Joule Thief 101

[MileHigh]

Brad:

Still hung up on this nonsense are you?

Here is the question:

You have an ideal voltage source and an ideal coil of 5 Henrys.  At time t=0 seconds the coil connects to the ideal voltage source.  For three seconds the voltage is 4 volts.  Then for the next two seconds the voltage is zero volts.  Then for two seconds the voltage is negative three volts, and then for the next six seconds the voltage is 0.5 volts.  Then after that the voltage is zero volts.

The question is what happens starting at t = 0

Let's change it up and make it more difficult, and revamp the question:

You have an ideal voltage source and an ideal coil of 5 Henrys.  At time t=0 seconds the coil connects to the ideal voltage source.  The voltage source waveform is 20*t^2.  So as the time t increases, the voltage increases proportional to the square of the time.

The question is what happens starting at t = 0

The answer:

The current through the ideal coil starts from zero at time t = 0 and then increases with this formula:  i = 1.33*t^3.

Time..........Voltage.........Current
0...............0.................0
1...............20...............1.33
5...............500.............166.67
10.............2000............1333.33
20.............8000............10666.67
50.............50000..........166666.7

There you go, harder question answered.

So why don't you go reassemble your device for John and even take a shot at the original question.

MileHigh

[tinman]

Brad:

Still hung up on this nonsense are you?

Here is the question:

You have an ideal voltage source and an ideal coil of 5 Henrys.  At time t=0 seconds the coil connects to the ideal voltage source.  For three seconds the voltage is 4 volts.  Then for the next two seconds the voltage is zero volts.  Then for two seconds the voltage is negative three volts, and then for the next six seconds the voltage is 0.5 volts.  Then after that the voltage is zero volts.

The question is what happens starting at t = 0

Let's change it up and make it more difficult, and revamp the question:

You have an ideal voltage source and an ideal coil of 5 Henrys.  At time t=0 seconds the coil connects to the ideal voltage source.  The voltage source waveform is 20*t^2.  So as the time t increases, the voltage increases proportional to the square of the time.

The question is what happens starting at t = 0

The answer:

The current through the ideal coil starts from zero at time t = 0 and then increases with this formula:  i = 1.33*t^3.

Time..........Voltage.........Current
0...............0.................0
1...............20...............1.33
5...............500.............166.67
10.............2000............1333.33
20.............8000............10666.67
50.............50000..........166666.7

There you go, harder question answered.

So why don't you go reassemble your device for John and even take a shot at the original question.

MileHigh

And there you go.
I knew you could not answer the question correctly--you are way off.

Care to have another go MH?.

Some hints for you.
1-An ideal voltage is one that dose not change in selected value.
2-An ideal inductor has no resistance nor capacitance.
3-What is the current value of 1 volt across a resister with a value of 0 ohms ?.


Brad

[MileHigh]

And there you go.
I knew you could not answer the question correctly--you are way off.

Care to have another go MH?.

Some hints for you.
1-An ideal voltage is one that dose not change in selected value.
2-An ideal inductor has no resistance nor capacitance.
3-What is the current value of 1 volt across a resister with a value of 0 ohms ?.


Brad

I am not sure what to say, I am not sure you correctly processed what I posted.  Perhaps try again tomorrow when the neurons will be firing differently?

<<< 1-An ideal voltage is one that dose not change in selected value. >>>

After six years you are lucky that you still have people that want to work with you and make attempts to help you.

[minnie]

   It's all got a bit too George Carlin methinks.
           John.

[Magluvin]

Magluvin:

There is an answer to the conundrum with ideal electrical components.  We do have to add the ideal diode, but what you missed was the impossibly infinitely fast switching function.  I will explain but first let's go back to the air tanks.

In the air tank example, and in fact using an unreal model where we ignore temperature for a second and have idealized components, when one tank discharges into the other tank, that spins up an ideal flywheel pump.   So you can set it up such that when the first tank drops to 70.7 psi, you then stop the air flow out of the first tank, and then the ideal flywheel pump takes over and pumps extra air into the second tank so that it also reaches 70.7 psi and therefore no energy is lost.  The net result is no energy was lost and extra air was pumped into the second tank by the ideal flywheel pump.

So, let's do that with ideal capacitors and ideal diodes and an ideal switch.

As Poynt said, an ideal capacitor connecting to an ideal capacitor is a no-no.  You were absolutely right though about the conundrum of "missing" electrons to get to 7.07 volts in each cap.

If you have two ideal caps, and an ideal inductor connecting between them, that is a manageable situation with no ripping of space-time.  In this case, Cap A discharges into Cap B via the coil and the charge goes back and forth forever.  There are no extra electrons and there is never a condition with 7.07 volts in each cap.  The best you can get is this in terms of equal voltage something like 5 volts in each cap, and current flowing through the coil.  I am not saying it has to be 5 volts in each cap either, just an equal voltage in each cap.  The coil discharges into Cap B and stops discharging when Cap A has zero volts and Cap B has 10 volts.  Then the whole process reverses.  During that infinite back-and-forth cycle there will be an instant in time where there is an equal voltage in each cap and the "missing" energy is in the coil.

I will do another post to explain the two caps with 7.07 volts each.

MileHigh

Ok. Thought about what it is I need to say to show my reasoning..... At least you said " You were absolutely right though about the conundrum of "missing" electrons to get to 7.07 volts in each cap."


What I had gotten from this discussion way back was that when we discharge a full cap into an empty directly, we lose half the energy 'because' of resistance. I frown on that. I can agree that some heat will occur because of the resistance and very high current flow during the transfer, but I fail to see that the heat generated caused the loss.

Like the ideal scheme. If we had 2 ideal caps, super conductive to say, one at 10uf at 10v, and the other at 10uf 0v, and we do the dump, Im wondering why there would be an unfathomable explosion or what ever when we hit the superconducting switch. In fact, there should be no heat generated at all with ideal caps and switch because there is no resistance. So if it says so in the book, Id like to read that book.

Now furthermore, if we have 2 ideal caps and dump the 10v cap to the 0v cap, there still should still be 5v each, considering the electron count measurements I described earlier. It seems the 'loss due to resistance' is made up for unknown reasons and we are just suppose to agree.  Well I cannot. Unless, whom ever made that statement back then 'meant' the energy lost is equal to the heat generated. But then there is still the blame on the resistance for the loss. So what Im saying is, it isnt the 'resistance that causes the loss', it is the fact that we expanded the electrical pressure into a larger container haphazardly, with which in the end, we wind up with only half of the usable energy. The energy wasnt simply used up as heat. it was reduced by letting the pressure change value and containment size get larger, and heat was generated because of the resistance.

So the energy that we lost in the transfer and equalization was due to stupidity of doing so by not using the flow of pressure from one cap to the other. Dumb, stupid, ridiculous thing to do.    

This may be showing that conservation of energy may not be all its cracked up to be. None of us(I dont believe) have ideal caps or switches or even inductors to test this. We are just told it is so. So there is no real reason why it cannot be questioned in such that I have.

Sure we can say that if we use an ideal inductor and ideal diode that we could convert all the energy from the 10v ideal cap to another 0v ideal cap and say that it would be a 100%transfer. But we can just about do that now with regular components, and eliminating the diode and replacing it with timed switching like I did in sim. Even with all the crappy resistance in the caps, the wires, the inductor and switches, we can come damn close to full transfer. So it leaves me to think that if the superconductor components are so superior, then why might we only be gaining that little bit we lost with the regular components? Maybe it can do better than that?? How would we know without access to these components. The best thing Ive seen with super conductors is we can float them above a magnet as long as we keep it super chilled. Probably wasting more energy chilling the thing than it would take to levitate a normal object of the same weight.


So really what Im looking for is the answer to the conundrum that you actually agree with me on. If you and I can agree that we cannot produce 2 caps with 7.07v each from a 10v cap 'by direct dump', then where would we lose half of the source cap energy in an ideal scenario by doing a direct cap to cap equalization? No heat. Where did it go? Excluding using an inductor and diode/timed switching, as it is another subject I will bring up after this cap to cap deal.


Mags