Thread: Energy density in supercapacitors

In battery we have (in ideal case) one electron per atom or three elecrons (in case of aluminum) which flow from anode to cathode.In activaded carbon there should be immensely smaller amount of free electrons than in metals.I do not have exact data for carbon, but for example in Germanium it is 2.1 x 10^12/cm-3 compared to 8.4 x 10^22/cm-3 for copper.Ten orders of magnitude smaller.I guess number of free carriers in carbon should be even lower than in Germanium.Usual discription of supercap contains phrase that electrons are transfered from one plate to other.I suggest you could transfer only free electrons?Therefore ultracap made of activated carbon shoud provide us 10 billions times smaller amount of electrons flowing from anode to cathode than battery with anode made of copper.Yet the best ultracapacitors could rival lead-acid batteries in energy density.How is it possible?

Originally Posted by Stanley514

In battery we have (in ideal case) one electron per atom or three elecrons (in case of aluminum) which flow from anode to cathode.In activaded carbon there should be immensely smaller amount of free electrons than in metals.I do not have exact data for carbon, but for example in Germanium it is 2.1 x 10^12/cm-3 compared to 8.4 x 10^22/cm-3 for copper.Ten orders of magnitude smaller.I guess number of free carriers in carbon should be even lower than in Germanium.

The situation is more complicated than you assume. There are two important factors: Dielectric breakdown field, and plate surface charge density. The latter is sort of what you were getting at, but surface charge density isn't so simply related to atom density. There are built-in potentials at the surface that alter the density dramatically.

As it happens, surfaces are messy things, analysis-wise, so capacitor designers often resort to quasi-empirical methods. Perhaps the most useful of these are Hall coefficient measurements, from which one may infer charge densities. What you typically find is that there isn't a dramatic difference in the maximum surface charge densities of most metals; they typically fall within a factor of ten or so (of course, having written those words, there will have to be a headline tomorrow announcing the discovery of a material with a record-breaking charge density). As a consequence, choice of materials is often driven by other considerations (cost, ease of processing, etc.).

The situation is more complicated than you assume. There are two important factors: Dielectric breakdown field, and plate surface charge density. The latter is sort of what you were getting at, but surface charge density isn't so simply related to atom density. There are built-in potentials at the surface that alter the density dramatically.

Well,but it seems to me that energy density of electric energy storage (either battery or supercup) should be anyhow related to amount of electrons that pass through external circuit and light a bulb,for example.More electrons = more current,roughly.
There is voltage too,of course,but voltage of supercaps is usually compareable to that of Li-ion batteries - 2-5 V.

What you typically find is that there isn't a dramatic difference in the maximum surface charge densities of most metals; they typically fall within a factor of ten or so

Carbon which is used in supecap plates is not metal.It is dielectric, in difference from Germanium which is a semiconductor.It should contain maybe hundreds of billions of free electrons less than majority of metals.

Originally Posted by Stanley514

Well,but it seems to me that energy density of electric energy storage (either battery or supercup) should be anyhow related to amount of electrons that pass through external circuit and light a bulb,for example.More electrons = more current,roughly.

You're really, very confused. Don't mix up "more current" with "higher energy." I gave the correct factors in my first post: Breakdown voltage of the dielectric, and maximum surface charge density of the plates.

There's always a limit to the charge density supported by a material. At some point, you've crowded so many electrons in a given area that they repel additional ones. The maximum density depends on lots of factors, like surface work function and topography.

You're really, very confused. Don't mix up "more current" with "higher energy."

Well, but is it true that supercups suppose to give you many billion times less current than batteries?

Originally Posted by Stanley514

You're really, very confused. Don't mix up "more current" with "higher energy."

Well, but is it true that supercups suppose to give you many billion times less current than batteries?

No. Where did you read that? How was the comparison made? Again, current by itself isn't relevant. I can get a huge current from either batteries or supercaps. I can configure the test to make either larger than the other. The key isn't current, it's energy. I can always use circuits to trade V for I at will.

Any other qualified mind on subject?

Originally Posted by Stanley514

Any other qualified mind on subject?

While you're waiting for ... a second opinion, just do some simple calculations yourself. Today's supercapacitors have resistances in the milliohm range. Anyone asserting that there is a factor of "billions" increase in current in the offing must explain how this can be achieved. The voltages cannot be increased by many orders of magnitude, nor can the resistance be practically reduced by many orders of magnitude. There's simply no practical way that factors of billions increase can be achieved. The flip side of the argument is that there is no way that supercapacitors are inferior to batteries by that same factor of billions. That's why I asked you to cite the source of your belief. You've either misinterpreted what was said, or that source is unreliable.

That's why I asked you to cite the source of your belief.

It"s quite easy to find info that carbon which is used as material of choice for supercaps is non-metal Carbon - Wikipedia, the free encyclopedia and its resistivity suppose to be higher than Germanium which is metalloid and who's resistivity is (20 °C) 1 Ω·m.
Germanium - Wikipedia, the free encyclopedia
Carrier concentration in Germanium is 2.1 x 10^12/cm-3 vs. 8 x 4^22/cm -3 for copper.
https://docs.google.com/viewer?a=v&q...Lc2UiIXUfxxwhQ
Page 8 , tab 1-1.
Obviously, you cannot transfer electrons from one plate to the other if they aren't free.You whould have to ionize carbon fist.
But even if carbon would have the same carrier concentration as metals, how many positive or negative ions per carbon atoms they have available?

Originally Posted by Stanley514

That's why I asked you to cite the source of your belief.

It"s quite easy to find info that carbon which is used as material of choice for supercaps is non-metal Carbon - Wikipedia, the free encyclopedia and its resistivity suppose to be higher than Germanium which is metalloid and who's resistivity is (20 °C) 1 Ω·m.
Germanium - Wikipedia, the free encyclopedia
Carrier concentration in Germanium is 2.1 x 10^12/cm-3 vs. 8 x 4^22/cm -3 for copper.
https://docs.google.com/viewer?a=v&q...Lc2UiIXUfxxwhQ
Page 8 , tab 1-1.
Obviously, you cannot transfer electrons from one plate to the other if they aren't free.You whould have to ionize carbon fist.
But even if carbon would have the same carrier concentration as metals, how many positive or negative ions per carbon atoms they have available?

Please (re)read my answer in post 2. You seem to be ignoring it. You really should read it, because it's basically right.

I suggest that to obtain sufficient power in electric circuit you need to have either high voltage or high current.
I do not know about any more factors.

Originally Posted by Stanley514

I suggest that to obtain sufficient power in electric circuit you need to have either high voltage or high power.
I do not no about any more factors.

Nope. It looks like it's time for EE101.

Voltage is not power. Power is not energy.

Power is the time rate of change in energy. A small amount of energy can still be used to generate a lot of power (just not for a long time).

In DC circuits, power = V*I, and energy is V*I*t.

For a capacitor, the energy stored can be written several ways. One is 0.5*C*V^2. Since Q=C*V, we can also write that as energy = 0.5*Q*V. From this last form, we can see that the energy stored by a capacitance -- any capacitance -- is maximized if we can maximize the product of the charge stored and the voltage. If we can maximize the breakdown voltage and the charge density, then we maximize the energy density. Now perhaps you can understand why I wrote what I wrote.

Next, how does one maximize charge density? You are making the assumption that charge density is somehow directly related to the density of atoms (or the electrons you can get from them). But it's much more complicated than that. Remember: Electrons repel each other. At some point, you pile up so many that the next one would rather just leave than get jammed in with the others. That's why I brought up the stuff about estimating charge densitiies from Hall measurements (or some other method, but that one works well).

The bottom line is that you can't just say that Metal X is better than Metal Y because it has more ionizable atoms per unit area. It's not nearly so simple.

Originally Posted by Stanley514

Any other qualified mind on subject?

I can't see anything wrong with what tk421 said.

I don't know a much about how supercaps work, but their main advantage is that they can be charged and discharged much more rapidly than batteries so they are used for temporary energy storage in electric vehicles, for example.

My understanding is that they do not have greater energy density than batteries but they have much higher power density because they can charge and discharge very rapidly. This means they can be used to provide high peak currents that would probably damage a battery.

In DC circuits, power = V*I, and energy is V*I*t.

Correct power=Voltage x current.
Voltage of supercap = 2-5 V. Comparable to battery.
Current is ??? Could you provide some estimation?
Sorry but to deny that current is flow of electrons through cirtain point at
certain time is meaningless.

In DC circuits, power = V*I, and energy is V*I*t.

Correct power=Voltage x current.
Voltage of supercap = 2-5 V. Comparable to battery.
Current of supercap is ??? Could you provide some estimation?
Sorry, but to deny that current is flow of electrons through a certain point at
a certain time is meaningless.

The bottom line is that you can't just say that Metal X is better than Metal Y because it has more ionizable atoms per unit area. It's not nearly so simple.

In a battery every metal that used on anod is typically solves in electrolyte.Therefore most of it ionizes regardless of metal kind.Some materials of course could have worse solubility.

Originally Posted by Stanley514

Current is ??? Could you provide some estimation?

Current depends on the load. It has nothing to do with the battery or supercap (other than limitations based on internal resistance, etc.)

Originally Posted by Stanley514

In DC circuits, power = V*I, and energy is V*I*t.

Correct power=Voltage x current.
Voltage of supercap = 2-5 V. Comparable to battery.
Current of supercap is ??? Could you provide some estimation?
Sorry, but to deny that current is flow of electrons through a certain point at
a certain time is meaningless.

You really, really, really need to study some basic electricity, and some basic logic.

First, you asked about energy density in supercapacitors. I answered, and also corrected several deep misconceptions in what you wrote.

Now you are complaining that I didn't talk about current flow. But as your question wasn't about that -- it was about energy storage -- I don't see a basis for your complaint.

Sheesh!

Obviously, the current delivered will be, as Strange said, a function of the load and of any resistances in the wiring (including in the plates of the capacitor). There's a famous formula that gives us the current: I = V/R. You may have seen this in the form V = IR. It's the same equation (it's called Ohm's law), but just rearranged to solve for current (I).

Now if you charge up a capacitor to some voltage V, and then connect a load to it (we'll use a resistor R to represent the collective action of the load and wiring resistance), then the current will start at a value V/R, and decay exponentially toward zero. If you want higher peak currents, charge to a higher voltage and reduce R. If you want all that to happen over longer times (to deliver more energy), then use more capacitance.

That's it.