To be deleted...

[nul-points]

@ nul-point and itsu:

May be you have two super caps instead of the two rechargeable batteries. The tests would be much shorter.

I attach the photo of a suitable super cap, but any similar type would work, more or less capacitance, even the 5 Volt types would be good.

The 1 F type super caps are low cost.

With a laboratory power supply or with a 1.5 V battery the super caps should be charged to 1.5 Volt before the test.

Greetings, Conrad

Conrad - you're a Genius!!

When we next meet at the Bierkeller, i owe you a large beer ...and a big kiss!  ;-) 

The supercaps are a perfect demonstration for two reasons:-

a)  you can run 2 tests with the same starting energy each time, one with the full circuit, swapping i/p & o/p and one with just the regular i/p, no o/p storage;

b) you can show clearly that there is no extra charge in the output cap, when you start

and of course the 2 tests only took a total of about 30 minutes, as you implied

I have to go to band practice now, but i hope to upload the 2 graphs later, if not now

I'll explain the details later, for sure, but the graph data shows that the same i/p energy only illuminates the LED for approximately half the time without the output storage swap system - approx 6 minutes compared to approx 12.5 minutes using the full circuit

More later, all the best
np
 

So - here is the result of your suggestion

[conradelektro]

The supercaps are a perfect demonstration for two reasons:-

Nul-Points you give me too much credit.

It was repeatedly suggested by multiple persons in the past to demonstrate OU with a big electrolytic capacitor or a super cap instead of using batteries, because one can see the result within minutes (instead of many hours or even days).

See for instance https://www.youtube.com/watch?v=JNBi6qoW5SI&feature=youtu.be, Laserhacker runs a pulse motor on a 1000 µF capacitor.

The capacitance of the super cap (1 F, 10 F or 100 F) will define the duration of the test. The smaller the capacitance the shorter the test. Things happen too fast if the test only lasts a few seconds. About two to  five minutes would be practical. Of course one usually does not have man different super caps at home or does nor want to spend too much money buying many different ones.

A OU set up must be able to run from a relatively small super cap (may be 10 F) or one suspects that the OU feature is in the battery (which would be chemistry not electronics).

Many experimenters can not imagine that a Joule Thief type circuit with one LED can run for months with a AA battery. I have a Joule Thief which runs up to 6 months with an AA battery. Of course the red LED is not very bright, but sufficient during the dark to be seen clearly from 10 meters away (in order to give directions).

I am looking forward to your explanations of the graphs.

Greetings, Conrad

[nul-points]

No, i would have continued with my 1 battery, 1 supercap testing, Conrad, if you hadn't suggested replacing both batteries with supercaps

I agree that there are certain aspects of a system which it is easier to quantify using just 'charge', rather than chemistry - and as i was thinking about the possibility of extending the work of which one battery was capable, i moved away from my initial curiosity about the behaviour of the circuit with capacitor storage on both i/p & o/p

So  - i'm very grateful that you made that suggestion ...and that i was also able to get the results so promptly because of it

The upper graph shows the terminal voltages of the i/p & output capacitors (both 1F); in this case the cathode of the LED 'load' is connected to circuit ground, or common, instead of the o/p capacitor, so the o/p current is only used to illuminate the LED - no storage occurs, so the output & input storage cannot be swapped as the test progresses

The blue trace shows the input capacitor get charged to approx 4.9V and as it discharges to 4.6V i connect it to the input of the circuit; the oscillator starts; the LED illuminates; and the input capacitor discharges down to approx 0.6V

At the point when the LED is no longer illuminated, i disconnect the input capacitor and start discharging it, to identify tbe end time of the test run - the duration of LED illumination, without the full circuit, is approx 6 minutes

The lower graph, then, shows the results achieved when the whole circuit is in operation  - the output is also a 1F capacitor, like the input, and the cathode of tbe LED is connected to its positive terminal

As before, the input capacitor gets charged to approx 4.9V and then connected to the input of the circuit when the voltage discharges to 4.6V; the oscillator starts; the LED illuminates; and the input capacitor discharges down to approx 0.42V, whilst the same current passing through the LED charges the output capacitor

At the point when the LED is no longer illuminated, approx 6 minutes after the start of the test run, i toggle switch S1 which swaps the input & output capacitors - the LED now re-illuminates, powered by the charge which has just been received by what was the output capacitor

This second amount of 'charge' illuminates the LED for an extra 4 minutes approximately, and whilst it does this the LED current is now charging the new output capacitor

When the LED is again no longer illuminated, i toggle S1 once more and the capacitors are now back in their original connection; the input capacitor has received some extra charge during the previous swap, the LED re-illuminates and the input capacitor discharges down to 0.6V approx

At this stage, the LED illuminates for a further 2.5 minutes approx  - the total duration of LED illumination, therefore, with this circuit arrangement is 6 + 4 + 2.5 = 12.5 minutes approx

So, this flashlight circuit is capable of extending the useful 'work' done by an input charge of up to 100% approx

Interesting!

Tests continue
np

[conradelektro]

So, this flashlight circuit is capable of extending the useful 'work' done by an input charge of up to 100% approx

Interesting!

Nul-Points, now I understand. Really nice this second graph, it shows the story very well. And it looks and sounds plausible. I have never seen before such a clear demonstration of a "battery (or cap) swapping circuit".


I would like to know:

Does the LED shine brighter in case "the cathode of the LED is connected to circuit ground, instead of the o/p capacitor" (first graph) in comparison to the the situation when the "whole circuit is in operation (second graph)?


I suspect:

You could make the circuit (first graph or second graph) run longer by dimming the LED a bit by setting the variable resistor VR1 to a higher resistance. And on the other hand, you could make it run shorter by letting the LED shine a bit brighter by setting the variable resistor VR1 to a lower resistance. All of course only works within certain limits.


Under normal circumstances I would replicate your "flash-light", but at the moment I am very busy with "electrostatic experiments" based on the electrophorus principle https://en.wikipedia.org/wiki/Electrophorus, which is known since 1762. The electrophorus is the intellectual root of all "electrostatic machines". The best web site about electrostatic machines I found is this one: http://www.coe.ufrj.br/~acmq/electrostatic.html .

Greetings, Conrad

[nul-points]

...
I would like to know:

Does the LED shine brighter in case "the cathode of the LED is connected to circuit ground, instead of the o/p capacitor" (first graph) in comparison to the the situation when the "whole circuit is in operation (second graph)?

I suspect:

You could make the circuit (first graph or second graph) run longer by dimming the LED a bit by setting the variable resistor VR1 to a higher resistance. And on the other hand, you could make it run shorter by letting the LED shine a bit brighter by setting the variable resistor VR1 to a lower resistance. All of course only works within certain limits.


...at the moment I am very busy with "electrostatic experiments" based on the electrophorus principle https://en.wikipedia.org/wiki/Electrophorus, which is known since 1762. The electrophorus is the intellectual root of all "electrostatic machines". The best web site about electrostatic machines I found is this one: http://www.coe.ufrj.br/~acmq/electrostatic.html .
...

Good question Conrad, difficult to answer objectively!  I'll set up a little DIY illumination tester and get back to you on that


This circuit, in common with most oscillators i suspect, will discharge a battery or capacitor source more or less quickly depending on the drive conditions of the active device - in this case i attempted to adjust the drive to produce the brightest output (since it is nominally a 'flashlight' application), so in one respect the operating conditions are 'optimum', rather than arbitrary

I think that the reported behaviour probably addresses your theory: both tests complete a discharge of the input capacitor (to the point of extinction of the LED) in a period of 6 seconds - i think that this tells us that both tests are having the sane effect on the input energy supply

The difference between the two tests is that in the 2nd test the energy being drawn from the input is diverted into another store and can then be re-used (and this process is repeated a couple of times)

Your own investigations sound interesting - i've seen mention of 'electrophorus' a few times recently so i will extend  my education and read up from those links which you provide - many thanks!

I'll report back if i can get a quantitive answer for you, about the relative illumination levels betwen the 2 tests

All the best
np