Author Topic: Why trying to store energy in a capacitor can be less efficient than you think  (Read 41415 times)

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Offline IanB

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It was suggested I start a new thread with this rather than having it lost in the noise of the free energy thread. It shows us that we can't "just store energy in a capacitor". There are important factors to consider if attempting to use a capacitor to capture energy is not to result in huge system losses.

We see that naively attempting to charge a capacitor from a voltage source is going to result in the loss of half the energy supplied (a 50% system efficiency):


« Last Edit: January 30, 2015, 02:21:36 am by IanB »
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Online Zucca

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Interestig!

BUT slow down a little bit....

I am still thinking the above is applicable only if R is different from 0 or infinite.
In theory with R=0 there is no lost energy, neither with R=infinite (here there will be no energy transfer at all).

Now let´s put an Rp in parellel to the C and call the R above Rs.
The lost energy will be:

1/2 C [V*Rp/(Rp+Rs)]²

{I did no calculations for what I wrote; just used the Thevenin/Norton theorem}

Now you can`t eliminate the Rs anymore.

So.... I would say since in reality there will be always an Rp in parallel to the C to optimize the energy transfer you want Rp as big as possible and Rs as small as possible. fuuu... the world makes sense again?

BTW the 1/2 max energy efficiency in an real power source (i.e. with Rs in serie in a V) is a common result. I think even with inductors or other stuff it would be the same. Maybe just generalize with an impedance and you got the above results too...

It could be I am wrong. In this case sorry in advance.

« Last Edit: January 29, 2015, 06:24:05 pm by zucca »
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Offline Galaxyrise

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When you posted that before, you added a comment about not charging from a constant voltage source if you want efficiency.  That got me curious what the efficiency is from a constant current supply.  Let's see if I can avoid careless mistakes...

Charging capacitance C to voltage V with current I will take CV/I seconds.  The power dissipated by the resistance is I2R, so energy is I2RCV/I = IRCV joules. 

So trickle charging a capacitor from a photo cell might be decently efficient?
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Online Zucca

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So trickle charging a capacitor from a photo cell might be decently efficient?

It depends on how you put on paper reality. Surely charging a C with a photo cell is not a stupid idea, because a photo cell can be modelized very well with an ideal I source...
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Offline Galaxyrise

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I am still thinking the above is applicable only if R is different from 0 or infinite.
In theory with R=0 there is no lost energy, neither with R=infinite (here there will be no energy transfer at all).
Yeah, it's got a R/R in it, so it's not defined with R = 0 or R = infinity.  I stopped and thought about R=0 for awhile, too.  It doesn't ever apply to reality, though: you can't have infinite current even with superconductors.

If you're going to add in the parasitics like parallel resistance, you should consider inductance and rise time as well; but they all only serve to make the loss larger; the moral of "less efficient than you think" still holds!
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Offline AG6QR

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So trickle charging a capacitor from a photo cell might be decently efficient?

It depends on how you put on paper reality. Surely charging a C with a photo cell is not a stupid idea, because a photo cell can be modelized very well with an ideal I source...

Assume the solar cell is an ideal one, a constant current source, at any voltage from 0 up to Vmax.  If you connect that to a capacitor and charge the capacitor, then yes, there will be no energy loss in the electrical circuitry between the solar cell and the capacitor.  All the electrical energy produced by the cell will go into the capacitor.  No loss.

Or is there loss?

The solar cell will have been producing half of its rated power over the charge cycle.  The power is V*I, so when V is low, most of the solar radiation falling on the cell is not being converted to electrical energy.  In order for the cell to produce maximum rated power, it must output its constant current at Vmax.

You have still lost half of the available solar energy.  It's just that it wasn't wasted in resistance of the circuit; it was wasted by having your constant current source operating at, on average, half of its rated voltage.
 

Offline dannyf

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Quote
While simplifies to:

No, it doesn't.

Quote
Therefore the energy loss from power dissipated in the system resistance is equal to the power stored in the capacitor, and is independent of the resistance

All you need to do is to step back and read that sentence of yours and think, big picture, what it means. It means that regardless of your component parameters, such a system is always 50% efficient.

Just think about that for sanity's sake and you will realize that something is terribly wrong here.
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Offline IanB

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All you need to do is to step back and read that sentence of yours and think, big picture, what it means. It means that regardless of your component parameters, such a system is always 50% efficient.

Just think about that for sanity's sake and you will realize that something is terribly wrong here.

Perhaps you can point to the mistake?
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Online Zucca

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No, it doesn't.

Please provide the correct result...
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Offline suicidaleggroll

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It's just an oversimplification of the problem.  It is only true if you have an ideal voltage source capable of infinite current with zero voltage drop.  In reality, as you start drawing current from the source, the voltage will dip and it will look more and more like a current source (at least until it blows), which breaks the assumptions made in the first post.

I agree though, that for an ideal voltage source, the system efficiency will be 50%.  A similar argument could be made for batteries as well though.  If you want to charge a battery efficiently, you don't drive it with a fixed voltage source through a resistor, you use constant current up to a set point and then CV afterward, same with a capacitor, if energy storage and efficiency is your goal at least.
 

Offline IanB

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It's just an oversimplification of the problem.  It is only true if you have an ideal voltage source capable of infinite current with zero voltage drop.  In reality, as you start drawing current from the source, the voltage will dip and it will look more and more like a current source (at least until it blows), which breaks the assumptions made in the first post.
Note that the dotted line box represents a "real" voltage source, where R represents the internal resistance of the source. Current flowing through R (non-zero) will cause the observed voltage to dip at the output terminals as expected. Since the resistance of the system is non-zero, the current will never be infinite.

Quote
I agree though, that for an ideal voltage source, the system efficiency will be 50%.  A similar argument could be made for batteries as well though.  If you want to charge a battery efficiently, you don't drive it with a fixed voltage source through a resistor, you use constant current up to a set point and then CV afterward, same with a capacitor, if energy storage and efficiency is your goal at least.
As shown in the analysis, the efficiency will be 50% even with real voltage sources.

Yes, a constant current source will improve matters considerably, but it must be an active source designed to minimize internal losses.
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Offline Hugoneus

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This is a very classic example of charging a capacitor. The system will dissipate as much energy as it will store. Nothing wrong with the calculations.
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Offline suicidaleggroll

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Note that the dotted line box represents a "real" voltage source, where R represents the internal resistance of the source. Current flowing through R (non-zero) will cause the observed voltage to dip at the output terminals as expected. Since the resistance of the system is non-zero, the current will never be infinite.
Sorry, hadn't had lunch yet, my brain was off in left field

Yes, a constant current source will improve matters considerably, but it must be an active source designed to minimize internal losses.
But this is no different than any other storage device.  You don't just dump a voltage source through a resistor into your storage system if you care about efficiency.  The efficiency for a capacitor will be 50% if you do this, but it's not going to be much better for a battery either.  The difference is a capacitor's voltage increases linearly from 0 with a CC input, while a battery's voltage increases rapidly from 0 to the nominal voltage, where it stays through the majority of the charge cycle, before increasing rapidly at the end.  So the average voltage for the capacitor during the charge cycle is half of the nominal voltage (meaning the other half is burned off in the resistor), while for a battery it might be more like 80% (meaning the other 20% is burned off in the resistor).  Either way it's crap if you care about efficiency.
« Last Edit: January 29, 2015, 07:41:02 pm by suicidaleggroll »
 

Offline Hugoneus

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But this is no different than any other storage device.  You don't just dump a voltage source through a resistor into your storage system if you care about efficiency.  The efficiency for a capacitor will be 50% if you do this, but it's not going to be much better for a battery either.

Yes, I think the reason this example is so popular is that even if you let R approach zero, the system would still lose 50% of its delivered energy (in that case mostly in the form of broadband EM radiation and sound).
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Online Zucca

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Or is there loss?

Yeah, it is basically right what you say. BTW I meant a solar cell can be modelized with a Is source and a Rp in parallel. The Vmax you mention is just Vmax=Is*Rp, in other words the open circuit voltage.

OFF TOPIC: This web site is fantastic to understand solar cells: http://pveducation.org/pvcdrom
« Last Edit: January 29, 2015, 07:46:36 pm by zucca »
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Offline IanB

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Yes, I think the reason this example is so popular is that even if you let R approach zero, the system would still lose 50% of its delivered energy (in that case mostly in the form of broadband EM radiation and sound).

I think this example is interesting because it may run counter to intuition and can present a trap for the unwary. For example, what will it do to your system if you try to put a capacitor across the output of a PWM to smooth the voltage...?
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Offline Hugoneus

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Yes, I think the reason this example is so popular is that even if you let R approach zero, the system would still lose 50% of its delivered energy (in that case mostly in the form of broadband EM radiation and sound).

I think this example is interesting because it may run counter to intuition and can present a trap for the unwary. For example, what will it do to your system if you try to put a capacitor across the output of a PWM to smooth the voltage...?

It will smooth the PWM alright!   :scared:
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Offline dannyf

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Two points for you to think about:

Quote
The capacitor and constant voltage 50% power loss is very classic in power electronics.

As the power loss is independent of the resistance in the circuit, what happens to the other 50% of energy when you charge up an ideal capacitor with an ideal power source?

Quote
The best way to charge capacitors efficiencly is constant current charging.

When you increase R, a voltage source approaches a current source. So if you believe that the 50% efficiency is independent of the resistance, charging a battery with a current source is every bit as inefficient as charging a battery with a voltage source.

Enjoy.
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Offline AG6QR

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Two points for you to think about:

Quote
The capacitor and constant voltage 50% power loss is very classic in power electronics.

As the power loss is independent of the resistance in the circuit, what happens to the other 50% of energy when you charge up an ideal capacitor with an ideal power source?

It's dissipated in the resistor.

Changing the value of resistance changes the rate at which the charging happens, as well as changing the rate at which energy is dissipated.  It changes both rates in equal proportions.  Since power is rate of energy (more precisely, it's the derivative of energy with respect to time), changing the resistance changes the power dissipated in the resistor, as well as changing the power delivered to the capacitor.  But changing the resistance does not change the ratio of those two powers, as integrated over the charge cycle.
 

Offline Hugoneus

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Two points for you to think about:
Quote
The capacitor and constant voltage 50% power loss is very classic in power electronics.
As the power loss is independent of the resistance in the circuit, what happens to the other 50% of energy when you charge up an ideal capacitor with an ideal power source?

Mathematically this would mean an infinitely high pulse of current which has an infinitely short duration. In real life the resistance can approach zero, but never reach zero. In the case of very small resistance, not all of the power loss would be thermal. Most of it would be in the form of broadband EM emission, light and sound.

Quote
Quote
The best way to charge capacitors efficiencly is constant current charging.
When you increase R, a voltage source approaches a current source. So if you believe that the 50% efficiency is independent of the resistance, charging a battery with a current source is every bit as inefficient as charging a battery with a voltage source.
Enjoy.

This is incorrect. A voltage source with a large series resistor is not a current source.
« Last Edit: January 29, 2015, 08:47:52 pm by Hugoneus »
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Offline c4757p

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:blah:

Would it kill you to quote with the author's name so it's a bit easier to piece together the conversation without re-reading the page every time somebody adds a comment?
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Online tggzzz

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I find it confusing that you intermix power and energy willy-nilly.

Why do you start with an equation of how much power is dissipated in a resistor? Starting with how much energy is dissipated is far more appropriate.
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Offline IanB

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The capacitor and constant voltage 50% power loss is very classic in power electronics.

I first became aware of this phenomenon yesterday, so I'm gratified to learn that it is "common knowledge" among those whose job it is to know such things.

It puts me in mind of this particular comic:

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Offline IanB

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I find it confusing that you intermix power and energy willy-nilly.

Why do you start with an equation of how much power is dissipated in a resistor? Starting with how much energy is dissipated is far more appropriate.

It is a document editing error. I will fix it up later. I was trying to say that total power loss is power integrated is energy, but the wording came out badly.
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Offline Galaxyrise

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When you increase R, a voltage source approaches a current source. So if you believe that the 50% efficiency is independent of the resistance, charging a battery with a current source is every bit as inefficient as charging a battery with a voltage source.
You've made an inefficient current source that way, but hey, at least there's almost no energy loss outside your current source ;)
« Last Edit: January 29, 2015, 09:04:42 pm by Galaxyrise »
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