# Free Energy | searching for free energy and discussing free energy

## Solid States Devices => solid state devices => Topic started by: dieter on February 14, 2014, 04:48:52 PM

Post by: dieter on February 14, 2014, 04:48:52 PM
Hi. As an amateur experimenter I have a simple question:
When I take two electrolyt caps, fill one to eg. 16 vdc, shorten the other so its empty, then I connect them +- and -+, to eachother, I get 8 v in both.

Since the energy in a cap is farad*v*v (pls correct me if I'm wrong), this means 50% of the energy has vanished. Where has it gone?
Title: Re: Silly question about capacitors
Post by: MarkE on February 14, 2014, 05:04:43 PM
Hi. As an amateur experimenter I have a simple question:
When I take two electrolyt caps, fill one to eg. 16 vdc, shorten the other so its empty, then I connect them +- and -+, to eachother, I get 8 v in both.

Since the energy in a cap is farad*v*v (pls correct me if I'm wrong), this means 50% of the energy has vanished. Where has it gone?
Almost of it goes into heat, the rest which is quite small goes into radiation.  Model each capacitor with a small series resistance, ESR.   All capacitors exhibit some ESR, and then of course the wiring has some resistance.  Most of the lost energy heats the wiring and ESR resistance.
Title: Re: Silly question about capacitors
Post by: vasik041 on February 14, 2014, 05:20:38 PM
Quote
Almost of it goes into heat, the rest which is quite small goes into radiation.

I agree.

Here an illustration of the process.
Title: Re: Silly question about capacitors
Post by: Cadman on February 14, 2014, 06:17:36 PM

@dieter
Since you asked, the formula for caps energy is joules = 0.5 * voltage^2 * C

Title: Re: Silly question about capacitors
Post by: Farmhand on February 14, 2014, 07:34:59 PM
I think there is some element of impedance matching involved, for instance if we use a very low resistance inductor in the positive line and no switch in the current loop between the capacitors as shown in my drawing, as well as we are doing it for a reason so there is a load involved as well. Then the efficiency of the transfer can be better I think.

So the idea is the system is charged to 10 volts to begin with, the inductior is considered ideal and we switch the RL load in for 1.5 mS only once then calculate the power dissipated by the resistor as legitimate load power and calculate or measure the voltage on the two capacitors and calculate the energy missing.
Anyone want to put that in a simulator ?

Poynt99 did a good paper on this kind of thing. But I cannot find it.  :-[

..

Title: Re: Silly question about capacitors
Post by: dieter on February 14, 2014, 08:01:56 PM
Cadman, so I was right with F*v*v, the *0.5 just uses the half, but the relation is the same.

Now, you guys say that, like a leathal strike of Watts within a cap is lost in heat creation? I never notized a cap getting hot by unloading it. Do they?

Even more absurd: when you add an LED, so cap A has to flow trough it to go to cap B, you still get 8v in both, but the LED was lit.

When the cap loses 50% when unloaded, what happens during loading? Or, does both lose 25% and when 2 caps are used as described, the loss adds to 50%?

Anyway it seems, using caps can be a very wasteful factor. At least, if you don't take care of what's in the previous posting. Well I  thought this is the proof for the conservation of energy being wrong  ;D   . Which would have been pretty weird.
Title: Re: Silly question about capacitors
Post by: Farmhand on February 14, 2014, 08:13:30 PM
It's good to know the formula for this and other things but I find it is simpler, faster and more accurate to just use a calculator that tells us the energy in a capacitor. When installing be sure to uncheck any toolbars ect. that the installation package may offer, unless you want them of course.

All you need to know is the voltage across the capacitor and it's capacitance, then the calculator will tell you the Charge in Coulombs and the Energy in and Joules.

..
Title: Re: Silly question about capacitors
Post by: mscoffman on February 14, 2014, 08:25:19 PM
Yes everyone is correct. When you connect two capacitors together in the real world one needs to reduce the complex
impedance to a resistive impedance. This real impedance will be some number because pure capacitive impedance
is only a theoretical construct. So even if this resistive component is a decimal with a number of zeroes after it is likely
that both identical capacitors have the same amount. The means that half the energy has been left behind dissipated in
the source impedance and half is dissipated in the destination impedance. The residual voltage is what happens when
capacitive impedance "dissipates" energy. Inductive impedance does the same but exchange word "current" for "voltage"
in the previous discussion.

:S:MarkSCoffman
Title: Re: Silly question about capacitors
Post by: TinselKoala on February 14, 2014, 11:31:38 PM
::)
Title: Re: Silly question about capacitors
Post by: Magluvin on February 15, 2014, 12:57:55 AM
Hi. As an amateur experimenter I have a simple question:
When I take two electrolyt caps, fill one to eg. 16 vdc, shorten the other so its empty, then I connect them +- and -+, to eachother, I get 8 v in both.

Since the energy in a cap is farad*v*v (pls correct me if I'm wrong), this means 50% of the energy has vanished. Where has it gone?

Lets say we have 2 equal size air tanks, 1 filled to 300psi and the other with 0 psi.

We connect the 2 together with a hose and we open the valves on the tanks. We should end up with 150psi in each. But we lost 50% of the total energy stored the first 300psi tank by the simple transfer, expansion and equalization of the 2 tanks.  Now, each tank contains half of the total air pressure that was in the 300psi tank. How did we lose 50%? The total amount of air(electrical charge) is still there in the 2 tanks, but we lost pressure. ???

We lost it stupidly. ;)   It wasnt lost due to resistance. It wasnt lost due to air leaking out. We lost it by reducing the pressure that was in the first tank and expanding it into 2 cambers, each the size of the first full tank.

Now, if we had a pump motor that uses air pressure to provide mechanical output(turbine may be a bit lossy due to blow by of the turbine fan) and we connect the air motor inline with the hose between the 2 tanks(1 full, 1 empty), then we are capturing that lost 50%(or so) of the total energy of the first tank during the transfer. So now we are not losing that energy stupidly. ;)

And in the end, we still have 150 psi in each tank, just like when we transfer the pressure from 1 tank to the other directly without the air powered motor, but without the huge stupid loss. ;)

So I dont believe that resistance is the issue with losing 50% of the initial energy of the full tank equalizing to the second tank.  Its just the fact that we reduced the energy of 300psi air pressure of 1 tank and expanded it into a tank that is twice as large, without doing anything with the actions of the transfer. And once the 2 tanks are equalized, there is no way that the 2 tanks of 150psi can be used to compress the total amount of air back into 1 tank at 300psi. But, if we mounted a large flywheel to the air motor, the energy that was lost in in the direct connect example is stored in the flywheel, and once the 2 tanks become equal(150psi), the flywheel is fully charged and would continue to pump the originally empty tank, pretty close to what was in the originally full tank, minus losses in the mechanical motor and resistances of the air being pushed through the hoses and fittings, etc.

Now with the caps, we can use an inductor as the air motor and get most all the energy of the full cap into the empty cap. Using a diode(check valve) to keep the previously empty cap, now nearly full, from going back to the first tank. ;) A bit of loss in the diode due to voltage drop inherent in the diode.

Mags
Title: Re: Silly question about capacitors
Post by: poynt99 on February 15, 2014, 01:45:28 AM
How about one more time.  :)

See the attachment for more info on capacitor energy transfer and how to maximize it.
Title: Re: Silly question about capacitors
Post by: Farmhand on February 15, 2014, 02:10:37 AM
Thanks Poynt99, I have that filed appropriately now so i can link it in future.

....

To anyone, does anyone have the time to simulate my circuit example drawing in Falstad simulator or something ? I am trying to learn LT spice, but I don't have a lot of time lately to learn new stuff with my medical issues and chores ect. If I had time to learn it I would simulate it myself. All we need is a DC one shot pulse to the mosfet or switch, the pulse width can vary, but should not allow full current through the series inductor, the 10 Ohm load would limit the current to 1 Ampere max anyway with the circuit pre-charged to 10 volts.

Maybe a longer pulse than 1.5 mS might work better or more or less load resistance to vary peak current

..

Mags, pretty good explanation, the inductor acts as a low loss load which when the switch is opened discharges it's stored energy into the smaller capacitor and raises it's voltage above that of the supply capacitor, that can happen due to the inductors properties.

Cheers
Title: Re: Silly question about capacitors
Post by: Magluvin on February 15, 2014, 02:11:59 AM
To expand on it a bit further, how can we get an actual 50% of the 300psi(total energy) into each tank? Meaning, the 2 tanks together equal round abouts the amount of energy of the original 300psi tank.

We would need the 2 tanks to round about 212psi each, and used together can do as much work as the single 300psi tank. Even though the pressures are different, 300psi(1 tank) and 212psi(2 tanks), the same amount of work can be done with each. The 2 tanks of 212psi may seem like less than 300psi, but we have more total air in the 2 tanks than the 1 tank at 300psi. Like the difference between a 12v batt at 12ah and a 24v batt at 6ah.

But how can this be done?   ??? ;)

We have our 2 tanks, 1 full(300psi) and the other empty, and we have our air motor with flywheel setup. Now we open the valves. The pressure from the full tanks drives the motor/flywheel and begins to pressurize the empty tank. But, we have an added redirect valve between the source tank and the air motor. When we see the source tank get down to 212psi, we turn the added valve. This valve closes off the source tank and opens the input to the air motor to the open air instead. Now, at this point, we stopped the source tank from depressurizing and it now contains 1/2 the energy of the original 300psi. But our flywheel is still going, and pumping the originally empty tank with outside air. And if there wasnt losses in the motor and such, it would be able to pump the second tank to right about 212psi. Now we have 2 tanks that have the same energy potential of the original single tank that was 300psi.   ;)   But one thing has changed. There is more total air in the 2 tanks than was in the single 300psi tank, because we pulled in air from the outside once we turned our newly added redirect valve to do so.

With caps, in this situation, we are basically doing the same thing, except once we turn our redirecting valve, the flywheel(inductor) is just transferring from one plate of the cap to the other.
So my analogy would be better to have 2 air tanks for the source and 2 tanks for the empty cap. The problem would be that the source tanks would need to be 1 full at 150psi and the other source tank at a negative 150 psi to have a clearer comparison between the 2 ideas. But the single tank to single tank gives a basic and simple understanding. ;)

Mags
Title: Re: Silly question about capacitors
Post by: Magluvin on February 15, 2014, 02:32:40 AM

Mags, pretty good explanation, the inductor acts as a low loss load which when the switch is opened discharges it's stored energy into the smaller capacitor and raises it's voltage above that of the supply capacitor, that can happen due to the inductors properties.

Hey Farmhand

The inductor/flywheel is a way of capturing energy that is lost because of a stupid way of losing it. ;)
Think. Do you believe that the energy lost in the direct transfer between 2 caps can be captured by storing the heat created during that transfer? Or by capturing the 'radiation' ? Maybe a little will be recovered, but definitely not all or near 50%, no way no how. Lets say the air tank analogy produces heat in the hose.  Does that in any way explain how we took 1 tank full, directly to 1 tank empty till they are equal, has anything to do with the 50% loss in initial source energy?? ;) ;D

There is something a miss there with the ideals of the so called losses of direct cap to cap transfer. I have some new ideas on this lately that may blow your mind, but I will reveal them when I have it all down pat myself. ;)   It took me a while the get this far to be able to explain what I have here.

Mags
Title: Re: Silly question about capacitors
Post by: Marsing on February 15, 2014, 03:00:25 AM
hi all..  nice explanation

what is formula the time needed to transfer from c1( full cap ) to c2 ( empty cap)?
according to VAsic041 pict, i took about 200ms
Title: Re: Silly question about capacitors
Post by: MarkE on February 15, 2014, 03:23:26 AM
Lets say we have 2 equal size air tanks, 1 filled to 300psi and the other with 0 psi.

We connect the 2 together with a hose and we open the valves on the tanks. We should end up with 150psi in each. But we lost 50% of the total energy stored the first 300psi tank by the simple transfer, expansion and equalization of the 2 tanks.  Now, each tank contains half of the total air pressure that was in the 300psi tank. How did we lose 50%? The total amount of air(electrical charge) is still there in the 2 tanks, but we lost pressure. ???

We lost it stupidly. ;)   It wasnt lost due to resistance. It wasnt lost due to air leaking out. We lost it by reducing the pressure that was in the first tank and expanding it into 2 cambers, each the size of the first full tank.

Now, if we had a pump motor that uses air pressure to provide mechanical output(turbine may be a bit lossy due to blow by of the turbine fan) and we connect the air motor inline with the hose between the 2 tanks(1 full, 1 empty), then we are capturing that lost 50%(or so) of the total energy of the first tank during the transfer. So now we are not losing that energy stupidly. ;)

And in the end, we still have 150 psi in each tank, just like when we transfer the pressure from 1 tank to the other directly without the air powered motor, but without the huge stupid loss. ;)

So I dont believe that resistance is the issue with losing 50% of the initial energy of the full tank equalizing to the second tank.  Its just the fact that we reduced the energy of 300psi air pressure of 1 tank and expanded it into a tank that is twice as large, without doing anything with the actions of the transfer. And once the 2 tanks are equalized, there is no way that the 2 tanks of 150psi can be used to compress the total amount of air back into 1 tank at 300psi. But, if we mounted a large flywheel to the air motor, the energy that was lost in in the direct connect example is stored in the flywheel, and once the 2 tanks become equal(150psi), the flywheel is fully charged and would continue to pump the originally empty tank, pretty close to what was in the originally full tank, minus losses in the mechanical motor and resistances of the air being pushed through the hoses and fittings, etc.

Now with the caps, we can use an inductor as the air motor and get most all the energy of the full cap into the empty cap. Using a diode(check valve) to keep the previously empty cap, now nearly full, from going back to the first tank. ;) A bit of loss in the diode due to voltage drop inherent in the diode.

Mags
Magluvin, what you propose will not fix the problem.  If we were to scale the capacitors or tanks so that they are not equal then we can greatly improve the efficiency going one way.  That is how charge pump power converters operate with relatively good efficiency for small loads.
Title: Re: Silly question about capacitors
Post by: Farmhand on February 15, 2014, 03:40:59 AM
There is no loss of "charge".

Did no one look at the "Charge" in "Coulombs", If we take 2 x 10 000 uF capacitors and we have one at 16 volts the electronics assistant tells me it has 160 Millicoulombs of charge.
Now if I was to connect the two caps together and equalize the charge which results in both having 8 volts across them then the electronics assistant tells me they will both have 80 Millicoulombs of "charge" in them so added together there is still 160 Millicoulombs of charge, so no loss.. Where is the loss. The energy in a capacitor is really only "potential energy". I think the capacitor actually stores "charge", not "energy".

The voltage increase on a capacitor has a non linear effect on the "potential energy". meaning from 0 to 8 volts is less "potential energy" than from 8 volts to 16 volts, the voltage increase is the same but the potential energy increase is not.

Cheers
Title: Re: Silly question about capacitors
Post by: Magluvin on February 15, 2014, 03:47:11 AM
Magluvin, what you propose will not fix the problem.  If we were to scale the capacitors or tanks so that they are not equal then we can greatly improve the efficiency going one way.  That is how charge pump power converters operate with relatively good efficiency for small loads.

Hey Mark

What problem are we talking about?  I thought the issue was described in the first post about the 50% loss when connecting a full cap to an empty cap. ;)   From what I understand, the size difference in caps still have losses due to the same issues, just different ratios of losses depending on the size differences in capacitors used. Like if we had a 10uf cap with 1000v and dumped it into a 10,000uf cap, the loss would be huge and beyond 50%, but the other way around the loss would be less. ;) What I am claiming is, I dont believe that the loss from cap to cap directly is related to resistance losses during direct transfer.  What hit me hard when I first encountered this problem was the fact that there is a 50% loss, no matter what the resistance is, whether it be high resistance or extremely low resistance, the loss is the same. It seems odd, no?  For some reason, resistance is the claimed cause of the loss, but there is no variance in loss with different values of resistance. Where in other circuitry, there is a difference when there is a difference. ;)

For example, say we have 2 caps that had no resistance, zero resistance, zero losses of any kind. When we connect them, one charged and one empty, and lets get rid of inductance for this for now to eliminate oscillations,  do you believe that when we discharge a 100uf cap with 100v into an empty 100uf cap, that there would be more than 50v in each cap????? ;)

Gota go pick up my pizza.

Mags
Title: Re: Silly question about capacitors
Post by: MarkE on February 15, 2014, 03:57:18 AM
Hey Mark

What problem are we talking about?  I thought the issue was described in the first post about the 50% loss when connecting a full cap to an empty cap. ;)   From what I understand, the size difference in caps still have losses due to the same issues, just different ratios of losses depending on the size differences in capacitors used. Like if we had a 10uf cap with 1000v and dumped it into a 10,000uf cap, the loss would be huge and beyond 50%, but the other way around the loss would be less. ;) What I am claiming is, I dont believe that the loss from cap to cap directly is related to resistance losses during direct transfer.  What hit me hard when I first encountered this problem was the fact that there is a 50% loss, no matter what the resistance is, whether it be high resistance or extremely low resistance, the loss is the same. It seems odd, no?  For some reason, resistance is the claimed cause of the loss, but there is no variance in loss with different values of resistance. Where in other circuitry, there is a difference when there is a difference. ;)

For example, say we have 2 caps that had no resistance, zero resistance, zero losses of any kind. When we connect them, one charged and one empty, and lets get rid of inductance for this for now to eliminate oscillations,  do you believe that when we discharge a 100uf cap with 100v into an empty 100uf cap, that there would be more than 50v in each cap????? ;)

Gota go pick up my pizza.

Mags
Magluvin, take a 100uF cap.  Charge it to 10V.  Connect that to a 1uF cap.  Calculate the net energy.  It is close to the original.  The worst case is always when the capacitors are the same value.
Title: Re: Silly question about capacitors
Post by: Marsing on February 15, 2014, 04:02:22 AM

Did no one look at the "Charge" in "Coulombs", If we take 2 x 10 000 uF capacitors and we have one at 16 volts the electronics assistant tells me it has 160 Millicoulombs of charge.
Now if I was to connect the two caps together and equalize the charge which results in both having 8 volts across them then the electronics assistant tells me they will both have 80 Millicoulombs of "charge" in them so added together there is still 160 Millicoulombs of charge, so no loss.. Where is the loss. The energy in a capacitor is really only "potential energy". I think the capacitor actually stores "charge", not "energy".

ok,  but you need energy to put back  only one cap  full of charge
the lose here is about to back to first condition.

btw, tesla have a great way to take advantages of this
Title: Re: Silly question about capacitors
Post by: Magluvin on February 15, 2014, 04:51:19 AM
Magluvin, take a 100uF cap.  Charge it to 10V.  Connect that to a 1uF cap.  Calculate the net energy.  It is close to the original.  The worst case is always when the capacitors are the same value.

Ugh, just wrote a long post and my laptop went for a shutdown.

Your right. And I was, well, half right. lol

"From what I understand, the size difference in caps still have losses due to the same issues, just different ratios of losses depending on the size differences in capacitors used. Like if we had a 10uf cap with 1000v and dumped it into a 10,000uf cap, the loss would be huge and beyond 50%, but the other way around the loss would be less"

Im trying to figure out what I was thinking when I wrote that. Was running late to pick up the pizza. I had gone over these things with Poynt a while back. I get what you are saying.

Anyways, thanks for catching that. ;D

Mags

Title: Re: Silly question about capacitors
Post by: Farmhand on February 15, 2014, 04:51:32 AM
ok,  but you need energy to put back  only one cap  full of charge
the lose here is about to back to first condition.

btw, tesla have a great way to take advantages of this

I guess the question is why ? Why would you create the second condition if you want to keep the first ? You've already done the work to put the potential energy in the first capacitor, why not just keep it there if that's where you want it ?

As I said the "energy in a capacitor" thing is really just "potential energy" it can be wasted, or used wisely.

Same thing with two 10000 liter water tanks side by side, one empty one full, if you let the full one equalize with the empty one, the same amount of water is there. None wasted. If you want all the water in the first tank just leave it there.

With no reason to do a thing it is always a loss. It's like making a big high stack of bricks then leveling it out into two shorter stacks it takes energy to make the big stack into the two small stacks. It cannot do itself, something must do it.

Cheers

P.S. Imagine making one tall stack of bricks by hand then changing that to two shorter stacks, does that take no energy to do ? No you must expend energy to get from state one to state two. Then if you want to make the two stacks back into one tall one you must expend more energy again. It's really quite simple. The amount of brick remains the same.

Also the higher the stack gets the harder it becomes to add another brick to the top of the stack.  ;) It depends on how you take them down and if you store the energy or not when you do it as to how much loss is involved.

...
Title: Re: Silly question about capacitors
Post by: Marsing on February 15, 2014, 04:57:58 AM

I guess the question is why ? Why would you create the second condition if you want to keep the first ? You've already done the work to put the potential energy in the first capacitor, why not just keep it there if that's where you want it ?

hey...    PM , oscillation .. etc
this is not about why,    But What and how

Did you forget this  forum name ?
;) ;) ;) ;) ;)
Title: Re: Silly question about capacitors
Post by: TinselKoala on February 15, 2014, 04:59:43 AM
There is no loss of "charge".

Did no one look at the "Charge" in "Coulombs", If we take 2 x 10 000 uF capacitors and we have one at 16 volts the electronics assistant tells me it has 160 Millicoulombs of charge.
Now if I was to connect the two caps together and equalize the charge which results in both having 8 volts across them then the electronics assistant tells me they will both have 80 Millicoulombs of "charge" in them so added together there is still 160 Millicoulombs of charge, so no loss.. Where is the loss. The energy in a capacitor is really only "potential energy". I think the capacitor actually stores "charge", not "energy".

The voltage increase on a capacitor has a non linear effect on the "potential energy". meaning from 0 to 8 volts is less "potential energy" than from 8 volts to 16 volts, the voltage increase is the same but the potential energy increase is not.

Cheers

Charge is indeed the conserved quantity here, and capacitors do "store" charge... and just like springs or containers of compressed air, the more charge you put into a capacitor the more _energy_ it takes to do it and the more energy will be returned when you let the charge back out. To say that this is not storing energy is playing with words, I think.
Also, this is real energy, not potential energy we are talking about. The formula for the energy on the capacitor is, as we have seen, 1/2(CV2) in Joules, where C is capacitance in Farads and V is voltage in Volts. Note that this is the same form as the expression for Kinetic Energy of a moving mass: 1/2(mv2) where m is mass and v is velocity. It's not the "potential energy" form like mgh.
The relationship between energy on a cap and the voltage is, as the formula shows, quadratic, not "linear".  Voltage is "charge pressure". You can think of individual negative charges all trying to repel each other, in a fixed volume. The more charges you stuff in there the harder it is to stuff any individual charge in there. By the formula, if you take a 100 uF cap and charge it from 0 to 6 volts, you have to put 0.0018 Joules of work (energy) into the charging. If you then charge it from 6 to 12 volts ... the same number of volts increase as before ... you have to put 0.0054 Joules in to do it ... because you are pushing against the charge pressure of the charges already in there.
If I did the math right, that is.
Title: Re: Silly question about capacitors
Post by: Farmhand on February 15, 2014, 05:03:43 AM
My undertanding is stored energy is potential energy and energy at work is kinetic energy.

..

I made a P.S. to my last post.
Title: Re: Silly question about capacitors
Post by: MileHigh on February 15, 2014, 05:38:23 AM
Farmhand:

Quote
Same thing with two 10000 liter water tanks side by side, one empty one full, if you let the full one equalize with the empty one, the same amount of water is there. None wasted. If you want all the water in the first tank just leave it there.

No water is wasted but you have lost one half of the GPE of the water.  It has all turned into heat.  There is no point in challenging the premise of the experiment, it is a given.

MileHigh
Title: Re: Silly question about capacitors
Post by: TinselKoala on February 15, 2014, 06:01:26 AM
My undertanding is stored energy is potential energy and energy at work is kinetic energy.

..

I made a P.S. to my last post.
Sure, OK. I was just trying to make the point that the equation has the same form (and for the same reason if you look at the full derivation) as the formula for KE of a moving mass. The energy on a capacitor can also be expressed as QV/2 where Q is the charge in Coulombs. This is more analogous to "potential energy", as it is linear in voltage.
Title: Re: Silly question about capacitors
Post by: Farmhand on February 15, 2014, 06:19:20 AM
Farmhand:

No water is wasted but you have lost one half of the GPE of the water.  It has all turned into heat.  There is no point in challenging the premise of the experiment, it is a given.

MileHigh

WHo is challenging the premise of the experiment. I helping to explain in layman's terms. Why energy is expended and where. How it can be a more efficient expenditure or not.

Look what I also wrote.

Quote
With no reason to do a thing it is always a loss. It's like making a big high stack of bricks then leveling it out into two shorter stacks it takes energy to make the big stack into the two small stacks. It cannot do itself, something must do it.

Cheers

P.S. Imagine making one tall stack of bricks by hand then changing that to two shorter stacks, does that take no energy to do ? No you must expend energy to get from state one to state two. Then if you want to make the two stacks back into one tall one you must expend more energy again. It's really quite simple. The amount of brick remains the same.

Also the higher the stack gets the harder it becomes to add another brick to the top of the stack.  ;) It depends on how you take them down and if you store the energy or not when you do it as to how much loss is involved.

So I'm not saying energy is not expended.

Tinsel, I get ya, energy is energy, I probably didn't word my post well. There is no assurance that the potential energy will be converted into "wanted work" at any specific efficiency.

Cheers
Title: Re: Silly question about capacitors
Post by: Farmhand on February 15, 2014, 06:26:19 AM
Farmhand:

No water is wasted but you have lost one half of the GPE of the water.  It has all turned into heat.  There is no point in challenging the premise of the experiment, it is a given.

MileHigh

Well yes of course. It would take an additional input of energy to equalize the water "up" to a level with the same top height as the first one. In other words equalize "up" (rather than equalize down) and the GPE would then be increased.  That's just common sense.

..
Title: Re: Silly question about capacitors
Post by: MarkE on February 15, 2014, 06:32:28 AM
Farmhand:

No water is wasted but you have lost one half of the GPE of the water.  It has all turned into heat.  There is no point in challenging the premise of the experiment, it is a given.

MileHigh
And this is of course the problem that our friend webby can't bring himself to see with the scenario he set up in defense of our favorite investor funds funeral director Mr. Wayne.
Title: Re: Silly question about capacitors
Post by: Farmhand on February 15, 2014, 06:51:27 AM
I fail to see how anyone could not agree that if you lower 5000 liters (kg's) of water closer to the ground you would not be lowering the (GPE) of it. Energy must be dissipated, but it
could be used partly for other purposes, exactly like the principal of the Hydro electric power generator, just one example.

As with the low loss series inductor the storing of some of the energy in the magnetic field could increase the efficiency of the transfer. Like storing momentum. Or as you like to say MileHigh "like a flywheel".  The thing is that when I transfer water and equalize rain water tanks so I can catch more water, I know I am reducing the "head" of the water in the fuller tank, but it doesn't matter because I only want the water. And the rain will refill the tanks and recharge the "head" of water for free.  ;D So it makes sense to "just do it" when the time is right, there must be a reason to do a thing or it is 100 % waste.

Cheers

Trust me, I don't pump water up high so I can let if fall under the force of gravity to use it.  ;) I catch it up high if I want to use it by gravity force.

..
Title: Re: Silly question about capacitors
Post by: MarkE on February 15, 2014, 07:12:41 AM
I fail to see how anyone could not agree that if you lower 5000 liters (kg's) of water closer to the ground you would not be lowering the (GPE) of it. Energy must be dissipated, but it
could be used partly for other purposes, exactly like the principal of the Hydro electric power generator, just one example.

As with the low loss series inductor the storing of some of the energy in the magnetic field could increase the efficiency of the transfer. Like storing momentum. Or as you like to say MileHigh "like a flywheel".  The thing is that when I transfer water and equalize rain water tanks so I can catch more water, I know I am reducing the "head" of the water in the fuller tank, but it doesn't matter because I only want the water. And the rain will refill the tanks and recharge the "head" of water for free.  ;D So it makes sense to "just do it" when the time is right, there must be a reason to do a thing or it is 100 % waste.

Cheers

Trust me, I don't pump water up high so I can let if fall under the force of gravity to use it.  ;) I catch it up high if I want to use it by gravity force.

..
Farmhand, if one wanted to extract the energy that otherwise goes to heat then one would need to devise a machine that has a highly variable impedance.  Throughout the transfer the machine needs to represent an impedance that is proportional to the difference in head between the two tanks.  At the start of the transfer, that is not too difficult as the difference is quite substantial.  Near the end of the transfer, the machine needs to present an impedance that is almost zero.  One might imagine a wide water wheel with a tapered diameter that slides along its axis, positioned by a lever connected to floats in each of the two tanks.
Title: Re: Silly question about capacitors
Post by: Farmhand on February 15, 2014, 08:12:29 AM
Farmhand, if one wanted to extract the energy that otherwise goes to heat then one would need to devise a machine that has a highly variable impedance.  Throughout the transfer the machine needs to represent an impedance that is proportional to the difference in head between the two tanks.  At the start of the transfer, that is not too difficult as the difference is quite substantial.  Near the end of the transfer, the machine needs to present an impedance that is almost zero.  One might imagine a wide water wheel with a tapered diameter that slides along its axis, positioned by a lever connected to floats in each of the two tanks.

I'm not wanting to design a machine. I'm just saying that for people to say half the energy is lost as heat and that is that is not telling the entire story I don't think.

Going back to the electrical model, how much energy is lost in the circuit I posted with a load when the switching is done so as to get good use of the inductor ? As compared to not using a load and an inductor ?

One of my points is that if there is a good reason to do it then the loss is regrettable but acceptable. If there is no good reason then it is just an experiment to show the loss of a pointless procedure. Like i said I only want the water but if I wanted to I could lower the water and cause it to do work on the way down but not much that would be worth the effort.

I made no claims, I'm simply trying to explain it in a way that people with a lesser training can understand better. Like the bricks analogy. Why make one big stack if you want two small ones anyway ?

..

A good example of Power is not Energy and Energy is not Charge ect. ect..

Title: Re: Silly question about capacitors
Post by: dieter on February 15, 2014, 10:34:56 AM
Wow, the lots of replies indicates that there is still a little bit of a paradoxon going on.

Personally I find it best to be compared to water. The deeper you get, the higher the pressure, in an exponentional amount, probably just like a cap: ^2.

Take a full waterbottle, make a pinhole near the bottom and see how the pressure is constantly reduced. Even atlhough there is more pressure-energy stored in the 8 to 16v, it can only flow to cap B until both have the same pressure.
Still having problems with the heat dissipation. Is gravity heat dissipation? Considering the need of flow from + to - (or in reverse, depending on popular naming) is just like gravity, I am not sure if you can explain gravity by heat dissipation. What is gravity anyway, as it is opposite to conservative models of centrifugal force.

As we see, we can utilize the energy flow with a load, but this will slow down the pressure equalisation. Now, does that mean we can lower the energy requirement to fill cap A to 16v when we allow more time to be "consumed"?
Title: Re: Silly question about capacitors
Post by: MarkE on February 15, 2014, 12:15:30 PM
I'm not wanting to design a machine. I'm just saying that for people to say half the energy is lost as heat and that is that is not telling the entire story I don't think.

Going back to the electrical model, how much energy is lost in the circuit I posted with a load when the switching is done so as to get good use of the inductor ? As compared to not using a load and an inductor ?

One of my points is that if there is a good reason to do it then the loss is regrettable but acceptable. If there is no good reason then it is just an experiment to show the loss of a pointless procedure. Like i said I only want the water but if I wanted to I could lower the water and cause it to do work on the way down but not much that would be worth the effort.

I made no claims, I'm simply trying to explain it in a way that people with a lesser training can understand better. Like the bricks analogy. Why make one big stack if you want two small ones anyway ?

..

A good example of Power is not Energy and Energy is not Charge ect. ect..
Farmhand you can and should answer your first question by replacing the load resistor with a capacitor of like size as the input capacitor and operating the circuit until the input capacitor depletes to 50% of its starting voltage.  For your machine more than 50% will be lost, because you will have the switching losses on top of the transfer losses.

If your intent is to retain the 50% energy, then you need a third energy store to hold that energy.

This particular problem is one of energy, so we have to stick with accounting for energy:  beginning, middle, and end.
Title: Re: Silly question about capacitors
Post by: MarkE on February 15, 2014, 12:24:29 PM
Wow, the lots of replies indicates that there is still a little bit of a paradoxon going on.

Personally I find it best to be compared to water. The deeper you get, the higher the pressure, in an exponentional amount, probably just like a cap: ^2.

Take a full waterbottle, make a pinhole near the bottom and see how the pressure is constantly reduced. Even atlhough there is more pressure-energy stored in the 8 to 16v, it can only flow to cap B until both have the same pressure.
Still having problems with the heat dissipation. Is gravity heat dissipation? Considering the need of flow from + to - (or in reverse, depending on popular naming) is just like gravity, I am not sure if you can explain gravity by heat dissipation. What is gravity anyway, as it is opposite to conservative models of centrifugal force.

As we see, we can utilize the energy flow with a load, but this will slow down the pressure equalisation. Now, does that mean we can lower the energy requirement to fill cap A to 16v when we allow more time to be "consumed"?
The water case is like a capacitor, but the pressure does not go up exponentially, anymore than the voltage on a capacitor goes up exponentially with stored charge.  What goes up quadratically in each case is the stored energy.

Gravity is not heat dissipation.  Nor is gravity analagous to heat dissipation.  Gravity is not opposite to centrifugal force.  Gravity is an acceleration between masses.  Gravitational fields have to the best of our ability to evaluate them always behaved in a conservative manner.  Centrifugal force, is a reaction force to centripetal force.  Neither is conservative.
Title: Re: Silly question about capacitors
Post by: dieter on February 15, 2014, 06:32:57 PM
I meant conservative in the way it is seen from science, a conservative view, contrary to  to be openminded for a new view, like, we thought we are attracted by the ground, but now the new view seems to be that we are repelled from the sky.

Not sure tho if gravity is yet officially understood by science. I mean, other than Newtons apple.
Title: Re: Silly question about capacitors
Post by: MarkE on February 15, 2014, 07:34:21 PM
I meant conservative in the way it is seen from science, a conservative view, contrary to  to be openminded for a new view, like, we thought we are attracted by the ground, but now the new view seems to be that we are repelled from the sky.

Not sure tho if gravity is yet officially understood by science. I mean, other than Newtons apple.
We float ever so gently in the sky as it buoys us slightly from the earth that so strongly attracts us.
Title: Re: Silly question about capacitors
Post by: Farmhand on February 16, 2014, 07:07:58 AM
Farmhand you can and should answer your first question by replacing the load resistor with a capacitor of like size as the input capacitor and operating the circuit until the input capacitor depletes to 50% of its starting voltage.  For your machine more than 50% will be lost, because you will have the switching losses on top of the transfer losses.

If your intent is to retain the 50% energy, then you need a third energy store to hold that energy.

This particular problem is one of energy, so we have to stick with accounting for energy:  beginning, middle, and end.

Fine by me, I was first to mention Poynt99's very good paper on the subject and in the actual post where I said there was no loss of "Charge" I was referring to the charge only as I also said that the capacitor actually stores charge and the charge was conserved. The wasting of energy by the exercise described in the first post is not contested by me.

When i said there is no loss I was referring to the charge and I never mentioned energy in that post. I don't think you are the maker of the rules as to what I can say in any given post.

The charge is conserved but the energy is lost. The fact that the energy is lost is obvious. I was merely pointing out that if done a different way it can produce less loss. A different way means a different setup.

The actual experiment described in the first post will always produce a loss of 50% of the potential energy, but the charge will be conserved. Happy.  Sheez.

..

When we release the pressure of certain pressurized gas vessels cold is produced rather than heat. True. This is the result of a loss of energy is it not ?

..

Here is the post below I think might be irking you, quoted in total.

Quote
There is no loss of "charge".

Did no one look at the "Charge" in "Coulombs", If we take 2 x 10 000 uF capacitors and we have one at 16 volts the electronics assistant tells me it has 160 Millicoulombs of charge.
Now if I was to connect the two caps together and equalize the charge which results in both having 8 volts across them then the electronics assistant tells me they will both have 80 Millicoulombs of "charge" in them so added together there is still 160 Millicoulombs of charge, so no loss.. Where is the loss. The energy in a capacitor is really only "potential energy". I think the capacitor actually stores "charge", not "energy".

The voltage increase on a capacitor has a non linear effect on the "potential energy". meaning from 0 to 8 volts is less "potential energy" than from 8 volts to 16 volts, the voltage increase is the same but the potential energy increase is not.

Cheers
..

I also stated to Tinsel that I may not have worded the post well. and my intention was to say that there is no assurance that the potential energy will be transformed into useful work at any specific rate of efficiency.

..

Title: Re: Silly question about capacitors
Post by: MarkE on February 16, 2014, 07:33:04 AM
Fine by me, I was first to mention Poynt99's very good paper on the subject and in the actual post where I said there was no loss of "Charge" I was referring to the charge only as I also said that the capacitor actually stores charge and the charge was conserved. The wasting of energy by the exercise described in the first post is not contested by me.

When i said there is no loss I was referring to the charge and I never mentioned energy in that post. I don't think you are the maker of the rules as to what I can say in any given post.

The charge is conserved but the energy is lost. The fact that the energy is lost is obvious. I was merely pointing out that if done a different way it can produce less loss. A different way means a different setup.

The actual experiment described in the first post will always produce a loss of 50% of the potential energy, but the charge will be conserved. Happy.  Sheez.

..

When we release the pressure of certain pressurized gas vessels cold is produced rather than heat. True. This is the result of a loss of energy is it not ?

..

Here is the post below I think might be irking you, quoted in total.
..

I also stated to Tinsel that I may not have worded the post well. and my intention was to say that there is no assurance that the potential energy will be transformed into useful work at any specific rate of efficiency.

..
I haven't seen anyone contest the conservation of charge.  The only issue to cover is whether and if so how to avoid the energy loss.  As you pointed out the energy loss appears in the resistance.  If one can make the resistance a reflection of a useful load, then the internal 50% energy loss is mitigated by some amount of useful external work done.
Title: Re: Silly question about capacitors
Post by: forest on February 16, 2014, 10:09:53 AM
what about the change of two charged capacitor connection from parallel to series ? how law of conservation of charge apply to that situation ?
Title: Re: Silly question about capacitors
Post by: dieter on February 16, 2014, 10:16:32 AM
But Resistance in series adds up to more resistance and therefor more loss, isn't it?

I'm back at square one, heat dissipation is no explanation to me. If there was really more "pressure" in the 8 to 16 volts, then the 2 caps simply should be equalized at 16- ((sqr(16)/2)^2)=12 volts. Because the electricity does not jump across the room and run away, nor have I ever seen a cap getting excessively hot simply by charging and uncharging it within its supposed specs.

You may call me silly , hence silly questions, but the paradoxon is yet unexplained to me. I'd rather see there are many paradoxons in established sience that are always "talked to death" in the sense of "science knows everything". No offence tho. Maybe I'm just not getting it.
Title: Re: Silly question about capacitors
Post by: dieter on February 16, 2014, 10:23:22 AM
forest, actually the initial experiment is in series, even a loop in series. Of course, the cap must have polarity, like electrolytics do have, otherwise it would be parallel and serial in the same time. (in this little 2 caps experiment
Or did I misunderstand your question?
Title: Re: Silly question about capacitors
Post by: MarkE on February 16, 2014, 10:29:00 AM
what about the change of two charged capacitor connection from parallel to series ? how law of conservation of charge apply to that situation ?
Charging up one capacitor with capacitance C and then connecting it in series with another discharged capacitor with the same capacitance C has the same effect on energy loss as the situation in the OP.  And as Dieter pointed-out once the switch is closed the two parallel capacitors are in series in the loop.
Title: Re: Silly question about capacitors
Post by: MarkE on February 16, 2014, 10:34:18 AM
But Resistance in series adds up to more resistance and therefor more loss, isn't it?

I'm back at square one, heat dissipation is no explanation to me. If there was really more "pressure" in the 8 to 16 volts, then the 2 caps simply should be equalized at 16- ((sqr(16)/2)^2)=12 volts. Because the electricity does not jump across the room and run away, nor have I ever seen a cap getting excessively hot simply by charging and uncharging it within its supposed specs.

You may call me silly , hence silly questions, but the paradoxon is yet unexplained to me. I'd rather see there are many paradoxons in established sience that are always "talked to death" in the sense of "science knows everything". No offence tho. Maybe I'm just not getting it.
What increasing the resistance does is increase the amount of time that it takes to transfer the charge.  It does not change the percentage of energy lost transferring the charge.
If you want to see capacitors get hot doing this we can set up an experiment that will heat them nicely by choice of capacitor and the test circuit.  Ceramic and film capacitors can handle very high ripple currents. Electrolytic and double layer capacitors are a different story.  Ripple current ratings and limiting internal heating of electrolytic capacitors is an important aspect of power converter design in converters that use electrolytics.
Title: Re: Silly question about capacitors
Post by: dieter on February 16, 2014, 10:49:03 AM
This is pretty much the opposite of what I've learned yet. Resistance causes heat loss but has no "slow down effect", and I've never heared of cooling elements for caps, tho they obviously are common for transistors or ICs...

The cap in a simple ac/dc supply that smoothens the rectified current is in heavy duty to do exactly what we discuss here, but they are as uncritical as a cap can be.

It just doesn't make sense, sorry.
Title: Re: Silly question about capacitors
Post by: MarkE on February 16, 2014, 11:03:13 AM
This is pretty much the opposite of what I've learned yet. Resistance causes heat loss but has no "slow down effect", and I've never heared of cooling elements for caps, tho they obviously are common for transistors or ICs...

The cap in a simple ac/dc supply that smoothens the rectified current is in heavy duty to do exactly what we discuss here, but they are as uncritical as a cap can be.

It just doesn't make sense, sorry.
Dieter, plug the circuit into any SPICE simulator as vasik041 did.  The amount of time that it takes for the second capacitor to come with 0.1% of 50% of the starting voltage is 6.9 RC time constants.  Increase C or R and the absolute time to stabilize to within any particular percentage of the 50% asymptote stretches out.

You can find capacitor with heat sinks easily.  Go take a look into your local power utility substation.

You are absolutely wrong about what you think about electrolytic capacitors.  Very often more capacitance is used than would be required to meet charge storage requirements in order to stay within ripple current requirements.  Ripple current requirements are set by internal temperature rise considerations.  There is lots of information on the web sites of the various electrolytic capacitor makers such as Nichicon and United Chemicon.  http://nichicon-us.com/english/products/alm_mini/pict_f.htm
Title: Re: Silly question about capacitors
Post by: dieter on February 16, 2014, 11:26:05 AM
Ok, I may be wrong there and yes, some caps may not be charged and uncharged too fast, otherwise they may be destroyed.

But I came to an other explanation that works for me:

Since we connect the 2 caps in series, the total capacitance will suddently be only 50% anymore. This happens faster than SOL, but immediately. The charge in cap A is now too high for its capacitance and quantum mechanics, those impatient dudes, can't wait til those lazy electrons have moved trough the slow wire. So they simply kick 50% of the charge "out" somehow. The remaining half, that now still fills Cap 1, will then slowly flow to cap 2 until they re equalized. Now they are still 50% full each, but as soon as you seperate them from eachother, they will double their capicitance, resulting in 8 Volts= 25% of the energy in each one.

Now I just have to find out, where the 50% has gone. So you say that's where the heat is caused, by forcing electron charges into the dielectric surroundings?

EDIT: okokok, but what happens when we increase the capacitance of an empty cap? Could't that be at least a good way to collect charges, eg. from the surroundings?
Ok, a silly question again but phew, I can test that right now. ::)

EDIT 2: Wait a sec, who said the initial charge of cap 1 (16v) is the max capacitance? Maybe the reaction is the same no matter how full the cap is and immediately altering the capacitence cannot be compensated by charge flow in the cirquit? When this causes heat, does then heat cause a charge?

I actually just reallize that I don't know anything about caps.
Title: Re: Silly question about capacitors
Post by: MarkE on February 16, 2014, 01:02:47 PM
Dieter, yes you are wrong as the linked reference explains in Item 2.  Electrolytic capacitors must be applied such that they do not suffer excess ripple current which in turn would cause excess internal heating and premature failure.

That is certainly a creative interpretation you offer.  The accepted theory previously explained is backed by careful experiments.

There are machines that vary capacitance.  If a voltage is placed across them, they emit E/M radiation.  More commonly, these machines act as stress or fine position sensors.

I don't think anyone said that the capacitor was charged to its upper voltage limit, only that it was charged to a particular voltage.

Title: Re: Silly question about capacitors
Post by: forest on February 16, 2014, 01:12:44 PM
sorry, I was not clear enough. Take two non-polar capacitors in paralell (maybe would be good to use two electrolytic to form non-polar to have bigger capacitance from start and lower leakage to air), charge them to the same voltage  (because of being parallel), then disconnect and connect them in series and measure. what is going on ? energy rise ????
Title: Re: Silly question about capacitors
Post by: dieter on February 16, 2014, 01:54:33 PM
Forest, that is most likely only the same like with batteries in series, double voltage, same current. As two caps in series are technically equal to doubling the gap between one cap's plates, it will reduce the Capacitance to 50%, but since the energy is the same, the voltage is forced to rise.
At least that's what I think, tho I might be rather unconcentrated and or confused ATM.

MarkE, they emit E/M Radiation ? How? RF? HV? But actually that's what I meant, losses trough radiation.

BTW. by "who said that" I meant "nobody said that", it's always tricky to translate sayings 1:1.
Title: Re: Silly question about capacitors
Post by: MarkE on February 16, 2014, 02:15:03 PM
Forest, no there is no energy gain.  Two capacitors in series is effectively just the previous case but charging both capacitors first.  No charge transfers and no energy is lost:

Original energy = 0.5*2*C1*Vcharge^2.  = C1*Vcharge^2.
Qtotal = 2*C1*Vcharge

Vseries = Qtotal/Cseries
Cseries = 0.5*C1

Eseries = 0.5*C*V^2 = 0.5*(0.5*C1)*(2*Vcharge)^2 = C1*Vcharge^2 = Original energy
Title: Re: Silly question about capacitors
Post by: MarkE on February 16, 2014, 02:17:06 PM
Forest, that is most likely only the same like with batteries in series, double voltage, same current. As two caps in series are technically equal to doubling the gap between one cap's plates, it will reduce the Capacitance to 50%, but since the energy is the same, the voltage is forced to rise.
At least that's what I think, tho I might be rather unconcentrated and or confused ATM.

MarkE, they emit E/M Radiation ? How? RF? HV? But actually that's what I meant, losses trough radiation.

BTW. by "who said that" I meant "nobody said that", it's always tricky to translate sayings 1:1.
Dieter anytime charge accelerates there is E/M radiation.  Connecting a full capacitor to an empty capacitor accelerates charge impeded only by the resistance and inductance.
Title: Re: Silly question about capacitors
Post by: poynt99 on February 16, 2014, 03:24:26 PM
Dieter,

As Mark said, a tiny bit of energy is radiated at the moment you connect the two capacitors together. Turn on your AM/FM radio and you may in fact "hear" it.

However, most of the energy is dissipated (lost) in the combined resistance in the capacitors themselves, and the conductors between them.

If one could replace the connecting conductor with a pure inductance, there would be no energy loss at all during the transfer (assuming ideal capacitors).
Title: Re: Silly question about capacitors
Post by: dieter on February 16, 2014, 03:27:06 PM
Surrounded by scary formulas, I made more simple experiments. The two caps, empty, were connected in a loop, then gently heated to about 45 deg. C., then I doubled their capacitance by unconnecting them, still warm. Now one has 12 mvdc, the other -12mvdc charge.

Of course I don't think that the thousands of intelligent and educted scientists made a bad job so that I have to reveal the truth about caps after 10 Minutes of experiments  ;) , it's just that science is very if not only profit-oriented, so certain aspects are never really examined. Example given the back emf is usually seen as a nasty, dangerous thing that needs to be eliminated, eg. by heat dissipation, instead of  harvesting it to save energy. After all, they want us to consume electricity since one reactor gives about 1/3 billion \$ in a year.
Title: Re: Silly question about capacitors
Post by: MileHigh on February 16, 2014, 03:41:28 PM
Dieter:

If it made economic and/or technical sense to harvest back-EMF it would be done.  But in general it doesn't make sense so it is not done.  To say that something like back-EMF is "never really examined" is not true at all.  There is a huge amount of material to learn if you want to learn about electronics and a wise person would not make judgements or pronouncements without first understanding all of the issues.  I am not directing this comment specifically at you, it is a comment that applies to everyone.

MileHigh

P.S.:  And I will defer to comments by MarkE below about how back-EMF is being harvested and recycled in a lot of applications nowadays.  I am also on a learning curve that I will never get off of.
Title: Re: Silly question about capacitors
Post by: MarkE on February 16, 2014, 03:41:43 PM
Surrounded by scary formulas. I made more simple experiments. The two caps, empty, where connected in a loop, then gently heated to about 45 deg. C., then I doubled their capacitance by unconnwcting them, still warm. Now one has 12 mvdc, the other -12mvdc charge.

Of course I don't think that the thousands of intelligent and educted scientists made a baf job so I have to reveal the truth about caps after 10 Minutes of experiments  ;) , it's just that science is very if not only profit-oriented, so certain aspects are never really examined. Example given the back emf is usually seen as a nasty, dangerous thing that needs to be eliminated, eg. by heat dissipation, instead of  harvesting it to save energy. After all, they want us to consume electricity since one reactor gives about 1/3 billion \$ in a year.
BEMF is recycled in virtually all high efficiency power converters and motor drives.  It is a necessity of smaller packaging where we can't afford the space needed to dissipate the wasted heat of dissipative BEMF discharge circuits.  40 years ago people threw away a lot of power in order to save component cost.  That all changed about 20 years ago.  Only in the lowest end consumer products do we see low efficiency power converters and motor drives.
Title: Re: Silly question about capacitors
Post by: MarkE on February 16, 2014, 03:44:38 PM
Dieter:

If it made economical and/or technical sense to harvest back-EMF it would be done.  But in general it doesn't make sense so it is not done.  To say that something like back-EMF is "never really examined" is not true at all.  There is a huge amount of material to learn if you want to learn about electronics and a wise person would not make judgements or pronouncements without first understanding all of the issues.  I am not directing this comment specifically at you, it is a comment that applies to everyone.

MileHigh
It really is done everywhere now.  Virtually all stepper motor drives that are in everything from 3D printers to robotics are full bridge drivers that recycle BEMF.  All high efficiency motorized appliances use one form of bridge driver or another that recycles BEMF.  Virtually all switching DC-DC, and DC-AC converters to recycle BEMF.
Title: Re: Silly question about capacitors
Post by: dieter on February 16, 2014, 04:06:01 PM
Today probably, yes, but for some 30 years it wasn't an issue. Would it be wise to say, Science is not financed by economy? Conclusion?

Anyway, what I have said about the -12 and +12 Mvdc charge may be caused by my body, acting as an antenna. Since I had them in my pocket while I was sitting in front of a heater fan.

Basicly I was referring to a document, I think it was "the secret life of capacitors" from the download section, that discovers a relation between self-charging and wheather changes, in pressure or temperature.
BTW Milehigh, I didn't say "something like a back-emf is never really examined", as you impressed. The idea of harvesting it wasn't really examined, at least for a long time. Thats a fact, just watch some older schematics, how they used to deal with bemfs between some cmos chips.
Title: Re: Silly question about capacitors
Post by: Vladokv on February 16, 2014, 09:01:30 PM
Energy is lost as heat. Condesators heat up dramatically at high current. Look at induction heating - condesator  in LC tank heats up dramaticaly. Polyester condesators with no electrolyte heats up! Currents in that very moment are enormous and so energy losses are. Your setup looks like Colpitt oscillator with very low inductance(and straight wire have inductance). Current jumps back and forth maybe and million times(some miliseconds) before come to stop.
Title: Re: Silly question about capacitors
Post by: Magluvin on February 16, 2014, 10:30:42 PM
Energy is lost as heat. Condesators heat up dramatically at high current. Look at induction heating - condesator  in LC tank heats up dramaticaly. Polyester condesators with no electrolyte heats up! Currents in that very moment are enormous and so energy losses are. Your setup looks like Colpitt oscillator with very low inductance(and straight wire have inductance). Current jumps back and forth maybe and million times(some miliseconds) before come to stop.

If you read my previous post with the analogy of 2 air tanks, then consider this....

2 caps of the same size, 1000uf with 5v and another 1000uf cap with 0v. We connect them together till each has 2.5v, and we have lost half the energy, and of the loss, it is said to be lost in heat.

Now lets take the same 2 caps, one with 5v and one with 0v. Now we set it up with a very low ohm switch that is controlled by a circuit that opens the switch when a voltage gets to 2.5v or lower.
So between our 2 caps, we include our switch device and an inductor of say 10mh(which can be had with pretty low ohm resistance) in series with the switch.

The circuit that detects a voltage at or below 2.5v will monitor the source cap. We close the switch and when the circuit opens the switch, we will find that the 2 caps have very close to 2.5v each, with near 0v across the inductor just before the time of switch off. But the inductor has a stored amount of energy that is near equal to what looks like we have lost during the transfer from on cap to the next.

In the end, the 2 caps have the same voltage using the switch and inductor, as we did if we just connected the caps directly.  Now, there is resistance in all things, the caps, the coil, the switch, the wires. Well where are our heat losses when using the inductor? Sure the inductor slows down the transfer, so less current, but over a longer period of time using the inductor. So there must be heat losses while using the inductor, right?

Well after all that, say the 2 caps have no resistance, and say no inductance, just to avoid ringing, what voltage would you say would be in each cap when we connect them together? More than 2.5v each????  No losses would mean what? 3.5v in each?  I cannot agree.  Lets say we consider electron charge.  2 empty caps, 0v each. The have say an equal amount of electrons on each plate. We take 1 cap and we pull electrons from one plate and push them onto the other plate. For simplified example we say 5v has pulled 5000 electrons from one side of the cap to the other. So now 1 plate is missing 5000 electrons and one plate is over loaded with 5000 electrons. When we connect the empty cap to the charges cap, how do you think that imbalance of electrons is going to be divided up between the 2 caps? Do we lose any electrons in the exchange?? Would we gain any electrons using an inductor?????? ::)   Can we not see that in order for there to be a balance in the end, the empty cap, one plate will take on 2500 electrons from the overfilled plate of the source cap, and the other empty cap plate will give 2500 electrons to the source caps depleted plate. So now each cap has a plate with an over flow of 2500electrons and the other 2 plates have a depletion of 2500 electrons, giving each cap 2.5v. Why is that so hard to comprehend? Any other way would have to add or take way electrons from the caps from an outside source. But no, we end up with half the voltage in each once the imbalance, balances out.

And my contention is that we lose that energy between cap to cap by simply not putting it to work during the transfer. Heat, what ever. Lol, that heat is a spike, just like the spike that many claim is just about nothing when it comes to a field collapse of an inductor, when we consider the time period that it happens. So why is this spike such a great loss, when they consider the field collapse to be such a little thing. ;)

And the reason why there is less loss with a direct transfer from a tiny cap to a large cap is the fact that once things balance out, the amount of energy in the tiny cap is mostly all in the large cap, and can be used, as such from the large cap, just a different voltage range. Less current, longer time.

Along with the large cap to the smaller cap, the large cap doesnt lose much because the tiny cap only took little.  Same can be said with the air tanks. ;)

The only heat that I would consider the loss would be the heat generated by charging that source cap to full, then wasting it by depressurizing into another container of equal size without taking advantage of the action of the transfer by using it to do something of value. ;) ;D In the end, there is no more or less total air in the tanks(total), nor is there in the caps(total), at any given point. ;)

Mags
Title: Re: Silly question about capacitors
Post by: Vladokv on February 16, 2014, 11:15:08 PM
I understand what you say and will try to elaborate where you are wrong.
I don't like air tank analogy because is lossy, so I will stick to electronic.
First assume that LC tank have zero resistance. Gives implication ->
1. Then we can not avoid ringing. Not possible.
2. Where is energy? -> 25% one cap, 25% another cap and 50% in inductor field.
both caps have 2,5V and there is present magnetic field in and around inductor.
3. What happens when inductor return field to zero? -> one cap have 5V and another 0.
After full cycle again first cap have 5V and another 0.
Also to mention when caps have equal voltage it is 2,5V and field of inductor

Offtopic note: inductor will slowly lose energy by emitting radio wave, if everything else is from superconductive material
Title: Re: Silly question about capacitors
Post by: Farmhand on February 16, 2014, 11:42:08 PM
No one mentioned charge till I did.

And since the total charge is conserved if the action were to conserve energy rather than charge then the total charge would need to increase. As Mags I think pointed to. Just because the first post was talking of energy being lost does not mean someone cannot point out that the charge is conserved. Just because the first post was about lost energy does that mean some people seem to think that mentioning that the charge was conserved is making some claim or something ?

I think the fact the charge was conserved is relevant as for the reasons I gave above. If the action described in the first post did anything less than lose half the energy then charge would have to be increased I think and that is a relevant factor towards explaining why it must be that way in that example.

I still say if you want half the charge in each cap then just charge the two caps, if you want the potential energy in one cap then do that.

Like the brick analogy, if we make one big stack then decide we want two small ones we can't just push over half the stack and hope it lands in a neat stack half the height, bricks will break, so we must remove them and make the second stack by hand the same way the first on was made, and it takes energy to undo it, then re-stack the second small stack.

So the moral of the story is if you want two caps charged to 8 volts then just do that. Don't push stuff up hill then carry it back down again.

Cheers
Title: Re: Silly question about capacitors
Post by: Farmhand on February 17, 2014, 12:32:18 AM
sorry, I was not clear enough. Take two non-polar capacitors in paralell (maybe would be good to use two electrolytic to form non-polar to have bigger capacitance from start and lower leakage to air), charge them to the same voltage  (because of being parallel), then disconnect and connect them in series and measure. what is going on ? energy rise ????

To explain it without all the hoopla, if we have one 10000 uf cap at 16 volts it has 160 Microcoulombs of charge and 1.28 Joules of energy.

Now to the question I think Forest asked.

A 10000 uf cap at 8 volts has 80 Microcoulombs of charge and 320 Millijoules of energy.

If we put two 10000 uf caps in series the capacitance is then halved so we end up with a 5000 uf capacitor charged to 16 volts, which has 80 Micocoulombs of charge and 640 millijoules of energy.

Charging the two caps in parallel then putting in series conserves both charge and energy because no actual transfer was done, the connection is assumed to cost us no energy, as in it cost no energy to just put them in series. However the cap is half the size, double the caps to 20000 uF so we end up with a 10000 uF capacitor charged to 16 volts and we need to double the work we need to do to charge them.

A 20000 uF capacitor charged to 8 volts has 160 Microcoulombs of charge and 640 Millijoules of energy, two in series is a 10000 uf Cap charged to 16 volts and has the original 160 Micocoulombs of charge and 1.28 Joules of energy.

Cheers

I guess it is always very good to know the formula for things, but I have to say that programs like the "Electronics Assistant", are powerful tools if used intuitively, they save time and don't usually lie. They can even teach us if we use them and practice the calculation till we get it right.

EDIT: Fixed typo.
..
Title: Re: Silly question about capacitors
Post by: forest on February 17, 2014, 09:50:35 AM
sorry, I'm not good with formulas I tried but without assuming  both laws of conservation (energy and charge) I got silly results. When you assume energy is conserved then something weird is going on with charge when such capacitors are re-arranged from parallel to series connection according to maths IF you do not assume conservation of charge but simply the capacitance values halved , probably the other way is what Farmhand presented - if you do not assume energy must be conserved but charge is conserved then you got weird results with energy stored and so on..... only when you assume both and disregard possible capacitance change you got the same charge and energy in both situation. Sorry, it's probably due to my lack of deep understanding of electrical science....

Please , can you find a bug in my conclusion ... :
Let Ci be a capacitance of one capacitor (both are identical)

When connected in parallel and charged to U1:
Q =Ci*U1 (of one capacitor)
C=2*Ci
E=0.5*(2*Ci)*U1^2=Ci*U1^2
Q=2*Ci*U1 (total charge stored)

when you change connection to series  capacitance is changed and I only assume that energy is the same and look for the final charge :

now capacitance is:

1/C=1/Ci + 1/Ci => C=0.5*Ci
Energy is the same so

0.5*(0.5*Ci)*U2^2=Ci*U1^2 => U2=2*U1 (as we can predict voltage rised 2 times)

Q=(0.5*Ci)*U2 =(0.5*Ci)*2*U1=Ci*U1

but originally that was Q=2*Ci*U1

where is the error ?  :(

There must be an error somewhere ...  ???
Title: Re: Silly question about capacitors
Post by: MarkE on February 17, 2014, 12:00:10 PM
Forest, what is throwing you off is that when you connect the two capacitors in series, charge does not transfer unless there is a current path around the loop.  So, if the first was charged before, it retains that charge and voltage difference, and ditto for the second capacitor, including for 0V charge on either.