Switching energy from one cap to another
Switching energy from one cap to another
Hello there,
It's interesting that if you try to switch energy from one cap
to another with a 'low resistance' switch there is always a
loss of energy involved. In particular, if you try to switch
the two caps in parallel when one cap has voltage V and the
other cap has zero volts across it you loose *half* the total
energy! The energy is lost in the resistance of the switch,
no matter how low it is, and decreasing the resistance doesnt
help save energy.
Here is the diagram:
http://www.smpstech.com/charge.htm
(see fig 1)
Take care,
Al
It's interesting that if you try to switch energy from one cap
to another with a 'low resistance' switch there is always a
loss of energy involved. In particular, if you try to switch
the two caps in parallel when one cap has voltage V and the
other cap has zero volts across it you loose *half* the total
energy! The energy is lost in the resistance of the switch,
no matter how low it is, and decreasing the resistance doesnt
help save energy.
Here is the diagram:
http://www.smpstech.com/charge.htm
(see fig 1)
Take care,
Al
LEDs vs Bulbs, LEDs are winning.
Hi,
I think this is rather natural.
If you compare the capacitors with two buckets, one full and one empty, you end up with two half full buckets if you 'connect' them. The total amount of water is the same but the volume is twice as big.
TOK
I think this is rather natural.
If you compare the capacitors with two buckets, one full and one empty, you end up with two half full buckets if you 'connect' them. The total amount of water is the same but the volume is twice as big.
TOK
Gorgon the Caretaker  Character in a childrens TVshow from 1968.
There is a similar relationship when you charge a capacitor through a resistor from a fixed voltage. Half of the energy removed from the source is dissipated in the resistor and the other half remains in the capacitor. Again, this is true for all values (except zero) of resistance and capacitance.
The energy loss is less if the capacitor is initially charged to a voltage near the the supply voltage. This is what allows charge pumps to run at reasonable efficiencies. The trick is to use large enough capacitors so that the load current only discharges them slightly during a cycle. The down side is that the current flows in short pulses and can be a significant noise source on a circuit board.
The energy loss is less if the capacitor is initially charged to a voltage near the the supply voltage. This is what allows charge pumps to run at reasonable efficiencies. The trick is to use large enough capacitors so that the load current only discharges them slightly during a cycle. The down side is that the current flows in short pulses and can be a significant noise source on a circuit board.
An interesting proposition ? _ I may not be seeing the point here but I thought all of this to be simple i.e. if you switch charge between two capacitors then what remains constant is not the energy (In Joules) but the charge (In coulombs) whatever you do or whatever resistance you put in circuit  you will lose the difference between CV^2/2 (the stored energy) and CV (The stored charge)  Is this not so ?
BB
Hi again,
Yes, and also the case where one cap is larger than
the other. If the 0v cap is 2uf and the charged cap is 20uf then
the energy transfered is over 90 percent. The voltage doesnt
drop as much either, to approx 90 percent.
I guess what is not intuitive about the equal caps case is, it seems
that if you reduce the switch resistance to zero you get perfect
transfer...only problem is, in real life there is no such thing as
a perfect switch so there will always be loss. Interesting also
is that that loss is always 50 percent...you would think that the
more you reduce the switch resistance the less loss there would
be, but the loss is constant; 0.1, 0.01, and 0.001 ohms all give
the same results
Take care,
Al
Yes, and also the case where one cap is larger than
the other. If the 0v cap is 2uf and the charged cap is 20uf then
the energy transfered is over 90 percent. The voltage doesnt
drop as much either, to approx 90 percent.
I guess what is not intuitive about the equal caps case is, it seems
that if you reduce the switch resistance to zero you get perfect
transfer...only problem is, in real life there is no such thing as
a perfect switch so there will always be loss. Interesting also
is that that loss is always 50 percent...you would think that the
more you reduce the switch resistance the less loss there would
be, but the loss is constant; 0.1, 0.01, and 0.001 ohms all give
the same results
Take care,
Al
LEDs vs Bulbs, LEDs are winning.
Further thoughts
Is that really the case ?? In physics/mechanics a similar case exists for bodies in motion i.e. if two bodies (Moving in exactly the same direction, one faster than the other  makes a better example) collide then the resultant combined velocity comes somewhere tne Momentum (MV = lbf * ft/sec = ft.lbf/sec = power = the rate of doing work) ) and the Kinetic Energy (CV^2/2 = ft.lbf = energy). The 'somewhere between' is dependant upon the Coefficient of Recovery which is dependant upon the elasticity of the colliding bodies  If the elasticity is perfect the KWE survives  if the bodies are totally rigid then only the momentum survives. That begs the question  'Is there a similar 'somewhere in between' for an electric charge being shared between two capacitors ?
BB
Charge is conserved in this case.
Paralleling a charged capacitor with an equal uncharged capacitor results in half the charge in each capacitor. This results in half the voltage, which means each capacitor has one quarter of the original energy, for a total of one half the original energy. The other half of the original energy is dissipated in the switch.
Consider three capacitors, one charged and the other two discharged. Paralleling these three capacitors results in one third the voltage, and one ninth the original energy in each capacitor for a total of one third the original energy. In this case, the loss is two thirds.
If the charged capacitor is C1 and it is paralleled by a discharged capacitor C2, the the fraction of the energy lost is;
J[lost]/J[original] = C2/(C1+C2)
Even for the case of two equal capacitors, the energy lost is double the energy transferred.
Paralleling a charged capacitor with an equal uncharged capacitor results in half the charge in each capacitor. This results in half the voltage, which means each capacitor has one quarter of the original energy, for a total of one half the original energy. The other half of the original energy is dissipated in the switch.
Consider three capacitors, one charged and the other two discharged. Paralleling these three capacitors results in one third the voltage, and one ninth the original energy in each capacitor for a total of one third the original energy. In this case, the loss is two thirds.
If the charged capacitor is C1 and it is paralleled by a discharged capacitor C2, the the fraction of the energy lost is;
J[lost]/J[original] = C2/(C1+C2)
Even for the case of two equal capacitors, the energy lost is double the energy transferred.

 Posts: 1752
 Joined: Fri Aug 22, 2003 1:01 am
 Location: Izmir, Turkiye; from Rochester, NY
 Contact:
I'm pretty sure zero energy is lost with perfect capacitors, and a perfect switch (infinite Ohms off, zero Ohms on).
Total charge remains the same, and voltage is half as said above. But the RC time constant is doubled relative to a particular resistive load. All the energy is still there, it just takes longer at a lower power to flow through a load.
I once worked out an inductorless stepdown switching supply where 2 caps where charged in series, then switched to parallel to charge a load's filter cap. Worked, but wasn't worth the trouble with losses in steering doiodes and transistors, etc.
Cheers,
Total charge remains the same, and voltage is half as said above. But the RC time constant is doubled relative to a particular resistive load. All the energy is still there, it just takes longer at a lower power to flow through a load.
I once worked out an inductorless stepdown switching supply where 2 caps where charged in series, then switched to parallel to charge a load's filter cap. Worked, but wasn't worth the trouble with losses in steering doiodes and transistors, etc.
Cheers,
Dale Y
A lossless solution does not conserve charge. If the initial energy was divided equally between the two capacitors, each capacitor would be at 70.7 percent of the original voltage. Since charge is proportional to voltage, the charge in each capacitor would have to be 70.7 percent of the total original charge. This would result in a charge 41.4 percent higher than the original charge, violating the principle of conservation of charge.
When you connected the ideal capacitors together, the voltages would have to change in zero seconds, due to a zero ohm resistance and a zero henry inductance. The current between the capacitors would be infinite. the rate of current rise would be infinite, as would the rate of voltage rise. This would radiate, except that the ideal parts also have zero physical dimensions, which would imply infinite energy density. As a ideal problem, there are too many zeroes and infinities to make much sense, unless you are trying to trigger a nucleur fusion reaction.
When you connected the ideal capacitors together, the voltages would have to change in zero seconds, due to a zero ohm resistance and a zero henry inductance. The current between the capacitors would be infinite. the rate of current rise would be infinite, as would the rate of voltage rise. This would radiate, except that the ideal parts also have zero physical dimensions, which would imply infinite energy density. As a ideal problem, there are too many zeroes and infinities to make much sense, unless you are trying to trigger a nucleur fusion reaction.

 Posts: 1752
 Joined: Fri Aug 22, 2003 1:01 am
 Location: Izmir, Turkiye; from Rochester, NY
 Contact:
True. Tomarrow is Saturday, I'll try work out an example with real/typical values. Based on MrAl's first post, it looked like he's confusing half voltage after connecting 2nd cap with half the energy being gone. It's not gone, it is spread out differently; of course with some disapated as heat of the real switch and wire.
Dale Y
There is an interesting symmetry if the two capacitors are connected by a finite resistor. At time zero, the middle of the resistor is at half of the initial voltage on the charged capacitor. As charge is transferred from the initially charged capacitor, the voltage decreases. The same charge is added to the discharged capacitor, so its voltage increases the same amount. The center of the resistor thus remains at one half the initial voltage. The final condition, when time is infinite, has both capacitors charged to one half the initial voltage. At this point, both ends and the center of the resistor are at one half the initial voltage.
Since the center does not change voltage, the resistor may be split in half and the center attached to a voltage source of one half the initial voltage. The uncharged capacitor then charges as in a normal RC circuit, with a voltage of one half the initial voltage and a resistence of one half the total resistance. In this case, the energy disssipated in one half of the resistor is equal to the energy transferred to the capacitor at time equal to infinity.
Since the current is the same in the other half of the resistor, it will also dissipate the same energy. At the end, the resistor has dissipated twice the energy transferred to the second capacitor, and the first capacitor contains the same energy as the second capacitor. The total energy is four times the energy transferred to the second capacitor and is equal to the initial energy in the first capacitor.
This might even work with unequal capacitors, but I'm not sure of that right now.
Since the center does not change voltage, the resistor may be split in half and the center attached to a voltage source of one half the initial voltage. The uncharged capacitor then charges as in a normal RC circuit, with a voltage of one half the initial voltage and a resistence of one half the total resistance. In this case, the energy disssipated in one half of the resistor is equal to the energy transferred to the capacitor at time equal to infinity.
Since the current is the same in the other half of the resistor, it will also dissipate the same energy. At the end, the resistor has dissipated twice the energy transferred to the second capacitor, and the first capacitor contains the same energy as the second capacitor. The total energy is four times the energy transferred to the second capacitor and is equal to the initial energy in the first capacitor.
This might even work with unequal capacitors, but I'm not sure of that right now.
Hello again,
Will:
I missed your first post because i was writing when you posted i guess.
Your second post is interesting...but this is 'really the case' because
there is energy lost due to the resistance, even though it's small.
To work up an elastic equivalent i think we would replace the resistance
(the switch or just a plain resistor like 0.01 ohms) with an inductor.
If the inductance is ideal, the circuit oscillates, with the energy
being transferred from C1 to C2 and then from C2 to C1, and this repeats.
With ideal elements, there is no loss (of course), but we have to
remember that this isnt possible in the real world. In the real world
we always have some resistance, or some energy take off which is still
loss due to resistance. This means we would at least have to insert
a small resistance in series with the inductor, which brings us back
to square one: we loose half the energy in the resistor.
dyarker:
You may be right in that zero energy is lost with perfect capacitors and a
perfect switch, but the problem is there is no such thing. Even if there
was, closing the switch would lead to an infinite current peak which also
is not possible. There has to be something limiting the current, and whatever
limits it, eats some of it. This is why we end up losing half the energy
even though the resistance (and cap esr=0) is very very small like 0.01 ohms.
rshayes:
Yes, exactly! The infinite current alone is impossible.
dyarker:
No im not confusing anything I can provide the proof i guess if i sit
down and write out the equations. There's only two caps and one resistor,
and at t=0 only one cap has a voltage across it and the other is zero.
rshayes:
Yes that's an interesting view because it forces us to look at what is
happening in the resistor itself...i would even use two resistors:
both of which are 0.01 ohms (for a total of 0.02 ohms, still very low)
and look at the voltage at the junction of the two resistors.
This helps understand the loss in the resistor...as the resistance goes
down the charge is transferred faster, but the peak current goes up so
the loss stays the same.
Take care,
Al
Will:
I missed your first post because i was writing when you posted i guess.
Your second post is interesting...but this is 'really the case' because
there is energy lost due to the resistance, even though it's small.
To work up an elastic equivalent i think we would replace the resistance
(the switch or just a plain resistor like 0.01 ohms) with an inductor.
If the inductance is ideal, the circuit oscillates, with the energy
being transferred from C1 to C2 and then from C2 to C1, and this repeats.
With ideal elements, there is no loss (of course), but we have to
remember that this isnt possible in the real world. In the real world
we always have some resistance, or some energy take off which is still
loss due to resistance. This means we would at least have to insert
a small resistance in series with the inductor, which brings us back
to square one: we loose half the energy in the resistor.
dyarker:
You may be right in that zero energy is lost with perfect capacitors and a
perfect switch, but the problem is there is no such thing. Even if there
was, closing the switch would lead to an infinite current peak which also
is not possible. There has to be something limiting the current, and whatever
limits it, eats some of it. This is why we end up losing half the energy
even though the resistance (and cap esr=0) is very very small like 0.01 ohms.
rshayes:
Yes, exactly! The infinite current alone is impossible.
dyarker:
No im not confusing anything I can provide the proof i guess if i sit
down and write out the equations. There's only two caps and one resistor,
and at t=0 only one cap has a voltage across it and the other is zero.
rshayes:
Yes that's an interesting view because it forces us to look at what is
happening in the resistor itself...i would even use two resistors:
both of which are 0.01 ohms (for a total of 0.02 ohms, still very low)
and look at the voltage at the junction of the two resistors.
This helps understand the loss in the resistor...as the resistance goes
down the charge is transferred faster, but the peak current goes up so
the loss stays the same.
Take care,
Al
LEDs vs Bulbs, LEDs are winning.
Interesting !  Without disagreeing with anyone specific the facts are (i) Energy is always lost in such transfers  more below on that (ii) Irrespective of series resistance, inductors etc (They only alter the shape/time of equilibration) the eventual charge is always the same (same no. of electrons) as the original charge and, since V = Q/C the voltage is reduced per the ratio c1/(c2 + c1) and, since E = CV^2/2 then, if c1 + c2 = n*c1, V2 = V1/n and E2 = n.C1.(v1/n)^2/2 = C1.V1^2/2n then, for (C2 + C1)/C1 = 3  E2 =C1.v1^2/6 = E2/3
BB
 Chris Smith
 Posts: 4325
 Joined: Tue Dec 04, 2001 1:01 am
 Location: Bieber Ca.
With a charged cap we have
U = 1/2 C (Vsquared) ....jeez with the fancy new board, you'd think we could have superscripts.....
assuming the second cap is the same value and they are connected by a super conductor..... the capacitance doubles and the voltage is cut in half (as I recall...somebody check me on that) so, after the connection.....we have:
U = 1/2 (2C) ((1/2 V) squared) = C((1/2 V) squared) = 1/4C(Vsquared)
which as the original poster indicated, is half the energy.....
The question is, where the hell did the other half of the energy go..... not into heating the super conductor.... there is no resistance, therefore, no heating.....
I suspect the instantaneous and infinite current flow causes a magnetic field so strong that there is a rip in the space time continuum and changes the cosmological constant accelerating the expansion of the universe to the point where the weak force is weakened so that every atom in the experimenters body is ripped apart by the electrical attraction of protons and electrons.....
he wont do that again any time soon.
U = 1/2 C (Vsquared) ....jeez with the fancy new board, you'd think we could have superscripts.....
assuming the second cap is the same value and they are connected by a super conductor..... the capacitance doubles and the voltage is cut in half (as I recall...somebody check me on that) so, after the connection.....we have:
U = 1/2 (2C) ((1/2 V) squared) = C((1/2 V) squared) = 1/4C(Vsquared)
which as the original poster indicated, is half the energy.....
The question is, where the hell did the other half of the energy go..... not into heating the super conductor.... there is no resistance, therefore, no heating.....
I suspect the instantaneous and infinite current flow causes a magnetic field so strong that there is a rip in the space time continuum and changes the cosmological constant accelerating the expansion of the universe to the point where the weak force is weakened so that every atom in the experimenters body is ripped apart by the electrical attraction of protons and electrons.....
he wont do that again any time soon.
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