Renewable Energy Questions/Discussion > Renewable Energy Q&A
dis -charging capacitors
oztules:
"I have 2 identical caps paralelled....12000uf, 56wv....Once charged to 40vdc,..then discharged into a 12 volt battery,...
The caps drop to battery voltage....so I used 28 volts....so is it E=.5 x 24000uf x 28vdc squared,.....or the 40vdc squared.....
just not sure....artv "
My bad Artv. The uf needs to be converted to farads, not uf (millionths of a farad) so your 24000uf = .024 farad.
And yes, as i understand it, it is the differential that we can use, not the absolute.... so 28v^2, even though the cap is charged to 40v plate to plate.... the pressure below 12v cannot be utilised, so does not exist for that circuit.... but the electron count is the same.
Looked at another way, we have the same electron count for the 24000uf, so that remains the same. But we now only have working pressure of 28v.... so thats all that will count for the energy available to do work......
So E=.024F x 28^2 in joules... or 18 joules.......... there are 4.2 joules per calorie. 4200 joules per Calorie (normal diet calorie) etc...
Now a watt is a joule per second... so if it took 1 second to use up your 18 joules, (18/1) we should have 18 watts of power for 1 second .
"
So the higher the uf rating on a cap the longer it takes to charge?, because the more uf means more available electrons, which in turn mean more amps??"
Ross has answered this well, but I think you need to ask some things a little clearer. In this case the answer is yes.... but only if you qualify the question with.... for a given voltage and resistance in the circuit... so the higher uf rating on a cap, the longer it takes to charge.
Not being pedantic, but we could change the charge rate by raising the volts or changing the resistance, or do both, and then the statement is not right..... but all things left the same it is true, and yes the more uf, means more room for more electrons.
If we think of a capacitor as two plates separated by a dielectric say a piece of insulated paper, then the bigger the plates (keeping the dielectric the same thickness), then there is more surface area to accommodate more electrons...... so it will take the same amps for longer or more amps for the same time for the larger plates (uf) to fully charge it up......
It's really just a numbers game after all. The uf being the number of electrons, the voltage being the pressure your going to push them with, and the resistance being how many hinderences you leave scattered in the path of those electrons that they must overcome with the applied pressure.
clear as mud again??
..................oztules
artv:
Thanks for these , I've reread many times...
Ross ....why introduce resistance.........seems the less resistance the faster the caps can charge,..the faster you can discharge into the battery......gaining more power......just need some kind of switch to keep up??
Oz....you lost me with the calorie thing :-[
Is that 18 watts going into the battery every second??
If thats right then you also have to add the 3 lights that are lit at the same time.....artv
rossw:
--- Quote from: artv on January 26, 2012, 05:26:45 pm ---Ross ....why introduce resistance.........seems the less resistance the faster the caps can charge,..the faster you can discharge into the battery......gaining more power......just need some kind of switch to keep up??
--- End quote ---
3 inevitable things. Death, taxes and resistance.
More seriously: if you have an large capacitor, and a generator capable of making a lot of power, and you put a diode between the generator and the cap (and you must do if you want DC that you can charge the cap with and then dump into your battery) - without any resistance, the capacitor is going to attempt to pull a virtually infinite current.
Your diode(s) will last less than one cycle before they're toast.
I also mentioned resistance because resistance, capacitance, current, voltage and time are all inextricably interlinked.
For any given capacitance and voltage difference:
- Resistance goes up, current goes down, time taken to charge increases.
- Resistance goes down, current goes up, time taken to chage decreases.
- With infinite resistance, zero current flows.
- With zero resistance, infinite current flows.
For any given resistance and voltage difference:
- Capacitance goes up, current remains the same, time taken to charge increases.
- Capacitance goes down, current remains the same, time taken to charge decreases
Etc, etc.
WooferHound:
Pulse Charging
http://www.sklaic.info/forum/index.php?topic=28.0
http://www.high-voltage-lab.com/75/pulse-charger-for-reviving-tired-lead-acid-batteries
http://forum.allaboutcircuits.com/showthread.php?t=54560
http://forum.allaboutcircuits.com/showthread.php?t=54682
oztules:
"Oz....you lost me with the calorie thing :-[
Is that 18 watts going into the battery every second??
If thats right then you also have to add the 3 lights that are lit at the same time.....artv "
I think we need to distinguish between energy and power or joules and watts, so you can see the difference.
If we were to take a hand grenade, we may find about 150 grams of tnt inside. We know it goes bang, and is deadly, and appears to release a lot of energy.... but does it really.
1 gram of TNT is worth about 4kilo joules of energy.. so 1 grenade stores about 150x4= 600 kilo joules of energy.
1 gram of chocolate is worth about 22kilo joules of energy... so 150 grams of chocky stores about 150x22=3300 kilo joules of energy.
So 150 grams of chocolate is over 5 times the energy of the grenade..... but you don't explode when you eat it.... why not?
Well power is energy expended over time. If we expend it very quickly like the grenade, we develop a lot of power for a very short time.
With the chocolate, we use a much larger amount of energy over a longer time.... small power for long time.
It is the same with capacitors, we can discharge them quickly or slowly. The energy will be the same, but the power will be completely different.
A practical example
I build low impedance electric fencing units sometimes, and we charge up a 50 uf capacitor to 600v and then release the energy into the primary of a transformer very very quickly via a SCR..... lets look at the figures.
The joules will be (50uf capacitor) 1/2 x .000050 x 600^2 or about 9 joules.
When I measure the output on the secondary, we find about 40 amps@8000volts using the fence tester into a load.
If voltsxamps=watts then 40x8000=320000 watts.... and that includes the transformer and switching losses.
So we know how much power it put out, but over what time did this happen..... well if a watt= joule per second, then it follows that joules/watts= time so 9/320000=.000028 seconds... or 28 millionths of a second.
It is actually faster than that, as the transformer is saturated to hell, and pulled our output watts down considerably, so in reality, it is probably closer to a 10 millionth of a second..... but you can see that we put out probably close to half a million watts for about 10 millionths of a second in reality.
If we were to discharge that same capacitor over a 1 second period, the power would be 9 watts.... big difference... just like the grenade and the chocolate.
So you must get these two concepts separated into what they really mean. I hope this helps do that with a real world example.
I mentioned calories as your from USA. We use joules, I thought you would be more familiar with calories.
From a practical standpoint, 1 calorie can heat up 1cc of water 1 degree Celsius. that small c calorie is equal to 4.2 joules. The big C calorie is 1000 small calories, or 4.2kj. If your wife diets she will be familiar with the big C calorie.
I don't know how fast your capacitor is discharging into your battery, so I can't give you your wattage. The 18 watts was just a simple example.
Is the mud any clearer...
.................oztules
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