Big Joule Theif

[Guruji]

Joule thief

Hi Slayer I did that and saw that video too it worked for me too.
The thing on the big toroid ferrite core I did it with just coil wire maybe that was wrong I had to use normal wires.
If it's because of that I will do it again it's ok.

@ Guruji


Here is a video on youtube that will help you with making a basic joule thief.

YouTube - Make a Joule Thief

Any npn transistor should work.
The led will only light one way to.
The long side of the led goes to the collector and the shorter end goes to the emitter or ground.

Once you get the basic JT going you will have no problem with the other circuits.

[lanenal]

A quick enhancement: By using sqrt(Vmax*Vmin), the relative error can be reduced by half. Since a discharging battery should drop in voltage, so Vmax is the runtime voltage at time zero, and Vmin is the runtime voltage at time of finish.

So, here are the things to measure:
1. starting rest voltage: Vrest0
2. starting runtime voltage: Vrun0
3. ending runtime voltage: Vrun1
4. ending rest voltage: Vrest1
5. runtime duration T

Suppose you know the resistor R. Let Vrun = sqrt(Vrun0*Vrun1). Then you have: Output(Vrest0,Vrest1)=T*Vrun*Vrun/R=T*Vrun0*Vrun1/R.
The maximum relative error of the estimated Output above should be (Vrun0/Vrun1 - 1), as the actual output is within [Vrun1*Vrun1/R, Vrun0*Vrun0/R].

@shlodo:
I would suggest a very rough way of doing this, without much expensive equipment, but the measurement might just be good enough given that the COP is significantly larger than 1.

Simply use a resistor (make sure it has the right wattage rating) as a load to the battery -- this way you can measure the voltage over the resistor every 15 min, then use an excel sheet to find out the work output -- the relative error in this method should be all right: given that the voltages over time (not the rest voltages, but the runtime voltages) does not change more than 10%, the maximum relative error would not exceed (1+10%)^2-1=21% (as Watt=V^2/R) even if you just measure runtime voltage only once. So, if your measured COP is something like 5, you know that the actual COP would be better than 5*(1-20%)/(1+20%) = 3.33 in the worst case (that is, output is off by 20% more and input energy is off by 20% less).

Another easy thing to do: with lead acid batteries, as long as you don't go below 50% of its capacity, the rest voltage changes are almost in linear relationship with its output work/energy, then your initial way of doing business is quite OK to my eyes. See my reference page in the previous post for lead acid batteries. So, if you are using lead acid batteries, play with them using my method above, to validate the linear relationship between voltage changes and energy output.

Hope that helps.

lanenal

[lanenal]

@shlodo:
I would suggest a very rough way of doing this, without much expensive equipment, but the measurement might just be good enough given that the COP is significantly larger than 1.

Simply use a resistor (make sure it has the right wattage rating) as a load to the battery -- this way you can measure the voltage over the resistor every 15 min, then use an excel sheet to find out the work output -- the relative error in this method should be all right: given that the voltages over time (not the rest voltages, but the runtime voltages) does not change more than 10%, the maximum relative error would not exceed (1+10%)^2-1=21% (as Watt=V^2/R) even if you just measure runtime voltage only once. So, if your measured COP is something like 5, you know that the actual COP would be better than 5*(1-20%)/(1+20%) = 3.33 in the worst case (that is, output is off by 20% more and input energy is off by 20% less).

Another easy thing to do: with lead acid batteries, as long as you don't go below 50% of its capacity, the rest voltage changes are almost in linear relationship with its output work/energy, then your initial way of doing business is quite OK to my eyes. See my reference page in the previous post for lead acid batteries. So, if you are using lead acid batteries, play with them using my method above, to validate the linear relationship between voltage changes and energy output.

Hope that helps.

lanenal

@lanenal
You are quite right, I couldnt believe I overlooked that! I realised not long after I made the post. But I have done many tests and they have all been positive. I will have to repeat the test with the same batteries...
I like your idea of finding emperical functions between V0 and V1. Ill have to look into it. Ive been trying to come up with a proper testing method.
I noticed Bedini uses a battery capacity meter which measures in % how many amp hours are in the batteries.
I saw a simple one at Jaycar.com.au for cheap too, but it reads in increments - 70-80%, 80-90% etc.. which I thought wasnt very accurate.

-Has anyone come across these??
-Can anyone out there suggest a proper testing method to compare the work we can do with the charge battery vs run battery??
-To accurately measure current consumed would i need to use a true RMS meter?

Any help would be much appreciated

Edit: explained the math in more details.
Edit: Sorry, seems I have to repost this to get my edit through.

[slayer007]

@ Guruji


Here is a video on youtube that will help you with making a basic joule thief.

YouTube - Make a Joule Thief

Any npn transistor should work.
The led will only light one way to.
The long side of the led goes to the collector and the shorter end goes to the emitter or ground.

Once you get the basic JT going you will have no problem with the other circuits.

[Guruji]

Thanks for helping guys. I did the 20 turn coil with two coils twisted then going around the torroid. Is that ok what it should be done?
Thanks

Guruji. if it's not working at all you might have your coil wired wrong.

Or I should say the wires comming off your coil might be wired in the wrong way.

Did you notice there should be a dot or a letter by one end of the coils in the circuit.

The dot should be the beginning of the coil the other side with no dot is the end.

You will notice the beginning of one coil will go with the end of the other coil.

So coil 1 with the dot should go to positive of the battery.
Then coil 2 with no dot should go to the positive also.
Coil 2 is the one that should have the 1K resistor on it.

I hope that will help I don't know if I explained to well.

EDIT.

Thanks for showing your circuit Tectalabyss it looks interesting.

[Xenomorph]

[slayer007]

Hi Slayer007 thanks for response. My JT did not work don't know why. I'm using a 2" torroid as you did but nothing happened.
I was using electrolyte polorized 10uf but nothing happened. Maybe it's not the capacitor problem it's the winding problem for JT.
Any help please?
Thanks

Guruji. if it's not working at all you might have your coil wired wrong.

Or I should say the wires comming off your coil might be wired in the wrong way.

Did you notice there should be a dot or a letter by one end of the coils in the circuit.

The dot should be the beginning of the coil the other side with no dot is the end.

You will notice the beginning of one coil will go with the end of the other coil.

So coil 1 with the dot should go to positive of the battery.
Then coil 2 with no dot should go to the positive also.
Coil 2 is the one that should have the 1K resistor on it.

I hope that will help I don't know if I explained to well.

EDIT.

Thanks for showing your circuit Tectalabyss it looks interesting.

[Guruji]

Big joule thief

Hi Slayer007 thanks for response. My JT did not work don't know why. I'm using a 2" torroid as you did but nothing happened.
I was using electrolyte polorized 10uf but nothing happened. Maybe it's not the capacitor problem it's the winding problem for JT.
Any help please?
Thanks

[slayer007]

HI Slayer I did a Joule thief with your setup about that 10uf cap is it electrolyte or ac capacitor?
Thanks for your circuit.

Hello Guruji

All the capacitors in the circuit are DC capacitors.

The ones I'm using are small electrolyte non polarized caps.

Also look at Lidmotors Inverted Joule Thief circuit.
He added a cap to the base of the 2n3055 and also one at the 1k resistor of the Joule Thief.

These will make a differance when running your JT at higher voltage.

[Guruji]

Slayer JT setup

HI Slayer I did a Joule thief with your setup about that 10uf cap is it electrolyte or ac capacitor?
Thanks for your circuit.

[shlodo]

Just a couple comments -

1. Heavily sulfated batteries (from my experience) tend to drop voltage under load and then slowly recover.
2. Adding distilled water to a completely dry gel battery seems to have gotten it back to life. Completely!

ABC

Yes I agree. a dead battery will see maybe 30V when u start charging and drop down as the impediance in the battery changes.
Normal Current charging will see an infinite resistance in the battery, but Radiant charge doesn't.

Ive brought 12V 7AH batteries back to life and they now accept normal charge

@Redeagle - thanks for the insight

@Xenomorph - I know that voltages doesnt tell us everything but there is definitley a lot of energy going in. In the test I did I did load tests and even charging for 30s-1minute would show an increase in voltage under load. A 20 Min charge will show clearly a lot more nergy in because the motor i use to load goes faster and harder.

-shlodo

[ABCStore]

Just a couple comments -

1. Heavily sulfated batteries (from my experience) tend to drop voltage under load and then slowly recover.
2. Adding distilled water to a completely dry gel battery seems to have gotten it back to life. Completely!

ABC

[redeagle]

@shlodo---don't forget that LABs also have a water cap effect. IF the charge pulse is switched of after the water was ionized into H+ and OH- but before the rest of electrolysis occurred, then they will recombine into H2O and release that energy that was sent to them to break them up. Thus it becomes a water fuel cell as well as a Lead/Sulfuric acid galvanic cell.

Conventional charging is only concerned about reversing the external electron flow and says to go ahead and electrolyze the water out of the battery. Between that and heat induced evaporation, you have to add water.

The bedini or joule thief style of charging reduces heat and the electrolyzing effect in the battery.

Relative conclusions:

1. Batteries charge faster

1. Batteries are less likely to be damage in charging

1. Sharp gradients pulsed across the battery organize the ions for maximum surface area. That explains the higher capacity

1. The water fuel cell effect explains why larger batteries are more efficient in charging

[Xenomorph]

@Shlodo,
i can only confirm what Lanelal said,
you have to measure the actual work output of the batteries (Rc discharge etc.) to make an accurate statement about the actual charge in the batteries.
Experimenting with a Tesla Switch my batteries show an average 1.2 V increase resting overnight, which i consider a ghost charge. Under load they will tend to return to the previous voltage. Cold charging is more diffuse than hot charging.
I have come to the conclusion that pure voltage measurements on batteries can in some cases be insufficient and misleading especially in regards to COP determinations.

[shlodo]

shlodo:

You are definitely heading toward the right direction, but I would like to give you some advice. Since the run and charge batteries are not of the same type, it is not very reliable to draw conclusions yet -- and even if they are of the same type, you may not draw convincing conclusions directly on voltage changes. It is better to do load tests to find out the relationship between the voltage change and the work output. For example, you can repeat your tests to find out the relationship between Energy output (in joules or Watt*Hours) and rest voltages (V0, V1), which are the initial and the final voltages. So, once you got an empirical function of Output(V0, V1), then you can easily calculate the COP. I don't know, maybe if you search on the net, you might find something ready for use.

Good luck!

lanenal

Edit:

Here is a reference about Lead=acid batteries:
Charging the lead-acid battery

@lanenal
You are quite right, I couldnt believe I overlooked that! I realised not long after I made the post. But I have done many tests and they have all been positive. I will have to repeat the test with the same batteries...
I like your idea of finding emperical functions between V0 and V1. Ill have to look into it. Ive been trying to come up with a proper testing method.
I noticed Bedini uses a battery capacity meter which measures in % how many amp hours are in the batteries.
I saw a simple one at Jaycar.com.au for cheap too, but it reads in increments - 70-80%, 80-90% etc.. which I thought wasnt very accurate.

-Has anyone come across these??
-Can anyone out there suggest a proper testing method to compare the work we can do with the charge battery vs run battery??
-To accurately measure current consumed would i need to use a true RMS meter?

Any help would be much appreciated

@Redeagle
I am noticing a definite change taking place in the batteries that im charging with the Potential charge. As I stated before, there is still some current coming out the back end. But I could use a larger pot to limit it.

"The trick is to engineer effects not causes. Charging batteries by moving lead ions instead of moving electrons only is a much more efficient method of charging and the spikes have been said to create a crystalized surface on the face of the lead plates which increases capacity." - Redeagle

That's interesting. Im noticing an inertia effect of the lead ions as you say. Its kind of like the battery keeps charging after u switch it off. Like it takes a while for them to be able to stop and flow the other way... anyway I heard Bedini talking about something like that once i dunno if its right..

@Tectalabyss
I like your setup, its pretty ambitious! Running a 555 > JT > JT > Ignition Coil > CFL + Charge! lol! Sounds like something Lidmotor would do! Its cool seeing people try seemingly "crazy" combinations.
Im wondering about your 555 timer.. Is it to reduce the on time of the CFL to reduce power? If you're interested in pure charging I would remove it.. But if its for a light source I can see why
Also, your results show an increase in the Source battery. Is this correct?
What kind of batteries are u using?

-shlodo