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Light LED with 4 nanoamps

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Newton Brawn

Sat Jul 05 2014, 01:40AM

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

Zamboni

Here some news, and RESULTS:

I just get a 1200pF polyester capacitor and charge it with 340volts using a diode and 220V line.

Then I disconected the capacitor from the diode and line,

And connected the cap with a NE2 neon bulb....

The result: I got a really visible red flash from the lamp !

The conlusion is that the Zamboni battery cam be connected with a 1000pF cap and a pendulum in such way that (when the cap is charged) the mobile blade of pendulum make a contact with one pole of the bulb, transfering the energy storaged in the cap to the bulb.
After the contact done, the cap voltage is reduced and the blade returns to initial position and starts to be charged again.

Here the proposed arrangement:
B= batery that provides high voltage,
C= capacitor that is carged by the battery and discharged throgh the pendulun to the bulb
NE2 = neon bulb 70V, 1/8W
P = pendulum, that work as a electroscope, opening the blade when charged

Note: the pendulum cam be assembled in the horizontal position, using the vertical shaft. for more sensibility.

]zzaamm_model_1.pdf[/file]

Regards

Newton

EDIT - fixing the typing

Zamboni

Mon Jul 07 2014, 03:45PM

Registered Member #2836 Joined: Fri Apr 30 2010, 01:24PM
Location:
Posts: 41

The conlusion is that the Zamboni battery cam be connected with a 1000pF cap and a pendulum in such way that (when the cap is charged) the mobile blade of pendulum make a contact with one pole of the bulb, transfering the energy storaged in the cap to the bulb.
After the contact done, the cap voltage is reduced and the blade returns to initial position and starts to be charged again.

*****I believe your proposed circuit will work. I am trying to engineer a set of conductive leaves that will allow your circuit to work. Thanks for your idea.

Paul

Newton Brawn

Tue Jul 08 2014, 10:23PM

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

Hi Zamboni,

1-Refining my experiments, this time I have charged the 0.001uF capacitor with 170V and discharged through the neon bulb.
THE RESULT: I got a really visible red flash from the lamp !

2- The calculed time for 3 nanoamper charge a 0.001uF to 500V is about 167 seconds, (considering no leaking through the surfaces or insulation creepage) That means, you can get a light flash at every 2.8 minute.

3- The electrostatic switch may take some energy from the battery in order to move itself. The blade mass, self capacitance, friction will increase the one minute charge time.
A more efficient switch with the blade performoing a push-pull arrangement may be more suitabe.

Let us know your experiments

Regards

EDIT " 167 seconds"

Antonio

Thu Jul 10 2014, 12:54AM

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

You can obtain more efficiency by discharging the capacitor through a transformer, reducing the high voltage in the capacitor to the level required by the neon lamp. This also allows the use of a smaller capacitor, using better the available energy. Try first the direct version. Leakage in the capacitor may be a problem. Use a high-voltage capacitor (or make one) and keep it dry and clean. The mechanical assembly uses very little energy, just the amount required to charge its capacitance just before the discharge, probably negligible in comparison with the energy required to charge a capacitor of even a few tens of pF.

Newton Brawn

Tue Jul 15 2014, 07:50PM

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

Hi:
Iam trying to calculate a small transformer to couple the capacitor with the lamp.
The primary voltage could be 500V , secondary 83V, providing voltage to a NE2 neon bulb (or 2.5V secondary feeding a LED)
So far, I do not have problems. The transformer ratio could be 6:1 or 200:1 in the case using a LED.
Now lets see the primary turns:
Input data:
Core = Ferrite , 16 mm diameter, area 0.0002m2 , max induction 0.3T.
Primary wave shape = sine half wave pulse.
Pulse lenght = 0.005s,
Bmax max = Epk x 2 x t / pi x N x A. >> N= Epk x 2 x t/ pi x Bmax x A
N= 500x2x0.005 / 3.14x0.3x0.0002 = 26526 turns.
It is too much turns. My limits are 600 turns, 32awg or thicker wire.
Did I make a mistake ? Or any wrong approach ?
One way to reduce such primary turns, could be a reduce the pulse length, that could set to 0.00005 seconds. Them the primary will be 265 turns, very easy to wind in a confortable way. BUT now a low power converter is required to shop the 500V to 20kHz.
Any suggestion for such converter?
Regards

Dr. Slack

Tue Jul 15 2014, 10:35PM

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

Can I suggest that instead of a complicated transformer, you go for something simpler, with one winding. This is basically an inverting buck converter. Your low current pile charges the capacitor, stores energy in it. Eventually, the switch closes, connecting the capacitor to the inductor. Current builds and voltage drops as the energy stored as voltage in the cap is transferred to energy stored in the inductor as current. The top terminal of the inductor sinks current from the capacitor.

Eventually, after 1/4 cycle of the resonant period of L1 and C1, the voltage will reach zero as the current peaks. Now the inductor will continue sucking current out of its non-grounded terminal. Without the lamp+diode there, the voltage would ring negative to the inverse of the previous +ve voltage (less losses) as it recharged the capacitor -ve. However, when it gets to -2v (LED) or -100v (neon) (round figures, don't quibble) it will suck charge out of the lamp instead of the capacitor. As the voltage across the inductor now is clamped to a low value, the current will decay only slowly, and the inductor will deliver its energy in a long low voltage pulse to the lamp.

This ought to simplify your calculations of the inductor. The core has to store the total energy of the cap without saturating. Once you've done that, the pulse length can be whatever it comes out to be, as long as it's not silly figures for current. The peak current from the cap / into the lamp is of course given by the energy that the cap dumps into the inductor, 0.5CV^2 = 0.2LI^2. The arrangement dumps essentially all of the capacitor energy into the lamp, and the choice of inductor and cap allow you to choose the peak current and pulse duration more or less freely, so there is no advantage in using a transformer per se.

Tony Matt

Thu Jul 17 2014, 11:41PM

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

Hi Dr Slack,
Now it looks easy, nice converter.
Do you mind say the amount of henryes that such inductor has to have?

I have experimented charge a 0.001Uf with 170V and dischage it in a 6mm LED.
I got a very small flash, difficult to see.

Another experiment I replicate the Newton experiment I got good and visible flash on a neon
C=0.001Uf, V=170V, neon 10mm lenght.

Let me know the inductor volts seconds per amper

Cheers

EDIT : Please read 0.001uF in the place of 0.001Uf

Dr. Slack

Fri Jul 18 2014, 07:50AM

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

Do you mind say the amount of henryes that such inductor has to have?

It's more complicated than that.

Decide what energy you want to dump into the lamp in a single shot.

Add a bit for losses, this is the energy that the inductor will need to store. If you are using an ungapped low permeability core, this will define the minimum core volume. If you are using a gapped high mu core, it will define the minimum volume of the air gap. If you are using an ungapped high mu core, then read up about energy storage in inductors.

Add a bit more for losses, this is the energy the capacitor will have to store. Choose a voltage level and capacitance jointly to give you this energy.

Now decide what peak current you want to send into the lamp. Finally the inductance so that 0.5*Ipeak^2*L gives your chosen energy.

Tony Matt

Sat Jul 19 2014, 02:02AM

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

Hi Dr SlacK

Please correct me if I am dreaming or writhing
something crazi...

available capacitor = 0.001uF, charged with 500volts

available energy: E= 0.5 x V x C^2

E= 0.5 x 0.001 x 500^2 = 125 microjoules

Adding 30% energy extra = 156 microjoules

using a tv flyback ferrite hi mu, 16mm diameter.>> core area = 0.0002m^2

how I can get the air gap volume based on the 156 microjoule ?

Dr. Slack

Sat Jul 19 2014, 01:50PM

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

how I can get the air gap volume based on the 156 microjoule ?

try this link, for magnetic energy stored per unit volume.

You will need to choose a Bmax. For ferrite, 0.4T may be pushing it, 0.2T might be more conservative.

However, if you are starting with a flyback core, that's designed to work in this energy storage mode, so unless it is actually gapped, it's likely to be low u to start with. However, do the sums assuming high u, extra air will merely mean you need a few more turns to get your inductance, and will reduce the likelyhood of saturation.

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