Stainless
Flashlight Enthusiast
I'm intrigued by the potential of capacitors/supercapacitors. Can aynone point me to an online source for learning the fundamentals? Thanks
Where to learn basics of capacitors?
- Thread starter Stainless
- Start date
highlandsun
Enlightened
My favorite site on the web is the HyperPhysics stuff: HyperPhysics
Great explanations, and cool javascript calculators for fast answers to quick questions...
For supercapacitors, PolyStor used to have good info but I think they spun that off into its own company. Look for PowerStor on Google.
Great explanations, and cool javascript calculators for fast answers to quick questions...
For supercapacitors, PolyStor used to have good info but I think they spun that off into its own company. Look for PowerStor on Google.
Stainless
Flashlight Enthusiast
Thanks. I'll check your leads.
Tomas
Banned
Then there's this guy:
PowerLabs
Who uses a little capacitor array to pump out a quick burst of "about" 3500 volts at 90,000 amps (about 315,000,000 watts) to run his toy.
Honestly, one of my thoughts was "How bright a flash would about 3 hundred million LS's make all at once when powered by something like this?"
I'm not sure I can conceive it, really.
Ya know what's really frightening about it, though? The powerpack is only about the size of a suitcase.
Fascinating project, really.
_ ^..^ _
PowerLabs
Who uses a little capacitor array to pump out a quick burst of "about" 3500 volts at 90,000 amps (about 315,000,000 watts) to run his toy.
Honestly, one of my thoughts was "How bright a flash would about 3 hundred million LS's make all at once when powered by something like this?"
I'm not sure I can conceive it, really.
Ya know what's really frightening about it, though? The powerpack is only about the size of a suitcase.
Fascinating project, really.
_ ^..^ _
Floating Spots
Enlightened
Two good references:
http://www.ibiblio.org/obp/electricCircuits/
http://www.web-ee.com/primers/primers_frm.htm
Specifically on caps:
http://www.ibiblio.org/obp/electricCircuits/DC/DC_13.html
Hope this helps.
Jon
http://www.ibiblio.org/obp/electricCircuits/
http://www.web-ee.com/primers/primers_frm.htm
Specifically on caps:
http://www.ibiblio.org/obp/electricCircuits/DC/DC_13.html
Hope this helps.
Jon
MrAl
Flashlight Enthusiast
Hello there,
An interesting formula is the formula for
calculating the 25% discharge ampere hour rating
of a capacitor so you can get an idea how a
capacitor of a given value compares to a
battery.
The formula is:
ah = C * (0.25) * V/3600
where
ah is the ampere hour rating
C is the value in farads
V is the original voltage of the equiv battery
So, for a capacitor of 1 farad and comparing this
to a 3 volt battery:
ah = 1 * 0.25 * 3 / 3600
So ah = 0.000208
which is not even one milliampere hour!
One the other hand, if we compared this to
a 100 farad capacitor charged to 4.5v we would get
ah = 100 * 0.25 * 4.5 / 3600
or
ah=0.031
which is 31 milliampere hours
which would, ideally, run a 30ma LED for a full
hour!
An interesting side note is that if the particular
(super) capacitor would allow a current as high
as 3 amperes to flow though it, you could charge this
last example back up in less then two minutes!
Good luck with your LED circuits,
Al
An interesting formula is the formula for
calculating the 25% discharge ampere hour rating
of a capacitor so you can get an idea how a
capacitor of a given value compares to a
battery.
The formula is:
ah = C * (0.25) * V/3600
where
ah is the ampere hour rating
C is the value in farads
V is the original voltage of the equiv battery
So, for a capacitor of 1 farad and comparing this
to a 3 volt battery:
ah = 1 * 0.25 * 3 / 3600
So ah = 0.000208
which is not even one milliampere hour!
One the other hand, if we compared this to
a 100 farad capacitor charged to 4.5v we would get
ah = 100 * 0.25 * 4.5 / 3600
or
ah=0.031
which is 31 milliampere hours
which would, ideally, run a 30ma LED for a full
hour!
An interesting side note is that if the particular
(super) capacitor would allow a current as high
as 3 amperes to flow though it, you could charge this
last example back up in less then two minutes!
Good luck with your LED circuits,
Al
Stainless
Flashlight Enthusiast
Thanks guys, I'm on my way.
MrAL:
Can you (roughly) compare the voltage/current requirements of a "standard" white led vs a LS led vs a one watt LS. (It will probally take me many weeks of learning before I can dig out such info on my own.)
Thanks again.
Stainless
MrAL:
Can you (roughly) compare the voltage/current requirements of a "standard" white led vs a LS led vs a one watt LS. (It will probally take me many weeks of learning before I can dig out such info on my own.)
Thanks again.
Stainless
INRETECH
Flashlight Enthusiast
highlandsun
Enlightened
Al, something's missing from your quickie formula - where do you factor in the capacitor's voltage?
On a separate question, you see the properties of a capacitor described when discharging through a resistor. Should one expect a cap discharging thru an LED to follow the same curve? If I have a red Luxeon Star with 2.95V forward voltage and 350mA current, plugging this into V=IR says this is equivalent to a 8.43ohm resistor. So would you expect the LED to discharge a cap at the same rate as an 8.43ohm resistor?
On a separate question, you see the properties of a capacitor described when discharging through a resistor. Should one expect a cap discharging thru an LED to follow the same curve? If I have a red Luxeon Star with 2.95V forward voltage and 350mA current, plugging this into V=IR says this is equivalent to a 8.43ohm resistor. So would you expect the LED to discharge a cap at the same rate as an 8.43ohm resistor?
MrAl
Flashlight Enthusiast
Hello there,
Stainless:
A 'standard' white led is probably taken to be a white Nichia, at 20ma nominal current.
An "LS" is usually taken as a 1 watt LS, and "LS-5W" or "5W LS" means a five watt LS.
The average current for a 20ma led is 20ma, for the LS is 350ma, and for the
5W LS is 700ma.
I hope this clears up the differences in white led's.
INRETECH:
Wow thats some big cap :-)
If you use that with a small step up circuit
i bet you could keep an led running quite a long
time!
highlandsum:
The formula comes from a basic formula for a capacitor:
dv/dt=i/C {v in volts, t in seconds, i in amperes, C in farads}
Since we are solving for a relatively short period of time
(relative to five time constants)
we let
dv=v1-v2
where
v1 is the max voltage, and
v2 is the min voltage.
Since we want to know the -25% point (v2)
this means v2 is 75% of v1, or the difference (v1-v2)
is 25% of v1, always, so we can take
dv=0.25 * v1 (or 25% of max voltage).
Thus we take take dv to be 0.25 times Vmax.
Substituting into the original eq and multiplying both sides by dt we get:
0.25 * Vmax = i * dt / C
So you see the capacitors voltage gets factored in there :-)
Now this last eq is in units of volts, amps, seconds, and farads.
To change the seconds into hours (so we can multiply by C and get the
right side to be in ampere hours) we have to multiply dt by 3600.
Doing this leads us to:
0.25 * Vmax = i * 3600 * dt / C
Now simply multiply both sides by C and we get:
C * 0.25 * Vmax = i * 3600 * dt
and since dt=t1-t2=tp we get:
C * 0.25 * Vmax = i * 3600 * tp
and now divide both sides by 3600 and we get:
C * 0.25 * Vmax / 3600 = i * tp
and now the right side of this last eq is in ampere hours.
This means the approximate ampere hour rating of the capacitor
discharged down 25% of it's original value is equal to
ah=0.25 * C * Vmax / 3600.
Since i is assumed linear, the formula is approximate, but it does
give you some guideline to show an estimation of what you might
get from a given capacitor.
A more exact analysis would lead to a more complex equation which
most people wont want to use anyway.
To find the approximate 20% discharge ampere hour rating, you would
simply substitute 0.20 for the 0.25. The smaller the percentage,
the less the error.
Of course there are more things to take into
consideration sometimes, but then there are when
you use a regular battery sometimes too, such as
series resistance.
Another use, besides trying to find the run time
of a particular LED with a particular value cap,
is to use the formula to get an 'overall' idea
what a particular cap might give you as compared
to a battery of known ampere hour capacity.
For example, you can quickly (in your head)
realize that a 100uf capacitor charged up to
a low voltage (say 5 to 10 volts) wont give you
very much time at all, even with a small 20ma
led, while a 100 Farad capacitor starts to show
some significant run time possiblities with the
smaller LED's (20ma), and this is true no matter
how you connect it to the cap.
Good luck with your LED circuits,
Al
Stainless:
A 'standard' white led is probably taken to be a white Nichia, at 20ma nominal current.
An "LS" is usually taken as a 1 watt LS, and "LS-5W" or "5W LS" means a five watt LS.
The average current for a 20ma led is 20ma, for the LS is 350ma, and for the
5W LS is 700ma.
I hope this clears up the differences in white led's.
INRETECH:
Wow thats some big cap :-)
If you use that with a small step up circuit
i bet you could keep an led running quite a long
time!
highlandsum:
The formula comes from a basic formula for a capacitor:
dv/dt=i/C {v in volts, t in seconds, i in amperes, C in farads}
Since we are solving for a relatively short period of time
(relative to five time constants)
we let
dv=v1-v2
where
v1 is the max voltage, and
v2 is the min voltage.
Since we want to know the -25% point (v2)
this means v2 is 75% of v1, or the difference (v1-v2)
is 25% of v1, always, so we can take
dv=0.25 * v1 (or 25% of max voltage).
Thus we take take dv to be 0.25 times Vmax.
Substituting into the original eq and multiplying both sides by dt we get:
0.25 * Vmax = i * dt / C
So you see the capacitors voltage gets factored in there :-)
Now this last eq is in units of volts, amps, seconds, and farads.
To change the seconds into hours (so we can multiply by C and get the
right side to be in ampere hours) we have to multiply dt by 3600.
Doing this leads us to:
0.25 * Vmax = i * 3600 * dt / C
Now simply multiply both sides by C and we get:
C * 0.25 * Vmax = i * 3600 * dt
and since dt=t1-t2=tp we get:
C * 0.25 * Vmax = i * 3600 * tp
and now divide both sides by 3600 and we get:
C * 0.25 * Vmax / 3600 = i * tp
and now the right side of this last eq is in ampere hours.
This means the approximate ampere hour rating of the capacitor
discharged down 25% of it's original value is equal to
ah=0.25 * C * Vmax / 3600.
Since i is assumed linear, the formula is approximate, but it does
give you some guideline to show an estimation of what you might
get from a given capacitor.
A more exact analysis would lead to a more complex equation which
most people wont want to use anyway.
To find the approximate 20% discharge ampere hour rating, you would
simply substitute 0.20 for the 0.25. The smaller the percentage,
the less the error.
Of course there are more things to take into
consideration sometimes, but then there are when
you use a regular battery sometimes too, such as
series resistance.
Another use, besides trying to find the run time
of a particular LED with a particular value cap,
is to use the formula to get an 'overall' idea
what a particular cap might give you as compared
to a battery of known ampere hour capacity.
For example, you can quickly (in your head)
realize that a 100uf capacitor charged up to
a low voltage (say 5 to 10 volts) wont give you
very much time at all, even with a small 20ma
led, while a 100 Farad capacitor starts to show
some significant run time possiblities with the
smaller LED's (20ma), and this is true no matter
how you connect it to the cap.
Good luck with your LED circuits,
Al
INRETECH
Flashlight Enthusiast
The problem with the Cap is its very low voltage, there isn't many semiconductor circuits that work well under 1v, so that portion of the storage is pretty much wasted
highlandsun
Enlightened
Thanks for the detailed explanation, Al. Now it all makes sense...
Stainless
Flashlight Enthusiast
Mr Al:
Thanks.
Like I said "It will probally take me many weeks of learning before I can dig out such info on my own."
Stainless
Thanks.
Like I said "It will probally take me many weeks of learning before I can dig out such info on my own."
Stainless
MrAl
Flashlight Enthusiast
Hello again,
INRETECH:
Yeah i see what you mean, and that voltage rating is pretty low, but
perhaps you can take heart in the fact that 84% of the energy is
available between the voltages of 1v and 2.5 volts, so you only
loose about 16% of the capacitor's 'ability' by not being able to
use it at voltages less then 1v. Note also this doesnt effect the
charge cycle efficiency either, as the energy not removed below
1v does not have to be replaced during charge either. You probably
already looked at this, but i thought i would mention it anyway :-)
Would be nice if it had a 5v rating, but i wonder how much a thing
like that would end up costing?
highlandsum:
I hoped to clear up some of the issues a little to make it easier
to understand the whole idea behind trying to determine an approximate
ampere hour rating for a capacitor.
Stainless:
Im sure you will pick up on whatever you want to know about, and if you
have more questions you can always ask in these CP forums :-)
Good luck with your LED circuits,
Al
INRETECH:
Yeah i see what you mean, and that voltage rating is pretty low, but
perhaps you can take heart in the fact that 84% of the energy is
available between the voltages of 1v and 2.5 volts, so you only
loose about 16% of the capacitor's 'ability' by not being able to
use it at voltages less then 1v. Note also this doesnt effect the
charge cycle efficiency either, as the energy not removed below
1v does not have to be replaced during charge either. You probably
already looked at this, but i thought i would mention it anyway :-)
Would be nice if it had a 5v rating, but i wonder how much a thing
like that would end up costing?
highlandsum:
I hoped to clear up some of the issues a little to make it easier
to understand the whole idea behind trying to determine an approximate
ampere hour rating for a capacitor.
Stainless:
Im sure you will pick up on whatever you want to know about, and if you
have more questions you can always ask in these CP forums :-)
Good luck with your LED circuits,
Al
INRETECH
Flashlight Enthusiast
Its incredibly high cap is from the reason that its distance of the plates is only a atom thick caused by the oxide of the metal; if they were to increase the distance to raise the voltage up - it would seriously lower the cap
Floating Spots
Enlightened
The company I looked at about a year ago has a 5000F cap at 2.7V. They were looking for three volts last year, but didn't make it. They are:
http://www.nesscap.com/prod/ba3.htm
I'm pretty sure I quoted one at $240 (5000F, 2.7V).
Doesn't look like the voltage is going to go up much.
I'll have to reread the theory of operation.
http://www.nesscap.com/prod/ba3.htm
I'm pretty sure I quoted one at $240 (5000F, 2.7V).
Doesn't look like the voltage is going to go up much.
I'll have to reread the theory of operation.
INRETECH
Flashlight Enthusiast
In a normal cap, there is a insulator which seperates the plates allowing the charge to build on - positive and negative pulled towards each other
The greater the distance, the higher the voltage since the charges can not "punch thru" the insulator
But, as you pull the plates away from each other, you also decrease the storage cap (Farads) - like the flashlight equation
Voltage X Cap -> Size
Can't increase one without decreasing the other, or increasing the size
In the "SuperCap" the companies were able to place the conductors (plates) INCREDIBLY close to each other by just using the 1-atom layer Oxide of the actual plates, since the Oxide doesn't conduct electricity
But, unfort - by bringing the plates so incredibly close to each other, they brought down the working voltage, my 4000F cap is only rated for 2.5v
Unless they find another material that will form a slightly thicker oxide, the working voltage is bascially going to stay in the 2.5-2.7v range
The greater the distance, the higher the voltage since the charges can not "punch thru" the insulator
But, as you pull the plates away from each other, you also decrease the storage cap (Farads) - like the flashlight equation
Voltage X Cap -> Size
Can't increase one without decreasing the other, or increasing the size
In the "SuperCap" the companies were able to place the conductors (plates) INCREDIBLY close to each other by just using the 1-atom layer Oxide of the actual plates, since the Oxide doesn't conduct electricity
But, unfort - by bringing the plates so incredibly close to each other, they brought down the working voltage, my 4000F cap is only rated for 2.5v
Unless they find another material that will form a slightly thicker oxide, the working voltage is bascially going to stay in the 2.5-2.7v range
Clouddancer
Newly Enlightened
I've tried to follow you MrAl, but I'm a newbie at this and my mathematical ablities are way below you. I have three 2.5V 100F ultracaps on order, planning to use them in series which will result in as I understand it 5V 50F for two or 7.5 at 33F for three. Based on your calculations then what will those combinations result in for run times of LEDs?
MrAl
Flashlight Enthusiast
Hello again,
INRETECH:
That's pretty amazing that they could make a cap that high in value.
I was looking at the relationship between energy storage and plate
distance and it looks like as the plate distance doubles, the voltage
rating doubles, the capacitance goes down by 4, and the energy storage
stays the same. I guess that's what you meant by "size" ?
Floating Spots:
Wow that's quite expensive for one cap. I wonder what the guarentee is
on one of those?
Clouddancer:
Well, with a 2.5v cap you would have to use a step up circuit
to run the white led's.
If you had a circuit that could run from 1.5v to 2.5v, you could extract
the energy from the cap while it was discharging down to 1.5v .
Using the average of 2.0v, with an output of 3.5v for the white led
running at 20ma:
Current supplied by cap would be 35ma.
now we take again the simplified formula:
dv=i*dt/C
and solve for dt:
dt=dv*C/i
dv is 1 volt
C is 100F
i is 0.035 amps
so
dt=1*100/0.035
which equals
2857 seconds, which is about 48 minutes run time.
Unfortunately, if the circuit's efficiency is averaging about 75%
(typical) then the current draw on the cap is more like 47ma.
Using the same formula,
dt=1*100/0.047
which equals about 35 minutes run time.
Now connecting two caps in series increases the energy storage
because the capacitance only decreases by half as the voltage rating
doubles (unlike moving the plates apart).
This gives us a 50F cap at 5v.
This means if we can operate between 1.5v and 5v (average 3.25v)
the average current draw will be about 28.7ma (including circuit
inefficiency).
Now with the same formula:
dt=3.5*50/0.0287
which equals about one hour and 42 minutes run time.
Thus we have increased run time by a factor of about 3.
This is of course assuming we had a circuit that would
operate from 1.5v to 5v without any problems. Since the
output would be 3.5v, sometimes this is hard to do.
With two caps in series you could run the LED right off of
the caps with a series resistor, but the led would start to
dim as the cap got down near 4 volts.
With a constant current circuit and two caps in series you
could run a single 20ma LED also.
The voltage would range from about 3.6v to 5v with a current of
20ma so this would be a dv=1.4v and using the formula again:
dt=1.4*50/0.020
which equals about 58 minutes run time.
Using the three caps in series (assuming full 7.5v initial charge)
the dv would be 7.5-3.6=3.9 and the current still 20ma, so
using the formula:
dt=3.9*33.33/0.020
which equals one hour and 48 minutes (almost two hours run time).
Using a 75% eff step down circuit with three caps would give you a
little over two hours run time.
Good luck with your LED circuits,
Al
INRETECH:
That's pretty amazing that they could make a cap that high in value.
I was looking at the relationship between energy storage and plate
distance and it looks like as the plate distance doubles, the voltage
rating doubles, the capacitance goes down by 4, and the energy storage
stays the same. I guess that's what you meant by "size" ?
Floating Spots:
Wow that's quite expensive for one cap. I wonder what the guarentee is
on one of those?
Clouddancer:
Well, with a 2.5v cap you would have to use a step up circuit
to run the white led's.
If you had a circuit that could run from 1.5v to 2.5v, you could extract
the energy from the cap while it was discharging down to 1.5v .
Using the average of 2.0v, with an output of 3.5v for the white led
running at 20ma:
Current supplied by cap would be 35ma.
now we take again the simplified formula:
dv=i*dt/C
and solve for dt:
dt=dv*C/i
dv is 1 volt
C is 100F
i is 0.035 amps
so
dt=1*100/0.035
which equals
2857 seconds, which is about 48 minutes run time.
Unfortunately, if the circuit's efficiency is averaging about 75%
(typical) then the current draw on the cap is more like 47ma.
Using the same formula,
dt=1*100/0.047
which equals about 35 minutes run time.
Now connecting two caps in series increases the energy storage
because the capacitance only decreases by half as the voltage rating
doubles (unlike moving the plates apart).
This gives us a 50F cap at 5v.
This means if we can operate between 1.5v and 5v (average 3.25v)
the average current draw will be about 28.7ma (including circuit
inefficiency).
Now with the same formula:
dt=3.5*50/0.0287
which equals about one hour and 42 minutes run time.
Thus we have increased run time by a factor of about 3.
This is of course assuming we had a circuit that would
operate from 1.5v to 5v without any problems. Since the
output would be 3.5v, sometimes this is hard to do.
With two caps in series you could run the LED right off of
the caps with a series resistor, but the led would start to
dim as the cap got down near 4 volts.
With a constant current circuit and two caps in series you
could run a single 20ma LED also.
The voltage would range from about 3.6v to 5v with a current of
20ma so this would be a dv=1.4v and using the formula again:
dt=1.4*50/0.020
which equals about 58 minutes run time.
Using the three caps in series (assuming full 7.5v initial charge)
the dv would be 7.5-3.6=3.9 and the current still 20ma, so
using the formula:
dt=3.9*33.33/0.020
which equals one hour and 48 minutes (almost two hours run time).
Using a 75% eff step down circuit with three caps would give you a
little over two hours run time.
Good luck with your LED circuits,
Al
INRETECH
Flashlight Enthusiast
Size/Weight - Batteries are still more eff at storing energy than Caps
Every storage medium has its pros and cons; and the user must decide how best to suit their purpose
I recently picked up 3 LED "reader boards" and all three of them had NiCad batteries to store the text when power was off, and all three of the batteries had leaked on the circuit board and made a mess; I replaced all three of them with small SuperCaps
Every storage medium has its pros and cons; and the user must decide how best to suit their purpose
I recently picked up 3 LED "reader boards" and all three of them had NiCad batteries to store the text when power was off, and all three of the batteries had leaked on the circuit board and made a mess; I replaced all three of them with small SuperCaps
