Capacitor Questions (Voltage and Dielectric)

[Physicist97]

Ok so my question(s) has more than one part.
First, the energy stored in capacitor is $(1/2)CV^{2}$ , is the $V$ the voltage of the source that charged the capacitor?? If I had two identical capacitors, and charged one using a 9V source and one using a 12V source, would the one charged by a 12V source have more energy?

Second, what would be a material with a high dielectric constant, and how much would it cost by the kilo? I've been looking around and have found materials like Lead Magnesium Niobate, which has a high dielectric constant, but I can't find prices for these materials.

Thanks for any help!

 

[meBigGuy]

Can't help with the materials.

V is the voltage across the capacitor, period. It makes no difference how the charge got there. For example maybe you charge for X seconds through R, and then discharge Y seconds through R2. The voltage across the capacitor whan you are done is a measure of how much energy is in the capacitor.

 

[Jeff Rosenbury]

The source of the voltage is irrelevant once the capacitor is removed from the source. Thus 100 capacitors could be charged by the same 1V source, then disconnected and reconnected in series to give 100V. (Remember capacitors in series add reciprocally.)

The 1/2CV² sounds right to me.

My brother used to use granular titanium dioxide for its high dielectric constant, low loss coefficient at microwave frequencies, and low cost. But there are better materials if you ante up the big bucks. Solids are also generally better, but harder to work with.

Materials selection is all about the trade offs. Since we don't know your application, we can't really help much.

 

[sophiecentaur]

Apart from a bit of fun, is it really worth trying to manufacture a high capacitance Capacitor? You're talking in terms of needing some high tech equipment for this sort of thing. Price per unit could be pretty high. What sort of construction method were you planning on?

 

[Physicist97]

Hey guys, thanks for the responses! I asked about the voltage of a capacitor because I was looking at Kirchoff's Laws, and how the voltage drop around any circuit is zero, so, for example, a simple circuit with negligible resistance that just has a power source charging a capacitor would give you $V_{source}-V_{capacitor}=0$ . I was a bit confused but it does make sense now. Obviously the voltage of a capacitor is dependent on the amount of charge you can put on it, and a higher voltage battery would be able to put more charge on a capacitor, with every capacitor having a limit. I was doing this from mostly a design perspective, wanting to design a very high capacitance capacitor that would have enough power to drive a motorcycle. I don't actually know if I'll go through and build it, as sophiecentaur pointed out it'll become quite expensive. Also, always on the topic of a capacitor's voltage, if I did plan to build a parallel plate capacitor, is there any formula that would tell you how much voltage it can have given its dimensions (area and separation of plates) along with the material used? The amount of charge that could go on the plate would be unknown.

 

[Jeff Rosenbury]

Capacitance goes up as the distance between the plates shrinks. Unfortunately breakdown strength of dielectrics goes down with the shrinking distance. Thus there are limits to capacitors. It would be hard to beat commercially designed capacitors since the designers know all of this. Still, there are lots on the market with different voltages and whatnot.

Capacitors have about 1% of the energy density of batteries. There's something called a super-capacitor (more like a lame battery IMO) which has about 10% the energy density of batteries. So usually batteries are the way to go. Some of the newest batteries have an energy density of dynamite, but they are expensive and temperamental.

 

[NascentOxygen]

I was doing this from mostly a design perspective, wanting to design a very high capacitance capacitor that would have enough power to drive a motorcycle.

One of the foreseeable problems with using a capacitor to power anything is that its voltage is continuously falling as it's used. Whereas a battery holds its terminal voltage reasonably fixed, the voltage from a capacitor has fallen by 50% by the time it's half discharged. You can use electronics to boost the drive voltage back to a higher level, but that is one added complication that battery drive avoids.

 

[meBigGuy]

You would have to build super-capacitors to get competitive energy density.

This chart shows supercapacitors as competive with batteries in energy density.
https://en.wikipedia.org/wiki/Supercapacitor#/media/File:Supercapacitors-vs-batteries-chart.png

 

[Physicist97]

One of the foreseeable problems with using a capacitor to power anything is that its voltage is continuously falling as it's used. Whereas a battery holds its terminal voltage reasonably fixed, the voltage from a capacitor has fallen by 50% by the time it's half discharged. You can use electronics to boost the drive voltage back to a higher level, but that is one added complication that battery drive avoids.

This may seem like a silly question, but how is it that a battery can hold a reasonably fixed voltage? As it is used wouldn't the voltage go down and eventually get to zero, at which point it would need to be recharged?

The chart was very helpful, meBigGuy. So I guess a super-capacitor is what I had in mind, just didn't know it had a different name from capacitor.

 

[NascentOxygen]

A battery's voltage is determined by a chemical reaction. So long as those chemicals are available to continuously combine, the reaction continues and the potential exists.

 

[Physicist97]

http://www.ultracapacitors.org/index.php?option=com_content&Itemid=68&id=79&task=view
This site has a graph of voltage vs. time for capacitors and batteries. Is it correct? I'm not very good with chemistry but even if you have a chemical potential, wouldn't that potential drop as it is converted to electrical energy? The graph does show that batteries remain at somewhat constant voltage for a large portion of their life, if that is what you meant.

 

[Jeff Rosenbury]

This may seem like a silly question, but how is it that a battery can hold a reasonably fixed voltage? As it is used wouldn't the voltage go down and eventually get to zero, at which point it would need to be recharged?

The chart was very helpful, meBigGuy. So I guess a super-capacitor is what I had in mind, just didn't know it had a different name from capacitor.

You might want to read the wikipedia page the chart was attached to. The chart wasn't exactly fair.

That having been said, vehicle regenerative braking is one of their fortes.

 

[Physicist97]

You might want to read the wikipedia page the chart was attached to. The chart wasn't exactly fair.

That having been said, vehicle regenerative braking is one of their fortes.

Yea, I read the article and understand what you are saying. Most sources I've read usually agree that while super-capacitors have low energy densities compared to batteries, they tend to have higher power densities and can endure more cycles of charge and discharge (which explains why they are so useful in regenerative braking).

 

[thankz]

I tested a 100uf cap with a 9v batt and a 1.5v batt, potential difference is independent of coulombs as the cap held it's charge at the batt voltages.

 

[meBigGuy]

http://www.elna.co.jp/en/capacitor/alumi/principle.html

Building a capacitor requires large surface areas and very thin layers. For example, for tantalum (from https://en.wikipedia.org/wiki/Tantalum_capacitor):
"For example, a 220 μF 6 V capacitor will have a surface area close to 346 cm2, or 80% of the size of a sheet of paper (US Letter, 8.5×11 inch paper has area ~413 cm2), although the total volume of the pellet is only about 0.0016 cm3".

batteries vs. capacitors:
A battery's voltage is determined by electro-chemistry and may exhibit very little voltage change as it is discharged. (https://en.wikipedia.org/wiki/Electrochemical_cell)
It may, for example, increase its effective output resistance as it is drained, until it reaches a threshold where the voltage then drops quickly. It is difficult (with Li-ion anyway) to determine remaining energy based simply on a voltage measurement.

A capacitor's dynamic relationship between current and voltage is expressed by I = Cdv/dt. So, sourcing or sinking I will cause a dv/dt (delta-voltage per delta-time, or change in voltage per unit time) based on the size of the capacitor. That and E = 1/2 CV^2 which says the voltage is a function of the remaining energy ( or V = Q/C to relate it to charge). A capacitor has a pretty well defined and very direct behavior compared to a battery.

 

[sophiecentaur]

I tested a 100uf cap with a 9v batt and a 1.5v batt, potential difference is independent of coulombs as the cap held it's charge at the batt voltages.

I don't know what you meant here but it reads totally wrong. The charge is directly proportional to PD (Q = CV). How did you think you were testing the charge? Numerical examples are not as helpful as Equations because numbers are specific and you can easily lose the pattern.