Investigations into the stored energy of inductors. - Page 1
Forum section: electronics / projects

author: makerman
author_rank: Regular Contributor
timestamp: January 22, 2019, 07:45:23 pm
content: Hello, i'll be detailing my main project in this thread.I'm investigating the stored energy in an inductor, the standard equation says it's : W=1/2LI^2, implying that the two ways of increasing the stored energy are by increasing either the inductance or the current, i'm looking into what happens when we have two inductors, both with the same resistance but one with significantly higher inductance, will the latter store more energy given the same power input?I've done my first coil-pair test :Coil 1.Wire diameter : 0.53mmResistance per metre : 0.0775 OhmsInductor resistance : 24.5 OhmsSelf-inductance : 0.77 HenriesCoil 2.Wire diameter : 0.71mmResistance per metre : 0.0432 OhmsInductor resistance : 24.5 OhmsSelf-inductance : 0.232 HenriesI'm using a pulse width modulator to pulse the coil, and a pair of diodes on the coil output so that only the CEMF spike powers the load, the load being a Neon bulb for visual comparison of the coils' output. This isn't ideal since it's visual judgement so i'm formulating a better way to compare output now that i have a decent scope.In the test (pulsing at 300Hz with a 50% duty cycle) the first coil lit the Neon and the power usage of the PWM was 1.45 Watts.The first coil was then swapped out for the second, which lit the Neon visibly brighter and used 0.93 Watts.I'm not sure why the higher-inductance coil uses less power, i'm thinking it may be due to the larger time-constant ?The next step is to do a comparison of two coils with a much larger difference in inductance, i've got the large-inductance coil (over 3.5kM of 0.85 wire!) which has an inductance of 4.5 Henries and a resistance of 102 Ohms and i've ordered a kilo of 0.4mm wire which i'll cut to a resistance of 100 Ohms then perform the same experiment.Picture of the large inductor attahced, it's a beauty

author: duak
author_rank: Super Contributor
timestamp: January 22, 2019, 08:13:35 pm
content: The time constant of an R-L circuit is L/R. (This is in contrast to an R-C circuit where the time constant is R*C.)Are you familiar with exponential functions and how they apply to R-L and R-C circuits?If you go through all the math, given the same applied voltage and the same series resistance, the larger inductor will charge to a lower current for the same length of time. Since power is V * I, a lower current implies lower power.Cheers,

author: makerman
author_rank: Regular Contributor
timestamp: January 22, 2019, 10:09:45 pm
content: Thanks duak, exponential functions yes, how they apply to circuits no but i shall read up on it.I'm puzzled as to why the bulb shone more brightly with less power in?

author: IanB
author_rank: Super Contributor
timestamp: January 22, 2019, 10:22:35 pm
content: Quote from: makerman on January 22, 2019, 10:09:45 pm(Power Out) = (Power In) − (Power Losses)How might you account for and measure the power losses in each experiment?

author: makerman
author_rank: Regular Contributor
timestamp: January 22, 2019, 11:07:23 pm
content: Quote from: IanB on January 22, 2019, 10:22:35 pmThanks IanB.Well these are air-cored so no core losses, therefore only losses in the windings which is dependent on the resistance and both coils have the same resistance, hence my confusion as to it drawing less current but seemingly delivering more power to the bulb. Neons need a realtively high startup voltage but a smaller maintaining voltage and the current is on the order of a few mA or less.As for measurement, i just got my first decent scope a few days ago and will learn how to measure power in/out this week.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 22, 2019, 11:33:13 pm
content: PWM is not constant current. This doesn't seem to be a control method consistent with your premise!It seems you need a peak current mode controller, like UC3843 or its equivalent functionality which is easy enough to build from the block diagram. Got a few logic chips, comparators and the like?A more consistent load would be a good idea, too. Neon bulbs are notoriously nonlinear. How about a diode into a capacitor, and measuring the power delivered into that (say with a load resistor) -- a boost converter?Tim

author: makerman
author_rank: Regular Contributor
timestamp: January 23, 2019, 11:24:25 am
content: Quote from: T3sl4co1l on January 22, 2019, 11:33:13 pmYes but the current needs to be pulsed for CEMF to appear, thanks for the load suggestion.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 23, 2019, 12:20:20 pm
content: I should say: you won't get repeatable (constant) current while varying those conditions.Tim

author: spec
author_rank: Frequent Contributor
timestamp: January 25, 2019, 12:59:57 am
content: Hi makermanA real inductor first order equivalent circuit is a perfect inductor in series with a resistor, predominately the resistance of the copper wire winding.The energy stored in an inductor is, as you say (1/2)LI2. Notice that there is no resistance in that equation.The resistance only comes into play when you are inputting or outputting current into/out of the inductor, which is achieved by applying a voltage across the inductor.Another important point is that the inductance and resistance of an inductor change with frequency, although there are special inductors, particularly for switch mode power supplies and RF applications. Apart from having special cores, they also have special windings (Litz wire).

author: makerman
author_rank: Regular Contributor
timestamp: January 26, 2019, 05:02:12 pm
content: Thanks for advice all, i've simplified things by ditching the PWM and using a mechanical switch in the circuit, this way we have constant current to the coils.Coil 1 :R=102 OL=0.5 HCoil 2 :R=102 OL=4.6 HThe PSU is outputting 10VDC.When the switch is closed for coil 1 we use 93 mA, when the switch is closed for coil 2 we use 91 mA.When the switch is opened the CEMF from coil 2 lights up the Neon far in excess of coil 1, it's very visibly brighter.I'm getting better results with a switch than i did with a transistor in terms of CEMF, i'm not sure if mechanical switching is sharper than solid-state ?I can't check it with the scope since my x100 probe hasn't arrived yet but it's glaringly (pun intended!) obvious to the eye.I find it puzzling, i'm an electronics novice (11 years on and off as a hobbyist) but this implies that the higher the inductance, the larger the CEMF given the same power input, where is the extra potential coming from ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 26, 2019, 08:40:37 pm
content: Switches are very fast yes, in fact typically sub-nanosecond. In that incredibly short span of time, the contacts have barely opened up, so they break over very easily, and most of the turn-off time is spent sparking.With a neon lamp in parallel, the spark goes from air in the switch, to gas in the tube. The switchover happens when the voltage drop across the switch rises to, eh, probably 80-100V or so, the breakdown voltage of the neon lamp. (The voltage drop of a spark varies with its length, more or less, so it's rising while the distance between switch contacts is opening up.)The rest of the time, the neon lamp carries the current from the inductor, which decays according to (approximately) V = L * dI/dt, where V is set by the neon's voltage drop (about 60V usually).The approximation holds while dt << R/L.Since you've set initial currents equal, the neon light will be lit equally brightly (in the instant when it turns on).The EMF is the same, because the neon lamp is holding it constant.The current flow, and therefore neon bulb glow intensity, falls (approximately) linearly over time, until there's not enough current to ionize the neon bulb, and it turns off.Finally, when the bulb turns off, the voltage begins to settle out, and it may swing around a bit. If the voltage ticks back up, the neon may re-ignite, in a hammering sort of fashion, using up the last wisp of stored energy. (When this happens across a switch's air gap, it typically happens in more violent fashion, due to the higher voltage drop of the air gap, and the lower impedance of mains wiring -- the resulting rapid-fire ESD-like waveform can fry sensitive electronics!)What varies is, one simply stays on for longer than the other (about 9 times longer, given the values).To the unaided eye, all of this happens in an instant (namely, about 7 milliseconds; the eye's response time is on the order of 50ms), so you cannot perceive the duration of the flash, but rather one just looks as many times brighter than the other.To be able to perceive it, you could use a high speed camera, but that might be a bit difficult (to get a useful number of frames of the glow and decay, you'll need thousands of FPS); the next best thing would be a photodiode pointed at the neon light, so you can read out the intensity (as current flow through the photodiode) on an oscilloscope.HTH,Tim

author: makerman
author_rank: Regular Contributor
timestamp: January 26, 2019, 11:46:32 pm
content: Thanks very much Tim.

author: makerman
author_rank: Regular Contributor
timestamp: January 27, 2019, 07:26:12 pm
content: I managed to blow my multimeter the other day so ordered my first fluke (101), powered it on and when not measurnig anything it's showing anywhere between 6-20 mA then slowly drops down. The uni-t meter and other cheapies i have don't do this, should i be worried ?

author: MrAl
author_rank: Super Contributor
timestamp: January 27, 2019, 09:17:46 pm
content: Quote from: makerman on January 26, 2019, 05:02:12 pmHello there,All you really need to be concerned with is the (1/2)*L*i^2 part. However, that only comes into play AFTER the current has STABILIZED.I use caps for emphasis on the important points.Because the current is constant AFTER and the current is the same in each, we can change that to a constant:i=Iand squaring that we have a constant:I^2=Know we have:W=(1/2)*K*Land lumping both constants into K1 we have:W=K*Land it is clear that the bigger the L the higher the stored energy. End of story, well almost.What is not shown in W=K*L is the way the current acted during the first instants of operation.As the inductor charges, the current creeps up almost like a ramp. The size of the inductor tells us how long it takes for the inductor to charge up, and a bigger inductor takes LONGER to charge than a smaller inductor, hence it takes in more energy before the current stabilizes. Once it stabilizes, we have that W=K*L, but if you want to see why one has more energy then you MUST look at the current flow during the time it is charging. After it charges, it is too late to detect anything.The equation during the charge time for an ideal inductor would be:v=L*di/dtand solving for di/dt we get:di/dt=v/Land with constant v and constant ideal L we see this is a ramp with slope v/L.Now if we have small inductor L1=1 and large L2=5, and voltage source of 1 volt we have:di/dt=v/1=1/1and we have:di/dt=v/5=1/5and it is simple to see that when L got larger the SLOPE became much less steep, so it takes LONGER for the inductor to charge.As it is charging it takes energy and thus we end up with more once the current tapers off.Note we can not see the taper here but it is really like this:i=1/R-e^(-t*R/L)/Rand once that stabilizes the max energy is stored in the inductor.

author: makerman
author_rank: Regular Contributor
timestamp: January 28, 2019, 11:56:27 am
content: Thanks very much Mr AI.

author: intabits
author_rank: Frequent Contributor
timestamp: January 28, 2019, 09:00:40 pm
content: I'm confused.Quote from: makerman on January 22, 2019, 07:45:23 pmSame resistance in both coils, but L2 uses thicker, lower resistance wire.Doesn't that mean that it must have more turns?And if so, doesn't that mean that it must have a higher inductance?Assuming other dimensions are similar. (You don't specify those, and the pic shows more than 2 coils)

author: makerman
author_rank: Regular Contributor
timestamp: January 29, 2019, 11:54:21 pm
content: Thanks Intabits, i've posted the inductance values the wrong way around.

author: makerman
author_rank: Regular Contributor
timestamp: January 30, 2019, 12:05:18 am
content: If i connect the large inductor to the DC PSU, the PSU reads 18v , 0.17A so 3.06W, but if i do the calculation for the energy transfer after five time-constants :R = 102L = 4.65TC (L/R) = 0.045588235TCx5 (to reach 99.5% of charge) = 0.22549Energy = 0.5 x 4.65 x 0.17^2 = 0.067193Power = (energy/TCx5) = 0.29798439It indicates only 0.29 watts, what am i doing wrong ?

author: IanB
author_rank: Super Contributor
timestamp: January 30, 2019, 01:23:40 am
content: Quote from: makerman on January 30, 2019, 12:05:18 amEverything?Your power supply is indicating resistive power dissipation.Voltage = 18 VResistance = 102 ΩCurrent = 18/102 = 0.176 AContinuous dissipated power = 18 x 0.176 = 3.17 WSo far this has nothing to do with inductance. It is only about resistance.Then you try to do a calculation with inductance. You suggest the time constant is L/R = 4.65 H / 102 Ω = 0.046 s ? But I don't think that calculation works. Does henrys divided by ohms give seconds? (Yes, it does. However, you are looking at a transient accumulation of energy in a magnetic field compared to a continuous dissipation of heat in a resistor. You are not comparing like with like.)

author: makerman
author_rank: Regular Contributor
timestamp: January 30, 2019, 11:29:01 am
content: Thanks Ian, i have a lot to learn!

author: makerman
author_rank: Regular Contributor
timestamp: January 30, 2019, 11:32:23 am
content: But i wasn't trying to work out the transient then, just how much power was transferred, so it should be like-for-like.I needed the inductance for that calculation so that i knew the time it took for the transfer and hence energy transferre over time, what was i doing wrong?