Inductor principles - Page 1
Forum section: electronics / beginners

author: Trekgeek1
author_rank: Newbie
timestamp: July 18, 2012, 10:28:45 pm
content: Hello, everyone!New guy here and ready to annoy!Okay, I have experience in electronics, but I want to clear up some inductor questions once and for all.So from my schooling I know that you cannot change the current in an inductor instantaneously. This is because the energy stored in the magnetic field is a result of the flow of current. Changing the current instantly results in the stored energy changing instantly which requires infinite power. Therefore, it is not possible to instantly change the current flow in an inductor.So imagine we have an NPN BJT transistor with an inductor connected in series with the collector running with a Vcc of 10V. We turn the transistor on (let's just assume it is in active mode with some sufficient current flowing) and a current slowly begins to pass through the inductor. Slowly because the current cannot change instantly from zero to some non-zero value instantly. Now let's assume we do not have a flyback protection diode in parallel with the inductor. Next, we shut off the BJT (let's assume this takes near zero time). The current cannot stop instantaneously. To resist the shut off of the BJT, the inductor creates an EMF so as to preserve the current flow. So the inductor terminal connected to the collector becomes more positive than the terminal connected to the 10V Vcc supply. ( Everyone agree with this?)So now I see two possibilities:1.) The inductor EMF rises to such a magnitude that it exceeds the maximum voltage of the BJT and current flows through the BJT even though it is off, resulting in damage to the BJT.2.) The voltage rises, but doesn't exceed the forward voltage limits of the BJT, thus no current flows through the BJT. However, since the current MUST continue flowing, it essentially does by having the collector side of the inductor charge up like the plate of a capacitor. I imagine this would solve the problem since the energy in the inductor has been converted to energy stored in an electric field from the collector to the emitter of the BJT.So I guess my final question is:A.) Is it the case that an inductor ALWAYS SUCCEEDS in keeping current flowing (by breaking the BJT or arcing across an airgap in a mechanical switch) ?ORB.) The inductor ATTEMPTS to keep current flowing in the circuit but sometimes the best it can do is charge the terminals of the open switch like a capacitor?I personally have convinced myself that it is B, simply because the rule says the current through the inductor cannot change instantly, not the current through the circuit. Thus, if the circuit is configured in such a way that current cannot flow, the compromise is that the current will flow through the inductor but collect as a charge on some terminals of the circuit.Thanks to anyone with some input. I have electronics experience, but I constantly confuse myself with inductors.As a secondary thought:Am I correct in stating that the collector-side voltage of the inductor will become highly positive? I'm pretty sure that is the polarity that would attempt to keep conventional current flowing from collector to emitter.Alright, thanks to anyone and everyone who responds to this.Live long and prosper!

author: free_electron
author_rank: Super Contributor
timestamp: July 18, 2012, 11:08:36 pm
content: The only correct answer is A.at the moment you stop the current flow , the value of that current wants to keep going... ohms law applies. I is constant, R went infinite, so V goes infinite ( theoretical model )something has to give..You either fry the transistor (it may avalanche and survive a couple of times ), or the voltages become so large they jump from winding to winding inside the inductor and 'burn-off' that way.the capacitance in an open switch is very small. there is a lot of current in that inductor to be eable to hold in the capacitor :so the voltage across that cap goes up tremendously . the more electrons the higher the voltage ...

author: IanB
author_rank: Super Contributor
timestamp: July 18, 2012, 11:25:13 pm
content: (To the OP)The theory of an inductor does not say that the current cannot change instantly, it says that the current tries to resist instantaneous changes.If you establish current through an inductor and then isolate it by opening a switch then the current will necessarily fall to zero more or less instantly. In that case the energy of the collapsing magnetic field will be dissipated by other means such as by electromagnetic radiation or as heat.The "give" in the system that allows this to happen is the imperfect coupling between the magnetic field and the coil. An ideal perfect inductor does not exist.A high voltage may be produced across the inductor terminals at the same time and this may damage connected equipment.

author: Trekgeek1
author_rank: Newbie
timestamp: July 18, 2012, 11:42:18 pm
content: Quote from: IanB on July 18, 2012, 11:25:13 pmFor the most part I agree, but I do maintain that the current cannot change instantly because V=Ldi/dt which means an instantaneous change in current requires an infinite voltage. While it does not explicitly state it, I believe it is implied by the mathematics. I believe this statement is found in the majority of electronics textbooks as well. But as free_electron said, this is assuming a perfect model. In reality it seems that before the voltage reached infinite magnitude, it would arc somewhere, whether that was across a switch or across the room to a grounded object. Anything would be possible when approaching infinite voltage.

author: IanB
author_rank: Super Contributor
timestamp: July 18, 2012, 11:56:58 pm
content: Quote from: Trekgeek1 on July 18, 2012, 11:42:18 pmUnfortunately you have the cart before the horse. The inductor does not behave a certain way "because the equation says so"; rather the inductor behaves the way experiments show it to behave, and then the equation may or may not represent the experimental results with the accuracy you desire.If you do the experiments you will find the currentdoesgo to zero instantly, and the voltagedoes notbecome infinite.It follows that V = L dI/dt is an equation that describes an ideal perfect inductor, but that a typical real inductor is not ideal and does not perfectly follow that equation.(If you really got into it you would have a much more complicated and detailed set of equations to describe your inductor and these more realistic equations would more closely model the actual behavior of a real inductor.)

author: Trekgeek1
author_rank: Newbie
timestamp: July 19, 2012, 12:22:40 am
content: Absolutely fabulous point about equations being approximate models of behavior rather than dictating behavior.QuoteHowever, this would contradict the RL circuit response. If we have a single resistor attached to a voltage and supply a step input, the current will rise "instantly". If we have a resistor and an inductor in series, the current will rise slowly and the voltage across the resistor will start at the voltage supply and fall to zero. How would you explain the performance of an RL series circuit if the current does react instantly?

author: amspire
author_rank: Super Contributor
timestamp: July 19, 2012, 12:25:09 am
content: The inductor does succeed keeping the current flowing at the point the transistor turns off, and it flows into the collector capacitance, the self capacitance of the inductor, and any other circuit capacitance. The current continues until theALLthe energy in the inductor has been transferred to the capacitances minus the energy lost in resistances (the inductor resistance, the inductor core losses, the transistor collector resistance. etc) provided the voltage does not reach the transistor break down voltage.In real life, if you do not have a heavy duty negative voltage drive turning off the transistor (assuming a NPN transistor), the speed of the collector (or drain) voltage rise can turn the transistor back on via the collector to base capacitance a little causing the transistor to briefly conduct again which consumes a little more energy from the inductor. The base or gate drive may require a peak current capacity equal to the inductor current to stop the transistor/mosfet turning on again during the transistor/mosfet turn off.It all comes down to energy, and even with imperfect parts, the 1/2I2L energy in the inductor is real, and it absolutely has to go somewhere.If you replace the transistor with an ideal switch, the inductor voltage will not rise to an infinite voltage. It will rise to a voltage where the energy in the inductor's energy in the inductor's self-capacitance is equal to the energy that was in the inductor (less resistive/core losses). All wound inductors have to have a significant self capacitance as you have adjacent copper wires separated by a very thin insulating coating.If you have a 1mH inductor with 1A current and it has 100pF self capacitance, then if you turn off the current with an ideal switch, the peak voltage will beQuoteSome energy will be lost as EMF radiation, but in practice, that number will usually end up being bundled into the core loss figure from the datasheets. With a well designed core, the EMF losses are hopefully very small.In the case that the breakdown voltage of the transistor/mosfet is exceeded, it does not necessarily damage the transistor/mosfet. It is only damaged if the energy in the inductor is more then the device can cope with in avalanche breakdown mode.Richard.

author: Trekgeek1
author_rank: Newbie
timestamp: July 19, 2012, 01:09:57 am
content: Quote from: amspire on July 19, 2012, 12:25:09 amWhew! Well done, Sir.So let me summarize to see if I've got you right.1.) The current will NOT stop immediately.2.) With an ideal switch, the current will flow into the capacitance of the circuit, including the self capacitance of the inductor.3.) If the switch is not ideal, the rapid rise in inductor back EMF may arcs or cause avalanche in the transistor.4.) After the self capacitance is charged, it will behave as an LC oscillator, dissipating the energy in its internal resistance.Also, am I right with my polarity? The terminal of the inductor connected to the collector will rise highly positive with respect to the inductor terminal connected to the supply?I hope 1-4 are essentially correct, since this is what my intuition was telling me. I apologize if I've missed your point.

author: IanB
author_rank: Super Contributor
timestamp: July 19, 2012, 01:15:26 am
content: Ah, good point about the self capacitance.Any time you try to impose an instantaneous change on a system things will tend to vibrate or oscillate and this oscillation is how the system energy gets dissipated.There is an analogy here with the physical science of fluid flow. If you have liquid flowing through a long pipe the momentum of that liquid will try to resist instantaneous changes in flow. A long pipe with water flowing through it is a physical analog of an electrical inductor.So what happens if you force the water to stop flowing instantly, by suddenly closing a valve or tap? You get "water hammer". The force of the flowing water bends and expands the pipe causing it to jump and vibrate violently, possibly damaging something in the process. In this case the elastic expansion of the piping is the analog of electrical capacitance. The energy of the flowing water gets converted to oscillations in the piping until the energy dissipates as heat.Needless to say, water hammer is generally undesirable in piping systems and is best avoided.

author: amspire
author_rank: Super Contributor
timestamp: July 19, 2012, 01:27:50 am
content: Quote from: Trekgeek1 on July 19, 2012, 01:09:57 amCorrectQuoteCorrectQuoteOr as I described, more often then not, it actually turns the transistor back ON causing the transistor to dissipate the current in the inductor. The transistor has an internal capacitor from collector to base, so if you feed a big current into the collector capacitance, all that current goes into the base. If the base drive circuit cannot pull that current out fast enough to stop the base voltage rising to 0.7V, the transistor will turn on during the fast collector rise time period. If you are expecting to see 100V peak voltage and you only see 30V, there is a good chance the transistor is turning on again during the period the voltage on the collector rises quickly, and so the transistor is dissipating a large part of the inductor energy. This is also the reason that a switching transistor is often much hotter then you were expecting from your calculations. If you are not switching at a high rate, this may be a situation you are very happy to live with, as it eliminates the need for extra snubbing components.If you have an opamp driving a NPN transistor that is switching 1A, there is no way an opamp with a 10mA to 20mA output can sink enough base current from the transistor to stop it turning on during the period the collector voltage is rising at a very high rate.QuoteCorrectQuoteFor a NPN transistor, correct.QuoteI think your understanding is very good.As I said, the really key point is that regardless how imperfect components are, the energy in an inductor is real, and it always has to go somewhere. If an inductor is conducting 1A, it will continue to conduct 1A at the point the switch turns off. The energy will absolutely go somewhere, even if it is into its internal self capacitance.Richard

author: Trekgeek1
author_rank: Newbie
timestamp: July 19, 2012, 02:00:13 am
content: QuoteThank you. Hopefully my understanding is descent, I am an electrical engineer. My problem is my specialty really isn't inductors and I don't use them very often. Additionally, I think I have trouble with first order approximations in my mind. Inevitably I start second guessing myself. I'll know inductors one day, and then I won't the other. I always know the golden rule of current not changing instantly, but then I start giving myself too many "what if" questions that really stretch that knowledge. For example, when I ask "what if the switch is perfect". In that case I know that the current must keep going and that it cannot keep going!!!! I think I need to stop "what iffing" myself with ideal components. The only way I could seem to reconcile these two opposing ideas is that the current must "mass" or "collect" on the terminals of the inductor (or circuit in general). That way the energy was stored in an electric field and no current was necessary. Thank you so much for clearing that up.

author: amspire
author_rank: Super Contributor
timestamp: July 19, 2012, 04:24:29 am
content: I shouldn't have used the term "self-capacitance" for an inductors winding capacitance, as self-capacitance is actually a property of all objects and is different from winding capacitance.What we call capacitance in electronics is usually mutual-capacitance which is the capacitance between two objects. An isolated mass own its own also has a self capacitance that is a constant and does not depend on any second plate. The self capacitance of the earth is almost 1F and I think a human has about 100pF or something.The reason I am mentioning this is that even if you had an inductor that magically had zero winding capacitance, there is always self capacitance and so there will always be capacitance to limit slew rate and peak voltage from an inductor.Richard.

author: Kevin.D
author_rank: Frequent Contributor
timestamp: July 19, 2012, 07:00:31 am
content: Perhaps while the ampsire's and free electrons are answering questions like this I have a similar question to above but about capacitors. :)I never really understood it .If put a fixed charge in a capacitor to reach a certain voltage. If I then physically move the plates apart (Voltage between plates must rise since voltage is proportional to charge ,area of plates ,and distance between plates) .So how far can the plates be separated and how high will the p.d between them reach .? if I take one of the plates for a trip around the world and leave one behind will there still be the same charge between them when I bring them back together ?.If not wheres the energy gone ?

author: amspire
author_rank: Super Contributor
timestamp: July 19, 2012, 07:21:17 am
content: Say you have a capacitor in the form of two big parallel plates in space and you feed in 1A for 1 second into the capacitor. The negative plate has 1 coulomb of extra electrons, and the positive plate has a deficiency of 1 coulomb of electrons. If the capacitor was 1000uF, then it will have 1000v between the plates.Now pull the plates apart. As you do so, then number of excess and deficient electrons stays the same, but the as the capacitance between the plates drops, the voltage between the plates increases. It takes energy to pull the plates apart, so the energy stored in the capacitor, plus the energy required to pull the plates apart equals the energy in the original capacitor. Total energy is conserved.If you keep pulling the plates apart, the self capacitance of the plates limits the peak absolute voltage (yes - Voltage is an absolute measurement, not just a relative measurement).So if the self capacitance of each plate is 1nF, the voltage on the negative plate will peak at -1.0 GigaVolts assuming no leakage of electrons.The number of excess electrons on the negative plate is still 1 coulomb.Move the plates back together and you will end up with 1000V again across 1000uF.Richard.

author: amyk
author_rank: Super Contributor
timestamp: July 19, 2012, 07:27:24 am
content: One plate has a surplus of electrons and the other a deficit. Electrons from the environment will enter one and leave the other in order to neutralise the charge. There is no such thing as a perfect insulator.However, in the ideal case there will be the exact same charge once the plates return to their original separation. To move the plates apart you have performed work against the force created by the electric field between them, and when the plates return together they will perform work so the total energy is conserved. This is assuming they retain their charge and that the plates are large enough for edge effects to be negligible.You might also find this thread worth reading:https://www.eevblog.com/forum/general-chat/what-is-a-capacitance/

author: Kevin.D
author_rank: Frequent Contributor
timestamp: July 19, 2012, 09:48:50 am
content: Amspire ,guys thanks for your answers .so lets see if i got this straight .1 / ampspire ,In answer to my last question (taking the plate along distance and returning them together)the answer was yes ?. the same charge would be still there .2/ you say that the self capacitance of each plate is what will limit the voltageattainable by plate separation . This is the part I am not to clear on ,could you elucidate or give a link to a clear explanation .Just out of curiosity is there a device that ever used this principle to step up or down voltages ( A kind of Capacitive transformer you could say) ?. Charge a cap by some small voltage ,then somehow cause a slight plate separation ,Harvest your much higher voltage then repeat .I am curious about how to work out the distance above which further separation will no longer make a difference.

author: IanB
author_rank: Super Contributor
timestamp: July 19, 2012, 10:10:54 am
content: You could also do this thought experiment the other way round.Imagine a single metal plate sitting in free space, close to nothing.Now imagine you add a single electron to this plate. The plate will get a slight negative charge and therefore a slight negative voltage produced by this charge. If you keep adding electrons the negative charge will become greater and the magnitude of the negative voltage will increase. The magnitude of the voltage will depend on the size of the metal plate. If the plate is larger the voltage will be lower and if smaller the voltage will be higher (charge concentrated in smaller volume). This is the self-capacitance of the plate.Now run a parallel experiment with another very distant plate, only this time remove electrons from it. Consequently it will develop a positive voltage.To complete the experiment, imagine bringing the plates together. Now the negatively charged plate will be surrounded by a negative electric field, and the positively charged plate by a positive electric field. As the plates come together the two fields of opposite sign will tend to negate each other's influence where they overlap. In particular this effect will be strong in the space between the plates. As a result the potential difference between the plates will decrease the closer the plates get together. (I just tried to picture what this really means in terms of physics and realized I can't do it clearly. So my explanation is rather weak here.) At the limit the voltage difference would become zero if the plates touched and the charges equalized. So it kind of follows that the voltage difference would be nearly zero if the plates were nearly touching. The two charge collections are nearly intermingled and so they nearly cancel each other out.

author: amspire
author_rank: Super Contributor
timestamp: July 19, 2012, 01:49:13 pm
content: Quote from: kevotronic on July 19, 2012, 09:48:50 amYesQuoteThe mutual capacitance will drop to zero as the separation gets huge, but you cannot reduce the capacitance below the self capacitance. So as the distance increases, the voltage on a plate will rise towards the limit determined by the self capacitance.QuoteIt is theoretically possible, but when you do the numbers, it is probably not economical in any way. The cost for the mechanism to reliably move two plates between two very precise extremely small distances (probably in a vacuum) and the amount of energy required to move the plates will dwarf the very tiny amount of power you can transform.The numbers just are not good.QuoteThere is no simple answer. You will have to get the textbooks out and try an example.If you start with parallel plate capacitors, they will start with the capacitance being inversely proportional to the plate separation. Then as the distance between the plates becomes huge, it will act like two point charges. Not sure how that will fall off - inversely proportional to the 4th root of the separation? Not something I have needed to know in the last 30 years. The size of the capacitance between two point charges is so small it is not much commercial use.Finally when the separation is big enough, the limit is the self capacitance that is proportional to the square root of the surface area, I think from memory.If you are interested, it is not hard to google for the equations.Richard.