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Investigations into the stored energy of inductors.

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T3sl4co1l:
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

makerman:
Thanks very much Tim.

makerman:
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 ?

MrAl:

--- Quote from: makerman on January 26, 2019, 05:02:12 pm ---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 O
L=0.5 H

Coil 2 :

R=102 O
L=4.6 H

The 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 ?

--- End quote ---

Hello 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=I

and squaring that we have a constant:
I^2=K

now we have:
W=(1/2)*K*L

and lumping both constants into K1 we have:
W=K*L

and 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/dt

and solving for di/dt we get:
di/dt=v/L

and 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=1/1

and we have:
di/dt=v/5=1/5

and 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)/R

and once that stabilizes the max energy is stored in the inductor.


makerman:
Thanks very much Mr AI.

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