Capacitor discharge into inductor - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 22, 2017, 10:57:33 pm
content: I'm trying to find the peak inductor current when a charged capactor is discharged into the inductor.V = 300VC = 1uFL = 100uHBased on my searches on the net I have seen some folks do this by equation the capacitor energy and the inductor energy like so1/2 CV2= 1/2 LI2I would like to confirm if this is the right way to go?Because I fail to understand why doesn't the Inductor resistance figure in this equation? Will the inductor resistance not have any effect in determining the final peak current that will flow into the inductor?If the inductor dc resistance is 1ohm vs 2ohm will there be any difference in the max current that will flow into it?TIA

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: September 22, 2017, 11:24:52 pm
content: Quote from: ZeroResistance on September 22, 2017, 10:57:33 pmwell.... if you don't specify resistance, then that's the way to go. otoh, if you are concerned about inductor resistance, then you probably want to take into consideration capacitor resistance, capacitor inductance, inductor capacitance as well.take your pick, as to how far you want to idealize your circuit.if you just have resistance RL of the inductor L in mind, then max IL is of course V/RL

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 22, 2017, 11:42:55 pm
content: Many thanks for the reply.Yes ideally I should consider Cap ESR, ESL, wiring resistance, Inductor DC resistance etc... However I was concered only for the Inductor Resistance since it is quite significant in my case.You also said that the max current would be decided by V / RL. I don't quite understand this. Because when the inductor current is maximum the capacitor has discharged at this point of time so the voltage across the Cap is 0.So why then would the max current be decided by V / RL.Also would it mean the the max current should reach V / RL or does it mean that the max current would always be V/RL?

author: lordvader88
author_rank: Frequent Contributor
timestamp: September 23, 2017, 12:40:11 am
content: I'm not sure of the equations of the top of my head, or if u just want the ideal max current, I can;t remenber or look it up , too much alcohol and THC,V=L di/dt us a start and that capacior eq with time constanst 1/RC (Tau) and about 4-5x tau for the typical full charge/discharhe timebut with a sensitive enough ohm meter, u would get the DC resistance , and that would be used in a model of an inductor that includes its DC resistance.Depending on the frequency this inductor night operate at if u made this into an LC oscillator a model that includes the capacitance of the coil.Then its the easy to remember capacitive and inductive reactance equations

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 23, 2017, 08:46:59 am
content: Resonant impedance.When you put L and C together, you get an impedance and a frequency.The impedance is the characteristic or resonant impedance, Zo = sqrt(L/C).The frequency is the resonant frequency, Fo = 1 / (2*pi*sqrt(L*C)).If you have no losses, and you start with an initial (peak) voltage, then that voltage is transformed to the current Ipk = Vpk / Zo.How long does that transformation take? Exactly 1/4 wave!If there are losses, then the resonant frequency will be slightly lower, and the energy lost every 1/4 cycle means the peak will be lower, each time, and so on. In that case, using Zo (as if no losses) gives a worst case figure.Inductors and capacitors are really quite easy, once you know what ratios to look for.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 23, 2017, 08:47:35 am
content: Thanks Lordvader88, I appreciate your inputs and your time.Basically I want to calculate the following,1. Peak current in inductor.2. Rise time to peak.3. Power dissipation the coil for a single pulse.4. Temperature rise in the coil.I would like to consider the effects of inductor resistance also here.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 23, 2017, 08:52:24 am
content: Quote from: T3sl4co1l on September 23, 2017, 08:46:59 amBTW, I also have a diode in the circuit, will this still apply for a unidirectional pulse. Circuit is attached.

author: IanB
author_rank: Super Contributor
timestamp: September 23, 2017, 09:05:55 am
content: But if you can draw the circuit in spice, why don't you just simulate it and see what happens? It will be much easier than trying to calculate by hand.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 23, 2017, 09:12:59 am
content: Yes I tried simulating it, and also did some tests practically but I am getting variation in results, so just needed to brush up on the theory to validate the results I'm getting.

author: IanB
author_rank: Super Contributor
timestamp: September 23, 2017, 09:25:38 am
content: Quote from: ZeroResistance on September 23, 2017, 08:47:35 amYou can determine the maximum possible value by assuming ideal components and no resistance.The energy stored in the capacitor is:$$E={1\over 2}CV^2$$The maximum current will occur when the capacitor voltage is zero and all the energy has been transferred to the inductor:$$E={1\over 2}LI^2$$Equating the two and solving for \$I_\text{pk}\$:$${1\over 2}LI_\text{pk}^2={1\over 2}CV_\text{pk}^2$$$$I_\text{pk}=V_\text{pk}\sqrt{C\over L}$$This is the same as the answer provided by T3sl4co1l.QuoteT3sl4co1l answered that aboveQuoteYou can assume in the worst case that the entire energy initially stored in the capacitor will be dissipated in the inductor. This will be especially true if the inductor has significant resistance and if the coil does no work on the surroundings.QuoteIf you know the energy dissipated in the coil you can work out the temperature rise from the heat capacity of the coil (which you will have to figure out by weighing the coil and looking up the specific heat capacity of the materials it is made of).

author: IanB
author_rank: Super Contributor
timestamp: September 23, 2017, 09:31:59 am
content: Quote from: ZeroResistance on September 22, 2017, 10:57:33 pmIf the inductor has resistance then the peak current will be less. However, when designing things it is a safe approach to consider the largest possible value and design for that. That way the design will have a positive margin of error. The largest possible current occurs when the inductor has no resistance.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 23, 2017, 09:48:31 am
content: Thanks a lot IanB,that's a really comprehensive answer and I highly appreciate it.The coil might be doing work on the surroundings like generating a spark or magnetizing a magnetic material etc.In this case what determines the rise time of the current in the inductor. Is it only L and C? In non ideal conditions will the R of the inductor have a significant effect on the rise time.In the current case if I take a value of 1 ohm for the inductor what would be the implications?For power dissipation in the coilE of cap is 1/2 * 1uF * 300 * 300 = 0.045 Joules or 0.045 watt seconds or 45watt millisecondsnow if LTSpice says my total pulse duration is around 0.8mS.So would it be safe to say that the power dissipation over the entire pulse is 45 / 0.8 =56.25 Watts!!

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 23, 2017, 09:52:54 am
content: Quote from: IanB on September 23, 2017, 09:31:59 amThanks again IanB.I agree with you. I would have liked to know the effects of inductor resistance also from the point of heat dissipated in the coil, and also the % effect it has in peak current reduction.

author: IanB
author_rank: Super Contributor
timestamp: September 23, 2017, 10:04:10 am
content: Quote from: ZeroResistance on September 23, 2017, 09:48:31 amWhy the surprise? You can easily have high transient power levels in short pulses. But when considering heating of components this is less important then the average power levels during sustained operation.Quote from: ZeroResistance on September 23, 2017, 09:52:54 amThe resistance has no effect on the heat dissipated in the coil. All the energy of the capacitor gets dumped into the coil, no matter what the coil resistance.For the effect on peak current reduction, that's not something I know how to calculate. I would just simulate it. Since you know the theoretical answer with no resistance why not simulate that first and verify that spice gives the same answer as theory? Then introduce some series resistance in the inductor and simulate again to find out what happens.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 23, 2017, 10:13:27 am
content: Quote from: IanB on September 23, 2017, 10:04:10 amMay I humbly ask why the resistance has no effect in the heat disspated in the coil. I would have thought current flowing thru' a substantial resistance (1 ohm in this case) would dissipate some amount heat. In this case if the peak current is around 30A then it should have amount to a peak power dissipation of 30 * 30 * 1 =900W!! I don't know how to calculate this over the entire period of the pulse (800uS) though.QuoteAgreed!

author: IanB
author_rank: Super Contributor
timestamp: September 23, 2017, 10:22:01 am
content: Quote from: ZeroResistance on September 23, 2017, 10:13:27 amThe resistance will affect the instantaneous power levels and the time duration, but it can't affect the end result.The equation is a simple one:Energy stored in capacitor => Energy dissipated in inductorSince the value of the resistance does not appear in that equation, it cannot affect the outcome.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 23, 2017, 11:00:27 am
content: Can we say that :Energy stored in capacitor => Energy dissipated in inductor + Energy dissipated in internal resistanceAlso can we find out if we start of we 0.045 Joule of energy in the cap what would be actually transferred to the inductor and how much would be lost in the resistor?

author: IanB
author_rank: Super Contributor
timestamp: September 23, 2017, 11:24:28 am
content: Quote from: ZeroResistance on September 23, 2017, 11:00:27 amNo, we cannot. What we can say is this:Energy stored in capacitor => Work done by inductor on surroundings + Energy dissipated in resistancesThe energy dissipated in the system is by definition the energy converted to heat in resistances. The resistances either have to be the internal resistance of the inductor, or the resistance of the wires, switches and internal resistance of the capacitor.Now energy can leave the inductor by either means, if the inductor does work on the surroundings perhaps by moving mechanical objects, by inducing current in nearby objects, by radiating electromagnetic energy, or by magnetizing things.QuoteTo find out how much energy is transferred out of the inductor into other objects would require a detailed knowledge of the entire system beyond a simple electrical statement of the properties of the inductor. You need to know the size, shape, and geometrical arrangement of the inductor and its surroundings, or you need to make experimental measurements. So you cannot find out how much work will be done by the inductor with the information stated so far.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 23, 2017, 12:15:51 pm
content: Quote from: IanB on September 23, 2017, 11:24:28 amOk! got it!Now what about skin effect, does skin effect need to be taken into consideration for unidirectional pulses, or it only comes into play for ac currents?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 23, 2017, 07:39:00 pm
content: Skin effect is relevant for any change.If you prefer a transient (time) rather than Fourier (frequency) domain analysis, think of it this way: when a current is applied to a wire, at first the EM field around the wire carries the energy. Over time, the energy "soaks" into the conductor. At first, the conductor prohibits fields inside it, because it is conductive: a conductor shorts out the E field, drawing eddy currents on its surface, which produce a magnetic field which opposes the applied field.No conductor is perfect*, and the opposing magnetic field is not perfectly opposing. The outermost surface opposes just a small amount of the external field. Each layer down does its job as well, and eventually, the field is stopped. But each layer gives way over time, giving the effect that the applied field soaks in. Eventually, it soaks all the way through, and the full wire cross section ends up carrying a DC current.By "soak", I mean diffusion: much as smells diffuse through still air, electricity diffuses through conductive materials. Diffusion has a characteristic sqrt(t) or 1/sqrt(f) property. Hence, skin depth is proportional to 1/sqrt(f).*Except for superconductors, which exhibit skin effect (the outer layer of ~100 nm, depending on material) down to DC. Except for type 2 superconductors, which exhibit hysteresis loss (flux pinning). Though that's a topological change, not skin effect.This is all very low level and theoretical; what does it imply for a capacitor discharge? Well, if the coil has very little inductance, then the risetime will be fast, and only the outer layers of the coil will absorb the energy.FYI, merely shorting out a film capacitor can deliver a pulse of fractional megawatts!It is difficult for a beginner to understand just how sudden, how powerful, a spark is! One might be familiar with different kinds of shocks and surges seen in every day life: clapping hands, the ping of a hammer, the pop of a firecracker, or a gun. All of these events are faster than human experience can understand: they occur in milliseconds or less, while our senses can't resolve much better than 10 or 20 milliseconds. They are so fast, that they seem to happen instantly. They could happen in one microsecond or one millisecond, and we can't tell the difference!So it is for this reason, why it's hard to appreciate how destructive an electrical spark can be. An electrolytic capacitor can discharge in ten microseconds; a film capacitor in hundreds of nanoseconds; indeed, ESD -- the mere touch of a thin spark from your fingertip -- goes off in just a few nanoseconds, a million times faster than your senses!Now, that said: is there any hazard to the coil? No. Very doubtful anyway. Copper is very conductive, even on short time scales (microseconds). The skin depth might be on the order of 0.2mm (give or take). A larger wire will not get very hot, even at the surface, even for those microseconds. A smaller wire may get hot, but you need to compute the (energy / heat capacity) to tell how much.Can it ever be a problem? Absolutely! When fast capacitors are discharged into a small piece of wire, the wire will explode. The explosion happens in less than a microsecond, yet the speed of sound across the wire diameter might be tens of microseconds: the explosion expands faster than the speed of sound, in other words, it's created a shock wave! A shock wave is all it takes to detonate a high explosive, so this mechanism is very popular -- safe and effective -- for igniting them (exploding bridgewire).In summary: things in electronics can happen a whole lot faster than anything you might be familiar with; but don't worry, as the laws of physics stand firm. There is nothing that cannot be calculated here! And solid materials, like metals, are very robust, just mind their limits.Tim

author: IanB
author_rank: Super Contributor
timestamp: September 23, 2017, 08:01:42 pm
content: Quote from: T3sl4co1l on September 23, 2017, 07:39:00 pmThere is one significant exception to this. If a large enough capacitor bank is discharged into a coil, the coil may destroy itself due to the magnetic repulsion forces created inside the coil by the current. This is seen for example in coin crushers. The coil in such devices fragments into thousands of tiny pieces and must be surrounded by a safe enclosure to guard against the shrapnel hazard.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 23, 2017, 08:26:36 pm
content: Quote from: IanB on September 23, 2017, 08:01:42 pmTim

author: Cerebus
author_rank: Super Contributor
timestamp: September 23, 2017, 10:41:55 pm
content: Quote from: IanB on September 23, 2017, 09:05:55 amIf, as you try to sleep tonight, you hear the rattling of a soldering iron lead, that will be the ghost of Bob Pease coming to haunt you for suggesting trusting SPICE over thinking things out.

author: Dave
author_rank: Super Contributor
timestamp: September 23, 2017, 10:46:51 pm
content: Quote from: IanB on September 23, 2017, 08:01:42 pmActually, it's the opposite. The windings attract each other, because the current in adjacent windings is flowing in the same direction.

author: IanB
author_rank: Super Contributor
timestamp: September 23, 2017, 11:18:21 pm
content: Quote from: Cerebus on September 23, 2017, 10:41:55 pmWell yes, thinking things out cannot be emphasized enough. In this case I can write down two differential equations, one for the (ideal) capacitor and one for the inductor (with internal resistance):$$i = C {\text{d}v \over \text{d}t}\\v = L {\text{d}i \over \text{d}t} + iR_L$$Now to find the analytical solution to these equations will cause me to resort to a textbook, but since I know what the solution should look like (a decaying cosine wave when closing a switch on a charged capacitor as the initial condition), a numerical solution will be much more convenient.Years of engineering have taught me that I shall do the qualitative analysis in my head and the numerical analysis on a computer.

author: IanB
author_rank: Super Contributor
timestamp: September 23, 2017, 11:22:07 pm
content: Quote from: Dave on September 23, 2017, 10:46:51 pmBut what about all the parallel magnetic field lines running in the same direction down the center of the coil? Won't they cause an outward (radial) force on the windings, tending to increase the diameter of the coil?

author: Dave
author_rank: Super Contributor
timestamp: September 23, 2017, 11:30:55 pm
content: Quote from: IanB on September 23, 2017, 11:22:07 pmYou're right, completely forgot about that.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 23, 2017, 11:47:11 pm
content: Quote from: Dave on September 23, 2017, 11:30:55 pmYou're both right: the coil compresses axially and expands radially.Tim