So what advantage does having a high capacitance have. I was reading the capacitance section in the e-book, and I didn't quite get it. I am inferring that it allows more aperage, allowing it to fill up quicker, but I am not sure, so I figured I would ask

Also the C=eK (the e being the greek) for series circuits. if K is the dielectric for both capacitors, what then would would you do if the capacitors have two different dielectrics?

Thank you so much!!

 

Higher capacitances have higher energy density, have lower AC reactance (resistance to AC), and take longer to charge. One good use of big capacitors is in power supplies to smooth out the rectified AC. I'm sure theres more your inquiring about though.

 

A larger capacitance could be a disadvantage instead of an advantage, depending on application. When designing, we want to select the appropriate capacitance value, whether large, small, or medium.

A larger capacitance will charge more slowly, not more quickly. The forumla is T=R*C where T is time in seconds needed to achieve 36.8% of full charge, R is resistance in Ohms, and C is capacitance in Farads.

For two capacitors in series, Ct = 1/( 1/C1 + 1/C2 ). So if each capacitor had the same area, same distance between plates, and different absolute permittivity, Ct would equal 1/( 1/\(\epsilon\)1K + 1/\(\epsilon\)2K )

 

So if your capacitance was too low it would charge too fast, causing there to be too much current (when you turn on he circuit i.e. when Volts go from 0 to X) through the rest of the circuitry which could cause damage?

And for the capacitance, if let's say the coinciding areas were different, that would be in the denominator too, correct?

 

Please know there are a few inter-related measurements here. Charge, voltage, current, capacitance, and time.

We can charge a tiny capacitance quickly with a tiny current. We can charge a huge capacitance slowly with a big current. The problem is that "tiny" and "big" have no real meaning. An iron atom is "monstrously huge " compared to a quark. Jupiter is "microscopically dinky" compared to the sun.

Current will be inversely proportional to the resistance in series with the capacitor, in accordance with Ohm's Law. I = E / R. Current will be larger when the cap first starts to charge, and decrease as charging continues. At no point will current exceed what it would be if the cap were shorted.

Example One: 1 whole Farad in series with 1 megOhm with 5V applied. Current at the instant the cap starts charging is 5V over 1 megOhm = 5 microAmps. Time required to charge to 1.84V = 1 * 1000000 = about eleven and a half days. Time required for "full charge" (99%) = more than 8 weeks.

Example Two: 1 picoFarad in series with 1 Ohm with 500V applied. Current at the instant the cap starts charging is 500V / 1 = 500 Amps. Time required to charge to 184V = one picosecond. Time required for "full charge" (99%) = five picoseconds.

Play with some different values yourself, and see how the numbers change.

 

ok, i see. so 184 is 36.8 percent of 500, so I assume that the amp curve is logarithmic. Is there any mathematical reason why 36.8 percent is so significant. or is one of things that kinda "just is" (Well, nothing in math "just is" but should I just accept as fact for now?) so then how would you then derive the equation of time for the full 100% charge it obviously isn't (approx)3*36% as 11 isn't anything near 56 days, or eight weeks.

I just want to say thanks for your help, as well as everyone else's. I have posted so many easy and for you over the past week, obvious questions and you all had the patience to help. I sure hope all engineers are this friendly

 

Actually, it's how long it takes to charge up to 63.2% (~=63%) of the full charge, or discharge to 36.8% (~=37%) of it's initial charge.

 

a empty cap placed across a voltage diff will cause a short to appear on such supply (for a brief moment). the larger the cap the longer this short. taking a 1F empty cap and placing it across a benchtop power supply might blow a fuse in the supply, etc. such large brief currents can also melt the insides of the cap.

 

wookie, you mean 63% is that eight weeks? What is the formula for that? I assume it's logarithmic?

DC, how exactly would it short? is it because there is so much current that it would bypass the the dielectric? So is there a formula for field flux as compared to Amps, as in a given field flux can only withstand so many Amps before being bypassed?

Wow, this stuff is soooo interesting I love this

 

when you short a power supply the electrons rush from one side to the other. same thing happens when a empty cap is charged except the electrons never make it to the other side (but from the perspective of the supply there is still current flow which looks like a short), they simple fill up the cap as fast as they can. there are inherent limiters (such as cap resistance) which will limit the current, but a large cap can easily kill some supplies if not limited in some way, etc. it looks like a short because there is a huge amount of current flowing the instant the cap starts to charge up, just as there is when you simply take the + of a supply and touch it directly to the - side.

 

Ok I see what you mean, just a question on one of the things you said, for most projects is cap resistance considered negligable?

One thing I do not quite get. Let's say you have a circuit with 1000 volts of power, 1000 ohm resistor and a cap that will never arc or short or break in anway. so for the resistor (Etotal=1000 Itotal=1 Rtotal=1000) it would drop 1000 volts, so what then would the capacitor try to charge to? (also assume cap has no resisitance, if that makes sense)

also about that quick release from capacitors, that's like those elctric pens that zap people, right?

 

Your 1000 volts is a measure of potential energy. Power is the product of volts times amperage.

How is your hypothetical circuit arranged? If all elements are in series, the cap eventually ends up with all 1KV across it. If the voltage source and resistor are in parallel with the cap in parallel, then it depends on which side of the resistor it's on. On one side, it will also charge to source - 1000 volts. On the other, there is no potential, so the cap will never charge up to any voltage.

The first section of the Ebook should make all these concepts pretty clear. Same for the need to have a schematic to look at when discussing a circuit.

 

ok thank you, I will keep that in mind next time I ask a circuit question. By the way, you did answer my question- they are in series

So if I understand this, regardless of the existance of the resistor, the cap will still charge to 1000 V, just with the cap it will take longer.

Thanks

 

Actually, it's how long it takes to charge up to 63.2% (~=63%) of the full charge, or discharge to 36.8% (~=37%) of it's initial charge.

Dang. I hate it when I read the charts wrong...

MusicTech, the curve is indeed logarithmic. I don't remember the formula, but Euler's number gets raised to some function of T, R and C. The capacitor never actually charges to 100%. 5 time constants give a 99% charge. Since capacitors have as much as 20% tolerance, and resistors usually have 5% tolerance, 99% is more than "close enough."

Hopefully someone else will either remember the formula or be able to find it quickly.

 

In response to

I dug up this formula to explain the curve:

Q = Q_{max}[1 - e^{(-t/RC)}]

where Q is charge, C is capacitance, t is time, and R is resistance

I believe that the 63.2%/86.5%/95%/98.2%/99.3% charge per sequential time constant are simply because they wanted to keep the math simple, providing the time constant equal to the product of R and C.

Also, regarding high capacitances - I'm not sure if the capability of higher value capacitors to pass AC current has been emphasized enough. The construction of filters that will pass certain frequencies to ground while passing a desirable bandwidth to a load depends on specific relationships to be attained. Capacitance is a critical factor in these designs.

For example, if you wanted to remove a hum from an audio amplifier, you could use a RC filter. A higher value cap will allow higher frequency signals to pass right through it to ground while allowing you to use equal or smaller resistance values that could maintain circuit performance.

Capacitive coupling is another excellent example. But I have already written too much...

So, I'm just trying to point out that looking at capacitors as simple little energy buckets can pigeon-hole your understanding of electronics.

- Cap lover

 

Wow, I didn't know that was it. i read a book just last night and it gave me that formula, but explained what it did very oddly

and e being the base of natural logs, id est 2.17.....

 

So, I'm just trying to point out that looking at capacitors as simple little energy buckets can pigeon-hole your understanding of electronics.

Indeed! Water as a metaphor for electricity leaves one all wet & high & dry.

The best way to understand capacitors is to use the formulas with different values to get a feel for how they work. Back in the Disco Age when I was learning this stuff, I also found it helpful to use some algebra on the formulas themselves. This helped me understand how current, charge, field, permittivity, et.al. dance together.

 

hmm that's a very interesting way of looking at it. thanks, I am glad I can use math in this, math may be the only scholarly thing better than EE, however. Math=EE=Math.... haha transistive property

 

oh and by the way, just one question. in that equation a couple of post s up, Q, chargeis in volts, right?

 

Charge is in Coulombs. Current is in Amps, which is the same as Coulombs per second. Capacitance is in Farads which is the same as Coulombs per Volt. Wikipedia all of the units, and it will all put together better.