Joule Thief 101

[Magluvin]

C = Q/V

Q = CV

So if Q is conserved and you double the capacitance to 2*C, then for the equation to hold then V has to be halved to V/2.

However, you lose half the energy when this happens.  Some of the charge moves through the resistor and suffers a voltage drop and loses some of its "bang."

Ah yes. The voltage drop. I was just thinking about that. Are you reading my mind through my phone?  lol  None the less, its true that I was.

Its funny the voltage drop across any value resistor doesnt change the outcome. It just changes the time till equalization as you said earlier. But when you first consider it, it makes you think a bit, dont it? In this case, what I believe, is that the resistor does just that and isnt a loss in this cap to cap deal. But thats just me until convinced otherwise. As for the answers so far, Im still at point A.

Mags

[Magluvin]

Misposted

[Pirate88179]

I should be able to shed some light on this subject.  We often say "things go to infinity" whereas MarkE would use a more commonly used term in scientific parlance, "undefined."  I will make use of both terms here.

Simple thought experiment for the real-world shorting one cap to the other and losing half the energy.  If you use say a 100 kohm resistor say it takes five minutes for the two caps to be equal in voltage.  (There is semi-related conundrum about the two caps "never" attaining the exact same voltage.)  Then you switch to a 5 kohm resistor and it takes 30 seconds.  Then a 100 ohm resistor and it takes two seconds.

Here is the simple conclusion:  It doesn't matter what value of resistor you use you always lose half the energy.  Hence when you short two caps together with "no resistor" you still lose half the energy.

Now for the ideal caps.  You can't short one cap to the other because you get infinite current for zero seconds.  In other words it is undefined.  Ending up with the same voltage in each cap or 7.07 volts in each cap is a non-starter, because you can never get there.  So let's switch to plan B and put the ideal inductor between the two ideal caps.  Now the energy cycles back and forth between each cap forever.  As you lower the value of the ideal inductor in an attempt to simulate a short with a value for zero for the inductance what happens?  The cycling frequency gets higher and higher until it approaches an infinite cycling frequency as the value of the inductance goes to zero.  One more time infinity crops up, and hence you can say the solution for a value of zero for the inductance is undefined.

Going back to losing half the energy for a real-world shorting of two caps together, you are over analyzing the situation.  Voltage times current through the resistor equals heat power.  So you convert energy stored in the electric field in the capacitor into heat energy.  It's actually very mundane, nothing to do with expanding into a new capacitor or volume.

There is a good old visualization trick for this one.  When the 10-volt cap is shorted to the 0-volt cap you lose half the energy and both caps are at 5 volts.  You can visualize this like a totally inelastic collision.  That simply means that energy is burnt off when two things collide.  So imagine a stationary metal block being hit by a ball of putty with a velocity of 10 meters per second.  After they hit they move together at 5 meters per second.  Here is the thing:  The ball off putty deformed when it hit and stuck to the metal block.  The deformation process was resistive in nature, like bending a coat hanger, and therefore the metal block and the ball of putty heated up due to the collision.  There is the signature of the resistive losses for the capacitor example.

Here is another way to look at the same thing:  On an axle you have three things:  A flywheel, a remote controlled clutch, and then another flywheel.  The clutch is between the two flywheels.  You spin up the first flywheel to 100 RPM.  The other flywheel is not turning.  Then you press the button and the clutch engages and connects them together.  It takes one second for the clutch to fully engage.  The net result is that both flywheels are now spinning together along with the "weightless" clutch at 50 RPM.  When the clutch engaged there was friction between the clutch plates producing heat.

MileHigh

I actually understand these examples and it has helped me to visualize what is happening.  I really get the 2 fly wheels example...as soon as I saw where you were going with the clutch idea...I yelled at my monitor 50 RPM! 50 RPM!  Then I read on to the point where you said they were all turning at 50 RPM.  This makes total sense.

Thanks.

Bill

[wattsup]

@Mags

If you consider a battery could be an analog to your cap, batteries usually cannot handle a charge of greater then 20% of its amperage rating. So if you have a 100 amp battery and 40 amps of charge, you are better off using two 100 amp batteries to be charged in parallel instead of that one.

So the problem I see is right away the use of two identical capacitors. The charge capacity may not match the greater discharge capacity and could this explain the loss. So if you had a 10uf charged to 10 volts, then discharge it into 2 x 10uf caps, then try 3 x 10uf caps and so on until you find the best number of caps that can hold that full 10uf 10 volts of the first cap. If you were using 5 caps in parallel to receive that 10 volts discharge and you then put those 5 caps in series and measure across them, would you get 10 volts or still get 5 volts. I would bank on maybe not 10 but close to 10 would be better then the 5 your are getting now.

I think that is where the problem is. You can take all the time in the world to charge that first cap, but then you put it in parallel with the second identical cap and expect it to take it all in one instant. Why should it?

Just thinking out load here.

wattsup

[Magneticitist]

Mag thanks for further explanation, but at this point from what I gather it seems there's a confusion as to why there actually would be some 'loss' at all when charging one cap with another.

First, why is resistance not an option? we can call it a 'loss' due to heat, or a dissipation of heat energy, or whatever, but why would there be no heat generated in the transfer of current across a conductor? If we're talking about a 10v 10uF cap 'dumping' into another 10v uF cap, it seems the 'heat' would have been in the tiny spark you may see, because the ultra fast discharge of those amps would not persist long enough to generate a level of heat within the conductor you would notice any other way other than let's say dumping that same cap into an indicator filament. Everyone here has certainly welded the legs of a cap to some metal during a discharge before. Just saying, why is heat not really considered a factor?

Ideally, were we to create a '100% transfer', it would be more like instead of 2 caps, the one cap magically grew twice the size in an instant, thereby retaining it's charge but doubling in capacitance.. am I way off there?

Alternatively I imagine it like two cups, same height (voltage). One is exactly twice the volume of the other (capacity). What happens when we pour the same amount of liquid into each cup. Obviously the height of each cups liquid level will be different, analogous to the voltage level of the caps. So in that ideal scenario, a gain or loss in 'pressure' would also accordingly result in the loss or gain of sustainability/duration. same amount of water, like same amount of air. ultimately releasing either the air or the water from these varying containers may not net you the same pressure as before, but that just requires 'tuning' now because we still have the same total water and air.


the electrical representation of that tank with the venturi metaphor for me is like taking a large amount of energy to charge a cap up, then slowly dump the cap into another cap through a resistor, making up for any current dissipated as heat in the process by providing the remaining charge via a solar cell. So, to do that with our circuits, we just need to add a solar cell to grab some of that sun, or wind turbine to grab some air, etc..

OK, well, then what did we even do? other than just magically increase the cap size? we can't magically also double the power in this scenario so that just has to 'spread out' so to speak right?
why would the voltage remain the same if we didn't also magically double the energy in the cap?

If we are looking at pressurized air spinning a turbine, and opening a valve in the air line increases the torque on the turbine, then we have clearly found a way to increase the efficiency of our air pressure-to-turbine exercise where our goal is to torque the turbine. but does the venturi actually increase the air pressure through the pipe? If so, isn't that just like lowering the resistance in a circuit and cranking up the current, draining the source faster? if it does not increase the air pressure in the pipe, then does it not simply just create a better air flow for the fan right at the exit valve? I mean look no lie that's badass but isn't just sort of just like finding the right capacitor that made your Joule Thief 75% efficient instead of 40%?

I don't see the real point of the cap argument because it's like saying we have to charge a cap initially.. but after that, we can facilitate this ideal transfer between that cap and another cap in some perfect resonance, where that same amount of energy just sloshes back and forth, yet in order to facilitate that process, we are introducing an outside energy that could also be used to do work, thereby ultimately recooping the energy it took us to get that initial cap charge. That's great but that energy came from somewhere else, not from within the system, so what's the difference between that and adding a solar cell? well, for one, I guess the solar cell will only work in the sun..  but to say the air flow and venturi analogy is more efficient because of that would be to ignore the amount of mechanical energy it took to fill that first air tank compared to the mechanical energy you would get out of it.