Charge Pump Circuit

[dragon]

I realize some think that there is some detriment to charge and discharge simultaneously but I've never seen any problems caused by this. It would, however, become a problem if you were using HV spikes to charge a battery while connected to sensitive low voltage loads. It's not the case here. Also the series batteries are temporarily disconnected when the charge battery is in operation.

In reality we do it all the time - solar to battery to load - running accessories in the car while charging and maintaining the battery etc.

Quite often, with some systems, you wouldn't see a gain if the two were combined because the output of the charging portion is considerably less than the required input - in order to demonstrate the battery charging it has to be into a separate battery.

[RoliK]

Interesting thread. I want to make a replication. Because of the switch.
Is it possible to use a single change-over relay? Would be much easier.

Lg rolik

[SkyWatcher]

Hi dragon, thanks for sharing the information, i will definitely be trying the 2 battery version.
Like i was saying, Bedinis 1984 free energy generator, discharged and then charged the same battery, very similar to this, with a mechanical switch and seemed fine, though I recall they said something changed within the battery also.
Though his version had capacitors charged by a rotating generator and yet it still used capacitors dumped into the battery, just like we're doing here.

Hi Rolik, yes, wistiti posted a video here, where he used a relay driven by a 555 timer.
peace love light

[Wistiti]

Interesting thread. I want to make a replication. Because of the switch.
Is it possible to use a single change-over relay? Would be much easier.

Lg rolik

Hi Rolik. As Skywatcher say it is what I use for switching and it work really well!

All, here is a video of the last circuit Sky an Dragon have share:
https://youtu.be/5iybsIiUecE
My load is 300w and the starting voltage of the battery is:
B1: 12,59v
B2: 12,53v

Until now, the load is really slow to warm this little glass of water... for sure I will not have a tea from it!! maybe the voltage going to it is to small. I am thinking to adjust it with a dcdc converter as Sky have already suggest... (see the image)

Any comments are welcome!

[Wistiti]

Battery resting

Hey Sky!
When you leave your battery resting, do you disconnect everything or you keep them connect series....??
Thank you!

[SkyWatcher]

Hi wistiti, thanks for sharing the video.
I disconnect all connections from the batteries when resting.
I'm using tractor batteries, which can give high amps briefly, more efficiently than gel batteries, i would think.
Also, Can the relay efficiently pulse high amperage, I'm not so sure.

After resting more than 24 hours:
Battery A = 12.54 volts
Battery B = 12.52 volts
Battery C = 12.63 volts charge battery

Also, water has the highest heat holding capacity of all materials i'm aware of, which might explain why it's heating so slowly.
peace love light

[dragon]

Wistiti, have you measured the amperage moving through the heating element? I noticed in one of you other videos you have a clamp meter handy - set it on AC and clamp one leg of the element. That element, normally, would require somewhere in the range of 25 amps. Something in the range of 15 to 20 amps should be cycling through it with the pump circuit. If the amperage is low we need to find out why or what might be restricting it. I don't see a problem with the relay, I use a 30amp SPDT automotive relay in the demonstration video here and as long as it's running below its rated amperage it shouldn't be a problem.

SkyWatcher is right about the gel cells, they aren't as effective as the tractor battery - starting batteries have the ability to "give up" their energy very quickly... they can develop 100-300+ amps very quickly. Deep cycle batteries store a lot more energy but are designed to give it up at a slower rate. In any case, since you have the gel cells, they should be fine producing 5 to 10 amp bursts needed to drive the pump - I use them quite often in a lot of my tests even though they're a bit short winded ( 7ah ratings ), also, quite often because of their limited energy you can see if things are going good or bad much quicker than would be seen in a larger battery.

In any case, your looking to relieve any possible restrictions in the pump circuit to develop huge amp flows from the capacitive discharge.

[Wistiti]

Hi Dragon
yes I have already mesure the amp going to the load and it is in the range of 5 amp... That why it is not eating fast... The fact Sky and you point about the CCA of the "Starting battery" vs "Deepcycle" might be the case why I have so low amp reading...

About the relay switching, I also think mine is rate for 30amp. so I do not see why it should eventually fail.? Anyway it is still working fine after the few test I have done. We will see with the time.

Ciao!

[dragon]

Something a little more advanced running on 18650 li-ion cells using an IR2153 half bridge driver.

[dragon]

Wistiti, I don't believe your low amp problem is caused by the battery type. It has to be a resistance problem or voltage drop. If you connect the heating element directly to a charged cap it should instantly go to full amperage then taper off until the cap has been drained. Likewise, connecting to a battery - any battery should drive that element. I haven't tried the other arrangements charging a battery, I suspect there may be something there that is slowing the movement of current.

Additional Food for thought.... Let's say your using the double pass circuit ( fig 2 in the charge pump compare drawings ). The first cycle draws 25 amps from the battery through the load to charge the capacitor, the second cycle discharges the capacitor through the load. So you have a 25 amp load on the battery at a 50% duty cycle. If you run the system for 1 hour you are drawing 12.5 amp hours from the battery - 1/2 of the energy of a single pass circuit yet your still driving the load on every cycle. The charge pump circuit (fig 3) becomes more complicated because we are creating a voltage divider with the ability to double the current... a capacitive transformer of sorts. We then complicate it even more by charging the capacitors which increases the energy surges. Because of the voltage drop it becomes more important to use loads that are much lower in resistance ( hence my demonstrations of multiple loads in parallel ) or increase the working voltage.

Essentially, the single pass circuit drives a 300 watt load at a cost of 300 watts, the double pass circuit will drive a 300 watt load at a cost of 150 watts - the charge pump circuit does the same at a cost of 75 watts.

[Wistiti]

Thank you Dragon for the reply!
I will do more test in the next few day.

[SkyWatcher]

Hi all, have not been making any tests, not feeling too well, though my head is starting to clear up.

Thanks for sharing that food for thought dragon.
So what about the variation i've been testing, the drawing i posted.

If it works out the same, 300 watts output, while only using 150 watts, then what does charging the battery on the second pass contribute.
Maybe it works out the same, 75 watts input, if charging the battery is only 50% efficient.
Hmm, have to think more about this.
peace love light

[dragon]

Once again we jump to the conclusion that the output is higher than the input... I don't believe this is the case. When we charge a cap it takes a certain amount of time, during that time you have a peak energy exchange at the beginning and as the cap fills the voltage rises, reducing the potential difference and the current flow decreases - this gives us an average energy from start to finish.

Then we have the resistive load which is said to be "using" the energy. I tend to question the amount the resistive load actually uses or converts vs what is generally wasted during the neutralizing process of an energy source such as a battery. The ohms law and BTU standards don't always coincide with each other so in order to explain the phenomenon we introduce a COP value that allows us to acceptably bend the rules without really breaking them.

If you take a higher potential and dump it into a lower potential ( thus storing the energy not neutralizing it ) the resistance between them acts as a regulator of time - so then what is really lost?

If you take a fully charged capacitor of a given value and dump it into another capacitor of equal value that is not charged you loose 1/2 of the energy. But... if you charge both caps one higher than the other the losses are reduced drastically and the resistance in line plays a very small role in those losses while balancing the pair.

As an example I'll use the 1F caps - one charged to 24 volts and the other to 12 volts. The 24volt cap contains 288 joules the 12 volt cap 72 joules. If we connect them in parallel they will balance at around 18 volts each. Each now contain 162 joules. We lost 36 joules of energy during the balancing - with or without a resistive load this proves to be accurate. The exchange shows a 10% loss between them, considerably lower than the 50% loss of the first example - one charged one not charged.

Now consider as an (extreme) example the same caps charged to 112 volts and 100 volts using the same potential difference as the first example. We have 6272 joules in the higher potential and 5000 joules in the second a total of 11,272 joules stored. They would balance at around 106 volts each containing 5618 joules or a total of 11,236 joules, again a loss of 36 joules because of the 12 volt difference but the actual loss of energy is less than 1% - with or without the resistive load. 6272 joules - 5000 joules = an exchange of 1272 joules... is it possible to drive a 1200 watt resistive load with only a 1% loss?? This is what I'm trying to find out...

The circuit itself isn't anything special - not overunity nor even unity - simply a circuit that allows us a small view of the possibilities.... if your willing to look beyond what's on the surface and question how and why the accepted rules apply.

[bistander]

Cap energy

...
As an example I'll use the 1F caps - one charged to 24 volts and the other to 12 volts. The 24volt cap contains 288 joules the 12 volt cap 72 joules. If we connect them in parallel they will balance at around 18 volts each. Each now contain 162 joules. We lost 36 joules of energy during the balancing - with or without a resistive load this proves to be accurate. The exchange shows a 10% loss between them,
...

Hi dragon,

When I do the calculation of connecting two 1F caps (one charged to 12V and one at 24V), I get a resulting voltage of 18.98V. Assuming ideal capacitors and circuitry, there are no losses or missing energy. Assuming an average voltage when connecting the two caps is incorrect.

Regards,

bi

[ricards]

Then we have the resistive load which is said to be "using" the energy. I tend to question the amount the resistive load actually uses or converts vs what is generally wasted during the neutralizing process of an energy source such as a battery. The ohms law and BTU standards don't always coincide with each other so in order to explain the phenomenon we introduce a COP value that allows us to acceptably bend the rules without really breaking them.

If you take a higher potential and dump it into a lower potential ( thus storing the energy not neutralizing it ) the resistance between them acts as a regulator of time - so then what is really lost?

If you take a fully charged capacitor of a given value and dump it into another capacitor of equal value that is not charged you loose 1/2 of the energy. But... if you charge both caps one higher than the other the losses are reduced drastically and the resistance in line plays a very small role in those losses while balancing the pair.

As an example I'll use the 1F caps - one charged to 24 volts and the other to 12 volts. The 24volt cap contains 288 joules the 12 volt cap 72 joules. If we connect them in parallel they will balance at around 18 volts each. Each now contain 162 joules. We lost 36 joules of energy during the balancing - with or without a resistive load this proves to be accurate. The exchange shows a 10% loss between them, considerably lower than the 50% loss of the first example - one charged one not charged.

Now consider as an (extreme) example the same caps charged to 112 volts and 100 volts using the same potential difference as the first example. We have 6272 joules in the higher potential and 5000 joules in the second a total of 11,272 joules stored. They would balance at around 106 volts each containing 5618 joules or a total of 11,236 joules, again a loss of 36 joules because of the 12 volt difference but the actual loss of energy is less than 1% - with or without the resistive load. 6272 joules - 5000 joules = an exchange of 1272 joules... is it possible to drive a 1200 watt resistive load with only a 1% loss?? This is what I'm trying to find out...

The circuit itself isn't anything special - not overunity nor even unity - simply a circuit that allows us a small view of the possibilities.... if your willing to look beyond what's on the surface and question how and why the accepted rules apply.

I agree with dragon, this is not OU, you cannot get more energy than what you put in!! BUT!!! you can use the energy WISELY to do MORE "WORK" (which can also be to generate another potential to supply the source).

I guess this is what the circuit can teach us (the REAL TRUTH about energy).
its an experiment, a TEST.. to study "Energy" and its behavior more.
the "Usage" "Dissipation" "Consuming" "Transformation" "Conversion" of energy is it real or myth, not just believe what we are taught or what we learned..
Been doing some test lately...
..to see how much I can use the same "Energy",
..to find out If your really just "Transforming" energy into other forms,
.. to see If the "Resistance" really was dissipating "Energy"