Confused about Inductors - Page 1
Forum section: electronics / beginners

author: BlueText
author_rank: Contributor
timestamp: November 13, 2022, 02:52:29 pm
content: Sorry about this but I'm a bit of a noob when it comes to electronics. I'm not sure if solving this problem has any practical use but, more of a puzzle really.What I have is 2 Inductors, (copper coil around a ferrite material as normal), connected together so that when Inductor 1 is charging, no current enters Inductor 2. Then, when Inductor 1 no longer receives current and it discharges, it flows into Inductor 2, charging Inductor 2. I assume I would need an arrangement of diodes or transistors/mosfets to do this.The thing is, I can't really picture how current is flowing from one Inductor to the other. I'm assuming that current flows in one direction when an Inductor is charging and current reverses direction when discharging but then, when I look at how a flyback diode is used, it seems as if current continues to flow in the same direction and the voltage is reversed instead.I guess if I could see a circuit of the above example it would help me understand this a bit better.Thank you if you can help.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: November 13, 2022, 03:48:51 pm
content: You could start by posting a schematic.

author: BlueText
author_rank: Contributor
timestamp: November 13, 2022, 05:13:31 pm
content: Quote from: Doctorandus_P on November 13, 2022, 03:48:51 pmUnfortunately I don't know what the schematic would look like. It would be just 2 inductors in series but setup up so the first one charges while the second does not and then first one discharges into the second.

author: golden_labels
author_rank: Super Contributor
timestamp: November 13, 2022, 05:34:17 pm
content: Posting any picture would really help. You do have some mental image of that situation, but you did not convey it to us. And your description doesn’t invoke any obvious circuits.There are some issues with the description. The misunderstanding is probably due to the missing image, but in its absence nothing else can be said.If two elements (not only inductors!) are in series, there is no way current could be on one but not on the other. That’s by definition. The current can’t disappear. So no,in this layoutit’s not even possible. Of course you may be thinking of a different layout, but… well, you did not draw the picture.When an inductor charges, current through it increases, not decreases.(1)While an ideal inductor can’t be fully charged, a real one will; at this point the current will be at maximum possible for the given voltage, not zero.An inductor is an element that opposes changes to current. So, whatever happens, the absolutely last possible thing is current getting reversed.(1)Under constant voltage. Speaking in a more general terms, the ratio of current and voltage goes to infinity.

author: MrAl
author_rank: Super Contributor
timestamp: November 13, 2022, 06:11:50 pm
content: Quote from: BlueText on November 13, 2022, 02:52:29 pmThat's not exactly correct but you are on the right track.When an inductor charges the voltage across it causes a current to flow. When the inductor discharges the voltage may reverse or just go to a lower value, but the current does not reverse. In fact, the simplest view of an inductor makes it look like a current source that tries to maintain the current that was originally flowing through it. So if you pass 1 amp through an inductor and then remove the voltage source (with also some resistance across the inductor to prevent high spike voltages) the current will try to stay at 1 amp and in the same direction until the inductor starts to lose energy at which time the current will then start to decrease.The idea that it tries to keep the same current flowing all the time is the basis for its use in filter circuits.If you understand how a capacitor works in a filter circuit then you can imagine how an inductor works too because the inductor is said to be the 'dual' of the capacitor. What the capacitor does with voltage is what the inductor does with current.You probably know that when you charge a cap up and the voltage reaches some value such as 10 volts and you go to discharge the cap, the voltage stays nearly 10v for some time and gradually ramps down as time goes on. Well, the inductor does the same thing except it works on the current. If you charge an inductor with 10 amps and then go to discharge the inductor the CURRENT will start to decrease slowly and so it appears to be trying to keep the current the same (10 amps) for some time. Of course it wont stay at 10 amps but will drop down little by little over time. The amount it drops depends on the load you connect to it in the same way that a cap voltage drops as you connect a load.So the inductor works like a cap but instead of trying to keep the voltage constant it tries to keep the current constant.In most circuits used for filtering (DC that is) it wont be long before the cap or inductor is charged up again, but during the time it is discharging it helps to keep the output (current or voltage) constant. It's not perfectly constant, but it's much better than no filtering at all. This is how both the cap and inductor works in a switching power supply.For AC filtering it is a little different. There is also resonance to think about, but that's a different story.

author: IanB
author_rank: Super Contributor
timestamp: November 13, 2022, 06:31:00 pm
content: The best way to think about this is in conceptual terms is by thinking of energy rather than current.An inductor is an energy storage device, and when you "charge it up" it stores energy in its magnetic field. This is how DC-DC converters or flyback converters work. In these circuits an inductor is charged up by in input circuit to store energy, and then the inductor is discharged into the output circuit to release that energy, but at a different voltage from the input voltage, thus achieving voltage conversion.Now, conceptually, you could discharge the first inductor into a second inductor, and this would form something like a "tank circuit". This could be the basis of an oscillator, with energy sloshing back and forth between the two inductors. However, in practice it is much more convenient to to use a capacitor as the second energy storage device, and so actual tank circuits you will see consist of an inductor and capacitor in parallel.So, although in theory it might be possible to construct something like you imagine, in practice I don't think it is useful to do so.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: November 13, 2022, 06:46:42 pm
content: Two coils in series are the same as one larger coil

author: Kasper
author_rank: Frequent Contributor
timestamp: November 13, 2022, 07:15:12 pm
content: My power electronics prof said this many times and I still find it helpful:Capacitors resist change in voltage.Inductors resist change in current.When you stop supplying power to an inductor, the current through it won't instantly change direction. It will keep flowing for a bit while it ramps down and stops.When a circuit opens suddenly that current can cause problems. Sometimes flyback diodes are used to give it somewhere to go.

author: BlueText
author_rank: Contributor
timestamp: November 13, 2022, 07:21:59 pm
content: Wow, thanks for the answers, particularly comparing inductors with capacitors. I had no idea they were kind of opposites of each other, even though I've seen them both in a resonant circuit.What I actually wanted to do, (but wanted to keep the example simple), was to have 3 inductors in series, one charging, one discharging and one staying empty, with this arrangement going around in a circle, with only a small input current added to start this off. I assumed that some transistors could be used to control the flow of this but if the direction of the current doesn't change I'll have to think of some other method to do this.

author: golden_labels
author_rank: Super Contributor
timestamp: November 13, 2022, 08:41:15 pm
content: May I ask you once more to attach images to show, how do you imagine operation of that circuit? You do not need to make a perfect schematic inEDA, just anything to convey the conceptual model you have in your head.Also repeating myself: if inductors are in series, they all see the same current. They are all getting charged or discharged at the same time. This is not specific to inductors, but applies to any current-carrying path (component). Any components in series always see exactly the same current, any components in parallel always see exactly the same voltage. For components to observe different currents, they must be in some way in parallel paths. Simply parallel, or because there is a path that branches off, or anything else, but they can’t be in series.(1)They may be temporarily in series configuration (e.g. due to switching), but during that period they are observing the same current.(1)I may be on pedantic side here, but that’s because — without having any image — no other assumption can be safely made regarding your use of “in series”.

author: MrAl
author_rank: Super Contributor
timestamp: November 14, 2022, 05:04:04 am
content: Quote from: BlueText on November 13, 2022, 07:21:59 pmHi,Can i ask, what are you looking to find out or demonstrate? Perhaps we can draw a schematic that would help understand this.

author: james_s
author_rank: Super Contributor
timestamp: November 14, 2022, 08:13:50 am
content: Quote from: BlueText on November 13, 2022, 07:21:59 pmTogether they make up the class of 'reactive' components. You might want to pick up a book like The Art of Electronics, it will go over a lot of these basic concepts.

author: EPAIII
author_rank: Super Contributor
timestamp: November 14, 2022, 09:53:16 am
content: With any three components in a ring configuration and the needed switching circuitry to try to transfer energy (current) from one to the next and then to the following one, there will always be losses at each transfer (some of the energy always becomes heat which is lost to the ring). So, you can not just start it with an initial current and expect it to run forever.If you would just tell us what you are actually trying to accomplish with this, then perhaps we could be more helpful. And I do hope this is not some kind of perpetual motion machine.Just taking your idea a bit further, if two inductors were in series then any current would be flowing through both of them in an equal amount. And that means that equal amounts of energy would be stored in each of them. Now if they were disconnected and the second one were then connected to a third one, that second one would only have 1/2 of the original energy. The energy in the first one would quickly convert to heat and be lost. And the current in the second one would try to flow in the third one. But, since the inductance of the third one would oppose it, there would be an initial high resistance in the new, series circuit. This high resistance would reduce the amount of current overall and at the very best you would only have half the energy that was originally present divided between the second and third inductor. From the power equation (I^2 x R) the current would then drop to 1/4 the original value or less. Actually it would be less, never the full 1/4 value.The same thing will happen when the circuit tries to move the energy to the third/first combination: now you are down to LESS THAN 1/16 of the original current. And it will continue disappearing (being converted to heat which transfers to the surroundings) at each successive transfer. It will quickly die to an undetectable level.Something similar would happen if you tried to transfer energy from one capacitor to another in a ring. But with capacitors it would be the Voltage that quickly dies down to nothing.As one of my favorite SiFi authors said, "TINSTASFL" (There Is No Such Thing As A Free Lunch).So, do tell us what the ultimate purpose here is.Quote from: BlueText on November 13, 2022, 07:21:59 pm

author: wasedadoc
author_rank: Super Contributor
timestamp: November 14, 2022, 07:31:57 pm
content: Quote from: EPAIII on November 14, 2022, 09:53:16 amOnly equal amounts of energy if the two inductors are identical in some way(s).

author: MrAl
author_rank: Super Contributor
timestamp: November 15, 2022, 02:24:04 pm
content: Oh yes energy in an inductor is (1/2)*i^2*L so if 'i' is the same and L is different than each one in the series string could have a different stored energy. It might be interesting to look at the energy transfer as they both discharge.To further illustrate the inductor/capacitor duality, here are some defining expressions.Inductor:v=L*di/dtCapacitor:i=C*dv/dtIt's interesting that if you swap 'v' and 'i' and also 'C' and 'L' you get the expression for the other component.For energy:E=(1/2)*i^2*L E=(1/2)*v^2*C Again swap 'i' and 'v' and also 'C' and 'L' and you get the expression for the other component.In some cases you can even construct the dual of the circuit the component is in by swapping resistances with conductances and inductances with capacitances (or vice versa). That's how to construct the dual circuit but not every circuit has a dual.

author: BlueText
author_rank: Contributor
timestamp: November 15, 2022, 08:51:36 pm
content: Thanks all for the help so far.Quote from: EPAIII on November 14, 2022, 09:53:16 amLOL, don't worry, I expect to input some kind of current occasionally to keep the thing going. I came to the same realisations as you describe regarding the inductors and that is, (I think), although one inductor can energise another, all you will end up with is both inductors levelling out at 50% of their charge and both of them dissipating to zero.Given that, I've decided to go with a variant of an LC circuit with 2 inductors and 1 capacitor. I've come up with the following circuit diagram but I fully acknowledge it might be wrong.Both transistors are PNP. The only aim here is to create a resonant circuit where the ferrite in the inductors increases in magnetisation, decreases in magnetisation and then has a rest period. Would this do the job?

author: IanB
author_rank: Super Contributor
timestamp: November 15, 2022, 09:20:10 pm
content: If you simply put a capacitor in parallel with an inductor you get a resonant circuit automatically. No need for any other components.

author: BlueText
author_rank: Contributor
timestamp: November 15, 2022, 09:40:18 pm
content: Quote from: IanB on November 15, 2022, 09:20:10 pmTrue, but then you don't get a rest period on the inductors, which is what I'm looking for.

author: BillyO
author_rank: Super Contributor
timestamp: November 16, 2022, 02:13:16 am
content: Look up the Hartley oscillator.

author: MrAl
author_rank: Super Contributor
timestamp: November 16, 2022, 04:36:48 am
content: Quote from: BlueText on November 15, 2022, 08:51:36 pmHi,Why do you want to create such a circuit what is the purpose of it?Why would you want to have the inductors which have initial energy dump all their energy like that?We'd have to do a little circuit analysis to see if there is anything we can call resonance.Also, inductors always have some series resistance but theoretical inductors may not. Which type of inductor do you want to specify for this circuit, with some series resistance or with none.

author: james_s
author_rank: Super Contributor
timestamp: November 16, 2022, 07:06:00 am
content: Quote from: BlueText on November 15, 2022, 09:40:18 pmIf there's a rest period then it's not resonant. What exactly are you trying to accomplish?

author: EPAIII
author_rank: Super Contributor
timestamp: November 16, 2022, 09:03:40 am
content: Why do the inductors need a rest period? Inductors have run for over 100 years with no rest.And why are you avoiding telling us the ultimate purpose of this?Quote from: BlueText on November 15, 2022, 09:40:18 pm

author: MrAl
author_rank: Super Contributor
timestamp: November 16, 2022, 04:15:34 pm
content: Good question.Also, we could see resonance that damps out after some time. That would mean we could see a sine response for some time as it gradually dies down to zero.An example would be an ideal capacitor in parallel with an ideal inductor. That would theoretically resonate forever. Add some parallel resistance however and the oscillations die down after some time. The time it takes to die down depends on the value of the resistance. With a low value we will see the oscillations die down quickly, with a high resistance it could take years in theory.In most practical circuits it dies down in less than a second but before that we see a sinusoidal type response sometimes referred to as "ringing".

author: golden_labels
author_rank: Super Contributor
timestamp: November 16, 2022, 06:59:31 pm
content: In this circuit the transistors are not conducting. If a diode is biased forward, the BE junction is biased in reverse. So the current through CE is 0A. If the diode is biased in reverse, then trivially current is 0A. So in any case the only valid current in this circuit is no current.If we ignore diodes and transistors, we end up with two inductors and a capacitor in parallel. For the sake of a thought experiment, let’s assume there is current on one inductor, no voltage and no current on the other inductor and capacitor.To begin with, if we ignore the capacitor, there is little to think about. The situation is akin to the two-capacitor paradox, but with inductors. An infinitely large voltage must be assumed on the other inductor and, independent of that, we end up with a gibberish solution while trying to calculate how is energy distributed between inductors.If we ignore the other inductor, we end up with a pre-chargedtank circuit, that oscillates forever.If we add the other inductor, that remains an LC tank, just with different inductance (two inductors are in parallel).The difference is one of the inductors being charged initially. That creates atransientbefore the circuit settles for steady state. And this transient is where a simple, conceptual model ends and mathematical gore starts. Speaking honestly, I am not sure if I would be even able to derive the solution after so many years without putting a lot of effort into refreshing my memory. Even less likely I would be able to pass the answer to you in any form you could understand at this level. So for now I will stop, which I do with no regrets, as this is a purely hypothetical question after all.

author: BlueText
author_rank: Contributor
timestamp: November 16, 2022, 08:59:01 pm
content: Quote from: BillyO on November 16, 2022, 02:13:16 amAh thanks for that. I didn't know this existed!Regarding the circuit, I admit I may have it wrong, but here is what it is supposed to do. The capacitor would have a starting charge with it, say, discharging its current in a downward direction and charging the left inductor. Then the inductor discharges into the capacitor in the same direction so it ends up being polarised in the opposite direction it started in. Same as a classic LC circuit. Then, when the capacitor discharges again, as the current is flowing in the opposite direction it charges the right inductor instead, which discharges as normal and the circuit returns to its original state. The diodes and transistors were designed to only allow current to circulate in one half of the circuit at a time, dependant on it's direction.I am assuming, with the PNP transistors, that current circulating clockwise in the left hand part of the circuit would enter the top transistor and cut off the right hand part of the circuit from receiving any current?As for the purpose of the circuit, well, I am interested in learning electronics and I thought the best way to do it would be to set myself a challenge and having an inductor rest while another is doing something would be interesting and force my brain to work to come up with a solution. If the circuit works, it could be interesting to have different kinds of inductors on each side so it would have two speeds of resonance or something, maybe producing an asymmetric AC current. Not sure if there are any real world applications for that though.Regarding resistance, I would assume any actual resonant circuit built in the real world will have resistance and wouldn't run for ever, so would need an input current occasionally.

author: BlueText
author_rank: Contributor
timestamp: November 16, 2022, 09:04:11 pm
content: Quote from: golden_labels on November 16, 2022, 06:59:31 pmApologies, I may have used the wrong transistor symbol. I am assuming a current going into the left side of the top PNP transistor would block current from circulating in the right hand side of the circuit.

author: IanB
author_rank: Super Contributor
timestamp: November 16, 2022, 09:20:54 pm
content: For a transistor to turn on you have to apply a voltage to its base. If your circuit starts out with only a charge on the capacitor and both inductors "at rest", then where is the voltage to turn on either transistor? Both transistors will be off and no current can flow through them.

author: MrAl
author_rank: Super Contributor
timestamp: November 17, 2022, 05:29:04 am
content: Quote from: BlueText on November 16, 2022, 08:59:01 pmI think if you have initial energy in both inductors you can see some behavior like you are looking for which would be resonance for a little while then a drop off in energy. I dont think it serves a good purpose for learning about circuits though it's just too esoteric to be useful in the respect.What would be more useful would be a circuit that is more like a regular textbook circuit like with an inductor and cap and resistor, and then later a diode, then maybe later a transistor.But back to the first line above. What do you think you will get out of this if you prove that with initial energy in both inductors and a particular current direction for both you see some resonance followed by a complete damping out of all energy?I just feel you wont get anything out of it because it's just too unusual. Maybe in an advanced course in engineering you may, but even then you can probably do much better.I've seen other circuits like this that are so far out in left field that nobody agrees on exactly how they work or how they should work. It ends up being a never ending discussion on anything but electronics, but more on philosophy and how things with no parasitics work or how things with parasitics work or somewhere in between.There are also numerous programming examples where people try to write the most obscure lines of code to perform a simple function and then see who can figure it out. It ends up being a huge waste of time and is mostly just for the novelty of the thing.My advice would be to first study the more common circuits with LRC and maybe some other things and learn the three types of resonance and how they are related, and the three possible types of responses: underdamped, critically damped, and over damped. That will get you some really useful insight about LRC circuits and related that you will be able to use in many circuits that come at you in the future. The circuit at hand will only get you as far as that circuit, if that, and that's it, as you will probably never encounter a circuit exactly like this again.A this point i would have to ask if you studied the more common LCR circuits yet. One thing you dont do is jump into more complicated circuit before you fully understand the more common ones and how to analyze them fully.

author: BlueText
author_rank: Contributor
timestamp: November 17, 2022, 06:39:53 pm
content: Quote from: IanB on November 16, 2022, 09:20:54 pmThe transistors are the PNP type which allow current to flow through when the base signal is low and shut off the current when the base signal is high.Quote from: MrAl on November 17, 2022, 05:29:04 amYes, I would say I jumped in at the deep end here.I guess I'm more into digital electronics like logic gates and stuff where you only have to think about whether things are on or off. Still I feel I've learned a lot here.I still think the circuit would work as I advised. The top transistor cuts off current to the right hand side of the circuit when the left side has current and vice versa. A normal LC circuit with only 1 inductor will have AC resonating in it, where as this will have half the AC sine wave through the left inductor and half the AC sine wave on the right inductor and not at the same time. If that makes sense?

author: wasedadoc
author_rank: Super Contributor
timestamp: November 17, 2022, 08:35:19 pm
content: Quote from: BlueText on November 17, 2022, 06:39:53 pmYou have already been told that it will not. The circuit will not allow current to flow anywhere. Period.

author: IanB
author_rank: Super Contributor
timestamp: November 17, 2022, 09:22:30 pm
content: Quote from: BlueText on November 17, 2022, 06:39:53 pmAnd where will you obtain these magical transistors that behave this way?I suggest you get yourself a handful of NPN and PNP transistors, a breadboard, some potentiometers, resistors and LEDs, and try some experiments to see how transistors really work.The whole of science is about experiment. Things do not work the way you imagine they do, they work the way experiment shows they do.

author: BlueText
author_rank: Contributor
timestamp: November 17, 2022, 10:43:12 pm
content: Quote from: IanB on November 17, 2022, 09:22:30 pmMy apologies, clearly I thought I knew how a PNP transistor worked, when I didn't. Is there any kind of switch/transistor that closes when a current is sent to its base?

author: MrAl
author_rank: Super Contributor
timestamp: November 18, 2022, 11:41:04 am
content: Quote from: BlueText on November 17, 2022, 10:43:12 pmI am not sure why some members are claiming that there will be no current flow. If you set the initial current in the two inductors you get something like you wanted. It doesnt make sense why you would need to see this happen though.This is with each inductor being 1mH and the cap being 0.5uf, and the current in the left inductor being 1 amp flowing up, and the current in the right inductor flowing up and also at 1 amp initially.So doesnt make any sense that anyone would say it does not do anything at all.Also, an inductor with at least some initial current will force something to happen no matter what. That's because they force the voltage to go to a level which will force conduction even if it is for only a short time.In this snap shot the horizontal is set at 200us per division.Clearly there is some ringing and then it damps out.The circuit may make more sense to some people if you remove the cap then examine the circuit, then put the cap back in and think about what happens then. Maybe replace the cap with a resistor and examine that circuit too first.

author: wasedadoc
author_rank: Super Contributor
timestamp: November 18, 2022, 02:32:12 pm
content: Current in the direction of the arrow in the schematic can only flow through the inductor on the left if the bottom transistor is conducting. To be conducting that transistor needs some base current. That base current can only come from the right hand loop. For current to flow in the right hand loop the top transistor must be conducting. However, the current flowing in the left hand loop is because the anode of the top diode is more positive than its cathode. That means the base of the top transistor is more positive than its emitter. As it is a PNP transistor that means its base-emitter is reverse biased. No base current. No collector current. Which is the opposite of what has just been determined as its necessary state in the first sentence.

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2022, 03:20:29 pm
content: Quote from: MrAl on November 18, 2022, 11:41:04 amBecause the stated starting point is the capacitor charged and no current in the inductors. If you build the circuit on the bench and charge up the capacitor with an external power supply using flying leads, where will the current in the inductors come from?

author: BlueText
author_rank: Contributor
timestamp: November 18, 2022, 06:05:06 pm
content: Quote from: MrAl on November 18, 2022, 11:41:04 amOK, that's interesting, and weird. I'm not sure how the right inductor can have current flowing up when the bottom diode would prevent current flowing in that direction. Are you using some kind of circuit simulator? I wouldn't mind trying that as it might avoid me asking some embarrassing question in future.

author: golden_labels
author_rank: Super Contributor
timestamp: November 18, 2022, 06:06:17 pm
content: Quote from: MrAl on November 18, 2022, 11:41:04 amBecause there is no path for the charge to flow? Of course we may argue about some capacitance and leakage in real transistors and diodes, and that in a physical BJT the base-collector junction can actually conduct. Which should allow for some transient ringing. But for ideal components the situation is as in the picture I attached. If I am not wrong, of course — feel free to correct me.Can you draw a loop in this picture? I can’t.Note that I am already pretend I do not see the currents are opposing each other.

author: BlueText
author_rank: Contributor
timestamp: November 18, 2022, 07:29:02 pm
content: Sooo, I may actually understand how PNP transistors work now.Electrons build up and unless a small number of them are allowed to escape to ground through the base, no current can pass through the transistor. Given that, maybe the following will work?Or maybe I have just submitted more circuit gore.Not sure if the diodes are needed now though...

author: wasedadoc
author_rank: Super Contributor
timestamp: November 18, 2022, 10:21:37 pm
content: Still will not work. Assume current flowing in the direction of the arrow in the left inductor. It goes through the top diode. It cannot go through the base-collector of the bottom transistor. (The base-collector is a n-p junction.) Two other possible routes are into the emitter of the top transistor or into the capacitor and thence into the emitter of the bottom transistor. If it goes into the capacitor it charges the capacitor, its top plate becoming more positive than the bottom plate. That causes reverse bias of the base-emitter junction of the bottom transistor so it cannot conduct. But that capacitor voltage is forward bias of the base-emitter of the top transistor allowing it to conduct so let's imagine the current goes into its emitter, out of the collector and through the right side inductor. Then through the bottom diode. But there is no continuing path back to the left inductor because the bottom transistor is not conducting. It isn't possible to have both transistors conducing simultaneously- if one has its base-emitter forward biased the other is reverse biased.Taking the diodes out does not change the above,So again no current flowing.

author: golden_labels
author_rank: Super Contributor
timestamp: November 18, 2022, 11:44:49 pm
content: BlueText, please attach images to the posts. Do not drop them at random external servers. The forum has an attachment feature.The circuit is symmetric with respect to the capacitor. Not only components, but also conditions present. Which means current through that capacitor must be zero. If it wasn’t, one of the sides would need to observe different conditions. That contradicts the initial assumptions. It would also raise a question of which side it would be, which by itself is not a problem, until one asks why in a perfectly symmetric situation one option would be preferred over the other.If current through the capacitor is zero, it may be treated as open. The circuit becomes what is seen in the attachment. Let’s assume that current I1is present. That implies that there must be current at transistors’ bases (IB). There are two paths available. The more obvious is loop through base-emitter junctions of both transistors. But there is nothing to drive that current. The other is two loops: one through L1, D1, base-emitter junction of Q1 and then emitter-collector of Q2, and a symmetric one for L2. However, that would induce impossible voltages. Assuming(1)voltage across an inductor isxvand each P-N junction dropsv: D1 cathode is(x-1)·v, Q1 base is(x-2)·v, the rest is dissipated on Q2. But since the situation is perfectly symmetric, equivalent reasoning for current through L2 indicates that D2 cathode must be(x-1)·v. And that cathode is the same node as Q1 base, which leads to a contradiction. Of course voltages are relative and L2 is independent of L1, so we may neatly assume that voltage at D2 cathode is(x-2)·v, but then base of Q2 is at(x-3)·v, which is not the same as D1 cathode, leading to the same contradiction.Another take is not involving currents at all. FromKVLthe sum of voltages on the inner loop, formed from base-emitter junctions of Q1 and Q2, must be 0. However, if there is current through those junctions, both see some positive voltage. And a sum of two positive numbers is always positive, which contradicts KVL.(1)The argument itself doesn’t depend on P-N junction drops being equal. I usedvfor them merely for simplicity. You may conduct the same reasoning with any set of non-zero values.

author: BlueText
author_rank: Contributor
timestamp: November 19, 2022, 10:17:02 am
content: Thanks for the circuit analysis. It's very helpful.I've played about with designs on the lushprojects.com circuit simulator and it looks like I might not actually need any transistors or gates and that just having the diodes are good enough.Dual LC circuitI was worried that 1 inductor would discharge straight into the other inductor, by-passing the capacitor altogether, but it seems current much prefers to flow into the capacitor first, as long as it doesn't get too saturated. There is still some degree of inductor to inductor transfer of current though, which I guess is damping down the oscillations a bit too much. Maybe some JFET transistors might help in this regard.

author: BlueText
author_rank: Contributor
timestamp: November 19, 2022, 12:09:58 pm
content: Actually this is a better example of the circuit. The capacitor is precharged with 5v. It resonates, but dies down too quickly as sometimes one inductor discharges straight into the other.Dual LC Circuit 2

author: golden_labels
author_rank: Super Contributor
timestamp: November 19, 2022, 12:23:20 pm
content: To start with, be aware this is a simulator created byPaul Falstadand he maintains the up-to-date version of thiscircuit simulator webapp. It seems Lushproject hosts a copy endorsed by Paul, but Iain hisemlf notes he’s no longer able to maintain it.Remember that any simulation program is a glorified calculator. It calculates a mathematical model, not represents reality. The components you draw are visual aids to construct mathematical equations. But it’s up to you to ensure those equations make sense. If you create an invalid model, you get invalid results.In this case the model seems reasonable. Chiefly, because what you have created is a more complicated version of an LC tank. Let me redraw your circuit. See the attachment for the rearrangement. First two steps are not introducing any changes to the circuit: the diodes get shifted closer to inductors and then inductor-diode pairs are moved to one side. In the last step a change is introduced: D2 is swapped with L2. While that introduces topological change, the rest of the circuit will not see any difference. Can you now see how this is like a standard LC tank, with inductance on the left and capacitance on the right, but with each inductor handling current going into one direction only? The diodes, having non-zero forward voltage, will introduce some minor shifts in the behavior, but the general idea stays the same.

author: MrAl
author_rank: Super Contributor
timestamp: November 19, 2022, 03:34:44 pm
content: Quote from: wasedadoc on November 18, 2022, 02:32:12 pmWell actually we have not yet established what the assumptions are about the components or operating conditions.This is why i asked about the theoretical assumptions earlier in this conversation.I am attaching a drawing that illustrates SOME of the possible assumptions that are either left out, or alternately ignored. Depending on which you choose, the operation of the circuit will be entirely different. That's why listing the assumptions is so important in a circuit of this nature.Once we know the assumptions, we can come up with a definite conclusion. Without that, there will continue to be very large differences of opinion.

author: MrAl
author_rank: Super Contributor
timestamp: November 19, 2022, 03:38:13 pm
content: Quote from: IanB on November 18, 2022, 03:20:29 pmI think the more basic question was can the circuit show any kind of resonance, and to that end we try to find different ways that it may do that. In particular, the initial conditions.There still has not been any attempt to establish any assumptions so there can be no definite conclusions yet. In a circuit as unusual as this the assumptions are just as important as the components themselves because they are a large part of what makes up the components.I post an attachment to show some of the things left out.

author: MrAl
author_rank: Super Contributor
timestamp: November 19, 2022, 03:41:21 pm
content: Quote from: BlueText on November 18, 2022, 06:05:06 pmEverything will be weird until you list the assumptions about the components. You will continue to see wide differences of opinion among members until you specify every little tiny detail. That's because these details change the behavior dramatically.See attachment that shows some of the details left out, but not all. There is still reverse leakage in the diodes and BE diode also. There is also internal capacitances.A lot of people use the LT Spice simulator used to be called "Switcher Cad". It's a pretty good one.

author: BlueText
author_rank: Contributor
timestamp: November 20, 2022, 07:50:42 am
content: Well I would say I'm ready to finish this project. Thank you for all the help as I've learnt a lot from this!Although I couldn't get the circuit to perform with gates I did settle on this design as it pretty much does the job I was looking for:Parallel LC circuitBasically I dumped the right hand side inductor / diode combo and replaced it with a small dual direction inductor which seemed to perform better as there is far less current trying to bypass the capacitor from one inductor to the other and it allows the capacitor to reset itself to its original polarity. Then I added more smaller inductors in parallel so I could keep the frequency high but have enough enough power to cycle the larger inductor.I guess I'll have to build it now and see if it can work at all in the real world.

author: wasedadoc
author_rank: Super Contributor
timestamp: November 20, 2022, 11:14:40 am
content: Quote from: BlueText on November 20, 2022, 07:50:42 amIt appears that you have yet to learn that inductors in parallel behave as a single inductor.https://www.omnicalculator.com/physics/parallel-inductors

author: BlueText
author_rank: Contributor
timestamp: November 20, 2022, 12:22:44 pm
content: Quote from: wasedadoc on November 20, 2022, 11:14:40 amIf you replaced the parallel bi-direction inductors with a single bi-direction inductor of equal total strength, when the single bi-direction inductor is almost discharged you get one inductor feeding into the other, which is not good for keeping the osculations going. Making the bi-direction inductor a lot smaller reduces this. Then you just need to add more bi-direction inductors in parallel to increase the peak heights to something more desirable. At least that's what the simulation shows.

author: golden_labels
author_rank: Super Contributor
timestamp: November 20, 2022, 12:54:17 pm
content: I assume that “bi-directional inductors” is used to contrast with the inductor-diode pair. If not, then you got something wrong: inductors are not directional devices.A bunch of inductors connected between two nodes may always be replaced with a single inductor with no difference in behavior, no matter how they were originally arranged. It works the same as with capacitors or resitors.(1)If you swapped all those inductors with a single one and the behavior has changed in even a slightest manner, it means that you have calculated the value wrong.As it happens, Falstad’s simulator has examples regarding that:Inductances in series,Inductances in parallel.It doesn’t matter they push current into each other. All those currents sum to 0 and are not seen on the outside.(1)In fact that doesn’t even depend on element type. They are all impedances and may be simplified to exactly one impedance. Because no basic passive element provides both real and imaginary part of the impedance, using more than one element may be required. But that’s it. Absolutely any subcircuit consisting of passives only and spanning between two nodes may be simplified that way without any change in behavior.

author: wasedadoc
author_rank: Super Contributor
timestamp: November 20, 2022, 01:26:41 pm
content: Quote from: BlueText on November 20, 2022, 12:22:44 pmThat is nonsense. If the simulation is showing that then you are doing it wrongly.

author: BlueText
author_rank: Contributor
timestamp: November 20, 2022, 05:49:55 pm
content: Quote from: golden_labels on November 20, 2022, 12:54:17 pmYep, I'm calling the inductors not coupled with a diode bi-directional in this circuit.Quote from: wasedadoc on November 20, 2022, 01:26:41 pmNope. Swap out the 5 100mH inductors for 1 500mH inductor and you'll see the oscillations die out much faster.

author: IanB
author_rank: Super Contributor
timestamp: November 20, 2022, 05:58:29 pm
content: Quote from: BlueText on November 20, 2022, 05:49:55 pmYes, because five 100 mH inductors in parallel makes 20 mH, not 500 mH.

author: MrAl
author_rank: Super Contributor
timestamp: November 20, 2022, 07:00:42 pm
content: Hello again,This turned out a little interesting after all i just want to post a little note about the inductors.When you connect inductors in series the total inductance goes up. So if you connect 5 inductors that are 100uH each you get 500uH and of course if they are 100mH then the total is 500mH. The ESR goes up by 5 times though so if each inductor has ESR of 0.1 Ohm then the total ESR is 0.5 Ohms. The saturation current rating will stay roughly the same. The wire insulation of the wire used to wind the inductor may do better too because the voltage will be divided among the 5 inductors.When you connect inductors in parallel the total inductance goes down and the saturation current rating goes up. So if you connect 5 inductors each 100mH with a saturation current rating of 1 amp, then the total inductance is 20mH and the saturation current rating goes up to about 5 amps. I have to say "about" 5 amps because it depends in part how matched the inductors are as to series resistance and how well the cores are matched, etc. The ESR goes down by 5 times, so if the ESR of each inductor is 0.1 Ohms then the ESR then becomes roughly 0.02 Ohms, but when you get down that low there may be other things to think about like how well the connections are made and how much wire is used to connect them and is the same length wire used to connect them all together, etc.This means that connecting them in series might work well when you need higher inductance but can also put up with higher ESR.Connecting them in parallel might work when you need lower ESR and/or a higher saturation current rating, but can put up with the lower inductance.

author: BlueText
author_rank: Contributor
timestamp: November 21, 2022, 08:21:28 am
content: Interesting. All I was saying was that if you have 2 inductors in the arrangements I have been using and at least one of them is coupled with a diode, when one of the inductors has nearly discharged into the capacitor, which is nearly full, current starts to leak from that inductor to the other, which is not desirable as that is lost energy and the oscillations die off very quickly.

author: golden_labels
author_rank: Super Contributor
timestamp: November 21, 2022, 09:09:20 am
content: MrAl talks about inductors that have ESR. The model used in this thread and in the simulator uses ideal inductors, which have no ESR. If you have diodes, the oscillations die out because power is lost in the diodes. If in Falstad’s simulator, in the “Options” menu, you select “Show Power”, the simulator will display power losses in components. In red is power lost, in green is power dumped into the circuit.Example(code1)— an LC tank vs an LC tank with a pair of diodes, in opposite directions, between components. Note how the capacitor and the inductor oscillate between red and green, indicating that they both consume and dump energy into the circuit, while diodes are only consuming energy.(code1)Code:$ 29 0.000005 35.60246606707791 50 5 53 5e-11l 288 192 288 288 0 1 0.028443326579426772 0c 384 192 384 288 0 0.000009999999999999999 -4.3700863618075605 10w 288 288 384 288 0w 384 192 288 192 0w 448 288 544 288 0c 544 192 544 288 0 0.000009999999999999999 -3.784107606531265 10.6l 448 192 448 288 0 1 0.02834242727798885 0d 480 144 512 144 2 defaultd 512 176 480 176 2 defaultw 480 144 480 160 0w 480 160 480 176 0w 480 160 448 160 0w 448 160 448 192 0w 512 144 512 160 0w 512 160 512 176 0w 512 160 544 160 0w 544 160 544 192 0o 0 64 0 12289 0.0001 0.031622766653233596 0 2 0 3o 6 64 0 4097 0.0000675788527344716 0.0512 1 2 6 3

author: EPAIII
author_rank: Super Contributor
timestamp: November 21, 2022, 10:39:38 am
content: I suggest that the OP buy some inductors, capacitors, transistors, diodes, etc. and a breadboard and actually build some of the circuits he is suggesting. Oh, and have a oscilloscope on hand so you can see what is happening when the circuit is started.Then, try to make it work as you suggest is possible. It may just do so for some small fraction of a second. Emphasis on the word "small" in that sentence.Do post some scope photos.