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Light LED with 4 nanoamps

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Dr. Slack

Mon Jul 21 2014, 11:24AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Just think of it as inductor Q, energy stored divided by energy dissipated per cycle. The eddy currents in the iron put an effective resistance in series with the inductor.

Ash Small

Mon Jul 21 2014, 07:55PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Dr. Slack wrote ...

Just think of it as inductor Q, energy stored divided by energy dissipated per cycle. The eddy currents in the iron put an effective resistance in series with the inductor.

If I had an easy way to charge a 1nF capacitor to 450V, I'd connect an led to the inductor and cap and try it, but I don't think I have an easy way to get 450V.

EDIT: If I get a chance I'll rig up a mains powered doubler, that should get to something around 460-480, maybe? (I won't rush into anything, though, not where 240ish mains is concerned. I'd need at least a fuse and a resistor between the mains and the doubler.)

Tony Matt

Thu Jul 24 2014, 03:05AM

Registered Member #3700 Joined: Sat Feb 19 2011, 12:59PM
Location:
Posts: 107

Hi Ash !

Your ideia of " easy way to charge a capacitor " is not bad at all....

1- supose that the bonanza battery carge 500V to the 0.001uF cap. The energy stored inside the cap will be 0.000125J.

2 - Then the pendulun moves and connect the 0.001uF in parallel with another cap. (40uF)

3- The voltage at the 40uF cap will be:

E = 0.5C.V^2 >>> V^2 = 2E/C >>>> V = (2E/C)^0.5

V = (2 X 0.000125/0.000040 )^0.5 = 2.5V

As the LED is conneted to the 40uF cap,, the cap will supply 2,5 volt to the LED.

Any suggestion or comments ?

EDIT : I the above calculation I have used the principle of " conservation of energy " to convert 500V to 2.5V, that is not the wright way.

The correct way is using the principle of carge conservation as:

1- Charge in the 0.001 cap with 500V:
Q = C x V = 0.001 X 500 x 10^-6 = 0.5 x 10-6 coulomb

2 - A capacitor with a charge of 0.5 x 10-6 coulombs and a potential of 2,5v shall have :
C = Q/V = 0.5 x 10-6 / 2.5 = 0.2 x 10-6 farads or 200nanofarads (nF).

So , please my apologies for my mistake.

The problem of two capacitors may be found in the internet.

Cheers

Dr. Slack

Thu Jul 24 2014, 07:39AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Hi Tony,

What makes you think all the energy in the 1nF cap will end up in the 40uF one? Do the sums again for voltage on the second cap, but this time use a non-zero approximation for the residual resistance of the interconnect, the switch, and the capacitor ESRs. If you don't fancy the algebra or the Laplace stuff, then just run it through a simulator.

Suggestion, try 1 ohm. Then try 10 ohm. Any difference in the final voltage on the 40uF cap?

Now do the same again, and assume a non-zero residual series inductance in the wiring and the switch. Is there any difference in the final voltage for the two resistance cases? Think about the Q of the residual series inductance and resistance.

Now try again with an enhanced series inductance.

Is a pattern emerging?

Tony Matt

Thu Jul 24 2014, 07:16PM

Registered Member #3700 Joined: Sat Feb 19 2011, 12:59PM
Location:
Posts: 107

Hi Dr Slack !

Of course my suggestion is just a concept.

We need to work, develope, test, and probably we may achieve a solution.

So far what I have for sure is:

1 - a good and visible flash in a NE2 bulb during the discharge of 0.001uF charged with 340V

2 - a poor but visible flash in a LED bulb during the discharge of 0.001uF charged with 340V

The transfering charge of a cap to another allways lose some energy. Losses in the ESR, pendulun contacts and some losses transformed in electromagnetic waves too.

I do not have a simulator, and the Laplace transformations are rusted... It is more than 45 year that I have play with such stuff.

Maybe some one can simulate such energy conversion, and guive us one idea about the losses, the role of a inductance or inductor to be added in such schema.

Dr. Slack

Fri Jul 25 2014, 07:40AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

I'm not sure I'd flatter it by calling it a concept. Concept implies an idea for which the physics will work, but maybe the engineering hasn't been nailed, or the economics. The physics won't work. Connect a 400v 1nF cap to an uncharged 40uF one, and they both end up at around 10mV, regardless of the value of resistor used to connect them. Have a think about which conservation laws are used to arrive at that result. I love conservation laws BTW, as long as you pick the right one, it allows you to cut through all the mathematical cr@p and arrive at a limiting result with certainty.

Antonio

Fri Jul 25 2014, 10:11PM

Registered Member #834 Joined: Tue Jun 12 2007, 10:57PM
Location: Brazil
Posts: 644

I made some experiments with this circuit, shown in the LTspice simulation below. To charge the 1 nF capacitor C1 I used an electrostatic machine (applying current in parallel, not voltage in series as in the simulation). My machine has on it the high-voltage diode D1, with about 20 V drop, that completes a buck converter (with some loss). The switch simulates a spark gap, that I set to breakdown at 450 V. The LED pulses visibly in the experimental setup.
The inductor L1 and the capacitor C2 are not critical. They control the maximum current in the LED, producing a short pulse of high current if small or a wide pulse of low current if big. The result is practically the same, as the total energy is what matters. A short pulse works better because the LED is more efficient with high current, and the spark gap operates for just a few tens of us anyway. The actual waveform seen in an oscilloscope is practically identical to the simulation.

A video, using 1 nF and 10 nF charged to ~500 V.

Tony Matt

Sun Jul 27 2014, 04:39AM

Registered Member #3700 Joined: Sat Feb 19 2011, 12:59PM
Location:
Posts: 107

Hi Antonio !

Amazing progress in the Zamboni project !

Now we have a stable spark gap that is very simple contruction, avoiding all problems of a mechanical/electrical construction, and a inductor that has a low inductance, simple and easy constuction.

If you do not mind, please inform the diameter of the coil, the number of turns and wire gage.

Cherrs

Antonio

Mon Jul 28 2014, 01:33AM

Registered Member #834 Joined: Tue Jun 12 2007, 10:57PM
Location: Brazil
Posts: 644

I used a section of PVC tube with 8.8 cm of diameter. 430 uH requires 53 turns of #32 wire (My inductor is actually two in series, with the same effect). I obtained a significant improvement by adding another diode (1N4007) from ground to the node between the gap and the coil, making a true buck converter. I had to increase C2 to 680 nF to reduce the LED current to a less dangerous level. Note that my generator generates a few uA of current. 0.5 uA produces one flash per second.

Newton Brawn

Tue Jul 29 2014, 03:27AM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Hi Antonio,

So nice get some light of so small energy...

And the air spark gap solved the switching problem with precision. The old technology of spark gap exeeded the actual semiconductor devices, at least in this project.

Maybe Antonio may try to check the result using a inductance of 1 henry in the place of 430uH. For me it is easy to wind at home a inductor of 0.5 to1.0 henry.

I suspect that 1 henry will allows the LED bright for few miliseconds, providing a better visual sensation .

Regards.

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