply it willi-nilly to the circuit youwantit to apply to.For one thing, the resistance of a piece of copper wire (or any wire) depends on the material (the copper) on the cross section area, AND the length. There's a formula for that also. Thin gauge wires (small cross section) will have higher resistance than thicker ones. Longer ones will also have more resistance.I suggest that you try to understand the basics before you run wildly into left field with a "marvelous", new idea. Chances are you have not re-invented the wheel.Quote from: robowaffe on July 27, 2023, 05:47:05 pm

author: Nominal Animal
author_rank: Super Contributor
timestamp: July 31, 2023, 07:31:43 pm
content: Quote from: tggzzz on July 31, 2023, 06:38:34 pmI don't use autotranslation, only my weak poor brain.Let me sketch out how I use the term, and why.In discrete time simulations, time advances in steps, and each point in time is typically called aninstant.Aninstantaneousmodel is one that can be used in such a simulation.For diodes, the most used model is the Shockley diode equation, \$I_D = I_S \left( e^\frac{V_D}{n V_T} - 1 \right)\$, where \$V_D\$ is the voltage across the diode, \$V_T\$ is the thermal voltage (25.852mV at 300 K or 27°C, and more generally \$V_T = k T / q\$, where \$k = 1.380649\times 10^{-23}\$ [Joules per Kelvin] is the Boltzmann constant, \$T\$ is the temperature (in Kelvins), and \$q = 1.602176634 \times 10^{-19}\$ ), \$n\$ is the ideality factor, and \$I_S\$ is the reverse-bias saturation current, and \$I_D\$ is the diode current.See how there is no time dependence at all? That we do not use time as a variable in the diode equation at all?This works, because eachinstantis iteratively solved until everything matches, very much like it was a DC steady-state situation!Ohm's law, \$V = R I\$, can be used as an approximation in such aninstant, even for a Shockley diode: \$R \approx \frac{I_S}{n V_T} e^\frac{V_D}{n V_T}\$. Yes, this is the slope of the V-I curve at those parameters, and only reasonable if we're not too close to the knee. A simple check is to calculate (or remember) the "\$R\$" from the previous time step, and compare it to the value at the initial guess for the current time step. (Typically the time step is shrunk until values converge and the delta from previous time step is acceptable.)Note that this is not a definition of \$R = dV / dI\$; we simply obtained the "instantaneous resistance" –– so called because this is an approximation that allows us to use Ohm's lawfor the calculations in this time step only, and only if the other variables do not change too much –– so we can model even a diode, in a small range of parameters, like when converging a time step in a spice simulation, using Ohm's law.Do you see how this all ties together? Don't get tied up in exact connotations of the terms, because we'd have to define them all in context first, and that would be horribly tedious.Quote from: tggzzz on July 31, 2023, 06:38:34 pmRight, it means the exact operating point. Thing is, the properties and behaviour of the item are easily approximatedaround that pointvia extrapolation, just like the resistance above.Again, the abovementioned \$R\$ for a Shockley diode is only "correct" at exactly one point. However, if that point is not near the knee, it can be a good approximation for a range of values. Being linear, you can solve systems of such in matrix form very efficiently.

author: EPAIII
author_rank: Super Contributor
timestamp: July 31, 2023, 08:04:12 pm
content: I am sorry that my German is not up to the task so I will say this in as PLAIN English as I can.Some here seem to delight in bringing up graph(s) of components that have non-linear current vs. Voltage segments and even segments where the current decreases for increases in Voltage. They seem to be saying that these "prove" that Ohm's Law is not true.I say that those examples only show that the RESISTANCE of those items is not constant. And I further state that, at each and EVERY point on those graphs the three values of Voltage, current, and resistance DO result in a STRICT equality if substituted in the simple expression of Ohm's Law:I = E/R.This is because the resistance (R) is DEFINED by Ohm's Law. In other words, R, by it's very definition is:R = E/I.This supersedes the older definition of the Ohm in terms of a volume of mercury and a constant current. Today the Volt is defined in terms of the Joule and the Coulomb and the Ampere is defined in terms of the Coulomb and the second. That older definition of the Ohm probably existed mainly due to the difficulties of counting electric charges as they passed a point (the Coulomb). Today the Ohm, by definition is E/I and that is directly from Ohm's Law.Thus, I = E/R or Ohm's Law, will be correct at any point and EVERY point in any circuit, BY DEFINITION.I find it confusing that some authors say that a component with a reversed slope on the I vs E curve is exhibiting "negative" resistance. That statement threw me for a total loop the first time I saw it. Nothing could be further from the truth. Even in those components, the resistance value never actually becomes negative. It is always positive. The confusion is in mistaking the slope of the curve for the calculated value using the actual I and E values at that point. These two, sometimes conflicting values, the slope of the curve and the calculated value using I and E, are only equal when the curve is a straight line which also passes through the zero or origin point.That the actual resistance of a given element in a circuit can have different values at different points on the I/E curve is a fact that we simply must live with in electricity and electronics. Send your thanks and complaints to the organizations that maintain the international standards; this is their method, not mine.

author: TimFox
author_rank: Super Contributor
timestamp: July 31, 2023, 08:19:41 pm
content: A device that follows Ohm’s Law is called “ohmic”.This is not trivial, since a large number of things (like metal wires and packaged resistors) follow that law closely.The unit “ohm” (1 V/A) is useful for other devices (like diodes) to quantify the derivative dV/dIat a point of the nonlinear V-I curve.This is all one needs to know.

author: tggzzz
author_rank: Super Contributor
timestamp: July 31, 2023, 08:45:21 pm
content: Quote from: TimFox on July 31, 2023, 08:19:41 pmI'm not sure why other people find that so difficult to understand!

author: tggzzz
author_rank: Super Contributor
timestamp: July 31, 2023, 08:55:29 pm
content: Quote from: EPAIII on July 31, 2023, 08:04:12 pmOhm's law is valid for, wait for it, ohmic devices - i.e. devices that obey Ohm's law. Spot the circularity!Many materials are not ohmic!QuoteSo, whatexactlyis thenumericalvalue of R when a current of 20mA is flowing through this diode?QuoteOnly for ohmic devices. Not for non-ohmic devices.QuotePlease answer the very simple question: what is the value of the resistance for that tunnel diode when I= 20mA?Numbersplease

author: tggzzz
author_rank: Super Contributor
timestamp: July 31, 2023, 09:02:03 pm
content: Quote from: Nominal Animal on July 31, 2023, 07:31:43 pmIf we don't agree on the meanings of words and concepts, then we might as well be talking to Humpty Dumpty!“I don't know what you mean by 'glory,' " Alice said.Humpty Dumpty smiled contemptuously. "Of course you don't—till I tell you. I meant 'there's a nice knock-down argument for you!' ""But 'glory' doesn't mean 'a nice knock-down argument'," Alice objected."When I use a word," Humpty Dumpty said, in rather a scornful tone, "it means just what I choose it to mean—neither more nor less.""The question is," said Alice, "whether you can make words mean so many different things.""The question is," said Humpty Dumpty, "which is to be master—that's all.”https://www.goodreads.com/quotes/tag/humpty-dumptyQuoteIn English, "instant" means "a point in time". In particular, it cannot mean "a point on a VI curve".https://en.wikipedia.org/wiki/Instant

author: Zero999
author_rank: Super Contributor
timestamp: July 31, 2023, 09:12:27 pm
content: Ohm's law is definitely valid, when it comes to power. P = VI is always true.Quote from: robowaffe on July 31, 2023, 05:17:54 amOne already exists. It's called SPICE. A good, widely used and free to use variant, is LTSpice, which can be found using a search engine and freely downloaded.

author: tggzzz
author_rank: Super Contributor
timestamp: July 31, 2023, 10:20:36 pm
content: Quote from: Zero999 on July 31, 2023, 09:12:27 pmWhile P=VI is valid for ohmic and non-ohmic devices (that are not reactive), that's not Ohm's law.QuoteTrolled!

author: TimFox
author_rank: Super Contributor
timestamp: July 31, 2023, 11:22:26 pm
content: With respect to the negative resistance (negative value of derivative dV/dIor its reciprocal conductance dI/dV), as in the tunnel diode shown above, note that only a powered circuit (including my personal favorite the transitron tube circuit) can exhibit negative resistance at the terminals. In the tunnel diode graph shown, there is a region with negative resistance and conductance (properly defined as derivatives), but theratioI/Vis always positive.

author: KaneTW
author_rank: Frequent Contributor
timestamp: July 31, 2023, 11:26:42 pm
content: Quote from: tooki on July 30, 2023, 06:18:01 pmThis is also how I interpret it. Another point suggesting it could be a result of German education.That is: R = U/I, always. But it doesn't mean that R is identical over all voltages, currents, time, temperature, the phase of the moon, etc.It's up to the user to know the limitations of the relationship. A piece of tungsten wire at a constant temperature is an "ohmic conductor" by how the English wikipedia defines it. A piece of tungsten wire that heats up to several thousand degrees Celsius as a result of the current passing through it is not.tggzzz is being dogmatic and/or trolling -- not sure, and not really interested in finding out. The resistance at a current of 20mA, throughanyreal device, is not a single value.

author: Manul
author_rank: Super Contributor
timestamp: July 31, 2023, 11:42:10 pm
content: Quote from: tggzzz on July 31, 2023, 08:55:29 pmI really don't like this, it reminds me trick algebra problems when they perform seemingly correct simplifications and prove that 1=0 or something of that sort.That is the VI graph of a real device, but it does not matter. I can put voltage on x axis and draw any arbitrary graph of coresponding current. For example, if I would draw a periodic wave, we could say that same current corresponds to infinite points in voltage axis. Which is undoubtly correct. In that case your question obviously cannot be answered, R is ambiguous. By math it is all correct, but the problem is that we got detached from how real world works. I think ambiguity here comes from incorrect causality.Same current may correspond to different voltages, but can same voltage correspond to different currents? If the trace of your tunnel diode would start looping back to the left, so that 0.3V corresponds to 10mAand40mA (for example), does such a graph makes sense? I think the answer is no. I think the reason is that, fundamentally speaking, field causes charges to move.In conclusion, that is a nice VI graph of tunnel diode, all is fine. I just think that you formulate your question the wrong way around and therefore it is without answer and misleading. Note that you can calculate R for any V without ambiguity.

author: bdunham7
author_rank: Super Contributor
timestamp: August 01, 2023, 12:16:00 am
content: Quote from: tggzzz on July 31, 2023, 08:55:29 pmTwo answers:1. It depends.2. It is what it is.

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 01, 2023, 12:51:10 am
content: Quote from: tggzzz on July 31, 2023, 09:02:03 pmWe are talking about "instantaneous" instead:https://en.wiktionary.org/wiki/instantaneous, specifically "happening within an imperceptibly brief period of time". For example, within one time step in a spice simulation.Within that time step, you can approximate many nonlinear semiconductor devices using Ohm's law. That's why current spice engines are so darn fast: they first approximate (often with just such linear models), then refine with more detail iteratively, for each time step. Linearization allows more efficient matrix forms, too.

author: TimFox
author_rank: Super Contributor
timestamp: August 01, 2023, 01:58:20 am
content: The ratio of voltage to currentV/Ifor a non-linear device is rarely a useful variable, while the derivative dV/dIis often useful, especially in small-signal or AC circuit analysis.Of course, for an ohmic device, the proportional equationV=IRmeans that the ratioV/Iand the derivative dV/dIare both equal toR.

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 01, 2023, 04:21:32 am
content: If \$V = R I\$, then \$\frac{\partial V}{\partial I} = R\$. For any continuous and differentiable \$V(I)\$, you can examine the V-I dependence near \$I = I_0\$, \$V(I_0) = V_0\$ via a linear approximation \$V(I) \approx V(I_0) + (I - I_0) \left[ \frac{\partial V(I)}{\partial I} \right]_{I=I_0} = V_0 + (I - I_0) R\$. This is basiclinear extrapolationaround \$V = V_0\$, \$I = I_0\$.To repeat, linear approximations of the system behaviour lets us find a good candidate for further numerical iterative search for the exact solution, using linear algebra. That is not possible if you cannot represent the "instantaneous behaviour" -- i.e., the very small changes within a single time step or iteration –– in linear form. Intuitively, you can think of it as a way to obtain a really good initial guess with relatively few math operations, so you can refine the solution. (*)Also note how that \$V \approx I R + V_0 - I_0 R\$ (where \$V_0\$, \$I_0\$, and \$R\$ are constant for the current time step or iteration) is still invertible, \$I \approx \frac{V}{R} + I_0 - \frac{V_0}{R}\$, and that in the approximation, the nonlinear component is described as if it was a linear component with voltage and/or current sources in series. This is very much in line with lumped-element model andmodel order reductionfor increasing the speed of SPICE simulations.(*) Consider how one would use Newton's method to find the square root of some positive number \$C\$. If your initial guess is \$x_0\$, you can iteratively refine it via \$x_{i+1} = \frac{x_i}{2} + \frac{C}{2 x_i} = \frac{x_i^2 + C}{2 x_i}\$. Use whichever right side is easier to calculate for you, the two are equivalent in this case.For \$C = 5\$, the solution is \$x = \sqrt{5} \approx 2.23606797749979\$. Let's say you start with \$x_0 = 1\$. The sequence would be \$1\$, \$3\$, \$\frac{7}{3}\$, \$\frac{47}{21}\$, \$\frac{2207}{987} \approx 2.236068895643364\$, so fourth iteration only has six or seven significant digits, and two more iterations are needed to get to 15 significant digits.An often recommended initial guess (for this; not in general) is \$C/2\$. The sequence of \$x_i\$ is then \$\frac{5}{2}\$, \$\frac{9}{4}\$, \$\frac{161}{72}\$, \$\frac{51841}{23184}\$, \$\frac{5374978561}{2403763488} \approx 2.23606797749979\$, and so on. We only needed four iterations to get 15 significant digits.For rate of convergence, and the amount of computation you need to do to arrive at an acceptable solution, the initial value makes a huge difference. When you have a state vector (with many components), it becomes even more important. This is why even a silly-looking linear approximation can be so useful in numerical electrical circuit analysis, and elsewhere: it's like a shortcut to an educated guess as to the solution.

author: robowaffe
author_rank: Regular Contributor
timestamp: August 01, 2023, 06:46:13 am
content: All the maths in the world wont solve a problem for you, if it just takes you down the wrong track, but its kinda impressive seeming when no-one knows what ur on about with all the confusing numbers all over the place.See if you can understand this.Unless you take ohms law to more than one resistor, you cant complete a voltage and resistance sim for all possible circuits.Ohms law isn't enough by itself to do a full resistor sim, but the whole thing is ohmic, because you cant disobey ohms law, but ohms law requires to be added more context to, to describe the whole system.

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 07:10:15 am
content: Quote from: Manul on July 31, 2023, 11:42:10 pmThe question is perfectly valid. It may be inconvenient for you to think about the consequences, but does not invalidate the question.The question is not a mathematical fiction, not "incorrect causality", not "detached from how real world works". Quite the opposite, in fact.Their negative resistance in part of the VI curve is the whole reason that they weretechnicallyandcommerciallyimportant.As wackypedia explains, "The "negative" differential resistance in part of their operating range allows them to function as oscillators and amplifiers, and in switching circuits using hysteresis. They are also used as frequency converters and detectors. Their low capacitance allows them to function at microwave frequencies, far above the range of ordinary diodes and transistors."https://en.wikipedia.org/wiki/Tunnel_diode#UsesIn some areas tunnel diodes still have remarkable performance: classic Tektronix scopes' superb triggering relies on tunnel diodes, and the Tektronix 1502 emits a 50ps risetime pulse from a tunnel diode.For other examples of negative resistance devices, seehttps://en.wikipedia.org/wiki/Negative_resistance#List_of_negative_resistance_devicesI rather like usingdiodes and capacitorsto amplify weak microwave signals in radio telescopes and satellite comms systems - see the parametric amplifiers, or this advert...

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 07:16:25 am
content: Quote from: Nominal Animal on August 01, 2023, 12:51:10 amWe are talking about time-invariant VI curves, not an artefact ofone of the manyforms of circuit simulations.

author: EPAIII
author_rank: Super Contributor
timestamp: August 01, 2023, 07:48:00 am
content: I will say it again.Resistance is NOT a derivative. It is the simple quotient of E/I.And the Ohm is NOT a derivative. It is also a quotient.Ohm's Law gives you both in all circumstances, for all components, in all circuits. That is how the powers that be (international standards organizations) have chosen to make it. Read the references, please.

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 08:04:04 am
content: Quote from: EPAIII on August 01, 2023, 07:48:00 amI will ask it again....So, whatexactlyis thenumericalvalue of R when a current of 20mA is flowing through this diode? (All you have to choose is the voltage (or "E" if you prefer) corresponding to 20mA.)If it is as simple and easy as you claim, then it should be simple and easy for you to answer that question.Come on, let's hear your answer. hint: your understanding is too simplistic. In particular you don't understand when Ohm's law doesn'tby definitionapply. That point has been made by several people on this thread, and in the second part of the introductory paragraph pfhttps://en.wikipedia.org/wiki/Ohm's_law

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 09:25:46 am
content: Quote from: Manul on July 31, 2023, 11:42:10 pmNow reassessthose statementsfor another passive two-terminal device, the neon bulb operating at 100V (N.B. the VI axes are changed):Yes, that non-ohmic behaviour is important in real-world devices, e.g.:I have a high-end Solartron digital voltmeter where the reference voltage is implemented using the non-ohmic behaviour of a neon bulb (high end in the 1960s, of course!)the world's oldest operating computer has logic and memory circuits in which the negative resistance of a neon discharge is exploited to provide the necessary amplification. It really is well worth visiting TNMoC to seehttps://www.tnmoc.org/witchin actionmy geiger counter has neon dekatrons as the high frequency counters (mechanical for low frequency). Memo to self: get a GM tube and measure the radioactivity of LoSalt from a supermarket

author: MrAl
author_rank: Super Contributor
timestamp: August 01, 2023, 11:23:33 am
content: Quote from: Nominal Animal on July 31, 2023, 01:23:57 amHello again,Ok I can see you are interpreting the word "break" differently than I am.The way you see it is that if you try to apply the law to a diode, it doesn't work, therefore it is broken.The way I see it is if you can 'break' a law at a certain point, then that law already does not apply. If it does not apply at that point then it cannot really be 'broken'.For example, if I try to apply r^2=x^2+y^2 to a resistor, does it become broken because it does not work or is it that it just does not apply. A more reasonable equality however would seem to be more in question I understand that.So we just have different outlooks on this.

author: Manul
author_rank: Super Contributor
timestamp: August 01, 2023, 11:25:10 am
content: Quote from: tggzzz on August 01, 2023, 08:04:04 amWhat I wanted to say is that this tunnel diode graph isVI, notIV graph. Someone was stepping up voltage and graphing current, you can't treat it as IV graph, just because you wish so. So being VI graph, it only makes sense to ask current (and R if you wish) at specific voltage. Not backwards. I mean, it's really logical and intuitive, isn't it?Of course, if it would be a simple resistor VI graph, or standard diode VI graph, you could use it as IV graph, because the slope never goes negative. That is again logical and intuitive.I see no problem here except that your question is asking to use a given graph, but the graph is the wrong one to use.Not that if I would take a current source and step the current, above point A current control loop would start ramping up the voltage and skip the valley region all together. So from point A the graph would look something like a super steep voltage rise to ~0.53V. With that graph your question would be a valid one. Specific current would correspond to specific voltage, simply because that's the way the graph was created in the first place.

author: MrAl
author_rank: Super Contributor
timestamp: August 01, 2023, 11:41:36 am
content: Hello again,I think a couple of interesting statements have come up in this thread that are worth repeating. I dont remember who posted these so if you want to take credit please feel free to do so. I am posting this because i think these statements (paraphrased) are simple yet still they explain the very nature of what Ohm's Law is all about. I'll put the statements in double quotes along with some short notes. Please remember these might be paraphrased so the exact text would have to be looked up in this thread."It’s simply understood that Ohm’s law has no predictive value with non-ohmic components."That might be the best way to put it. For a single value R=V/I at a single point Ohm's Law can not PREDICT the value of resistance, and that is what Ohm's Law is all about really."You cannot write a program to calculate the resistance at a single point using Ohm's Law for a nonlinear device."This is also a good way to put it.Because it has no real use at a single point, you can't create a program with code to calculate the resistance using Ohm's Law at a single point (v,i) unless the resistance is constant.And, one last note on the word being used, "instantaneous".I believe that what is meant here is not really related to time, but the way we think of things that occur at a single point even though they don't, strictly speaking, have any relation to time. It simply means "at a single point" such as at one V,I point.An example using the circle again would be something like this...5^2=3^2+4^2Now isn't that nice? We have an equality here and it looks a lot like the equality for some circle. However, it just relates two numbers it isn't the equation for a circle. If we try to force this to be an 'equation' for a circle, then we might note that we could see the same thing with this:5^2=4^2+3^2where it looks like i switched the x and y parameters. The equality still holds, but it's still not a circle. The equation for a simple circle looks like this:R^2=x^2+y^2(or similar).THAT relationship holds over a wide range of x and y.

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 01:24:56 pm
content: Quote from: Manul on August 01, 2023, 11:25:10 amNo, that's not the case. It may be intuitive to you, but it is wrong.The operation, results, and questions are identical whether you are set the current and measure the voltage or set the voltage and measure the current.QuoteYour not liking/understanding the physics and graph doesn't make it "wrong".So, based on your definition/(mis)understanding of resistance in non-Ohmic device, what is the resistance at 20mA? Stop dancing around the question!QuoteThere are indeed hysteresis effects, and those are the reason the TD can be used as a switch, logic element, or threshold detector. None of those are possible in non-ohmic devices.Repeat after me: Ohm's Law (R=V/I) applies to ohmic devices, and does non apply to non-ohmic devices. Not difficult!

author: Zero999
author_rank: Super Contributor
timestamp: August 01, 2023, 01:51:39 pm
content: Quote from: tggzzz on July 31, 2023, 10:20:36 pmI thought it was, along with P = I2R = V2/R and V = √(PR), I = √(P/R) etc. Plugging Ohm's law into a search engine, also brings up the aforementioned formulae.QuoteAlong with yourself and everyone else who's responded to this thread and others by the same original poster.After witnessing this argument, several tims on this forum and others, I've come to the conclusion that whether Ohm's law is a universal truth, or not, is one of semantics and isn't worth debating.

author: bdunham7
author_rank: Super Contributor
timestamp: August 01, 2023, 02:04:44 pm
content: Quote from: tggzzz on August 01, 2023, 01:24:56 pmI'm not sure I agree that there isn't a distinction here. If you are determining V given I (the IV graph) then the result is path-dependent. For determining I given V (the VI graph),in this caseit appears to not be path dependent and there is a single I for every V.QuoteFor every steady state circuit there would be a V and an I. Anyone who wishes to can find the result of the expression V/I for this case, there's no law against that, not even Ohm's. So if I deem R to mean 'result' or 'ratio', then there is no problem with R=VI for any circuit. What Ohm's law specifically means for ohmic materials is that R (for resistance) is a constant, giving the same V/I ratio in all cases. Of course there are limits on that too, no material is perfectly ohmic under all conditions. So what you have left of Ohm's "Law" after you analyze it is that V/I is fairly (but not perfectly) constant for some materials over a limited domain of V, I, temperature, etc and we can deem this material to have a resistance R which approximates the ratio V/I over that limited domain.As an aside, I'd point out the operation of the resistance ranges in the 5500-series Fluke calibrators. They aren't resistors, they are non-ohmic circuits that are specifically programmed to pretend to be resistors by adjusting their operation so that the voltage drop across them is a precise ratio of the test current put through them. So there's a case where R (resistance) very closely tracks V/I--better than most resistors--in a non-ohmic system. So simple rules are nice, but real life is complicated.AsTim Foxhas been pointing out, when people use the term R (in quasi-DC circuits, lets leave out reactance for now) they generally mean one of two things. In the ohmic case , they mean V/I and assume it is constant. In other cases where the discussion is of 'negative resistance' or 'source impedance' or terms like that, they mean dV/dI whether they know it or not. And, of course, if R is a constant, the R= dV/dI = V/I anyway.

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 02:16:42 pm
content: Quote from: bdunham7 on August 01, 2023, 02:04:44 pmTrue, but it doesn't change the point.QuoteYou can deem it to be anything you want, e.g. a glory. (Apologies to Lewid Carroll!) But if you want to exchange information, you have to agree on the meaning.There's an analogy with voltmeter calibrations. It is trivial and cheap to have all your meters agreeing with each other. It becomes difficult and expensive when you want your meter to agree with someone else's meterQuoteSeveral of us have pointed that out; few have understood - even when given concrete real-world examples designed to expand their understanding

author: Manul
author_rank: Super Contributor
timestamp: August 01, 2023, 02:35:21 pm
content: Quote from: tggzzz on August 01, 2023, 01:24:56 pmI'm not dancing around a question, I explained previously why I consider your question as simply invalid given how this graph was created and the nature of the device. I'm not implying that you don't understand how things work, in fact I'm pretty sure you do. So do I. I see nothing surprising, that one can't calculate resistance of a tunnel diode for a given current using VI graph. This data is simply not enough. Yet every point of a graph still has a defined R, and it is completely meaningful to talk about tunnel diode resistance at specific voltage points as a ratio between V and I. That's all I'm saying.

author: TimFox
author_rank: Super Contributor
timestamp: August 01, 2023, 03:55:20 pm
content: Tunnel diodes are useful due to their operation at negative resistance, defined by theslope, not the ratio.In this graph, the ratiosV/IandI/Varealways positive.

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 04:48:16 pm
content: Quote from: Manul on August 01, 2023, 02:35:21 pmThoseemphasisedstatements are mutually contradictory.If youincreasethe voltage and the currentreduces, how can that happen with a positive resistance? For the tunnel diode shown V=100mV I=28mA to V=400mV I=5mA.Why are you (incorrectly) concentrating on resistance "at specific voltage points" for one device (tunnel diode), but having to flip that to "at specific current points" for another device (neon bulb). Doesn't that raise warning flags in your mind?Which would you choose for Gunn diodes and why?Which would you choose for varactor diodes, and why?Which would you choose for a fluorescent lamp, and why?... UJT, thyristor, triac, IMPATT diode, a battery ...Tunnel diodes are usefulprecisely becausethey have anegative resistancecharacteristic in one part of their VI curve due to quantum mechanical tunnelling.See any textbookon tunnel diodes, or neon bulbs, or Gunn diodes, as well as some other useful devices.If you don't have a textbook, use wackypedia.Using your numerology, in what region is that negative resistance apparent?Hint:Tim Fox's statements are correct, but are apparently too theoretical (or something) for some people to understand. Hence my giving concrete examples, to try to get those with a more empirical bias to realise the limitations of their understanding.Think about what these diagrams are demonstrating:(both fromhttps://en.wikipedia.org/wiki/Negative_resistance)

author: wasedadoc
author_rank: Super Contributor
timestamp: August 01, 2023, 05:32:44 pm
content: At any point on that plot of tunnel diode current and voltage, both of those parameters are positive and dividing the voltage at the point by the current at that point gives a positive value of "resistance". This means that in a circuit applying the same dc voltage to a resistor of that computed value would result in the same dc current flowing.However if one applies a perturbation to the voltage it is apparent that there is a region where an increase in voltage is accompanied by a decrease in current and vice versa. The excursions of voltage and current have opposite polarity and the division operation gives a negative number - the "negative resistance".Irrespective of how one defines resistance, the effect is real and has been put to practical use in real circuits.

author: Manul
author_rank: Super Contributor
timestamp: August 01, 2023, 05:47:54 pm
content: Quote from: tggzzz on August 01, 2023, 04:48:16 pmEmphasised statements are not contradictory, I think I provided enough explanations above.QuoteHow can that happen? Are you serious? If you increase the voltage and current reduces, it simply means that resistance increased, not that resistance is not positive. This phenomena is called negative differential resistance if you wish. But resistance of the tunnel diode is always positive at any point! The direction of the flow of energy never reverses. I have hard time to understand what we are even talking about. I'm done here.

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 07:43:22 pm
content: Quote from: Manul on August 01, 2023, 05:47:54 pmIt does happen. Look at the VI curve. Read the numbers from the graph.Your contention that V=I*f(V,I) is definitely nothing like Ohms law, except where f(V,I) is aconstantR.Look at the illustrations you snipped. They illustrate negative resistance.Did your education include any calculus whatsoever?In my state school everybody took national maths exams at 15yo including integration and differentiation of polynomials except 1/x.

author: Manul
author_rank: Super Contributor
timestamp: August 01, 2023, 08:18:14 pm
content: Quote from: tggzzz on August 01, 2023, 07:43:22 pmMiscommunication is rising to epic levels. Of course, I'm not claiming to be highly proficient in english, but it's a bit weird. When I wrote "How can that happen? Are you serious?", it was a response to your "how can that happen with a positive resistance?". It was meant to rethorically express my disbelief, that you are asking this question, not that I'm asking it myself and not that I can't see a VI curve or I don't understand it. Is this some form of gaslighting or what? (Rethorical)

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 08:36:37 pm
content: Quote from: Manul on August 01, 2023, 08:18:14 pmOK, we'll let that pass. having different first languages introduces difficult and subtle problems.What about the other points I made in my last post and previous post, and you snipped?(Calculus, annotated graphs of generic VI curves)

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 01, 2023, 09:32:16 pm
content: Does the following help or clarify anything? It is just me arguing with myself, as a rhetoric device.Quote from: tggzzz on July 31, 2023, 08:55:29 pmLet's say I have a circuit which requires a voltage-controlled current source (or sink) with current between 10 and 20 mA, but my control voltage only varies by 0.1 V, but in the opposite way (0V at 20mA, 0.1V at 10mA). I pick the above tunnel diode, and bias my control voltage by about +0.2 V, from a soft-start supply that does not overshoot when starting up.How would you model the behaviour of that circuit? Me, I'd approximate it asI= 20 mA +VR, whereR= -0.1 Ω. In lumped-element model, that corresponds to a current source and a negative-resistance linear resistor, making its behaviour trivial to examine within its validity region, especially as part of a larger circuit behaviour analysis.(I could model it somewhat more accurately by fitting a 3rd, 5th, 7th, or any higher-degree polynomial curve, plus the Shockley diode exponential model for larger voltages, though. For a third degree polynomial,I= 0.52V- 2.86V² + 3.81V³. However, for larger circuit analysis, systems of third-degree or higher polynomials or exponentials need a lot of work to solve. So, the above approximation is just the simplest working one possible, and the one most useful for numerical computation.)Do you call that approximation "applying Ohm's law"?I do, because of how I have always used (and learned to apply) Ohm's law. Others disagree, but I do not know why: to me, it is just a model, like the rest of physics, and I don't know of any better name to use for such approximations.What is the "resistance" when the above tunnel diode current is 20 mA?The question of what the "resistance" is when the current through the tunnel diode is 20 mA is interesting, because it depends on the forward voltage. The possibilities are approximately 0.04 V, 0.2 V, and 0.52 V, "corresponding to 2 Ω, 10 Ω, and 26 Ω" (if we attempted to describe all of the behaviour of the tunnel diode using Ohm's law). Plainly, we can see that the Ohmic model ofV=RI(or its mathematical equivalentsI=V/R,R=V/I) does not work. Even if we were to considerRa function ofV, it cannot easily give us the three possible answers simultaneously.Thus, we see that Ohm's law does not describe the general behaviour of nonlinear components like this tunnel diode.This is not in question, this has been proven by the existence of such components.Does this mean that Ohm's law cannot be used with tunnel diodes?Yes, and No.Yes, because it just does not describe the overall V–I curve a tunnel diode exhibits.No, because there are two or more nearly linear regions where Ohm's law is a good approximation for the change in behaviour, if we use a lumped-element model and split the diode into two: one constant, one linear part.The first linear region is where the forward voltage over the tunnel diode is between 0 and 0.05 V. Here,I≃V× 2.3 Ω.The second linear region is where the forward voltage is between 0.16 V and 0.3 V. Here,I≃ 45 mA -V× 1.3 Ω.Also, any relatively narrow voltage range above 0.5 V can be treated as linear region around that center voltage, because there the current follows the exponential Shockley diode equation. The accuracy of the approximation then depends on the width of the voltage range used: the narrower, the better; the wider, the bigger the error at the edges too.I disagree. What's your point?Ohm's law is a model, and a truly simple one, because it describeslinear behaviour(that passes through origin, i.e. is zero at zero).It does not describe theentirebehaviour ofanycomponent, just like Newton's laws of motion do not describe physical motion when velocities are very high or very large masses are involved. Even resistors can burn out or be damaged by very high voltages. Electrical components, even solid-state components, also age because their chemical composition and/or structure degrades.What it does describe, is the behaviour of a component within some region of parameters, where its behaviour is linear. Often, the real component is then modeled as a constant current or voltage, plus a linear component. This kind of model is only valid in that region, and its accuracy depends on the selection of that region. Right or wrong, doing this is often also called "applying Ohm's law".Why do you keep repeating 'linear', as opposed to say "Ohmic"?Because oflinear algebra, and the tools that branch of mathematics provides, for solving large systems of linear equations.Even when your system is nonlinear, linear approximations of the parts near some operating point will quickly give you a candidate solution, that you can refine with more detailed, and more computationally intensive methods, including exponential and logarithmic behaviours of parts.I showed a particular example of this earlier in this thread, on how to apply Newton's method to calculate a square root of some value iteratively. For calculating the square root ofC, the recommended method to reduce the number of iterations, is to start with the linear approximation ofC/2.(If one usesfloating-pointnumbers – them being of formm×Bxwheremis mantissa, B is the constant radix of your floating-point format (usually 2), andxthe exponent – the equivalent linear approximation is (m/2)×Bx/2. The "famous"fast inverse square rootis not this, but quite closely related, for those interested in computing history.)Hell, this is exactly what we humans do when we do on-the-envelope first-order approximations intuitively!The mathematical subfield that deals with obtaining such an approximate solution fast then refining it, is calledperturbation theory.

author: tggzzz
author_rank: Super Contributor
timestamp: August 01, 2023, 10:48:09 pm
content: That's averythoughtful and useful postI agree with a lot of what you have written, but there are exceptions.Quote from: Nominal Animal on August 01, 2023, 09:32:16 pmI disagree,for the reasons you give below(Short form: Ohm's law is linear)QuoteYes, precisely.Caveat: unlike you, some people cannot grasp that Ohm's law does not apply to the many devices and materials that do not follow Ohm's law. The consequences of that tautology are beyond some people's comprehension.QuoteSuch "piecewise linear" approximations can be useful, but have limits.Other equally useful approximations have different limitations.QuotePrecisely! That linear behaviour/model is absolutely key!I well remember my first week at university, when I tripped up over that - and so did my tutorMy (incorrect) definition of "linear" was based on school pure maths and applied maths, where y=mx+c was a line. It took me, (and my tutor!) a while to get my head around the importance of requiring c=0, i.e. just as you state.QuoteIt is wrong, because Ohm's law is a linear model.But that wrongness conceals a deeper truth: the benefits of finding ways to take a non-linear system and model it as if it is a linear system within tight boundaries. Those benefits are so profound that such models are given a specific name: small-signal models with differential/small-signal parameters.Such small-signal models/parameters describe only the behaviour "near" a specific point; stray too far and they become invalid.Such small-signal models/parameters are usuallyverydifferent to the large-signal models. Small-signal models/parameters are much simpler to understand and reason about, and capture important behaviour but not all behaviour.But you mention those points belowQuote... because it is vitally importantOhm's law is merely the codification of one instance of linear behaviour/model. The concept of a linear system/model is far more general than Ohm's law.QuoteMathematically small-signal approximations are directly related to differential calculus. In particular the small-signal model is valid over the region where the differential (i.e. slope of a graph) is suitably constant.And with that we are back to the points that Tim Fox has been making very succinctly. (Too succinctly for some people).Linear models (e.g. Ohm's law) cannot describe non-linear systems (e.g. tunnel diode or neon), and any attempt to do so is a fundamental error.Of course neither can small-signal models describe non-linear systems.

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 02, 2023, 02:11:01 am
content: I was utterly wrong in #61 and later: the term "instantaneous" has nothing to do with tggzzz-tooki disagreement.It is either A) that like me, Tooki uses "Ohm's law" for all linear approximations, or B) Tooki is wrong and thinks Ohm's law always does apply, if you consider resistanceRa function (that depends on bothVandI, as well as timetand the properties of the component itself).The latter fails, because the functionV(I) can be multi-valued for nonlinear components. Mathematically, it is possible ifRdepends onVandIin sufficiently complex way, say involving square or cubic root, but then calling the formV=IRa law or a model would be nonsensical:V(I) =IR(V,I, ...)says nothing about the voltage behaviour, unless the functionR(V,I, ...) is also defined: It literally says only that "voltage is a function of current and resistance, with resistance being a function of voltage, current, and possibly other variables". It's like a weather forecast that says it might or might not rain, be sunny, or there could even be weird weather. It's not a forecast; and such a law is not a law or model because it does not define anything in any way, just states that anything is possible.Quote from: tggzzz on August 01, 2023, 10:48:09 pmThanks! Because of your reply, I now understand. I think!Quote from: tggzzz on August 01, 2023, 10:48:09 pmWell, no, "linear" does not by definition require passing through zero; the origin-passing ones are simply a subset of all linear ones. (I mean, the parentheses were not a definition, but limiting to a subset.)All functions of formy= mx+ c are indeed linear, and the term "linear" itself does not require passing through zero.I do not know the exact correct term. I use "linearandpassing through the origin"."Zero-intercept linear"is one possibility. "Ohmic behaviour" is another.This is a language problem, because if you separate Ohm's law from linear approximations, then Ohm's law isn'tjustlinear; it also requires passing through zero,V=0,I=0, which linear approximations in general do not.It is very important to realize that in the linear approximation, a nonlinear component is not represented simply by a resistor (with possibly negative resistance), but also by that constant voltage or current as if part of that component. In that sense, a linear approximation is definitely not "Ohmic".(I can definitely see myself also tripping on not separating the two, and making a mistake because of that.)Quote from: tggzzz on August 01, 2023, 10:48:09 pmIn case I haven't made it clear, I fully agree: they are spot on.(I wish I could write as clearly and to the point as he does. I don'twantto be this verbose.)In particular, I thought hisfirst post in this thread, reply #78fully outlined the use cases and sources of misunderstanding.I'm very tempted to delete my posts #61, #65, #72, #84, and replace #85 with a post that points out that nods to TimFox first, then points out the disagreement in between calling all linear approximations as applications of Ohm's law (as below the line). That way, I'd help clear things, and not just cause more confusion. I will not, because that would be falsifying history, and hiding my mistakes. I will put that text below, though.If call linear approximations of nonlinear components as applying "Lanclec's law", to completely separate it from Ohm's law, we find the two areanalogous:Rin Lanclec's law approximates ∂V/∂Iwithin the region of interest, for all components.The difference between the two is that Lanclec's law describes linear dependence that does not need to pass through zero, and thus involves "unphysical" current or voltage sources or sinks (depending on which one is trying to solve), to model the behaviour of a nonlinear component within some operating region where its behaviour is sufficiently linear. The choice of the operating region for each component determines the accuracy of Lanclec's law.In other words, within regions where nonlinear components behave sufficiently linearly (the closer to linear they behave, the smaller the error), you can replace each nonlinear component with a linear one (meaning one voltage or current source, plus an "Ohmic" component, possibly with negative resistance), and then apply Kirchhoff's laws to solve this simplified system.For complex systems, nobody relies on that linear approximation itself, but it lets us describe very large and complex systems using linear algebra (and thus solve them fast and efficiently), and it also gives us a starting point for refining the solution using more accurate models and numerical iterative methods.Thus, Ohm's law should not be considered a separate thing that always applies, but alumped-element modelof specific types of situations, like "Ohmic" components in electrical circuits; and a companion to Kirchhoff's circuit laws. "Lumped-element model" does not refer to lumping electrical components at all: it is a method of simplifying the description of systems – all sorts of systems, not just electrical circuits –, a mathematical tool.Lanclec's law trades limiting ourselves to sufficiently narrow regions of current and voltage for each nonlinear component, to letting us use the same fast mathematical tools to find an approximate or initial solution to the system, as when using Ohm's law and Kirchhoff's law. Its precision and validity is pre-determined by the choice of the region.However, if you do conflate Ohm's law and linear approximations, communication will be difficult or impossible with those who consider them separate.Also, there is obviously no such thing as "Lanclec's law", that is just a made up name for linear approximation of nonlinear components in electrical circuits.

author: robowaffe
author_rank: Regular Contributor
timestamp: August 02, 2023, 05:59:33 am
content: On linear versus non linear.In falstad, when u add another voltage source to a fixed set of solid state resistors, you only get a linear modification to all currents in the system.I think this is actually wrong, and it is possible to get a non linear change from a linear voltage change, but falstad doesnt sim it.I dont think you can fully trust falstad, it wont take you all the way, it can be wrong.

author: MrAl
author_rank: Super Contributor
timestamp: August 02, 2023, 10:23:22 am
content: Quote from: robowaffe on August 02, 2023, 05:59:33 amHello,Show the circuit if you want someone to check this out. Without a circuit it is impossible to discuss.

author: MrAl
author_rank: Super Contributor
timestamp: August 02, 2023, 10:29:57 am
content: Hello again,I find it interesting that so many people are worried about the exact nature and interpretation of what we call Ohm's Law.I guess because it is strange to see Ohm's Law being written the same way as a general resistance calculation would be written, and many are passionate about the field of electronics and magnetism.Ohm's Law: R=V/IGeneral resistance: R=V/IAny wonder why there is often such confusion, discussions, and disagreement.Maybe the key factor is that rules like this don't come out of thin air, they come from a study of materials and how they work. Another way to put it is that the materials and their properties come before the equations or expressions, not the other way around. You don't get an expression and then try to force it onto different materials or different ways to use it, you start with the materials and then come up with an expression to describe what is happening. If one expression happens to be the same as another, even if that is rare, that could be just a coincidence.

author: tggzzz
author_rank: Super Contributor
timestamp: August 02, 2023, 03:29:42 pm
content: Quote from: Nominal Animal on August 02, 2023, 02:11:01 amOh, mumble mumble mumble....Thinkingagainof something that happened almost half a century ago has refreshed my thinking.You (and I) are right that mx+c is linear for c != 0.The c=0 is necessary forsuperpositionof responses to be valid. If superposition is valid then the effects of different inputs can be modelled independently and then simply added together. Simply suppress (i.e. set voltage sources to a short circuit, current sources to an open circuit) all but one inputs and see what the response is. Rinse and repeat with other inputs, and add all the responses together. That's rather useful when doing circuit analysis!The reason for my mistake was that I was posting too late at night. That reason doesn't excuse it, though.QuoteBrevity is useful when the audience understands the point being made. But if the audience doesn't understand then more verbose explanations can sometimes help.QuotePlease don't "revise history"Often thepathto enlightenment is as helpful to others as the actual endpoint.OTOH, adding an extra comment allowing other readers to - if they wish - skip to enlightenment can also he helpful to others.TLDR: we agree

author: tggzzz
author_rank: Super Contributor
timestamp: August 02, 2023, 03:33:50 pm
content: Quote from: MrAl on August 02, 2023, 10:29:57 amPartly that, partly because the rules/laws are formulated in such a way that you canmathematicalanalysis of circuits is possible. See my and Nominal Animal's recent posts on linear systems and superposition.Simulation is, of course, a poor substitute for analysis, since simulation is merely numerology on steroidsUnfortunately, there are many cases where simulation is the best available tool.

author: MrAl
author_rank: Super Contributor
timestamp: August 03, 2023, 10:07:38 am
content: Quote from: tggzzz on August 02, 2023, 03:33:50 pmHi again,I do not know what you mean by "mathematical analysis" or "simulation", I have nothing against either of these.This is about the application of these techniques, not about the fundamental properties. That is, when you can apply one technique or another, or rather, when you SHOULD apply one technique or another, or even when you should assume that you CAN apply one technique or another. This is a little more of a subtle point than we usually run into, and that is unfortunate because that means it is harder to understand the logic behind it.

author: tggzzz
author_rank: Super Contributor
timestamp: August 03, 2023, 10:37:16 am
content: Quote from: MrAl on August 03, 2023, 10:07:38 amExamples of mathematical analysis:solving algebraic equations, firstly to derive the general principles of KCL and KVL, and - separately - using KCL and KVL to "predict" what will happen in your circuit given a whole range of inputssolving algebraic equations to derive the principles of how poles and zeros and how they affect stability, and - separately - predicting the phase margin of your circuitExamples of simulation:number crunching using LTSpice to "measure" what happens in your circuit when you apply one specific input waveformnumber crunching using LTSpice to see whether or not your circuit will oscillate with specific component valuesIn order to make mathematical analysis tractable, it is necessary todefinecomponents in such a way that superposition applies in systems of those components. A lot of effort is expended, with greater or lesser success, to ensure that components'definition(e.g. R=V/I, Z=1/jωC) enables such analysis.Not all components have characteristics that are compatible with those definitions, e.g. non-linear components I=k1(ek2V-1) e.g. small signal analysis of systems containing non-linear components, where the "small" means over a sufficiently limited range that they can be considered linear and hence superposition does apply e.g. the small signal diode model where I=k3VConsult a textbook to find the values of k1k2k3

author: robowaffe
author_rank: Regular Contributor
timestamp: August 03, 2023, 11:10:00 am
content: Quote from: MrAl on August 02, 2023, 10:23:22 amno, u just didnt understand what I said.But that is perfectly ok! Its confusing understanding someone else when we are talking about all this tech confusion.But it doesn't mean its bullshit when u say something and I dont understand it!

author: tggzzz
author_rank: Super Contributor
timestamp: August 03, 2023, 11:31:59 am
content: Quote from: robowaffe on August 03, 2023, 11:10:00 amIt appears MrAl understood what you wrote better than you understand what you wrote.MrAl's request is reasonable. Simon's response in another of your threads is relevanthttps://www.eevblog.com/forum/beginners/propagation-delay-question/msg4993969/#msg4993969

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 04, 2023, 07:49:15 am
content: Quote from: tggzzz on August 02, 2023, 03:29:42 pmI never do, if it involves an error or mistake I've made: it isnecessaryinformation to others, and valuable to myself as an expression of honesty. Everyone makes errors, but only those with a working solid spine will admit to making any specific ones, even if it stings a bit, socially speaking.Quote from: tggzzz on August 02, 2023, 03:29:42 pmYep, except I don't really know what to say to clear up any confusion, as this thread is already pretty darn chaotic already!

author: tggzzz
author_rank: Super Contributor
timestamp: August 04, 2023, 08:21:56 am
content: Quote from: Nominal Animal on August 04, 2023, 07:49:15 amSpot on, in all respectsAnd it seems we won't be distracted by robowaffe any more.

author: MrAl
author_rank: Super Contributor
timestamp: August 04, 2023, 08:29:59 am
content: Quote from: tggzzz on August 03, 2023, 10:37:16 amHello,Not sure why you went through all that when i already said this is not about that.It's about the application of a technique or law or rule of some kind. That's entirely different than an analysis method.

author: MrAl
author_rank: Super Contributor
timestamp: August 04, 2023, 08:35:53 am
content: Quote from: robowaffe on August 03, 2023, 11:10:00 amHello,Who said it was? Something can be questionable but still not be BS.When you make statements out of the blue if you want someone to understand you and believe you, you have to show something concrete. A schematic is a concrete way of showing what you mean when it comes to electronics. If you have a question about a circuit or make a statement, especially a broad ranging statement, you have to show a circuit. This is what talk about circuits is all about.If you make statements about a circuit and don't show a circuit to prove or just talk about it, it's going to be hard to talk about it, and if you keep doing that you probably won't be taken seriously after a while. There's nothing hard about showing a circuit in this day and age especially if you are already familiar with simulators, which allow you to easily draw circuits. Not only that though, but if you like to talk about circuits then you might as well learn how to draw them. It also makes it more interesting to everyone.You made a statement about a linear circuit and a given simulator, so you must have already drawn the circuit so why not show it and explain what you did that made it act the way it did.

author: John Coloccia
author_rank: Super Contributor
timestamp: August 04, 2023, 09:15:29 am
content: Something I always tell my guys: names are important, and confusing names are just as bad as design mistakes...and often lead to design mistakes."Negative" resistance would mean that as your voltage went up, conventional current would flow in the opposite direction of your voltage. That's the only definition that could possibly be mathematically and logically consistent. It's a simple consequence from V=IR, and by definition that's what it means to have negative resistance. It's exactly the same as negative voltage, i.e. hook up your power supply backwards. It's just convention that we always consider resistance positive, because that convention kind of models the physical reality well."Negative differential" resistance is, by definition, a situation where increasing voltage decreases current. It has the word negative in it, but in the sense that the increase in current with increasing voltage has a negative coefficient, not positive. It's still positive resistance by convention because the direction of the current doesn't change. If you claim it's negative resistance, you also have to change the sign of the voltage. I don't think anyone is intentionally making that claim.Ohms law still applies in any case. It relates voltage, current and resistance. The fact that R for some devices is a function of voltage doesn't really change anything.V = IRV = IR(V)It's the same thing, just like ohms law applies perfectly well when R is a function of temperature, as R almost nearly always is:V = IR(T)But it's all just by definition. What it means to have resistance is that if you apply a voltage V across a thing with a given R, you get the relationship V=IR, whatever R happens to be. For more complicated relationships, you use different words and different techniques because it's not resistance anymore. Maybe there's the concept of an equivalent resistance just to makes things simpler, but it's a different thing.

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 04, 2023, 02:18:43 pm
content: Quote from: John Coloccia on August 04, 2023, 09:15:29 amI disagree,but onlyin that when R is a function of voltage and current, i.e.V=I·R(V,I), it makes no sense to call it applying Ohm's law; much less than calling linear approximations applying Ohm's law (which I did, before now).(And that means I do not believe Ohm's law applies in all cases.)Feel free to disagree, because this is a nuanced thing, but here is my reason for the above opinion:Ohm's law is primarily a tool for modeling components in an electrical circuit. If you let the resistance vary depending on the voltage and current,R(V,I), you aren't saying anything about the behaviour of the circuit, because it says absolutely nothing about the voltage–current relationship! As an example, considerR=C/(V·I), which leads toV=±√C, but no definition forI, i.e. defines a constant positive or negative voltage device, regardless of current!Your example showed only a dependence on voltage, i.e.V=I·R(V), which tells us there is a linear dependence on current, but a possibly nonlinear dependence on voltage. I could call that avariantof Ohm's law, but note that it is not sufficient to model all nonlinear components.Your third example,V=I·R(T), which applies to basically all pure resistors, I can accept as within Ohm's law, because itdoesdescribe the (linear) dependence between the three variables, with just the resistance depending onother, externalvariables.

author: TimFox
author_rank: Super Contributor
timestamp: August 04, 2023, 02:30:16 pm
content: I still maintain that theratioofV/Iis almost never useful for non-proportional (non-ohmic) devices.For ohmic devices, the ratio is the usual resistance value that is obviously useful.For other devices, such as the tunnel diode shown above with its negative-resistance region, the ratio ofV/Iis always positive, but the slope (derivative) dV/dIgoes negative and gives a useful result.

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 04, 2023, 03:01:15 pm
content: Quote from: TimFox on August 04, 2023, 02:30:16 pmI do agree.Before the discussion with tggzzz, I considered both "applying Ohm's law", but now call the latter a linear approximation of nonlinear components, for simplicity. The two are closely related, perhaps even analogous, but indeed one isV/Iratio, and the other the dV/dIslope (or partial differential ∂V/∂I, if other variables like temperature are also considered; those too will be considered constant within the range the linear approximation is applied).Above, I tried to show that turningRinto a function, does not help or make Ohm's law more general without losing its descriptive properties, because either one makes the function describe nothing (with all complexity hidden inR), or it cannot describe all non-proportional/non-ohmic/nonlinear components.On the other hand, a linear approximation can be done for any kind of component, but it will only work within a small voltage–current range, depending on the component properties. As a tool, linear approximations get one "pretty close" with little work, but require one to ensure one keeps within the valid range. Although it is analogous to Ohm's law, withR=dV/dI(at specific chosenV,Inear the center of the range, with constantRused within the range), it is better called a linear approximation, for clarity and avoiding misunderstandings.

author: tggzzz
author_rank: Super Contributor
timestamp: August 04, 2023, 03:28:59 pm
content: Quote from: John Coloccia on August 04, 2023, 09:15:29 amYes and no, respectively.Ohm's lawonlyapplies to ohmic devices. Ohm's lawdoes not apply to non-ohmicdevices. That is simple and unambiguous, and I would have thought uncontroversial.So, what's the distinction between ohmic and non-ohmic devices?Ohmic devices are those where there is a linear relationship between voltage and current, and with zero current there is zero voltage. In standard mathematical notation, V=f(I) where f(I) is mI+candc=0.Non-ohmic devices have arbitrary f(I), and those functions are indeed a very wide rance of functions. Some functions are constant, some are exponential, some logarithmic, some are polynomial of varying order, some are strictly monotonic, some are monotonic, some are not monotonic, some are multivalued.Negative resistance (differential or not) is where anincreasein voltage corresponds to adecreasein current - whatever the values of V and I.It is possible to have positive V/I and negative dV/dI, e.g. a tunnel diode amongst others.It is possible to havenegativeV/I andpositivedV/dI, e.g. a cell/battery.(fromhttps://en.wikipedia.org/wiki/Negative_resistance)

author: TimFox
author_rank: Super Contributor
timestamp: August 04, 2023, 04:32:04 pm
content: Reply to the previous two replies:The reason why the slope (units of ohms or siemens, depending on which derivative) is a useful function of the current (or voltage) of the device is for calculating small-signal (or AC) behavior: it allows you to calculate, as a function of bias (or quiescent) operating point what a small change in voltage or current will cause on the other variable.For example, applying the Schockley equation for a forward-biased BE junction to the emitter current as a function of BE voltage and taking the derivative thereof is how you can calculate the transconductance of a BJT from base voltage to collector current, correcting slightly for the smaller (non-linear) base current and its (still not constant) ratio "beta".

author: Nominal Animal
author_rank: Super Contributor
timestamp: August 05, 2023, 03:53:58 am
content: Quote from: TimFox on August 04, 2023, 04:32:04 pmExcellent, thanks!Small signal analysis was not part of my electronics courses, but I happen to have struggled with controlling 4×20mA common-anode LED backlighting; instead of doing analysis, I just ran a bunch of spice simulations to "chart" the parameter space, varying all of the properties...To anyone not familiar with the math, the Shockley diode equation describes the diode current \$I\$ as a function of diode forward voltage \$V\$,$$I = I_S \left( e^\frac{V}{n V_T} - 1 \right)$$where \$I_S\$ is the reverse-bias saturation current, \$n\$ is the the ideality factor (\$n \ge 1\$, but small; \$1\$ for ideal diodes), and \$V_T\$ is the thermal voltage (0.025852 V at 300 K or 27°C, \$V_T = k T / q\$ where \$k\$ is Boltzmann constant, \$q\$ is electron charge, and \$T\$ is temperature in suitable units, typically Kelvins).The derivative of the current with respect to forward voltage is$$\frac{\partial I}{\partial V} = \frac{I_S}{n V_T} e^\frac{V}{n V_T}$$which is a straightforward exponential curve.Interestingly, the partial derivative with respect to thermal voltage is$$\frac{\partial I}{\partial V_T} = -\frac{V I_S}{n V_T^2} e^\frac{V}{n V_T} = -\frac{V}{V_T} \frac{I_S}{n V_T} e^\frac{V}{n V_T} = - \frac{V}{V_T} \frac{\partial I}{\partial V}$$which means that one can make a quick estimate of how the current varies at a given temperature and voltage, by multiplying the \$\partial I / \partial V\$ slope by \$-V/V_T\$. (It is as good an estimate as the Shockley diode represents the behaviour.)Many sources mention an approximation \$I = I_S e^\frac{V}{n V_T}\$. When \$V / (n V_T) \ge 4.61\$, the error is drops to 1%; with \$\ge 6.91\$, the error drops to 0.1%. In general, the relative error is \$\log\left(\frac{V}{n V_T}\right)\$ (using natural logarithm, often calledln).Fitting real-world measurements to Shockley diode model has to be numerically, but using free software like Octave or Gnuplot it is trivial.For example, if you create a filediode.txtwith one measurement per line, voltage (in volts) and current (in amperes) in decimal separated with one or more spaces on each line, then runningCode:gnuplot -e 'f(V)=Is*(exp(V/nVt)-1); fit f(x) "diode.txt" using 1:2 via Is,nVt'will tell you the least squares fit for \$I_S\$ and \$n V_T\$ under headingFinal set of parametersnear the bottom of the output. (In Linux and BSDs, you can append2>&1 | tail -n 7 | head -n 2to the command to get just the interesting output.)The reverse bias saturation current \$I_S\$ is often tiny, values between1e-20and1e-6are typical for real-world diodes, and LEDs can have it as low as1e-30. Seeltwiki standard.diofor examples.

author: IanB
author_rank: Super Contributor
timestamp: August 05, 2023, 04:50:38 am
content: Quote from: Nominal Animal on August 02, 2023, 02:11:01 amYou could say it is a proportional relationship, or linearly proportional if you wish to be very precise. But "proportional" without any qualifiers usually describes a relationship of the form \$y=kx\$, where \$k\$ is the constant of proportionality.Thus Ohm's law is a proportional relationship of the form \$V=RI\$ with \$R\$ being the constant. You could equally express it as \$I=GV\$ where \$G\$ is the conductance, or the reciprocal of the resistance.

author: magic
author_rank: Super Contributor
timestamp: August 06, 2023, 04:41:56 pm
content: Quote from: Nominal Animal on August 05, 2023, 03:53:58 amIt is even more interesting that the resulting derivative is negative, which means that forward current decreases with temperature at any given forward voltage...

author: TimFox
author_rank: Super Contributor
timestamp: August 06, 2023, 05:28:05 pm
content: A side effect of that temperature dependence is that the input bias current of BJT op amps (forward-biased diode) decreases with increasing temperature, but the bias current of JFET op amps (reverse-biased diode) increases.At high temperature, the BJT device may have lower bias than the JFET; compare the LM308 series to the TL071 series over temperature.

author: RoGeorge
author_rank: Super Contributor
timestamp: August 07, 2023, 10:46:03 am
content: Quote from: Nominal Animal on August 05, 2023, 03:53:58 amWow, always wished to have a quick way of draft plotting something.It was already there:gnuplot -p -e 'plot "diode.txt"', waiting.I didn't know gnuplot can be that easy, thank you!

author: magic
author_rank: Super Contributor
timestamp: August 07, 2023, 08:15:29 pm
content: Quote from: TimFox on August 06, 2023, 05:28:05 pmFor the record, this has nothing to do with what I wrote previously, and I was being ironic. Contrary to the erroneous analysis, at a constant forward voltage, diode current obviously increases with temperature, as everyone knows.Meanwhile bias current of bipolar amplifiers really tends to decrease with temperature because β of transistors increases. But it's not so simple, because input stage collector current needs not be constant over temperature. Indeed, many opamps increase their input stage bias to counter the loss of transconductance due to increasing VT. Examples: LM358 advertises "temperature compensated bias current" and National's datasheets shows a wavy curve with one peak near room temperature and another at 125°C. NJM4559 increases with temperature. AD797 decreases but then increases near 100°C.

author: TimFox
author_rank: Super Contributor
timestamp: August 07, 2023, 09:51:49 pm
content: Yes, I stand corrected. For a given applied forward-bias voltage, the current in a PN junction increases with junction temperature.The increase in beta for a BJT with temperature is a separate, not a side, issue.Also, for a given applied reverse-bias voltage, as in the JFET, the current increases with junction temperature.I need to be more careful with agreeing with someone else's statement that ends with ellipses.

author: schmitt trigger
author_rank: Super Contributor
timestamp: August 11, 2023, 02:11:14 pm
content: I am so happy that this thread has drifted into something far more intelligent and interesting than the original post.