Charging 12V Battery with 10V Power Supply

[CANGAS]

Guys, seems that you did not understand the point
You are talking about voltage  step up conversion, but I do  not

Mancha, in your hubris you do not care that some of us fail to have nice computer systems that are fast enough, so, we never (can) do youtube crap.

I have primitive old charging equipment that can charge to any voltage from any voltage. The most I have done is to charge 15.5 top charge from 1.55 top charge.

Is it possible to describe yo youtube crap in verbal English?

In other words, if you have something important to say, then, say it. Don't just tell us that we are to dumb to get it.

Thank you so much.


CANGAS 90

[Mancha]

to Cangas:
It was not hubris, but  could be crap as you said 
My comment  was addressed to TinesKoala and Tito. Not to everybody .
Because they are talking about voltage step up conversion.
What I am saying that  power input 10 V x 1A to charge the  12V battery is more efficient than 13.5 or 15.5  x 1 A.
It means  no step up or step down conversion, the same current from the  source  and the same current  to battery.
I hope that it is  more clear now.

[MileHigh]

Cangas:

You have to be kidding with your comments about "hubris" and "fast enough computer systems" and "crap."  You are making it sound like Milan is doing a demo with a $50,000 digital storage oscilloscope and is enjoying making the rest of us look poor.  Nothing could be further from the truth.  You can buy a Pentium 4 box for $30 that will play YouTube videos just fine.  An eight-year-old laptop will play YouTube videos just fine.

Meanwhile Milan makes videos in English which is not his first language and does just fine.  You need to rethink your statements.

Milan:

If your clip was called "Charging 12V battery with 10V RMS AC Power Supply" it would have been more appropriate.  You make comments that voltage doesn't really count and current is the primary factor for charging batteries.  This is not correct.  Current is always linked to voltage to calculate charging power, you cannot separate the two.  If people talk about "the best charging current" and similar things for batteries, the associated voltage is always implicit and understood to be there.  It still comes down to power x time = joules.

Note this is a fundamental property of Nature:  The "through variable" (current) times the "across variable" (voltage) = power.  You see it everywhere:  force x velocity, torque x angular velocity, water flow x water pressure, etc.

Your setup is indeed a quick and easy way to charge a battery from a 10 VAC power source.  I will be honest and state that I don't play with 12-volt batteries because I don't have a requirement to do that.  But if I had a requirement, I would do my research on smart chargers and buy a good quality smart battery charger.  I believe that they use inductive charging with current pulses and they make tests to see if the battery is fully charged and then they stop charging automatically.

You may want to explore constant-current LED power supplies.  You can string multiple batteries in series and the LED power supply will charge them at a constant current.  There are also dimmable constant-current LED power supplies where you can adjust the current with a potentiometer.  Naturally, you would want to stop the charging manually after the battery is fully charged.

MileHigh

[Mancha]

MileHigh,
thanks for your comment. I have been surprise  also with  some comment from some meber and did not react much.
Your suggestion  about tittle is very good and I will change it right now.
Regarding  what I said about current and joules .
You are right that we can not separate  voltage and cuurent, they goes together. What I said is not my statements , it is  from electro chemistry . Maybe I did not formulate it well  even I tried to explain it with single cell of electrolyser. 
Let is try it again. If  will look any law , any formula in electro chemistry . all of them are related to number of coulombs  or amps,. Because they (coulombs or amps) making deposit on cathode and oxidation on anode. So no voltage in all formulas  about this subject. But we need voltage of course. It tell us that we can use  any voltage range (from minimum needed voltage to above)
the work will be determined by passing current.
I will use electrolytic cell for example again.
By Faraday law we know that 96485 Coulombs is needed to liberate 1 gram of mas on electrode.

Imagine that we have cell with two stainless electrodes inside of  20 % NaOH solution.
Breaking voltage of NaOH solution is 1.69 V, plus overvoltage (supertension) of electrodes gives minimum operating voltage about 2 V. So we have MINIMUM input voltage which is 2 Volts .
The current depends of distance between electrodes, size of electrodes and type of solution and input voltage.
I already said that 2 V input will be minimum voltage input, under it, the cell will works like capacitor.
By  calculation  we can see that about 1.6 Amps (formula do not ask for voltage value) is needed to get one liter of Hydrogen /oxygen mixture for one hour.
So if we tune distance between two electrodes in cell to draw 1.6 amp at 2 Volts input we will get 1 liter og  gas for one hour.
It gives 2 x 1.6= 3.2  Joules or Watts hour.
Next .. if we distance electrodes each other  too much, the current will drops  for the same voltage input. Now we need to find which voltage input will gives 1.6 Amps  between electrodes.
Let say that this new voltage valuse is 3.4 V
We have  now 3.4 x 1.6= 5.44 Joules of Watts hour to get ONE liter of gas.
So  gas production is the SAME because it is related to Amps not to Joules
But there are losses in this case. Amps makes Joules Heat which we can calculate by formula E(heat)= R x I^2 
It tell us that distanced electrodes will produces  more heat than close electrodes, but gas production will be the same.
It is the same for the battery, because  the same process is happen inside of them.
So if we go back to video which I made , it tell us that lover voltage input will be more efficient than higher voltage input. Battery to be charged from  starting Capacity  stage to 100 % Capacity needs  X amount of coulombs . If you make it with 10V RMS will be lower energy input than if you make with 14.4 or 15.5 V input.
But as  I said in video , you need longer time to do that with 10 V RMS than with 14.4 V RMS.
I am using this charger if I am not in a hurry.

All The Best,

[MileHigh]

Milan:

Thank you for your reply.  I am just a layperson when it comes to battery electro-chemistry but I can make a few basic comments.

When you separate the electrodes it takes more voltage to sustain the same current to produce the gas like you state.  However, we know that the electrolyte itself acts like a resistor.  Therefore as the current flows through the electrolyte there is a voltage drop.  The farther apart the electrodes are, the greater the voltage drop assuming the same current flow.  In other words, this is just basic electronics concepts where the "resistor" is a fluid.

Of course there is the production of gas.  That is a chemical reaction with an associated chemical formula.  When you look at the formula, you see the molecular reaction to produce the gas.  Some reactions are endothermic, and some reactions are exothermic.  In some chemical formulas associated with the equal sign there is either a requirement of a certain number of electron-volts to make the reaction go forward, or in other cases the reaction produces a certain number of electron volts.  So the point is that some chemical equations require energy in the form of electron-volts, or they release energy in the form of electron volts.  Therefore, the chemical formulas account for the addition or generation of electrons, and the associated electron-volts for the reaction.  i.e.; they are accounting for current and voltage for the reaction, even if in common everyday usage people only speak about the current.

So, if for example you put four volts across the electrodes to produce gas, you may have a three volt drop due to the resistance of the electrolyte, and a one volt drop to produce the gas.  In another configuration it may only require three volts to produce the gas where two volts are the voltage drop through the electrolyte, and one volt remains to produce the gas.

There is no issue with respect to talking in terms of current only for purposes of discussing reactions to produce gas.  The voltage is implicit and can't be forgotten when you start to do a serious energy analysis.

MileHigh