How to light an incandescent bulb using capacitors?

[J_Nichols]

I am just learning and playing with electronics and I want to light a bulb with one or more capacitors for a few seconds. The bulb would be 25 watts, 250 Volts. So I would like to know the values of the capacitor that I need and what kind they should be.

 

[shortbus=]

As a newbie/learner myself, don't think this will work if the voltage is AC. A smaller DC voltage and a car light bulb may work though.

 

[J_Nichols]

As a newbie/learner myself, don't think this will work if the voltage is AC. A smaller DC voltage and a car light bulb may work though.

I can start with a small car light bulb. In that case, how many uF do the capacitor needs to run the light for about a second?

 

[Externet]

For a 12Volt - 24 Watt bulb, current is 2 Amperes.
2 Amperes times 1 second = 2 Coulombs
For a capacitor charged to 12VDC, the capacity should at least Q/V = 2/12 = 0.17 Farads = 170,000 μF
----> Then, a 170,000μF capacitor charged to 12VDC will do the one second.

The brightness will decrease during the one second.

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For a 250V DC - 25 Watt bulb, current is 0.1 Amperes.
0.1 Amperes times 1 second = 0.1 Coulomb
For a capacitor charged to 250VDC, the capacity should at least be Q/V = 0.1/250 = 0.0004 Farads = 400μF
----> Then, a 400μF capacitor charged to 250VDC will do the one second.

The brightness will decrease during the one second.

 

[AnalogKid]

The total energy in watt-seconds (W-s) stored in a capacitor = 1/2 x C (capacitance in farads) x Vsquared. But that is the *total* energy. As the capacitor discharges into the load, its output voltage decreases. At some point the voltage is so low that it no longer is enough to run whatever the load is, even though the capacitor is not yet fully discharged. So the capacitor size needed to do a job is depends on the the operating voltage range of the job. Example:

12 V 2 A bulb = 24 W. To run the bulb for one second takes 24 W-s.
Let's state that when the voltage sags to 6 V, the bulb is too dim to be useful. So the useful discharge voltage range of the cap is from 12 V to 6 V. Subtracting the ending energy level (at 6 V) from the starting energy level (at 12 V) gives the amount of energy available to the load. Since we already know the amount of power we need, we can rearrange the equation to give us the capacitor size needed to meet any particular power requirement.
V1=12
V2=6
24 W-s = 1/2CV1^2 - 1/2CV2^2 = 1/2C(V1^2-V2^2)

C = 48 / (144-36) = 407,407 uF

ak

 

[alec_t]

In the interests of safety, don't play with capacitors charged to high voltages (e.g. mains voltage). They can give you a FATAL shock.

 

[J_Nichols]

For a 12Volt - 24 Watt bulb, current is 2 Amperes.
2 Amperes times 1 second = 2 Coulombs
For a capacitor charged to 12VDC, the capacity should at least Q/V = 2/12 = 0.17 Farads = 170,000 μF
----> Then, a 170,000μF capacitor charged to 12VDC will do the one second.

The brightness will decrease during the one second.

---------------------------------------------------------------------

For a 250V DC - 25 Watt bulb, current is 0.1 Amperes.
0.1 Amperes times 1 second = 0.1 Coulomb
For a capacitor charged to 250VDC, the capacity should at least be Q/V = 0.1/250 = 0.0004 Farads = 400μF
----> Then, a 400μF capacitor charged to 250VDC will do the one second.

The brightness will decrease during the one second.

In the case that I want to light 2 bulbs (in series) rated at the same 250V 25 watts each one. How do I should calculate?
500VDC - 25 Watt bulb (or 50?) = 0.1 Amperes?

 

[MikeMl]

Anybody who has to ask this question has no business screwing around charging capacitors off the 250V line

 

[Externet]

In the case that I want to light 2 bulbs (in series) rated at the same 250V 25 watts each one. How do I should calculate?
500VDC - 25 Watt bulb (or 50?) = 0.1 Amperes?

Each 250VDC rated 25 Watt bulbs use 0.1 Amperes; 25W/250V = 0.1 A
and their resistance is 250V/0.1A = 2500 Ω
Two in series present 5K Ω. Current on 500V for two in series is 500V/5000Ω = 0.1 Amperes.
0.1 Amperes times 1 second = 0.1 Coulomb

For a capacitor charged to 500VDC, the capacity should at least be Q/V = 0.1C/500V = 0.0002 Farads = 200μF
----> Then, a 200μF capacitor charged to 500VDC will do the one second to two 250VDC - 25Watt rated
bulbs in series.

The brightness will decrease during the one second.

I am just learning and playing with electronics ...

----> Not your best choice if just learning and want to play !
Take this as an exercise in arithmetics, and play with other safer circuits.

Print and hang this :
----> https://s588.photobucket.com/user/Innernet/media/Triangles.jpg.html?sort=3&o=99

 

[audioguru]

An old fashioned incandescent light bulb draws 10 times its normal current until it is hot. Then the huge expensive capacitor must be maybe 10 times bigger and 10 times more expensive than you calculate.
Why not experiment with a modern cheap low voltage and low current modern LED instead? The low voltage will not kill you. The capacitor to light it for one second will be small and inexpensive.

 

[shortbus=]

I can start with a small car light bulb. In that case, how many uF do the capacitor needs to run the light for about a second?

The main reason for my statement and one that no one else seemed to pickup on- a capacitor passes AC . It doesn't store it. when beginning in electronics always start with low voltages, like 12V and under.

 

[schmitt trigger]

The statement from J_Nichols strongly implies that he has plans to store enough energy to light up an incandescent lamp for 1 second.

I'm also with MikeMi on this one. If you don't know what you are doing, if you are a rank beginner, you have no business playing with this circuit.

 

[J_Nichols]

An old fashioned incandescent light bulb draws 10 times its normal current until it is hot. Then the huge expensive capacitor must be maybe 10 times bigger and 10 times more expensive than you calculate.
Why not experiment with a modern cheap low voltage and low current modern LED instead? The low voltage will not kill you. The capacitor to light it for one second will be small and inexpensive.

That is a good ideas. I thought about an incasdencent bulb because it's the one I have.

 

[MrAl]

Hello,

Yes, the warning about high voltage caps is a good one. They can also explode like little bombs if faulty or are charged to excessive voltages.

One kind of equation for the light bulb if it was a pure resistance would be:
a*Vs=Vs*e^(-t/RC)

which of course simplifies greatly to:
RC=-t/log(a)

where:
RC is the resistance of the bulb times the capacitance of the capacitor to be used, and
'a' is the percent of the source voltage that once reached causes the bulb to be considered too dim, expressed as a fraction.

For example if the source voltage is 100 volts and 80 volts is considered a voltage that causes the bulb to become too dim, then the equation becomes:
RC=-t/log(0.8)

and this simplifies approximately to:
RC=4.48*t

So if t=1 second then we have simply:
RC=4.48

so if R=12 then the approximate value for C is 373000uf, which is a lot of capacitance.

Note this kind of result is independent of the supply voltage as we only need to know the percent decrease that causes the bulb to become too dim.

Also, as audioguru rather nicely pointed out (+1 Like), with a REAL bulb there will be some extra energy required to cause the bulb to heat up first before it actually even lights up. This could be small or large depending on the physical construction of the bulb. We would have to add that into the equation to get a more exact result as the result above assumes the bulb is a pure resistance.

I have some data somewhere we could probably use for this but i bet there is something on the web. Seems to me i ran across this before on a forum.

 

[Colin]

He really wants to connect the capacitor to a door-knob and light-up someone "like light globe."

 

[gary350]

When I was in high school 48 years ago I use to light up a 4 foot long Fluorescent bulb with a D size battery. That was too long ago for me to remember details but you need enough volts to make the filaments inside the bulb glow you might need more than 1 battery. Parallel the 2 pins on each end of the bulb with 2 wires. Connect the battery when both filaments glow the light comes on. The old alkaline D batteries are 8 amps.

 

[audioguru]

When I was in high school 48 years ago I use to light up a 4 foot long Fluorescent bulb with a D size battery. That was too long ago for me to remember details but you need enough volts to make the filaments inside the bulb glow you might need more than 1 battery. Parallel the 2 pins on each end of the bulb with 2 wires. Connect the battery when both filaments glow the light comes on. The old alkaline D batteries are 8 amps.

You mean the filaments at each end glow dim orange? That is not lighting the Fluorescent tube.

Oh, maybe you had a very high power AM radio station or high voltage electricity power lines next door that ionized the fluorescent coating inside causing the tube to light properly? Then it can light properly without the batteries heating the filaments.

 

[J_Nichols]

Thanks for all the information.
I am reading all the information little by little to better understand it.
When I have a question or answer I will reply.

 

[gary350]

You mean the filaments at each end glow dim orange? That is not lighting the Fluorescent tube.

Oh, maybe you had a very high power AM radio station or high voltage electricity power lines next door that ionized the fluorescent coating inside causing the tube to light properly? Then it can light properly without the batteries heating the filaments.

The fluorescent bulb did light up bright just like being in a light fixture. Must be something else I am forgetting. It was a hand held portable light. The filaments glowed orange from the battery. I may have had a buzz box auto transformer running on the battery too. That was 1967 or 1968 I just don't remember anymore.

 

[MrAl]

Hi,

Back in the 70's i experimented with fluorescent bulbs too. No problem lighting the two filaments for starting the tube, but I burnt one tube out before i discovered that the higher voltage power source had to be current limited

With one, i could not stand the noise the old fashioned ballast made so i connected it to the bulb with long wires so the ballast could be located remotely. It's amazing how differently we do things today.