Even with a lossless conductor, you will still have inductance, which, with the capacitors, forms an LC resonator which can radiate the energy.
Also, remember that superconductors lose that property in a strong enough magnetic field, and then you are back to the low but finite resistance.
This type of capacitor switching is used in charge pump arrangements using switches that definitely have finite resistances that are not in the fractional ohm range. The interesting thing is that the efficiency of these circuits does not depend strongly on the switch resistance.
Switching energy from one cap to another
It can't resonate if you open my super conducting "switch" after the initial transfer of charge (or after the first have cycle)..... now where did the energy go....rshayes wrote:Even with a lossless conductor, you will still have inductance, which, with the capacitors, forms an LC resonator which can radiate the energy.
back to my time space continuum rip me thinks...
Will:
Yeah geeze, i'll have to remember not to charge one cap with another
to half the voltage level in the future unless i dont mind reducing
efficiency to less than 50 percent
Chris:
Funny thing is, if the resistance is reduced to zero how do we deal
with the resulting infinite current peak? I would think that we
couldnt get to abs zero so we would always have 'some' resistance,
and that means we're back to square one again: loss in the
resistor. Agree?
cato:
Wow yeah, i'd like to see more fonts available like that too.
On your observation about the super conductor, again, if there
is no resistance we would have an infinite current, which is not
possible so that tells us something is missing. The missing part
is (as someone else brought up) the physical distance between
parts which means there is going to be inductance, which means our
circuit defaults to 'two caps connected by an inductor', where
we get (with a perfect inductor made from a superconductor and
somehow no magnetic radiation) a perfect oscillator where the
voltage transfers from cap 1 to cap 2 and back again over and
over, and the total energy is divided up between the two caps
*and* the newly introduced inductor, and always stays at 100
percent of the original energy.
Pretty amazing i would say, but then again there is no way to
build an inductor without radiating some energy during the
energy transfers (that i know of).
If we assume that there is *no* distance between the caps, and
there is no lead length either, then we no longer have two caps,
but only one, and that means we cant charge one up independently
of the other.
So i think to summarize, if there is *any* resistance then there
is at least some loss, and if there is *no* resistance then either
we have an inductor in series with the two caps or we have only
one cap, and the case with the inductor means a perfect oscillator,
and the case of one cap we cant charge one cap without also
charging the other too so we cant even do the experiment.
On the lighter part of your reply: ha ha. I hope that doesnt happen
though if i try this he he
Take care,
Al
Yeah geeze, i'll have to remember not to charge one cap with another
to half the voltage level in the future unless i dont mind reducing
efficiency to less than 50 percent
Chris:
Funny thing is, if the resistance is reduced to zero how do we deal
with the resulting infinite current peak? I would think that we
couldnt get to abs zero so we would always have 'some' resistance,
and that means we're back to square one again: loss in the
resistor. Agree?
cato:
Wow yeah, i'd like to see more fonts available like that too.
On your observation about the super conductor, again, if there
is no resistance we would have an infinite current, which is not
possible so that tells us something is missing. The missing part
is (as someone else brought up) the physical distance between
parts which means there is going to be inductance, which means our
circuit defaults to 'two caps connected by an inductor', where
we get (with a perfect inductor made from a superconductor and
somehow no magnetic radiation) a perfect oscillator where the
voltage transfers from cap 1 to cap 2 and back again over and
over, and the total energy is divided up between the two caps
*and* the newly introduced inductor, and always stays at 100
percent of the original energy.
Pretty amazing i would say, but then again there is no way to
build an inductor without radiating some energy during the
energy transfers (that i know of).
If we assume that there is *no* distance between the caps, and
there is no lead length either, then we no longer have two caps,
but only one, and that means we cant charge one up independently
of the other.
So i think to summarize, if there is *any* resistance then there
is at least some loss, and if there is *no* resistance then either
we have an inductor in series with the two caps or we have only
one cap, and the case with the inductor means a perfect oscillator,
and the case of one cap we cant charge one cap without also
charging the other too so we cant even do the experiment.
On the lighter part of your reply: ha ha. I hope that doesnt happen
though if i try this he he
Take care,
Al
LEDs vs Bulbs, LEDs are winning.
Can't have infinite current?.....pfff...they used to say we couldn't have super conductors.....now you can buy them on the internet:
http://www.oxinst.com/wps/wcm/connect/O ... 3Sn-Bronze#
:-p
besides..... if you have infinite current, you only need it for an infinitely short period of time....so, did you really have it in the first place?
http://www.oxinst.com/wps/wcm/connect/O ... 3Sn-Bronze#
:-p
besides..... if you have infinite current, you only need it for an infinitely short period of time....so, did you really have it in the first place?
Oscillation has nothing to do with it. The Energy loss occurs during the first half cycle....when the charge is transferred between c1 and c2...swap in an infinite resistor as soon as the first half cycle is completed....the charge is trapped in the two caps....there is no oscillation....MrAl wrote:Will:
the case with the inductor means a perfect oscillator,
and the case of one cap we cant charge one cap without also
charging the other too so we cant even do the experiment.
On the lighter part of your reply: ha ha. I hope that doesnt happen
though if i try this he he
Al
oooo..... I just figured it out......I bet what happens is... at such a high current....there is like a venturi effect type suction thingy....from the collapsing infinite magnetic field that was generated from the infinite current flow ....and ALL the charge is transferred to the second cap....thereby conserving energy....and avoiding that nasty space time tear.....
Hi again cato,
Well, if you have infinite current for an infinitely short period of time (0)
you have (if you recall) an 'impulse', which only exists in pure theory, not
in real life.
Im not sure what you mean by "Oscillation has nothing to do with it".
I was talking about the case where there are two caps connected by an
inductor (the other ends of caps grounded) and one cap charged up to
some dc value (like say 10 volts) at t=0. Once the circuit is allowed
to function (perhaps with a perfect switch) the energy transfers from
the first cap to the second cap via the inductor, then back again, then
back again again, etc. The waveforms are sinusoidal, with the first
cap at zero degrees and the second cap at 180 degrees (ie out of phase
completely). The energy is divided up between the two caps and the
inductor, with the total energy sometimes being in cap 1 and sometimes
in cap 2, and when the two cap voltages pass through their 50 percent
voltage points the inductor has 50 percent of the energy stored and the
two caps have the other 50 percent (recall when the two caps are at
half voltage they each contain 25 percent of the original energy).
Pretty cool huh? I think so
About the switching supplies...
I'd like to take another look at some of the capacitor voltage booster
chips and see how much they loose when they convert one voltage up to
another voltage. Unless it's done right there could be a lot of energy
loss.
Well, if you have infinite current for an infinitely short period of time (0)
you have (if you recall) an 'impulse', which only exists in pure theory, not
in real life.
Im not sure what you mean by "Oscillation has nothing to do with it".
I was talking about the case where there are two caps connected by an
inductor (the other ends of caps grounded) and one cap charged up to
some dc value (like say 10 volts) at t=0. Once the circuit is allowed
to function (perhaps with a perfect switch) the energy transfers from
the first cap to the second cap via the inductor, then back again, then
back again again, etc. The waveforms are sinusoidal, with the first
cap at zero degrees and the second cap at 180 degrees (ie out of phase
completely). The energy is divided up between the two caps and the
inductor, with the total energy sometimes being in cap 1 and sometimes
in cap 2, and when the two cap voltages pass through their 50 percent
voltage points the inductor has 50 percent of the energy stored and the
two caps have the other 50 percent (recall when the two caps are at
half voltage they each contain 25 percent of the original energy).
Pretty cool huh? I think so
About the switching supplies...
I'd like to take another look at some of the capacitor voltage booster
chips and see how much they loose when they convert one voltage up to
another voltage. Unless it's done right there could be a lot of energy
loss.
LEDs vs Bulbs, LEDs are winning.
MrAl wrote:Hi again cato,
Well, if you have infinite current for an infinitely short period of time (0)
you have (if you recall) an 'impulse', which only exists in pure theory, not
in real life.
Im not sure what you mean by "Oscillation has nothing to do with it".
I was talking about the case where there are two caps connected by an
inductor (the other ends of caps grounded) and one cap charged up to
some dc value (like say 10 volts) at t=0. Once the circuit is allowed
to function (perhaps with a perfect switch) the energy transfers from
the first cap to the second cap via the inductor,
Stop right there. Open the switch. Whats the energy? Is there a loss? Where did it go? Thats what I meant.
However, as I mentioned, I solved the problem. When half the charge is transferred, the magneticfield of your inductor begins to collapse and forces current to continue to flow, thereby draining the first cap completely. Open the switch at that point and you trap the charge in the second cap. ALL the energy is now stored in the second cap....think of those snow board dudes on the half pipe...down one side and up the other and stopping ....
close the switch and leave it close and the current rushes back and forth (as you described) for ever...since we are talking about a lossless system....think of the snow board dudes going back and forth....
then back again, then
back again again, etc. The waveforms are sinusoidal, with the first
cap at zero degrees and the second cap at 180 degrees (ie out of phase
completely). The energy is divided up between the two caps and the
inductor, with the total energy sometimes being in cap 1 and sometimes
in cap 2, and when the two cap voltages pass through their 50 percent
voltage points the inductor has 50 percent of the energy stored and the
two caps have the other 50 percent (recall when the two caps are at
half voltage they each contain 25 percent of the original energy).
Pretty cool huh? I think so
About the switching supplies...
I'd like to take another look at some of the capacitor voltage booster
chips and see how much they loose when they convert one voltage up to
another voltage. Unless it's done right there could be a lot of energy
loss.
If you have a lossless inductor, after one half cycle all of the energy is in the second capacitor. If you try to break the circuit before the zero current point, the voltage across the switch becomes infinite.
To get zero inductance will require that the capacitors be of zero size, which for any finite amount of energy storage will require infinite energy density. The dielectric will have an infinite voltage gradient across zero thickness. Also, being of zero size means that it doesn't exist, so it can't store energy and the problem can't be posed.
It is more reasonable to allow few microohms in the shorting device. This gives a realizable problem and a realizable solution.
To get zero inductance will require that the capacitors be of zero size, which for any finite amount of energy storage will require infinite energy density. The dielectric will have an infinite voltage gradient across zero thickness. Also, being of zero size means that it doesn't exist, so it can't store energy and the problem can't be posed.
It is more reasonable to allow few microohms in the shorting device. This gives a realizable problem and a realizable solution.
-
Robert Reed
- Posts: 2277
- Joined: Wed Nov 24, 2004 1:01 am
- Location: ASHTABULA,OHIO
- Contact:
OK
We have established ideal circuit components in an ideal world.When the switch is closed, there is a transient produced by a current inrush, but it cannot be infinate because we have already established an ideal inductance in the circuit (TC = LC ). Now with these ideal components , we also have the ideal tank circuit. As long as the switch were held closed, the circuit would ring forever(oscillate) at a frequency determined by its LC time constant. In this case that would be an extremely short time. Resonance high up in the GigaHz range probably. In the the ideal tank circuit, there is no loss,so energy transfer would be 100% ( no loss) . The only loss that would occur would be radiation losses due to the RF phenomina of the frequency involved and the associated connector lengths. Since at this point,we do not know the frequency or the connection lengths,this is impossible to determine.I dont know if there would be some inherant energy loss in the constant inductor energy conversion from magnetic field to electric current, but this is more food for thought.
We have established ideal circuit components in an ideal world.When the switch is closed, there is a transient produced by a current inrush, but it cannot be infinate because we have already established an ideal inductance in the circuit (TC = LC ). Now with these ideal components , we also have the ideal tank circuit. As long as the switch were held closed, the circuit would ring forever(oscillate) at a frequency determined by its LC time constant. In this case that would be an extremely short time. Resonance high up in the GigaHz range probably. In the the ideal tank circuit, there is no loss,so energy transfer would be 100% ( no loss) . The only loss that would occur would be radiation losses due to the RF phenomina of the frequency involved and the associated connector lengths. Since at this point,we do not know the frequency or the connection lengths,this is impossible to determine.I dont know if there would be some inherant energy loss in the constant inductor energy conversion from magnetic field to electric current, but this is more food for thought.
Hello again,
cato:
Oh ok i see what you mean now. You're thinking that if you let the
inductor+two caps circuit only oscillate for 1/2 cycle and then open
the switch (switch in series with inductor) then ALL of the energy is
transferred to the second capacitor.
All i can say is: Good thinking there! I guess that means you've
solved the problem of transferring ALL of the energy with no loss
(using ideal elements of course).
The nice thing is, when we introduce a small but measureable resistance
in series with the inductor we only get a small loss at the end of
only 1/2 cycle.
rshayes:
Well, i think cato has solved the problem because guess what? It just
so happens that after 1/2 cycle there is no energy left in the inductor
(as well as none left in the first cap) which means two things:
1. All of the energy is now in the second cap
2. Since there is no energy in the inductor the current must be zero, so
there's no problem with opening the switch (as long as it's timed perfect).
Very interesting right?
ADDED LATER:
I missed your post Robert...i use a 4.0528uH inductor with two 20uf
caps, which works out to an osc frequency of 25000 Hz...
F=1/(pi*sqrt(2*L*C))
cato:
Oh ok i see what you mean now. You're thinking that if you let the
inductor+two caps circuit only oscillate for 1/2 cycle and then open
the switch (switch in series with inductor) then ALL of the energy is
transferred to the second capacitor.
All i can say is: Good thinking there! I guess that means you've
solved the problem of transferring ALL of the energy with no loss
(using ideal elements of course).
The nice thing is, when we introduce a small but measureable resistance
in series with the inductor we only get a small loss at the end of
only 1/2 cycle.
rshayes:
Well, i think cato has solved the problem because guess what? It just
so happens that after 1/2 cycle there is no energy left in the inductor
(as well as none left in the first cap) which means two things:
1. All of the energy is now in the second cap
2. Since there is no energy in the inductor the current must be zero, so
there's no problem with opening the switch (as long as it's timed perfect).
Very interesting right?
ADDED LATER:
I missed your post Robert...i use a 4.0528uH inductor with two 20uf
caps, which works out to an osc frequency of 25000 Hz...
F=1/(pi*sqrt(2*L*C))
LEDs vs Bulbs, LEDs are winning.
This is cool ! - Here is my simplistic (possibly over-simplified) There is always resistance present - even if the circuit were connected by super-conductors, the capacitor connections would still have finite resistance. If no inductance were present then the lost energy will be dissipated as heat (It always will be) - but inductance will be present (Even in a superconductor) - each time a current flows in a conductor a magnetic field is formed around the conductor (If it were not then generators and motors would not work because multiple turn cioils would have nothing to concentrate) According to Lenz' Law the nature of the magnetic field is such that it 'Opposes the force producing it ' so that this would limit the current from reaching anywhere near infinity. With finite capacitance and inductance there will be a real resonance frequency at which the circuit will oscillate until all of the energy is dissipated as thermal energy. (Think of a normal resonant 'tank' circuit being maintained by a supply voltage - when the supply voltage is cut off the circuit continues to resonate until all of the energy is dissipated. IMHO E&OE LOL
BB
If you assume the original problem, where charge is transferred from a charged capacitor to a completely discharged capacitor, the efficiency of energy transfer is 33 percent, which is pretty poor. After the transfer, three quarters of the energy originally in the first capacitor is removed, and only one forth of it is in the second capacitor.
The trick is to not remove all of the energy from the second capacitor. Assume the two capacitors are equal (they usually are in most of the charge pump circuits). Before the transfer call the voltage on the first capacitor V1 and the voltage on the second capacitor V2 (less than or equal to V1). After the transfer, the voltage on both capacitors is the average of the initial voltages, (V1+V2)/2.
The efficiency of the transfer is the energy added to C2 divided by the sum of the energy lost and the energy added to C2.
E= J[transferred]/(J[transferred]+J[lost])
E= (V1+3V2)/(3V1+V2)
set k= V2/V1
E= (1+3k)/(3+k)
If k = 1, then E = 1.
This is where the capacitors are equally charged to begin with, there is no energy transfer, and no loss, so the efficiency is 100 percent.
If k = 0, then E = 1/3.
This is the case where the second capacitor is completed discharged, and agrees with our previous result.
k = .95, E = .975
k = .90, E = .949
k = .85, E = .922
k = .80, E = .895
It appears that the process is over 90 percent efficient provided less than 20 percent of the charge on the second capacitor is removed before repeatinc the cycle.
The trick is to not remove all of the energy from the second capacitor. Assume the two capacitors are equal (they usually are in most of the charge pump circuits). Before the transfer call the voltage on the first capacitor V1 and the voltage on the second capacitor V2 (less than or equal to V1). After the transfer, the voltage on both capacitors is the average of the initial voltages, (V1+V2)/2.
The efficiency of the transfer is the energy added to C2 divided by the sum of the energy lost and the energy added to C2.
E= J[transferred]/(J[transferred]+J[lost])
E= (V1+3V2)/(3V1+V2)
set k= V2/V1
E= (1+3k)/(3+k)
If k = 1, then E = 1.
This is where the capacitors are equally charged to begin with, there is no energy transfer, and no loss, so the efficiency is 100 percent.
If k = 0, then E = 1/3.
This is the case where the second capacitor is completed discharged, and agrees with our previous result.
k = .95, E = .975
k = .90, E = .949
k = .85, E = .922
k = .80, E = .895
It appears that the process is over 90 percent efficient provided less than 20 percent of the charge on the second capacitor is removed before repeatinc the cycle.
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