A resistor that never gets hot? - Page 1
Forum section: electronics / projects

author: John Heath
author_rank: Frequent Contributor
timestamp: January 12, 2018, 05:22:39 am
content: I am going to take a cd4016, quad FET switch and tie 2 switches to one side of a 1 u Henry coil and the other 2 to the other side. I will then drive the cd4016 at 100KHz in such a way that the 1 u Henry coil is changing polarity 100,000 times a second. A DC current will always see this 100,000 per second polarity switching 1 u Henry coil as reactive limiting the flow of DC current. As it is reactive there is no heat. A kind of endless perpetual counter EMF. Would this opposition to DC current be a resistor that never gets hot?

author: xavier60
author_rank: Super Contributor
timestamp: January 12, 2018, 05:49:25 am
content: When an AC voltage is applied to an inductor, the inductor alternately stores energy from the voltage source, then returns it back to the source, less losses. Twice per cycle.

author: AG6QR
author_rank: Frequent Contributor
timestamp: January 12, 2018, 06:26:29 am
content: Quote from: John Heath on January 12, 2018, 05:22:39 amNo.You mention switching the polarity at 100,000 Hz, so you're not talking about DC anymore.Reactance is not the same as resistance. A resistor always has its voltage and current in phase. A reactance does not. Ideal reactance doesn't dissipate energy, and doesn't get hot, but its impedance is non-resistive, so it's not a "resistor that doesn't get hot".In addition to the non-resistive reactance, there will, of course, be at least some parasitic resistance, which will dissipate energy and cause the resistor to get warm.The only resistor that doesn't get warmed up is one that has no voltage across it, and no current passing through it (one of those conditions implies the other).

author: AndyC_772
author_rank: Super Contributor
timestamp: January 12, 2018, 09:12:41 am
content: This doesn't sound like a question about parasitic components, but about pure theory.John, can you please post a rough schematic of what you have in mind? How exactly is your coil connected, and where does the current flow?Your assertion that "A DC current will always see this 100,000 per second polarity switching 1 u Henry coil as reactive limiting the flow of DC current" doesn't make any sense to me. A DC current flows through a pure inductor just like it would through a straight wire; there's no resistance at all.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 13, 2018, 12:38:08 am
content: Quote from: AG6QR on January 12, 2018, 06:26:29 amI was not as clear as I could have been when describing this circuit. The applied power to this circuit xould be a simple DC 5 volt source from a 7805. The only thing that is switching at 100,000 times persecond is the 1 u Henry coil. You may assume idea condition of no capacity and no resistance.Another way to describe this circuit would be the exact opposite of a class D amplifier. A high frequency square wave can drive a inductor without high current as the reactance of the inductor prevents it. In this case it is the opposite with a DC source that is driving it with the only high frequency AC being the inductor switching it polarity using a cd4015 quad FET to accomplish this.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 13, 2018, 12:44:36 am
content: Quote from: AndyC_772 on January 12, 2018, 09:12:41 amCan not give a diagram as I have not built it. It is a thought experiment only. It does not have to be a cd4016 quad FET switch. Anything that flips the polarity of the inductor 100,000 times a second will do.Would this be a resistor that never gets hot?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 13, 2018, 12:58:22 am
content: Quote from: blueskull on January 12, 2018, 05:26:30 amYes there would be efficiency issues in the real world where switches are not perfect nor are inductors. Too soon for this. First does it work as a heatless resistor in theory assuming idea condition. If so then worry about building it in the real world.

author: wraper
author_rank: Unknown
timestamp: January 13, 2018, 01:05:47 am
content: Quote from: John Heath on January 13, 2018, 12:58:22 amIt does not work as resistor. Energy that is apperantly consumed at one moment of time is returned back at another. If you take an ideal inductor without losses, net energy consumed will be 0.

author: westfw
author_rank: Super Contributor
timestamp: January 13, 2018, 01:57:53 am
content: This is essentially how fluorescent ballasts work. The Tube itself, once running, would happy pass enough current at normal mains voltage to damage itself, and needs a current limit (like an LED.) A "magnetic ballast" is essentially a huge inductor that appropriately limits the current at the 60Hz line frequency. An "Electronic Ballast" (also as used in CFLs) steps up the frequency and can thus use a smaller inductor to produce the same AC current limit.(A ballast also contains some magic to handle startup, when the tube needs the filament heated, and/or a temporary higher-voltage supply to start the tube discharging. But once operating, the ballast is essentially an "low-loss AC resistor"...)

author: retrolefty
author_rank: Super Contributor
timestamp: January 13, 2018, 02:07:08 am
content: Quote from: John Heath on January 13, 2018, 12:44:36 amGenerally it works better if you design and draw a schematic before you attempt to build a circuit.I really don't understand clearly what you are proposing. The reason that schematics work so wellin electronics is that's it's true that one schematic in worth a thousand of your typed words.

author: DaJMasta
author_rank: Super Contributor
timestamp: January 13, 2018, 02:25:20 am
content: I don't see how you'd pass current through a resistor and prevent it from heating up - if you were oscillating both ends in phase, then you'd never have a potential across it, no current would flow, and you'd have effectively no loss, but if you've got current - AC or DC, positive or negative - you're going to have power dissipation in the resistor.Just because the average voltage is zero doesn't mean the resistor won't act as a load - the potential would have to be the same on both sides all the time for no power to be lost, but then no current would be flowing through it, and why would you bother putting the resistor in, it would have no effect (in the limited scope of the exercise, of course it could be used for abnormal conditions or fixed impedance at different frequencies or what have you).

author: Marco
author_rank: Super Contributor
timestamp: January 13, 2018, 02:29:33 am
content: Just run it through a simulator with ideal components, your impedance will approach infinity.If you want to emulate a resistor with a switching circuit you have to dump the energy somewhere, if you want to avoid generating heat the most likely suspects are stored charge in a capacitor or chemicals in a battery.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 13, 2018, 02:31:49 pm
content: Quote from: westfw on January 13, 2018, 01:57:53 amInteresting. Yes I see your point. I put a small neon bulb on a variac one day then turned the variac up until the the neon voltage break down was met , around 70 volts. When this happened the neon bulb blew a 2 amp fuse on the variac. The little neon bulb never worked the same after that. It was a little blackish inside. I can conclude from this that a larger fluorescent tube is using the inductive " magnetic ballast" as it's current limiting resistor.This could be considered proof of concept that a heatless type of resistor is possible. However one would have to do it the opposite way to know for sure. By opposite I mean 150 volts DC supply for a standard fluoresent tube then switch the polarity of a standard ballast transformer 60 times a second. Is alternating polarity of the basllast transformer with a 150 volt DC power source the same as alternating alternating the power source , AC power , and not changing the polarity of the ballast transformer? If so then a general heatless type of resistor should be possible with a polarity switching inductor.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 13, 2018, 03:22:32 pm
content: Quote from: retrolefty on January 13, 2018, 02:07:08 amI understand your point of no diagram. A 5 volt reg 7805 with the 5 volts going to a 1 u Henry inductor then a 5 ohm resistor to ground. That's it. Turn it on and the resistor should have 5 volts at 1 amp , 5 watts of heat as the inductor has 0 ohms resistance to DC current. Now switch the polarity of the inductor back and forth. The faster you switch the less current to the resistor as inductance is like a flywheel resisting a change in direction. The frequency of the 1 u Henry polarity switching should set the effective resistance of the inductor. A heatless resistor of sorts.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: January 13, 2018, 03:54:33 pm
content: A pure reactance, capacitive or inductive, will have no losses. Unfortunately, there is no way to supply and switch the current with zero ohms in series. Look up the 2-capacitor problem. Now, there are other ways to do things. Seehttps://www.av8n.com/physics/capacitor-transfer.htm

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 13, 2018, 04:53:37 pm
content: Reactive power at DC can be a useful concept, but strictly by definition, that means nonsense.Reactive power is most commonly introduced as an AC steady state property. At 0Hz, the average value of the imaginary component is always and forever zero. So, there cannot be reactive DC power.Let's take a more general definition:Instantaneous power is p(t) = v(t) * i(t).Real power is the time average of p(t).Reactive power is the averaged difference between p(t) and the average. (I forget exactly how; RMS I think?)When only DC is present, real power is simply the product of V and I, whether you take the average or RMS figures. (Same thing.)When AC and DC are present, you must take account of them separately. You can have real DC power, but AC reactive power -- like in a switching supply, where the AC power is ~zero (because the ripple voltage (for a CV supply) is negligible) but real reactive power is being cycled back and forth (the inductor in a typical converter handles reactive power about 1/4th the DC power converted).Tim

author: AG6QR
author_rank: Frequent Contributor
timestamp: January 13, 2018, 06:31:05 pm
content: Quote from: John Heath on January 13, 2018, 12:38:08 amI think you were quite clear. But what's important isn't the nature of the original power source, what's important is the nature of the power presented to the load. AC can be converted to DC, and DC can be converted to AC. If you switch the polarity of electricity presented to the coil, no matter how you do it, the coil is not seeing DC anymore, even if the original power source was a battery.Putting aside parasitics for the moment, an inductor is not a resistor. The voltage across a resistor is proportional to the current. The voltage across an inductor is proportional to the derivative with respect to time of the current (that derivative, and thus the voltage across an inductor, is always precisely zero by definition under DC conditions). These two are not the same, and understanding the difference is fundamental to understanding general electronics and AC circuit theory.This inductor is not at all resistive, so how could you consider it a resistor that doesn't get hot?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 03:52:05 am
content: Quote from: Conrad Hoffman on January 13, 2018, 03:54:33 pmCool link. There is a less complicated way to transfer 99 % of the energy from one condenser to another. Let us start with 100 uf condenser "A" charged to 150 volts and condenser "B" 100 uf 0 volts. One diode in series with a inductor from conderser A to B. The inductor avoids the infinity current problem of just shorting the condensers together. The diode prevents the current from going into resonance from A to B then B to A.With all this in place 99 % of the energy from condenser A can be moved to condenser B.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 04:00:38 am
content: Quote from: AG6QR on January 13, 2018, 06:31:05 pmYou are correct. A "resistor" that does not get hot was a poor choice of words on my part. I will rephrase. A dynamically controlled reactive inductor as a replacement for a resistor to reduce heat loss.

author: kalel
author_rank: Frequent Contributor
timestamp: January 14, 2018, 04:01:26 am
content: Is not that in some way what e.g. a step down converter does (I could be wrong, still newbie)?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 04:06:00 am
content: Quote from: Marco on January 13, 2018, 02:29:33 amDo you have a link to a freeware simulator?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 04:17:18 am
content: Quote from: kalel on January 14, 2018, 04:01:26 amI think you are right on the money. The principle is the same with a twist of flipping the inductive polarity around instead of the current. I guess you could say a mirror image of a inductive voltage down converter.

author: timb
author_rank: Super Contributor
timestamp: January 14, 2018, 05:45:46 am
content: Quote from: John Heath on January 14, 2018, 04:06:00 amLTspice

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 06:37:42 am
content: Quote from: timb on January 14, 2018, 05:45:46 amthanks

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 07:35:15 am
content: Quote from: blueskull on January 12, 2018, 05:26:30 amYes yes very interesting blah blah blah.There is no problem with multi tasking a different subject within a thread especially if it is welcomed by the one that started the thread , myself. With this in mind I and I suspect many other members would like to know how you manage to build your own home brew mass spectrometer out of off the shelf parts? How could you miss with americium 241 ? How does this business of a ion breeze work. The questions are endless. If you happen to come back and read this please knock yourself out sparing no details on the home brew mass spectrometer.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 14, 2018, 08:21:25 am
content: FYI, LTSpice is free but not freeware. You might be looking for NGSpice instead..?LTSpice is supported in WINE, FWIW.Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 08:27:12 am
content: Quote from: T3sl4co1l on January 14, 2018, 08:21:25 amIs LTSpice crippleware , or stops working after a while? What is the difference between LTSpice and NGSpice?

author: Marco
author_rank: Super Contributor
timestamp: January 14, 2018, 12:58:57 pm
content: Quote from: John Heath on January 14, 2018, 04:06:00 amAnother free but not freeware simulator, which I use, is the free version of Simetrix (limited in circuit nodes). It has far better convergence. When you are just throwing unrealistic circuits at a simulator to get a feel for things that's helpful.I only know of one guaranteed convergence simulator though, NL5 (maybe Falstad too?). Which I see now also has free software to try it out. Never used it but might be interesting :https://en.softonic.com/download/idealcircuit/windows

author: timb
author_rank: Super Contributor
timestamp: January 14, 2018, 02:19:48 pm
content: TINA TI is also free (it’s a limited version of the full TINA simulator) and has an easier to use interface than LTspice (which is admittedly terrible). However, LTspice is the most powerful (and most accurate) free simulator you can get. The fact that Linear Tech uses it internally to develop their analog and power ICs should tell you everything you need to know. It’s well worth it to learn the UI.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 03:13:23 pm
content: Quote from: blueskull on January 14, 2018, 08:31:24 amWell LTSpice is now installed but it does not care for my user32.dll in windows vista. It seems there is always another problem. Maybe NGSpice or XSpice.I do not see a thread called "bump that thread "?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 03:48:18 pm
content: Quote from: timb on January 14, 2018, 02:19:48 pmIt this point I would be happy to have any simulator that will install and work. What happened to taking pride in writing code with people like Norton. Norton software would run on anything as he would write the code on the oldest computer he could find to make sure it was backwards compatible. And he had a night mare of mono , cga , ega and vga video cards to deal with.

author: timb
author_rank: Super Contributor
timestamp: January 14, 2018, 03:54:53 pm
content: Quote from: John Heath on January 14, 2018, 03:13:23 pmWindows Vista is pretty old and isn’t supported with the current Windows version (which is for 7 and above). You can grab the older version (LTspice IV) here:http://ltspice-iv.en.lo4d.com(It was replaced by LTspice 17 last year, so it’s still rather current.)If that doesn’t work, try TINA TI fromwww.ti.com, it’s easy to use.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 04:53:16 pm
content: Quote from: timb on January 14, 2018, 03:54:53 pmThank you for the helpNow that I look the one I installed is XVII or 17. You said TINA easier to use. That is polite way of saying LTSpice is not easy to use. I will give TINA a try.While I have your attention I was looking for a PIC series simulator. that can run off a hex file to save time. r as I realy do not need advanced simulator just quick and dirty for the meat ball st

author: AndyC_772
author_rank: Super Contributor
timestamp: January 14, 2018, 08:34:35 pm
content: LTSpiceiseasy to use. Slightly unconventional, admittedly, but that doesn't make it hard provided you can RTFM.IMHO, it's a very rare thing indeed when it comes to free software. It actually worth more than you paid for it.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 10:03:28 pm
content: Quote from: John Heath on January 14, 2018, 04:53:16 pmThis is not going well and I need to ventilate so that everyone else shares in my misery. The TINA solution plan "A" did not pan out as the download was not 1 not 2 but no less than 29 files of which I am supposed to know which files are needed. What happened to 1 file called TINA.exe that takes care of itself. Plan "B" the the older version 4 of LTSpice.exe not LTSpice17.exe has it has a problem with user32.dll. The download of LTSpice.exe ver 4 went well delivering the goods of 1 file LTSpice.exe clearly marked version 17 not version 4 . Why would a download button marked LTSpice ver 4 download a file marked LTSpice ver 17 that has the user32.dll problem? Plan "C" google the user32.dll problem for a possible fix. Found it. A rather long 3 paragraphs of unnecessary information ending with no fix yet for user32.dll problem . Why not " no fix " first then 3 long paragraphs of unnecessary information? Plan "D" return here to ventilate my misery. I could not even have that as the net when down before sending so all was lost. I had to wait 3 hours for the net to come back up then rewrite everything. Frustrating and still no simulator software. This is a rhetorical post.

author: timb
author_rank: Super Contributor
timestamp: January 14, 2018, 10:27:26 pm
content: Uh, TINA TI should be a single installer file, not 29. Where did you see 29 files at? Because I’m not seeing that here:http://www.ti.com/tool/TINA-TI(You will have to create a free account with TI before downloading.)Here’s a direct link to LTspice IV:http://ltspice.linear-tech.com/software/LTspiceIV.exe(FYI, there is no fix for the user32.dll problem with LTspice 17; it’s simply not designed to work on anything older than Windows 7. You really should think about upgrading to at least Windows 7 as Vista is increasingly becoming a security risk, due to it being EOL by Microsoft. Programs are also increasingly requiring a minimum of Windows 7 to function, so you’re going to see this more and more.)

author: John Heath
author_rank: Frequent Contributor
timestamp: January 14, 2018, 11:59:08 pm
content: Well that was easy. It was right there in front of my nose? The lesson here is do not press the button that says software and tools. That one takes you to a selection of 29 files not 1 friendly file Tina90.zip. Zip is unfortunate as my unzip is corrupted with another malware unzip that wants money before unzipping. It is clever malware that makes it impossible to uninstall with any of the anti malware software that I have used. My solution was to use Explorer as the default unzipper or manually with an old dos version of unzip. When I try the run Tina90.zip it tries to open explorer which trigger malware to think unauthorized redirect so it closes it. The end result is an oscillation of Tina90.zip opening explorer to unzip competing with auto malware shutting it down. Control alter delete for task manager does not work to shut it down as it keeps opening up again. The only way out was to pull the plug and reboot from scratch. I am not the only one with this problem so why would they use a zip format. Band width is not the problem it used to be so why use a zip file format. There is a light at the end of the tunnel. I am now registered at Tina with Email conformation and I have the file Tina90.zip. Just need to find a way to unzip it without the computer going crazy.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: January 15, 2018, 12:31:52 am
content: One of our local computer guys is quite famous and I'd suggest following his clean-up procedures here-http://securitytango.com/tango.phpI've never had a problem with LTSpice on any machine or OS. The learning curve is easy and the results 1st rate. But you need a clean machine to start with, for that or anything else.

author: kalel
author_rank: Frequent Contributor
timestamp: January 15, 2018, 01:24:08 am
content: You could try scanning the file with a service like this, and if it finds something, report it to the developers:https://www.virustotal.com/#/home/upload(it's not an ad for the service, it's just one of those multi-scanners that will detect more things than a single one)In any case, you could/should report your inconvenience to them, so that they can address the issue.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 15, 2018, 04:02:55 am
content: I have Tina9 installed and running. It was a matter of changing the window association for zip files from explorer to and old DOS zip.exe. Thanks to all for the help. I should leave something on the table. A scientific spectrum lab that is freeware. It is a camel not a race horse as people from around the world have been adding to it. A few screen shots to get a feel for it. Some unusual options such as quad antenna stereo input to detect the direction an EM wave is coming from , Real time internet streaming time stamped the universal time to share live data with others around the world. Auto logged audio wave files for later analysis also time stamped and much more. I use it for Van Allan belt monitoring , VLF monitoring lighting strikes from half way around the world and sun plasma bursts. Description and download athttp://www.qsl.net/dl4yhf/spectra1.html. The download is about 80 percent down the page.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 15, 2018, 05:23:57 am
content: Where are my ICs in the Tina9 software? I see a 555 timer and an op-amp. Where is the 4000 series and 74hc series of ICs. Do I have to down load these separately?

author: timb
author_rank: Super Contributor
timestamp: January 15, 2018, 01:17:08 pm
content: Quote from: John Heath on January 15, 2018, 05:23:57 amThese type of SPICE simulators are primarily for doing analog simulation, not digital. (Not to say you can’t; I know there are limited sets of 4000 and 74 series models for LTspice, for example, but these type of packages don’t excel pure at digital simulation, since they will be simulating the analog characteristics of each transistor and FET of the part.)You should also know that SPICE simulators like this aren’t intended to simulate an entire large circuit in a single go. You’re meant to break it up into smaller chunks. (If you need to use the output of one chunk as the input of another, you can save the output as a waveform (PWL file) and then use it as an input later.)Anyway, based on your OP we thought you needed a good analog simulator. If you want to do fast digital simulation we can recommend some software for that.o

author: John Heath
author_rank: Frequent Contributor
timestamp: January 15, 2018, 01:59:47 pm
content: I watched a few videos on Tina and LTSpice and I see what you mean. Analog for amplifiers , oscillators and such , cool. Easier to make mistake in Tina than the real world. Hard to burn resistors or over heat a transistor if they are only virtual. Good stuff. Should be able to simulate a polarity switching inductor with power FETs to see how it performs for DC currents.

author: IDEngineer
author_rank: Super Contributor
timestamp: January 15, 2018, 05:24:44 pm
content: Quote from: Conrad Hoffman on January 13, 2018, 03:54:33 pmThere's also the speed of light issue... even if all else is "perfect", the magnetic field in the inductor does not form and collapse instantaneously. Its inductive reactance is not constant with time which leads to very real effects that even theoretically perfect components cannot eliminate.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 15, 2018, 11:10:22 pm
content: Quote from: IDEngineer on January 15, 2018, 05:24:44 pmFascinating. I have been stewing this and I think there is a way to prove that Tina would screw it up just as you say. An example. I will construct a cable that has a series inductance .5 u Henry per meter with a shunt capacity of 9 p Farad per meter. If I ask Tina what the propagation speed of such a cable would be she would calculate 2 c ! Information must never travel faster that light so Tina is wrong. The natural properties of a vacuum are 1.2 u Henry by 9 p Farad per cubic meter which when calculated sets the speed of light at c. I suspect Tina is not going to stop and ask me how long my .5 u Henry inductor is to stay obedient to relativity. This being the case Tina could screw it up for real life simulation.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 15, 2018, 11:22:00 pm
content: Propagation delay is not a loss, indeed it works the other way: propagation is the general form, and capacitance and inductance manifest at low frequencies as an approximation.Consider that all inductors are just lengths of wire, coiled up somehow or another, and all capacitors are two conductors very nearby, wadded up somehow or another. These are just transmission lines -- not very good ones (typically dispersive, or interconnected in weird ways), but the essence is there.Inductor losses arise from well understood low frequency phenomena, so that we very rarely have to analyze problems as transmission lines, rather it's enough just to reflect on where these things are coming from.Other wave mechanics apply -- skin effect primarily (and the analogous effect in lossy permeable cores, whether the loss is hysteresis or eddy currents).Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 15, 2018, 11:26:21 pm
content: Quote from: John Heath on January 15, 2018, 11:10:22 pmIt is a poor craftsman who blames his tools.The model is only as good as the squishy meat-computer commanding it. If you told it to model an all-pole system, you should expect it to tell you nothing other than the properties of that system; you certainly should not expect an ideal transmission line, as that would be expecting the computer to do something completely different from what it was told!"Pray tell, Mr. Babbage; if one should enter incorrect numbers in the Engine, can it still produce the correct result?"Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: January 16, 2018, 12:07:45 am
content: Quote from: timb on January 15, 2018, 01:17:08 pmI was looking at some more videos on Tina. Apparently there are Tinas and there are TINAS . I took a picture off the videos so you can see the difference between Tina and TINA TINA gets optoelectronics , your choice of logic ICs , a bunch of MCUs , A/D D/A converters , the list go on and on. What did I get in the "Tina package" one stinking 555 timer ? What am I supposed to do with that blink a LED on and off? What am I chicken liver here? Why does TINA get all the good stuff while I am stuck with one stinking 555 timer? Pardon the humor as I need the practice. I am sure it is a Tina library add on. Just wondering where to find this Tina library add on for those MPUs and such?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 16, 2018, 12:26:46 am
content: Quote from: T3sl4co1l on January 15, 2018, 11:22:00 pmI think I have a handle on domain hysteresis and eddy currents but skin effect I do not understand. Why do electrons prefer scooting alone the skin of copper? If I hammer a copper round wire flat does it have less resistance as there is now more surface skin that the electrons like? If you can shed some light on this I would not mind hearing it.

author: timb
author_rank: Super Contributor
timestamp: January 16, 2018, 12:31:29 am
content: Quote from: John Heath on January 16, 2018, 12:07:45 amThe full version of TINA is a commercial package and costs several hundred dollars to start. However, TI offers models that will work in their cut down version of TINA. Also, most other standard Berkeley SPICE models will work, but these are full analog simulations of the digital parts, so it will be slow.To be honest, the digital simulation in TINA isn’t that great to begin with, so you’re not missing much. If you want basic digital simulation, check it iCircuit. It costs like $10, but it works on Windows, Mac, iOS and Android, is easy to use and includes a comprehensive library of digital building blocks (it also does basic analog simulation as well). There’s also the free, browser based Falstad Online Circuit Simulator that’s nice for sharing basic ideas with other people.By the way, did you see the working download link to LTspice IV I found for you? You really should give it a try. Watch some videos on it, play around.http://ltspice.linear-tech.com/software/LTspiceIV.exe

author: John Heath
author_rank: Frequent Contributor
timestamp: January 16, 2018, 12:51:17 am
content: Quote from: T3sl4co1l on January 15, 2018, 11:26:21 pmYes and no. I will tell Tina to drive a given transistor with a 10K resistor thinking the output on the collector output will be bigger but the same. Tina says no no no it will not be the same b/c , b/e capacitance , gain. Tina says you can not have the high frequencies therefore your output will look like this. Tina has taken on the responsibilities of thinking on her own using her data base of the specified transistor being used. The way I see it is in for a penny in for a pound. At high enough frequencies not only inductance of the circuit but the mass of the electron itself has to be added for total inductance. Will Tina compensate for this? In for a penny in for a pound so why not?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 16, 2018, 12:29:44 pm
content: Quote from: John Heath on January 16, 2018, 12:26:46 amAs you go up in frequency, the changing magnetic field around the wire affects the wire itself.Imagine a round wire (isolated in space, nothing near it) divided into concentric shells. The outermost shell encloses all the current carried by the interior, which therefore produces a magnetic field in the outermost shell (which is nearly the magnetic field at the surface of the wire -- just add the current of the outermost shell). This field is changing, so it induces a current in the conductor, which opposes the current flow. How much does it oppose? Depends on the resistance of the shell: circumference times resistivity of the material. As you go from inside to outside, the enclosed current increases, but so does the circumference; but area increases faster, so the current would actually be flipped, becoming negative on the outside -- as a result of assuming a constant (evenly distributed) current on the inside. This is a contradiction, so we know the current cannot be even!In fact, it's a feedback mechanism, where most of the current flows on the outside, and the current inside is determined by what current (and magnetic field) is left after passing through the outer shell, and so on.At very high frequencies, hardly any energy flows in the conductor itself, rather it's contained entirely within the space around the conductor. The conductor is a boundary condition, shaping the EM field, but not directly participating. This is true of ordinary TEM (stripline, coax, etc.) sorts of transmission lines, and even more explicit in waveguides (including the all-dielectric kind, and fiber optics), Goubau lines, and free waves.It's maybe more convenient to assume the energy is contained in the space around the conductor, and 'seeps' into the conductor. If instead of AC steady state, we think in terms of transient time, then: when a step change occurs, first the change propagates at the speed of light, in the space around the conductor. This imposes a current on the surface of the conductor, and an opposing magnetic field.Because the change is very rapid, the surface depth is very shallow (less than a micron, say). As time goes on, the rate of change outside the conductor remains zero (it's a step), but the rate of change within the conductor is nonzero: the magnetic field, and current flow, diffuse into the conductor.Still another way to think of it: the conductor has a very high index of refraction. This is nonsense, right? I mean it's a conductor... Well, refractive index is determined by permeability and permittivity; and permittivity is how 'conductive' a substance is to electric fields. Normally it gives capacitance, but if we merely expand this to the complex plane -- as is normal for AC steady state problems, anyway -- we can consider a complex capacitor, which gives real resistance. Thus, bulk resistivity is identical to complex permittivity. The magnitude of the index is large, but it's very lossy as well. Indeed, this tells us metals are good shields -- incident radiation is primarily reflected (because of the huge index mismatch), and what's left is quickly absorbed. If we think about how a wave propagates through such a material, its amplitude decays as it goes (because of the loss), while being phase shifted. If we look at the average effect, we see diffusion.Indeed, wave mechanics don't go away just because you've put down a conductor -- can you do standing waves in a conductor? Absolutely. For a moderately sized round wire (i.e., radius about 1-3 skin depths), the wave doesn't fully attenuate by the time it reaches the center; waves coming in from all sides means the center is a null. At certain frequencies, there are also zeroes away from center, so that current density is large at the surface, dropping to zero at some inner shell, then actually going backwards in the core (before again going to zero at the center).Or for a sheet, a particular thickness of foil/film can have a color. Hold a CD-ROM up to the light, and see that it looks not completely opaque, but slightly bluish. That's a thin (100nm??) film of aluminum metal you're seeing through.What does this mean for oddly shaped conductors? Transformers? Other structures? What if we want more current, or less (for that matter)?A ribbon shaped conductor, in isolation, is worse than a round wire. Consider where the magnetic field must go: for the round case, it is evenly distributed around the wire, and current density can't be any lower. For an oblong shape, the magnetic field will clip into the peaks, increasing current density there, and reducing efficiency.What if the conductor is not in isolation, but near other wires carrying currents? Well, all those currents act together, shaping the current distribution in each wire. A pair of wires carrying opposite currents (twin lead) has the magnetic field canceling out at far distances -- which means the back sides of the wires carry little current. The magnetic field is concentrated in the space between wires, and so the currents are concentrated on the facing sides of the wires.Likewise, two wires carrying equal currents (same direction) repel. (Kind of like skin effect all over again, just with the conductor broken up into pieces.)In a transformer, say you have a windup like this: 2 layers secondary, tape, 2 layers primary. Take a cross section, and you will see pairs of wires in the same direction (the two layers), and pairs opposing (the two layers separated by tape). The layers in the middle there have double trouble: they're being pinched by the currents outside of them, while also facing opposite currents. This double-pinches the current, so the inner layers have even worse distribution. This is proximity effect.You avoid proximity effect by interleaving opposing currents. Suppose the transformer were wound: (1 layer each) P, S, P, S. The middle layers are pinched the same way in two directions at once, resulting in more even current distribution.What else can we do? If we go back to the flat conductors, but use two carrying opposing currents, we can spread out the current by placing them broadside-facing. This works for transformers with single turn foil windings, PCB traces (particularly in multilayer boards when you need to move a lot of AC current around) and so on.So, is flat conductor better? Not alone, but when used correctly -- near opposing currents, flat to flat, it's much better.By the way, the space between wires -- that's carrying all that close-up magnetic field -- is the leakage inductance in a transformer. The interleaved case has less leakage, which is a big help for most switching converters.Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: January 16, 2018, 03:40:15 pm
content: Quote from: T3sl4co1l on January 16, 2018, 12:29:44 pmThank you for a quality post. I can see you are using different analogies hoping one of them will get the point across. I got some of it but not all.A few questions.A] 100 amps filtered DC going through a wire. No skin effect zip nada zero. Yes / no ?B] For AC current. Less inductance on the outside , skin , of the wire so more current on the skin . Yes / no?C] A and B are correct but it is not that simple. Yes / no ?If I have A and B right then it only leaves C "not that simple"If your answer to A is yes and B is yes is It is mcounter EMF

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 17, 2018, 12:22:03 am
content: Quote from: John Heath on January 16, 2018, 03:40:15 pmNone, current density is evenly distributed (according to resistivity and Ohm's law in bulk material).QuoteIs is?There's always EMF, but tracking the EMF around and within a wire doesn't seem as useful to me.Inductance isn't the motivating factor here -- it's not like, say, an equivalent circuit where there's a bunch of independent current paths wired in parallel, and current distributes according to their impedances. It's more like a ladder diagram, where there's two terminals (the external fields at the surface, where you can probe / connect to it), then a chain of alternating series and parallel components (namely, L and R), where the further down the chain you go, corresponds to depth beneath the surface. Only at DC do the resistors all connect in parallel (X_L = 0), while at higher frequencies, the L's cut off some of the resistance, causing R_AC to rise and concentrating current flow towards the surface.Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: January 17, 2018, 03:55:31 am
content: Quote from: timb on January 16, 2018, 12:31:29 amI did follow that link. The download for LTspice IV ends up being LTspice XVII ?? I have already installed LTspice XVII and it hangs up when I run it with error user32.dll. I have a picture of the error. Maybe you can make some sense of it.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 17, 2018, 04:54:27 am
content: Quote from: T3sl4co1l on January 17, 2018, 12:22:03 amYour words are similar to what Wikipedia says. Wikipedia is a reliable source for information so you are in good company ,, most of the time.https://en.wikipedia.org/wiki/Skin_effectIn the Wikipedia article they give an example of a coax cable and how the skin effect increases with frequency. My anntenna was up as it smacked of a theoretical model prediction not a measurement made. There is a difference between theory and a measurement made. There is a working example where an extreme dc current will cause negative charges moving within to converge to the center. This is a working example used in real life therefore a measurement made not a theoretical model. The working example is called a lithium lens. A 6 cm long round tube of lithium. They shoot anti protons through it. This in itself does not do much however when combined with 500,000 amps of current the negative charges , anti protons , converge to the center thus the name lithium lens to converge the negative anti proton negative charges to a focal point. I trust Wikipedia however when a measurement made is in conflict with Wiki ?. There is a fork in the road. Which path to takeI am on the fence on this one . I will provide a link to a video describing a lithium lens. I would be interested in your take on this. Is it in conflict with Wikipedia for the skin effect ?https://en.wikipedia.org/wiki/Skin_effect

author: John Heath
author_rank: Frequent Contributor
timestamp: January 17, 2018, 05:17:08 am
content: Quote from: blueskull on January 17, 2018, 04:41:17 amThat's what I like about eevblog. There is always someone that knows. It never fails. Win 7 Hmm. That could be done. I will copy LTspice XVII instal to a win 7 computer and try.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 17, 2018, 11:59:35 am
content: Quote from: John Heath on January 17, 2018, 04:54:27 amHave you tried to measure it?Have you formulated a problem which would be able to confirm or deny the frequency dependence of skin effect?A good example is measuring the series resistance component of a straight wire. Somewhat tricky to do as inductive reactance dominates (above ~10kHz depending on size) but a good exercise in nulling techniques and low impedance measurement.The reason the theory is so good is because it's been tested.QuoteYeah, so what?I highlighted the key word here.Note that they have to maintain that current for tens of microseconds, otherwise the beam will only be focused to (diameter - skin depth) and not the maximal extent. But they cannot maintain it indefinitely, as the lithium will fuse in a fraction of a second (which would be rather messy in the lab or in the beam line!).They have to use lithium because it is the first element that is conductive. Low atomic number --> small cross section to particles. It's not the most conductive element, so it takes quite a lot of voltage, and energy, to achieve that current. The rod is about 0.25mohm, so the drop is 127V and power 64MW; a 100us pulse requires 6kJ, probably stored in a somewhat larger electrolytic capacitor bank. The peak temperature rise will be around 100C, so it won't be destroyed by a short pulse, and can be cooled back to ambient temperature with an outer cooling jacket. (These figures are around room temperature; it may be cooled to LN2 or even LHe temperatures in the beamline, where conductivity will be even better.)Anyway, note that electrons are confined to the metal. The mean thermal energy is about 26meV (at room temperature), well within the conduction band (which, as this is a metal, we know the Fermi level is above the band gap or the band gap is nonexistent, and so there is a ready supply of freely mobile electrons in the material). The mean drift energy of those electrons, during the pulse, will be, probably fractional meV at best. That is, thermal energy dominates, and the electrons hardly budge when the pulse happens. Just offhand, I don't think this is anywhere near enough current to run into current saturation problems -- metals contain a truly massive number of conduction electrons. It would be an interesting calculation to see where that threshold is (left as an exercise to the student).Electrons are not free-as-in-vacuum, which requires 3eV or more. You would need a truly extreme electric field to achieve that in a bulk metal (or a little UV light shining on the surface).But whatisfree-as-in-vacuum? The antiprotons. Those are moving, certainly fast enough that they aren't dwelling in the lithium for long enough to annihilate significantly (there will be a fraction lost due to collision, though, and the debris from those collisions will contaminate the beam, and have to be separated -- perhaps by charge/mass separation and some apertures).This is not a device that you can simply plug in your cooled antiproton gas at one end, and have it bleed out the other side. No, that would completely annihilate all those hard-won antiprotons! They must be going very fast indeed, maybe not relativistic but many MeV certainly. This is well beyond ordinary electrical conduction and steady state thermalized matter, this is hugely concentrated, high energy, high temperature particle physics.On a related note, the LHC's beam stop (for protons, antiprotons and heavy nuclei, whatever they're shooting that day) is something like two meters of solid graphite. All two meters are needed. And the conductivity of graphite is well needed, too (the beam is swirled over the cylinder to better distribute the energy, rather than just ionizing a freaking hole in it). A beam of that energy has serious penetration depth!On that note, I wonder how much beam luminosity and energy can be handled by that lithium lens; at what point does the beam energy loss exceed electrical (conduction) loss? Hmm...Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: January 18, 2018, 09:12:28 am
content: Quote from: T3sl4co1l on January 17, 2018, 11:59:35 am

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 18, 2018, 01:10:19 pm
content: Big quoteTim

author: Marco
author_rank: Super Contributor
timestamp: January 18, 2018, 01:17:27 pm
content: I don't really understand the point of all this ... knowledge of EM won't allow you to break conservation of energy

author: John Heath
author_rank: Frequent Contributor
timestamp: January 18, 2018, 02:36:03 pm
content: Quote from: T3sl4co1l on January 18, 2018, 01:10:19 pmYes I see your point . I do not have the hang of EEVblog editor yet. I start talking after you say " The reason theory is good is because it has been tested."

author: John Heath
author_rank: Frequent Contributor
timestamp: January 18, 2018, 02:44:39 pm
content: Quote from: Marco on January 18, 2018, 01:17:27 pmI admire succinctness however in this case maybe too succinct. Can you put a finger on where you saw a violation to energy conservation laws?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 18, 2018, 08:29:42 pm
content: Quote from: John Heath on January 18, 2018, 09:12:28 amBut you don't need any of them to understand how to throw a softball.Only use the model level you need to, no more. Check the higher levels for relevance, keep them in mind for possible gotchas -- but don't use them if you don't have to.You literally cannot throw a softball in QM, not today without quantum computers at least. There are too many particles to keep track of, and too little value to be had from it. The collective motion of 10^24 particles is intractable. But averaging over those particles to approximate them as a continuous bulk solid, now you can start to do things. Acoustic waves, momentum transfer; aha, now we can hear the crack of the bat!If we never touch QM in the first place, that's just as well. Indeed, there's no way we can average over the materials of a ball and bat today, to determine the average properties of those materials; we're better off measuring it in the first place. Empiricism wins, here. (Well, something like an aluminum softball bat is probably solvable from first principles, today. Maybe a wooden bat too, but then we need to measure the microstructure -- the pores, crosslinking fibers, cellulose crystals and lignin macromolecules -- and things get exponentially harder, and less pure, more empirical. The ball itself, made from a variety of packed materials, will be similarly difficult.)Rest assured, even if you invoke QM to solve a classical problem the hard way, it can never give an answer greatly different from what you were already expecting, using  classical methods. The fancier model always reduces to the classical, under the constraints that apply to the classical model. (In the case of QM and Relativity vs. Newtonian mechanics, that's v << c, N(particles) >> (a lot) and T >> 0.)QuoteI'd be quite surprised if your Hall effect probe can resolve microamperes, let alone quantized charge!Riddle: if a resistor in series with a vacuum tube cathode shows a constant current flowing, yet the plate voltage is being toggled between two voltages (say, 1000V and 2000V), what's happening to the electron beam?(For argument's sake, let's say it's a CRT with a thin neck, so we can clamp a current meter around it, too.)QuoteYou're really jumping the shark here!What assumption is necessary for the classical approximation (bulk conductivity) to hold?N(particles) >> (a lot).And I suppose some things about thermal energy, thermalization time and such. In short, it's a quasi-static equilibrium, as far as the current is concerned.You don't start to violate that in metals until the high THz, where the conductivity takes on reactance due to electron motion, and eventually (even for single photons) that motion must exceed the plasma frequency or thermal energy in the material (and then you get photoelectric emission and weird quantum stuff like that).Thus, skin effect is a collective, bulk effect, which breaks down on very small scales and high energies.Note that it works on the margin, because high energies (i.e. ~optical photons) only interact with the first few layers of a metal anyway (more or less the Debye scattering distance, which is derived from the physics of electrons in a periodic crystal). The quantum version of skin effect is a parameter like this.Why doesn't the antiproton beam spread out and conduct through and around the solid, like the cold, obedient electrons do? Because there's nothing to stop it and move it around such a path. It feels its own field (self interaction), and it feels the field from the current flow, but at the velocity it's going, the cross section for particle interactions is pretty small. The relative momentum of any pair of particles (lithium nuclei and antiprotons, electrons and antiprotons) is way in the non-bonding range, so they certainly won't just snap in line and spread out. With a relativistic velocity (if they're that fast, I don't know), length contraction further reduces the cross section, giving low beam loss.The current from the beam is very small in relation to the total current, but it too has a magnetic field; that field is trapped within the lithium cylinder, just as you would expect. It will act as a tiny line-like current through the solid, not carried through the solid as a classical ohmic conductor (and so, there's no equivalent circuit between the cylinder's resistance, and external supply, and the beam), but nonetheless present, and its field will be shielded by the surrounding conductor according to the response you would expect. (Namely: presumably the antiprotons are in bunches, so that the skin depth will be very shallow, microns maybe. By the way, this has the convenient feature of shorting out most of the magnetic field of the beam, keeping it small, making the focusing that much more effective, I think?)This is an example of a problem where a mixed, semi-classical solution is possible (and necessary!). There's nothing wrong with using classical results in non-classical physics, as long as the assumptions remain correct -- QM happily uses the Coulomb formula to calculate energy potentials, for example. (Now, QED underlies EM & QM, but you don't need to invoke that except for very precise atom and particle interactions.)Tim

author: Marco
author_rank: Super Contributor
timestamp: January 18, 2018, 08:33:35 pm
content: Your original theory of the heatless resistor, if the supply is providing power it either needs to go into heat or some kind of storage medium. You lacked a storage medium, hence you're trying to break conservation of energy.I was wrong to suggest a simulator. It's better to just run it through in your head.- The current in the inductor will be a symmetrical triangle wave, possibly with some DC bias due to the way it started up.- The supply would see that wave, but with every other half of it negated (including the DC bias).- Draw it out in your head or on paper and determine the average current the supply sees.Hint, it's zero. The power supply is delivering no energy. Your circuit is simulating an infinite resistor, which indeed doesn't get hot ... but isn't very useful.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 19, 2018, 12:24:02 am
content: Quote from: Marco on January 18, 2018, 08:33:35 pmWe all have our own way of stewing on a problem. For myself it is usually using Newtonian analogies then pacing the floor for a few nights before any prototyping of ICs on a board. Most of the time the floor pacing has already sorted out what ICs , transistors will be used and how it will look.Back to your point. I have a light bulb 1 volt , 1 watt and a power supply 1 volt. When connected 1 amp will go through to light bulb for 1 watt. I want to dim the light to 1/2. I will switch the light on and off with a 50 percent duty cycle. Now my l watt light is only on for 50 percent of the time so the average wattage is only 1/2 watt not 1 watt. This is how a dimmer switch works with triacs to dim lights in a house.The idea presented here in this thread is to use an inductor instead of a triac to vary wattage by switching the polarity of the inductor. Theoretically it will work but is it practical ? Is there a market for it ?You are a 100 percent correct that there can not be a heatless resistor without violating energy conservation laws. However an inductor is not a resistor , more like a flywheel where energy can be shifted in time.

author: Marco
author_rank: Super Contributor
timestamp: January 19, 2018, 07:52:17 am
content: 60 Hz is pretty much DC in this case.Think of your circuit as a black box for a moment. It has a load below and a power supply up top, if it was passing a "DC current" to the load with a DC voltage drop, it would by necessity be absorbing power. V*I will not be denied. If it acts like a resistor it burns power or stores it, no way around.Which is why it actually has an infinite DC resistance, it doesn"t get hot because it passes no current. Simply chopping the input voltage through an inductor works of course, but then it's a simple buck.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 19, 2018, 09:22:22 am
content: Quote from: T3sl4co1l on January 18, 2018, 08:29:42 pm

author: John Heath
author_rank: Frequent Contributor
timestamp: January 19, 2018, 09:48:31 am
content: eebblog says I can modify a give post. That means I can go back in time and fix mistakes. So if anyone reads this post you are delusional as this post never happened in the new time line I will create by modifying and correcting mistakes made in the post in the past. Now I am getting the hang of eevblog editor.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 19, 2018, 12:07:05 pm
content: Quote from: Marco on January 18, 2018, 08:33:35 pmYou can get a lot of milage out of I*V=watt. However when it comes to AC reactance has to be considered. I have a video link that describes it better than I would.

author: Marco
author_rank: Super Contributor
timestamp: January 19, 2018, 01:50:48 pm
content: The AC reactance is neither here nor there, you want it to conduct a "DC current" ... it won't.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 19, 2018, 03:18:53 pm
content: Give that editing another go, it looks like you got the quote tags reversed??Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: January 20, 2018, 02:35:35 pm
content: Quote from: timb on January 16, 2018, 12:31:29 amI manage to build the cold resistor in Tina. It is not Tina Turner as it takes about 20 minutes to work up the transient simulation. Is there a way to speed it up a bit?

author: John Heath
author_rank: Frequent Contributor
timestamp: January 20, 2018, 08:56:30 pm
content: The cold resistor appears to work according to a Tina simulator. A 10 m Henry inductor switching it's polarity 10 time a second with 10 volts applied to it. Not sure what it's net average resistance would be as it is both charging and discharging the 10 volt source?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 20, 2018, 11:07:35 pm
content: Relays are a very strange component choice, for several reasons...Is that not shorted to ground on one side?Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: January 20, 2018, 11:32:13 pm
content: One side of the relay is not shorted to ground but there is a wire that could give that impression. I am still on a learning cure with Tina so it is not the best. The inductor can float relative relative the the relay coil driving stuff. Interesting that Tina would have nothing to do with it unless I put in the free wheeling diodes on the relay coils. She does not care for high voltages.I did ground the 10 volt power supply going to the 10 m Henry switching inductor. The only reason was to make measurements from a common ground. As you can see I am having a lot of fun with the Tina simulator. It is surprisingly sophisticated analyzing things down to the n second!I have a better picture of it to see that the side of the relay is not grounded despite it looking so.

author: Marco
author_rank: Super Contributor
timestamp: January 21, 2018, 10:30:46 am
content: If it was acting as a resistor VF4 wouldn't keep swapping polarity ...

author: John Heath
author_rank: Frequent Contributor
timestamp: January 21, 2018, 12:57:34 pm
content: Yes not a resistor in the real sense. A resistance to current that is 100 percent reactive therefore no heat.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: January 21, 2018, 01:21:38 pm
content: I think this is a tiny little bit like a current-mode analog to switched-capacitor circuits.And the voltages and currents are all over the place. Just because there is "a voltage" and "a current" at certain instances does not make it a resistor, and most definitly not a resitor that doesn't heat up.

author: Marco
author_rank: Super Contributor
timestamp: January 21, 2018, 01:49:51 pm
content: Quote from: John Heath on January 21, 2018, 12:57:34 pmIf it acted like a resistance VF4 would not keep swapping polarity either.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 21, 2018, 02:41:36 pm
content: Quote from: Marco on January 21, 2018, 01:49:51 pmI hear you. It is not resistance. I want to stop my car so I put the breaks on. The car stops but the break pads are hot. This would be resistance and the resistor , break pads , heat up. Another way to slow a car down is to put it in a lower gear and drop the clutch. This is reactance with the energy going into spinner the motor up not heat.Opps with all these car experiments I broke the breaks. I now have no breaks going down a really steep hill towards a speeding train. We will use reactance by gearing down to spin up the motor. This works until we are down to first gear but still moving at 5 MPH down hill. Now what ? And the big nasty train is getting closer! I will slam the car in reverse and spin the motor the opposite way around like a flywheel. We are now down to 2 MPH still going down hill and the speeding train is only 10 feet away. With the motor spinning backwardsI will slam it into first gear again to spin the motor the other way then reverse , first gear , reverse , first gear , reverse ... slowing us down to 1 MPH enough for the speeding train to pass before hitting it.First gear to reverse is VF4 flipping polarity with relays and the flywheel inertia of the spinning car motor is the inductor 10 m Henry. Not the best analogy but you can see why VF4 is flipping polarity. Battery V2 10 volts is the energy going into the car from going down hill.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: January 21, 2018, 02:51:17 pm
content: Quote from: John Heath on January 21, 2018, 02:41:36 pmThe problem with your analogy is that when you use the engine to break, you are still turning all that energy into heat. It's just being done in the engine now (or the clutch in the case of using flywheel inertia), and the advantage of engine breaking is that engines have cooling systems made to take a way a lot of heat, so you don't notice it. You can't just take energy and make it dissapear, which is what you are suggesting you would be doing.I think you just really don't understand enough the core concepts of what are going on here, so I'm not gonna bother anymore. Good luck.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 21, 2018, 03:45:58 pm
content: Do not go away mad. I did not mean to annoy you. I can explain this 10 different ways using 10 different analogies. If you hang in there I promise you will end up agreeing with the idea of real vs imaginary power.

author: Marco
author_rank: Super Contributor
timestamp: January 21, 2018, 04:25:26 pm
content: What it is is a poorly controller DC to AC converter.

author: John Heath
author_rank: Frequent Contributor
timestamp: January 21, 2018, 04:49:25 pm
content: Glad you came back. Yes agree a poorly design DC to AC converter. In reality relays would not be practical. The only reason for relay polarity flipping was a quick way for Tina software to simulator to test the concept with least amount of parts. The real deal would have to be power cmos FETs , 10 m ohms on , back to back with over voltage protection. Somewhat pointless to build as a simple triac power control for heatless power control is easier and has stood the test of time. I do not think there is a practical application for the flipping inductor approach for power control. However it is fun to to think of alternatives and free to build in a virtual Tina simulator to see how it performs.