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Offline artv

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dis -charging capacitors
« on: January 23, 2012, 08:54:36 pm »
hi all, if i charge caps to 40 vdc ,then connect them to a battery..........the caps will drop,  to battery voltage...                                     on a 12volt battery....does this mean 28 volts were dumped into the battery??...........but the amps are limited to the "uf" capacity???    the more "uf" the more amps???                                                                                                                                                      KVL ....from what i understand ....says there is only x amount of voltage ,....the amps only come from the resistance...........??              If I can charge caps in the blink of an eye and discharge them into a battery, in the blink of an eye........but stay at ~40vdc ..... does this mean i am dumping 28 volts into the battery ,on a contiuious basis??..............but limited to the "uf" ...amount of amps????...............maybe i should get another wobbly pop..........lol..........artv           

Offline WooferHound

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Re: dis -charging capacitors
« Reply #1 on: January 23, 2012, 10:31:14 pm »
A pulse charger does what you are talking about.
It's a different way of charging but it isn't better
Charging with pulses is the method that desulfators use
Try not to charge a battery any faster than 10% of it's capacity
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Offline oztules

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Re: dis -charging capacitors
« Reply #2 on: January 24, 2012, 05:16:23 pm »
Artv, there is some confusion here.
"hi all, if i charge caps to 40 vdc ,then connect them to a battery..........the caps will drop,  to battery voltage...                                     on a 12volt battery....does this mean 28 volts were dumped into the battery??...........but the amps are limited to the "uf" capacity???    the more "uf" the more amps???  "

We cant dump volts anywhere... because the volts are only a pressure figure. We need to drive/dump electrons into the battery. Each electron will give us 1 single molecule of change on the plates.... so the more electrons pushed into the battery by the voltage, the better.... it's not the voltage that does the charging. It delivers the pressure to push the electrons where we want them... but it is not what charges the plates in the battery.

The higher the uf, the more charge we can accumulate. It wont matter if it is at 1 volt or 100000 volts, the charge will still be 1 farad of electrons for a 1 farad capacitor. I think a farad of charge is about 6x10^23 or near 100000 coulombs. A coulomb is 1 amp of electrons flowing past a point in one second... about a 6x10^18 figure.

So can you see that the capacitor simply "stores" charge.... and that charge is just electrons.

The voltage gives it pressure to move those electrons. This voltage and charge can be equated to energy by
E=1/2 Capacitance X voltage squared.


Now, if we charge a 1 farad capacitor, we know how many electrons we have ready to go somewhere....... but we need pressure to push them around our circuit. So if we charge at 1farad at 1 volt.... they won't run up hill and try to get into the battery. If we charge it up at 40 volts, we have 40 ^2 (1600) times more energy stored..... but the same number of electrons..... but now they have a pressure value which if it is higher than your battery means they will flow.

We can see how many amps will flow by your ohms law..... where the amps = voltage/resistance in the circuit.

So it is not the uf that will drive higher current, it will be the volts/resistance that will determine amps.

So your 40-12v gives us a 28volt differential to work with.... so 28/resistance of the circuit is the current.

Of course, as we discharge the cap, we not only use up our electron store, but our pressure will drop too. So as it discharges, it will be like a balloon with a leak, not only do the air molecules get out and leave less in the balloon as a time function, but the pressure will drop as well, until we end up with no air in the balloon, and no pressure either.

The higher the resistance, the longer it takes to discharge....

Looked at another way, the uf tells us how many folks are in the room, and the voltage differential tells us how fast they can get out of the room in a stampede... more voltage , the more frenzied the stampede.



Does that help at all to see what happens?


..........oztules
Flinders Island...... Australia

Offline artv

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Re: dis -charging capacitors
« Reply #3 on: January 24, 2012, 06:24:18 pm »
Thanks for the replies...
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??
I'm not sure if that even makes sense :-[
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 :-\

Offline rossw

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Re: dis -charging capacitors
« Reply #4 on: January 24, 2012, 06:57:54 pm »
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??

They will all try to charge at a (nearly) infinite rate. The current could be scary though. So usually there is (some) series resistance, by design or just by nature. It is this resistance that limits the current that in turn affects the charge time.



Offline oztules

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Re: dis -charging capacitors
« Reply #5 on: January 24, 2012, 08:48:13 pm »
"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
Flinders Island...... Australia

Offline artv

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Re: dis -charging capacitors
« Reply #6 on: January 26, 2012, 05:26:45 pm »
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

Offline rossw

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Re: dis -charging capacitors
« Reply #7 on: January 26, 2012, 05:53:51 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??

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.


Offline oztules

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Re: dis -charging capacitors
« Reply #9 on: January 26, 2012, 07:03:20 pm »
"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
Flinders Island...... Australia

Offline artv

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Re: dis -charging capacitors
« Reply #10 on: January 28, 2012, 04:05:08 pm »
Hi Oz,...starting to clear
In your example ,you discharge 320000 watts in .000028 secs....so .5 sec=18watts...,   .25 sec=36watts and so on (correct)?
"very very quickly via a SCR"...this sounds like what I'm looking for...as it is I just connect and disconnect the caps to the battery by hand, but I can't switch fast enough ,probably 2-3 dis-charges/sec.
Can I measure the current between the, cap + out, and the battery +, but it would be a peak reading that would drop to zero very fast,...would that peak reading be accurate??
Also to find the circut resistance, connect to the common ground, but where should I connect the red lead,at the + caps out???
Is there a rule about ,time to charge and discharge,...ex. 1sec charge 1sec dis- charge??

In the fencer ..how long does it take to charge the cap? and what is the source??
It just got cloudy again......artv

Offline WooferHound

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Re: dis -charging capacitors
« Reply #11 on: January 28, 2012, 04:56:18 pm »
Didn't you look at the Pulse Charger links that I posted above ?
Didn't you search google for more information about Pulse Chargers ?
You are talking about Pulse Charging, there is tons of information out there  ! !
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Offline bvan1941

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Re: dis -charging capacitors
« Reply #12 on: January 28, 2012, 05:59:50 pm »
ARTV,
Capacitor don't charge or discharge in a linear line. They do charge and discharge over a period of time, in whats called TC'S (time constants). While the time constants are equal in time, the chrg/d'chrg per each TC varies in the following percentage rates: 
TC # 1 =   63%
TC #2 =    86%
TC #3=     95%
TC #4=     98%
TC #5=     99%

So to summarize, each TC period is a combination of the capacitance and the resistance in that circuit combined. Normally, the period is in milliseconds/ microseconds. It is a formula that can be found online. It's use is by electronoic techs / engineers to compute.
If you new the TC period, you would know then that in the first TC, the charge on that capacitor would be either 63% charged or discharged.
Hope that clears up some confusion and give you a perspective of that characteristic of capacitors.
Bill





















 

Offline artv

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Re: dis -charging capacitors
« Reply #13 on: January 28, 2012, 06:32:26 pm »
Hi Woofer,...I did look at the links, and thank-you very much for those..Just skimmed them so far...not alot of time for the internet
I don't want to charge batteries............I want to pulse electromagnets
The more volts ,more capacitance,.....the stronger the electromagnet???
I need to get more caps ....real caps...
You always give the best links......any suggestions where to get them, high uf, high voltage..
Right now the stored caps will bunt ...the alt ,..after shut-down..
I need bigger caps though.........artv

Offline oztules

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Re: dis -charging capacitors
« Reply #14 on: January 28, 2012, 08:28:22 pm »
"In your example ,you discharge 320000 watts in .000028 secs....so .5 sec=18watts...,   .25 sec=36watts and so on (correct)?"

Thats what the equations say. It is not linear, ( more like exponential) but the energy dissipated over those time frames give us the power. It will start out hard and dribble to an end. If the load is a LED, it will stop discharging before zero volts (led will drop out before then) or for  a battery, will only discharge to that voltage.


""very very quickly via a SCR"...this sounds like what I'm looking for...as it is I just connect and disconnect the caps to the battery by hand, but I can't switch fast enough ,probably 2-3 dis-charges/sec."

Yes you can use an SCR..... but it only switches off when it drops below it's turn off threshold. So with AC no problem ( it disappears through zero every cycle, with DC as in the fencer, the driving input impedance is only mild, so when the SCR shorts, it sees almost zero volts for a finite time, and can turn off. If the source is too strong, you may find turning the scr is not so simple as "discharge and go again".. like I have done in this instance. Driving a battery it won't turn off, as the volts don't drop to zero.


"Can I measure the current between the, cap + out, and the battery +, but it would be a peak reading that would drop to zero very fast,...would that peak reading be accurate??"

The peak reading will be just that... a peak. The output will decay from that peak at the start to the dribble at the end.



"Also to find the circuit resistance, connect to the common ground, but where should I connect the red lead,at the + caps out???"

I am not familiar with your circuit, so can't really comment sensibly.

 

"Is there a rule about ,time to charge and discharge,...ex. 1sec charge 1sec dis- charge??"

Bvan1941 covered this..... but maybe we should see how to make an equivalent water solution you may find easy to understand what is going on.

If we think of a capacitor as a water tank with a value of the capacitance as the square footage/inches of the base of the tank X the voltage as the height of the tank, and gravity as a constant.... we may be able to describe this.

As we fill the tank, the volume will be a factor of the floor area times the height.... but the height will also give us water pressure.. dependent on height. A High V capacitor would be tiny surface area and very tall height, and a low voltage one ... a bigger floor area and low height for the same capacitance.

Now as we fill the tank, the volume fills, but the back pressure attributable to the height will resist the incoming water... the higher it gets, the more pressure it will exert against the incoming water force... until the pressure of the height of the water equals the driving force behind the filling water.... were fully charged.

Now remove the filling force (hose) and let the water run out. When the tank is full, it will run out quickly.   But as the level drops, so does our volume, and our pressure.... so the flow rate slow down some.... this continues until we get to the bottom, where the pressure is now zero, and only the water laying on the bottom of the tank remains.... and there is no more pressure to push the dregs out.... unti it drips out, or evaporates over time.

The fill time and empty time will also depend on the pipe size on the input and output if they are not the same. (if not using the inlet for the outlet???) For the fencer, it comes in through a small hole, and to empty, we pull down an entire wall to empty it almost instantly.... close it back up and fill it again... etc.

Now I'm not an electrical engineer, but thats as good as I can do in the water analogy stakes....

Anyone else  have a better idea for the water analogy?


"In the fencer ..how long does it take to charge the cap? and what is the source??"

 This breaks a lot of the rules Bvan1941 mentioned.

It runs off 12v DC. It is a little oscillator driving a small ferrite transformer. The transformer aims for about 800v.... but never gets there.

When the cap is empty, it charges up to about 400v rapidly, and then eases off as the differential between the transformer output gets nearer to the cap voltage... but as the cap gets fuller, the oscillator runs harder, and so it does not follow the exponential time frame that you would think.... because the voltage is not a constant source. The terminal voltage is the cap voltage, but the driving EMF is dependent on the oscillator.

It takes about 1 second to charge up to 600v, at which time the SCR triggers into a very very low impedance (9 turns of 13 gauge wire) which pulls the cap down to zero (near enough) in a very very short time. This will start as an exponential  curve thing..... although the transformer saturates very very heavily, and that will effect the discharge curve. One moment it drives an inductor, and next a virtual short when saturated. This brings the voltage way below threshold... SCR releases, and the whole thing starts the slow recharge.

It is the saturation that lowers the output wattage so dramatically, and threw the time calcs out. The input power is much higher then the recovered output power.

It has to take at least a second to charge...... thats an occupational safety thing.... so is the very short duration pulse.



...............oztules



Flinders Island...... Australia