Light bulb not lit up by inductor - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: November 21, 2024, 12:02:16 am
content: I'm trying to demo how an inductor stores energy by having it power a DC light bulb. After charging up by being connected to a DC power supply, the current in the inductor builds up generating a magnetic field until the current through the inductor is constant. Then, when the switch flips, due to the inductor's nature, the current doesn't change instantaneously. The induced field maintains the current through the inductor, and it flows through the light bulb until the field is gone and the current is depleted. I think I might have missed some things, but I believe that that is the gist of it. The problem is, when switched, the light bulb doesn't light up even though the current and voltage at time of charging should be enough to power the bulb. Why could it not work?I'm working with a 1 uH inductor, and the bulb has a voltage rating of 6.3 volts and rated for 0.22 amps. I also put some resistors so that I can achieve a higher voltage across the inductor, and I tried putting one in series with the LED so as to help delay it a bit. In theory, it should work, but either the current is depleted too fast or the light bulb isn't turning on at all. What could be the issue?

author: Analog Kid
author_rank: Super Contributor
timestamp: November 21, 2024, 12:06:22 am
content: Quote from: LoveLaika on November 21, 2024, 12:02:16 am1. You're mixing up units used for capacitance (μF, or microfarads) and those used for inductance (Henrys, usually micro- or millihenrys.2. You guessed right: the current is depleted too fast for you to see it. But it is present in that circuit when you flip the switch.You'd need something like an oscilloscope to actually see the inductive current flowing.

author: Benta
author_rank: Super Contributor
timestamp: November 21, 2024, 12:11:24 am
content: Actually, a mains transformer primary should be able to do your experiment.Around 50 VA is fine, they have high inductance, and are at your voltages undangerous, But it might blow the bulb.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: November 21, 2024, 12:12:42 am
content: Thanks. Yeah, that's my bad. Slip of the tongue.With the resistors that I mentioned, I could use them to measure the voltage (and thereby view the current) flowing through the bulb. But, according to my supply, I'm reaching 30 volts, and it's drawing ~300 mA. How can I delay the depletion of the current? Would higher value resistors help?

author: PGPG
author_rank: Frequent Contributor
timestamp: November 21, 2024, 12:14:24 am
content: Quote from: LoveLaika on November 21, 2024, 12:02:16 amif it has 1uF than it is capacitor and not inductorQuote from: LoveLaika on November 21, 2024, 12:02:16 amIn theory it should not work.When you switch that way the capacitor it preserves stored energy because during switching I=0 as it is temporarily disconnected and current can't flow.To do the same with inductor you have to ensure U=0 during switching. So it have to be shorted during that time and it would be good to have its coil being made of superconductor to not loose energy during switching.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 21, 2024, 12:43:55 am
content: Three observations. Firstly, as PGPG says, the switch disconnects the coil briefly as it transits from one contact to the other. There's probably a spark inside the switch, dissipating some of the power.Secondly, are you using a filament lamp? If so, the thermal inertia of the filament is probably too high to show you the brief flash. You could try an LED instead.Thirdly, 1uH is a tiny inductor. I strongly recommend that you read up on the basic physics. Calculate how much energy the inductor will store, and what that would look like when delivered to a load like an LED. Also, confirm that the inductor will not saturate with that current.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: November 21, 2024, 01:10:25 am
content: I corrected my error regarding the appropriate units.When the inductor is switched, instantaneous current remains but voltage can drop to 0. Voltage can change instantaneously. If there's no voltage, the bulb won't turn on. Is that what you're saying? I was thinking about that when I saw your message.

author: RJSV
author_rank: Super Contributor
timestamp: November 21, 2024, 02:39:08 am
content: The typical garden light is a good example, although they use an oscillator, try search on 'Buck Boost converter' for background on that.Likely that you could simply create an LED flash, easy to see, but still too brief to read on your multimeter. An oscilloscope, sure, but it will be a one-off pulse, not a continuous thing.The way I like to consider the theory, is that the inductor field has been built up, and when that field collapses, it starts generating more and more voltage. That would make around 3V for the LED that is in the path to ground, or negative, driving something like 3 volts, into your LED and series resistor (try 200 ohms).You'll get a brief single pulse of light, if you skip the oscillator. But with a storage scope you could see the waveform.

author: fzabkar
author_rank: Super Contributor
timestamp: November 21, 2024, 02:51:12 am
content: Connect a 1A 1N4007 reverse biased diode between the rightmost switch terminal and the pole. The pole should connect to the cathode. When you open the switch, the inductor's current keeps flowing through the diode like a flywheel. You shouldn't need R2 (because you would be dissipating the majority of the inductor's energy in this resistor).

author: RJSV
author_rank: Super Contributor
timestamp: November 21, 2024, 02:58:40 am
content: After some thought, I realized there might be some, er, pesky complexities; In order to set up your schematic you first would do better with a push button, having the two switch terminals ACROSS your LED, literally shorting it out.With the push button down, you are charging up the magnetic field in the inductor (might want a series resistor).Now, a buck-boost circuit would have that inductor 'dump' into the LED, every time you 'open' the contacts of the push button. I guess your knife switch would work, using two connections, just like a push button.Complicated due to the fact that a buck boost circuit uses something like 1.5 V battery, so we might have to walk you through the details there.But your idea should work, just maybe as a brief transient flashy.- Rick

author: Andy Chee
author_rank: Super Contributor
timestamp: November 21, 2024, 03:18:38 am
content: Use a neon bulb, not an incandescent filament bulb.

author: vk6zgo
author_rank: Super Contributor
timestamp: November 21, 2024, 08:49:33 am
content: The inductor is way too low in value to store more than a miniscule quantity of energy, & on top of that, you have resistors in series with it to further reduce the amount available for storage.Then when you operate the switch you effectively place a short circuit across it, with the very low resistance of the bulb.The energy stored in the inductance would all be dissipated in far too short a time for the filament to heat to operating temperature.

author: Zero999
author_rank: Super Contributor
timestamp: November 21, 2024, 09:55:26 am
content: Try this instead. Note V1 must be below 5V, otherwise it might damage the LED.

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: November 21, 2024, 11:13:36 am
content: Quote from: LoveLaika on November 21, 2024, 12:12:42 amno, a higher resistor does not help:the time constant for an L-R combo (your incandescent lamp plus 1µH inductor, while having R2 = 0 Ohms instead of 100 Ohms) istau = L/R. hence, to get longer time constants, you either reduce the resistance or increase the inductance.(for a C-R combo, tau = C +R)in your example the current through the lamp will vanish after approx.5 * tau = 5 * 1uH/59mOhms = 84 µs. (again, R2 = 0)so if you want a 1 sec decay time, tau has to be approx. 1/5 sec.thus, L > 0.2 * 0.059 = 12 mH. as the lamp filament will heat up, its internal resistance will increase to approx. 10 times of its cold value. therefore, L = 120 mH is a more realistic value in your circuit.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 21, 2024, 12:28:25 pm
content: Quote from: Zero999 on November 21, 2024, 09:55:26 am@LoveLaika: take note of this post from @Zero999! ^^^^^^^^^^This is the correct way to observe the energy stored in the inductor. No sparking at the switch contacts or interruption to the current, which is the main problem with your original circuit.Is it clear to you how it works?

author: Siwastaja
author_rank: Super Contributor
timestamp: November 21, 2024, 12:34:32 pm
content: Thought experiments / counter questions:How much energy is stored by the inductor? You can look the formula up.For how long should this energy light up the bulb?How fast is human vision, or reaction time of the bulb (filament heating up)? Can it light up at all?What does the inductor do to maintain current going through it when the switch is mid-way switching, and thus open?

author: BILLPOD
author_rank: Frequent Contributor
timestamp: November 21, 2024, 01:59:00 pm
content: Good Morning Lovelaika, substitute an analog voltmeter for your bulb and if you've got the meter on it's lowest range, you 'should' see the needle twitch, but just once.

author: Picuino
author_rank: Super Contributor
timestamp: November 21, 2024, 03:34:01 pm
content: Try with a LED, bigger inductor and without right R2 resistor (only has the effect of dissipating the energy of the coil without producing light):https://www.picuino.com/circuits/?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgAbENPbhQ7vhDYEKKFGgRUpGbOJ582FDShsAbiAXDRmqmGJiqVGlVzij0BGwDKg3in53wB8yABmAQw4BnBuJRsAO66zmL6YS6QbABOISJiPNqGyFHBKMqhwtgC4apghGLY2ZlFAvHcIAAmDJ4ArhwALkwcDJXiUhKQ7GlC5aVJqti8VABiHTBd4LAQLCAAkgB2lbUAxg0eCysMbEA

author: RJSV
author_rank: Super Contributor
timestamp: November 21, 2024, 03:35:19 pm
content: Amusing to read the varied approaches. We have TWO different supply voltages, in the discussions.Similarly, TWO different light generators; filament and LED types. We could temporarily call them the low voltage, LED type, and the moderate voltage, filament type.Note that the filament has drastic variations in the internal resistance, over about a 200 mSec heat-up period. The LED type set up would have a flash visible for perhaps 100 uSec, as also noted in a previous reply, here.In my opinion the filament type of indicator is more current and more trouble. If not for some decorative uses, the LED type indicator is currently replacing the older tech filament.If you do things with LED, I'd estimate you will get a single flash, visible at 100 uSec duration. With a storage scope, you can observe the waveform of voltage across a resistor, to derive the initial current (during the 100 uSec).The newer schematic shown recently should have some resistor, in series with LED, and that's often at 59 ohms, in some devices, flashlights, etc.Using ohms law, V÷R = I although that's going to be transient voltage, as your storage scope will show.Thanks for your patience, it's a great question.

author: Picuino
author_rank: Super Contributor
timestamp: November 21, 2024, 03:42:09 pm
content: Resistor R2 does not reduce the led current (which is given by the coil current), it only increases the voltage drop of the coil and, therefore, the speed at which the coil is discharged.

author: RJSV
author_rank: Super Contributor
timestamp: November 21, 2024, 03:46:08 pm
content: Sorry Picuino, but you might be fast enough, on that knife switch movement, to take, maybe 300 mSec. and by the time that moving knife moves even 1% of the travel, the LED will have finished the visible pulse.Folks have to decide, incandescent or not, rather than mixing both together (using 30 volts would make more sense for the higher power incandescent filament lamp).

author: Zero999
author_rank: Super Contributor
timestamp: November 21, 2024, 05:53:46 pm
content: Quote from: SteveThackery on November 21, 2024, 12:28:25 pmI got the decimal point wrong, in my mental arithmetic. The inductor should be 100mH, not 1mH.Here's a simulation. The LED will only be on for 1ms, but that will be enough to see a flash, given it's driven hard at 30mA. The human eye can respond quite well to a short flash, as long as it's very bright.Note to those circuits with a DPST switch: it has to be break before make, otherwise as soon as the switch is opened the inductor will discharge into the switch, causing arcing across the contacts. Unfortunately all DPST switches I've seen are break before make.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 21, 2024, 06:21:27 pm
content: Quote from: Zero999 on November 21, 2024, 05:53:46 pmI've altered your quote to what I think you meant. I agree completely, but unfortunately if it were make before break the power supply would light up the LED. That's why the OP's circuit won't work as a demonstration.

author: RJSV
author_rank: Super Contributor
timestamp: November 21, 2024, 11:44:45 pm
content: My problem is visualizing what happens during the knife switch travel time, which is a century compared to time constants.Moving that switch in 300 mSec. is approx. 300 X the 1 mSec. flash time.So my curiosity is about what happens when the circuit is just 'open' sitting there with magnetic field at max.I recall that you would use parasitic capacitance, to get a 'tau' describing decay.You have decay of that magnetic energy, right ?Even if totally open circuit, I believe it was the parasitic capacitance that aids in the drain-down. I'm saying that occurs way, way before your knife edge does it's travel over to make contact.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: November 21, 2024, 11:56:12 pm
content: Quote from: SteveThackery on November 21, 2024, 12:28:25 pmThanks for replying. Yes, I believe I understand how that works. Reminiscent of how one should have a diode to dissipate the energy from a relay coil. Makes total sense to me. Thanks. This might be what I need with some tweaks. Just gotta test it and see if it is visible enough.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: November 21, 2024, 11:57:04 pm
content: Thanks so much. I think this is what I need, and it makes sense.

author: vk6zgo
author_rank: Super Contributor
timestamp: November 22, 2024, 01:33:54 am
content: Quote from: BILLPOD on November 21, 2024, 01:59:00 pmNot with the OP's original values.The inertia of the meter movement will prevent any twitch.

author: vk6zgo
author_rank: Super Contributor
timestamp: November 22, 2024, 01:45:23 am
content: Quote from: RJSV on November 21, 2024, 11:44:45 pmAs soon as the switch removes the supply, as the magnetic field can't be maintained without current from the supply, it starts to collapse, inducing reverse current into the coil.It can't flow anywhere, there is a build up of voltage across the coil, which can be discharged through any load now switched into circuit.With a very low inductance value, the voltage will be small to start with, & probably will decay rapidly.through the stray capacitance.Storage in an inductor is a real thing---Google Kettering ignition.

author: RJSV
author_rank: Super Contributor
timestamp: November 22, 2024, 02:43:23 am
content: Ahh; thanks, nice to ask my own basic physics question, as I feel 'irritated' and obsessed...good physics explanation.Anyway that drain-down isn't going to leave much. (I'm going to go search on that ignition.)OP: Do you really want an incandescent (filament) type, as LED component seems easier with less components ?

author: coppercone2
author_rank: Super Contributor
timestamp: November 22, 2024, 02:58:57 am
content: i think you need a bigger inductor so it explodes the light bulb

author: LoveLaika
author_rank: Frequent Contributor
timestamp: November 22, 2024, 06:35:28 pm
content: Testing with an LED seemed to work a lot better than a bulb, as in, the LED lit up briefly while the bulb did not.

author: RJSV
author_rank: Super Contributor
timestamp: November 22, 2024, 08:38:54 pm
content: Could assume, use the 30 volt power input, if you need to experiment with a filament.For an LED, a low voltage probably 1.5 volt,would do. I'm never comfortable with leaving out a series resistor, like 59 ohms.That drops LED voltage by roughly 500 mV for every 10 ma flowing (at peak).Then, I'd use the knife switch but only TWO of the terminals, as the travel time is way, way too long...unless you try some ballistic method of super fast (and ridiculous) action.You get the one flash, with LED, and can put a storage scope onto the series resistor to convert to current flow for a reading of peak current.A filament has that messy change in impedance, as it heats up. Plus needing that 30 volt supply, then.

author: RJSV
author_rank: Super Contributor
timestamp: November 22, 2024, 11:21:05 pm
content: Reply #12 has the good schematic, thanks to Zero999 for posting that.I would add the 59 ohms series resistor, for limiting any excess LED current.Sorry about this taking time to resolve, I've got a todo stack bigger than my own ego.

author: Picuino
author_rank: Super Contributor
timestamp: November 23, 2024, 12:14:53 pm
content: Another practical circuit:https://www.picuino.com/circuits/?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgwM9wVvjCe3TgIEQYNKLDDYMdQpHmFMcjEhiQkYePEmQUFOLrBrYatgDcQTGkOL7r3G1Gc0qYfVU-QEbAO5WnMDtaWw8-EH5wDAFIoLD-BxDBJMhw90chXgiBVP8wGglhKNFoqDTSoqK8T3KBauyQetyG+sjI1LAlJLik+v0AEwYAMwBDAFcAGwAXJgmGfucxI3CnVrqa-0Dg2ODUgHtkRucADzwtVhA0SSX9LUWqenUNKxw2IAYou must use an LED with the same maximum current as the relay coil (or a combination of LEDs in parallel that has a maximum current equal to or greater than the relay coil current).

author: RJSV
author_rank: Super Contributor
timestamp: November 23, 2024, 01:54:00 pm
content: Thanks. But I'm only partially following the oscillating circuit.What's the longer time period, 130 milli-secinds ? That seems like the relay 'buzzing' at around 8 hz, which would be about right. Of course that's not a very robust oscillator, as the moving contact just barely gets into an 'open' state, but not certain if that is any problem.The short time periods makes sense, also, but I'm having difficulty relating the two, looking at that pulse waveform shown.I think you just would get an illuminated LED similar to the solar garden light situation.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 23, 2024, 04:59:04 pm
content: Quote from: RJSV on November 23, 2024, 01:54:00 pmMe too. I cannot see why that circuit should oscillate UNLESS the lamp is a device showing negative resistance - perhaps a neon lamp. But neon-basednoscillators normally just use a capacitor.

author: Zero999
author_rank: Super Contributor
timestamp: November 23, 2024, 08:17:25 pm
content: Quote from: SteveThackery on November 23, 2024, 04:59:04 pmI can confirm the relay circuit will oscillate. Relays have hysteresis. It's a Schmitt trigger oscillator. I made one when I was a child, which I used as a shocker. I didn't put anything in parallel with the coil, just the terminals to zap people with. It's an old circuit, which has been used as a boost converter many years ago.

author: RJSV
author_rank: Super Contributor
timestamp: November 23, 2024, 09:12:27 pm
content: Yeah, I mean, (just so other readers are following this), you can work out the behavior, mechanically:With voltage to the coil, that relay contact should start opening up the circuit, but that's going to cut the power to itself, soon.But, mechanically there will be delay, as the contact is on a springy metal, which also has helped make good switch contact.I'd guess 10 or 50 mSec. to pull off so contact is broken. (That's the schmitt-like hysteresis).So, whatever that time takes, the LED is so much faster, you will get a pulse and the waveform has 6 uSec. duration.But then we have to wait, a bit, until the relay contact spring brings the contact back. That is, back to connect as the normally closed position it sits in, when un-powered.Thats some more milliseconds, maybe 100.I'd suggest, get rid of the oscillating relay, that could be all over the map anyway.(Can I use the relay from my scooter ? No.)Use your knife switch, instead of the normally open relay contacts. Then just open it, once.

author: RJSV
author_rank: Super Contributor
timestamp: November 23, 2024, 09:16:00 pm
content: Oops, I meant the normally open relay contact.

author: Zero999
author_rank: Super Contributor
timestamp: November 23, 2024, 10:33:20 pm
content: Quote from: RJSV on November 23, 2024, 09:16:00 pmThe circuit will only work with a normally closed relay contact in series with the coil.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 24, 2024, 01:38:12 pm
content: Oh, damn! I didn't realise it was a relay! What an idiot. I was fooled because the animation does not show the relay contacts moving.Sorry - bit of a brain fart there.

author: Zero999
author_rank: Super Contributor
timestamp: November 24, 2024, 02:08:39 pm
content: Quote from: RJSV on November 22, 2024, 11:21:05 pmR1 already limits the current through the coil and therefore the LED.