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Light LED with 4 nanoamps

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Zamboni

Wed Jun 18 2014, 08:09PM

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

Thank you all so much for your help.
Rich, you were right. Only experimentation will solve this issue. My end result was that I can't get a neon to light. I was able to get an LED to light...though so dimly that you need almost total darkness to see it.
I will update if I find a way.

Paul

klugesmith

Thu Jun 19 2014, 04:35AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

Paul,
You said your picoammeter indicates about 4 nA when connected directly to battery.
When the leads are reversed, do you observe similar magnitude with opposite sign?
This is analogous to checking the offset voltage of a sensitive voltmeter by reversing the leads.

What does the meter show when the neon lamp, or the LED, is in series?

What does it show when the 3.3 nF capacitor is in series? That ought to start out looking like the short circuit case. Then current should decrease as the capacitor charges, until we reach equilibrium between battery resistance and capacitor leakage. Time constant approx 150 Gohm * 3.3 nF = 495 seconds (no need to use a calculator or even pencil and paper for that one).

You could also monitor the capacitor voltage with your electrostatic voltmeter.

I've never personally worked with such low currents, but know that leakage over the surface of insulators (esp. if there's any ionic contamination) can be a problem.

ZakWolf

Thu Jun 19 2014, 06:56AM

Registered Member #3114 Joined: Sat Aug 14 2010, 08:33AM
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Posts: 608

darlington pair???

Zamboni

Thu Jun 19 2014, 02:34PM

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

Rich,
Thank you for your helpful suggestions. I had a couple of questions...

You said your picoammeter indicates about 4 nA when connected directly to battery.
When the leads are reversed, do you observe similar magnitude with opposite sign?
This is analogous to checking the offset voltage of a sensitive voltmeter by reversing the leads.

*****I was under the (perhaps mistaken) impression that you never hook up a battery with no load onto an ammeter. When I do my amp testing, it is always through a load. In my case, the load is my horizontal pendulum. You are suggesting that I hook up the battery directly to the pico-ammeter, with no load ?
I have switched the leads (by accident) and did get the negative, it was slightly different, but extremely close.

What does the meter show when the neon lamp, or the LED, is in series?

*****I have never tried that. I will give it a try and let you know. I found out that a neon bulb has about a megaohm of resistance, unfortunately, that is like a direct short to this type of battery. The LED would be a very interesting experiment. I will try it.

What does it show when the 3.3 nF capacitor is in series? That ought to start out looking like the short circuit case. Then current should decrease as the capacitor charges, until we reach equilibrium between battery resistance and capacitor leakage. Time constant approx 150 Gohm * 3.3 nF = 495 seconds (no need to use a calculator or even pencil and paper for that one).
You could also monitor the capacitor voltage with your electrostatic voltmeter.

*****I will try that also. Your figure of 495 seconds tracks very nicely with my experience. By the way...I would have needed paper, a calculator and the Internet to figure that out

I've never personally worked with such low currents, but know that leakage over the surface of insulators (esp. if there's any ionic contamination) can be a problem.

*****I can tell you that there are many interesting and very stressful problems with currents this low. Most people would have had the good sense to abandon this project years ago. I am just too stubborn (or maybe too stupid) to give up. Now that I have succeeded, the joy of watching my Devices operate makes it all seem worth it.

Paul

Newton Brawn

Fri Jun 20 2014, 03:47AM

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

Hi,

If the battery can charge the capacitor with 1500V, the energy storaged in the 0.0033 uF capacitor could be 0.5 . 0.0033 . 1500^2 . 0.000001 = 3.7mJ.

This energy may be switched by the pendulun, modified to act as a normaly open contact, that moves by the electrostatic force... Is the pendulun acting as a switch send the energy to a converter that provides low voltage to a neon or LED.

Such energy may fed a converter that provides low voltage to the neon Assuming that a led or neon consumption is in the range of of 0.005 watts, the lamp can glow 0.0037/0.005 = 0.7 seconds..

By construction, I believe that the pendulun has insulation very high, much better than solid state switches, neon lamp or LED...

It looks like a dream, please forgive-me.

EDIT "0.0033 uF"

Steve Conner

Fri Jun 20 2014, 10:19AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Buck converter with pendulum switch? Nice idea :)

Newton Brawn

Fri Jun 20 2014, 09:14PM

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

Hi Steve,

The idea is simple circuit where the battery charge the capacitor, and when the cap is full charge the eletrostatic force moves the pendulum (as a electroscope foil) that makes contact with the other terminal, discharging the cap to the converter.

The converter then start switching and feed the bulb for one second or less.

Someone may advise what is the minimal current that a NE2 or LED provides a visible flash ?

Regards

Antonio

Mon Jun 23 2014, 02:31AM

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

Some calculations to see if it's viable to light a LED using a DC/DC converter using the pendulum as switch:
Let's assume that a LED can produce a visible flash with 1 mA, a flash lasting for 0.1 s, and the voltage drop in the LED as 1.7 V. This requires the energy E=1.7*1e-3*0.1=170e-6 joules (170 uJ).
A capacitor storing this energy at 450 V shall have the capacitance C=170e-6/(0.5*450^2)=1.7e-9 farads (1.7 nF).
A pile producing 4 nA charges 1.7 nF to 450 V in t=1.7e-9*450/4e-9=191 seconds.
Too long time.
Reducing the time of the flash to 0.01 s reduces the energy to 17 uJ and the capacitor to 170 pF. This capacitor can be charged in 19.1 seconds. Less bad, but the flash would be not very visible.
The converter could be a buck circuit, using an inductor and a low leakage diode, powering directly the LED.

Bored Chemist

Mon Jun 23 2014, 07:46PM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

The easiest thing to look at would be to connect an LED (or several) in series with the neon lamp of a neon oscillator.
Compared to the non conducting state of the bulb, they will just look like a conductor. but, when the neon strikes they will carry the current so they should light up. The current spike will be quite large, but very brief so it might not damage the LED- in principle, a small resistor could limit the current to a mA or so. ( A megohm is a small resistance in this sort of circuit)
I strongly suspect that the neon will last longer than the LED in this configuration, but it would be interesting to find out if it works.

Another approach, albeit one that doesn't address the real question of what to do with "dud" batteries would be to include two LEDs between the capacitor plates and the moving contact so the swinging pendulum switches the current through the LED directly.

In any event, those a really cool items.

Newton Brawn

Fri Jul 04 2014, 11:08PM

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

Zamboni

See this thread

The coupling of the source and the lamp is made with a air gap. As the source has a capacitance, the discharge is made when there is enough energy in the source. and the energy storaged in the cap cam turn the lamp on.

Maybe you can try this arrangment to get some flash in the neon or led

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