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Wanderer
Guest
Wed Sep 07, 2011 4:00 pm
On Sep 7, 8:04 am, George Herold <gher...@teachspin.com> wrote:
Quote:
On Sep 6, 7:44 pm, Wanderer <wande...@dialup4less.com> wrote:
On Sep 6, 6:33 pm, Rui Maciel <rui.mac...@gmail.com> wrote:
John Larkin wrote:
The energy storage ratio is extreme, four or six orders of magnitude.
Yes, it appears that those values are coherent with the information which
has been presented in this thread. Yet, in some applications energy density
may not be the decisive property of a energy storage device. For example,
in applications such as small LED flashlights energy density may be less
important than recharging speed and even the ability to recharge them
through simpler means, such as mechanical generators. Is there any reason
that makes this sort of applications impractical?
Rui Maciel
Lets just do it.
Here's a supercap from digikey 1 Farad 6.3V around 4 dollars.
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=604-....
Lets say we can charge it to 6 volts and discharge it to 2 volts and
we build a constant current circuit to drive 10mA into a 2 volt Led.
Energy in a cap is (1/2)*C*V^2 so at 6 volts the cap has 18 joules and
at 2 volts it has 2 joules so we can supply 16 joules.
The LED at 10mA and 2 volts uses 0.02W.
A joule is a Watt*second so 2/0.02 = 100 seconds = 1.6 minutes
Opps, shouldn't that be 16 Joules/0.02 Watts = 800 seconds. (but
still much to short a time.)
Now a AAA alkaline has about 1000mAh at 1.5v. or 5400 joules. Let say
we boost that to 2v and we're 50% efficient. So, we supply 2700 joules
to the LED. 2700/0.02 = 135000 seconds = 37.5 hours.
Oh, you must use a AA battery or Miso will have another rant. :^)
George H.
So you will need a supercap with lots of Farads of capacitance and a
high voltage rating to compete with the battery.
Oops. Yes you're right. 800 secs =13.3 minutes
Rui Maciel
Guest
Wed Sep 07, 2011 4:18 pm
Wanderer wrote:
Quote:
Lets just do it.
Here's a supercap from digikey 1 Farad 6.3V around 4 dollars.
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=604-1018-
ND
Lets say we can charge it to 6 volts and discharge it to 2 volts and
we build a constant current circuit to drive 10mA into a 2 volt Led.
Energy in a cap is (1/2)*C*V^2 so at 6 volts the cap has 18 joules and
at 2 volts it has 2 joules so we can supply 16 joules.
The LED at 10mA and 2 volts uses 0.02W.
A joule is a Watt*second so 2/0.02 = 100 seconds = 1.6 minutes
Now a AAA alkaline has about 1000mAh at 1.5v. or 5400 joules. Let say
we boost that to 2v and we're 50% efficient. So, we supply 2700 joules
to the LED. 2700/0.02 = 135000 seconds = 37.5 hours.
So you will need a supercap with lots of Farads of capacitance and a
high voltage rating to compete with the battery.
I see what you mean. Yet, the capacitor you've shown is a somewhat smaller
than a AA battery and it appears to be possible to stack around 5 of those
to reach the same height. If I'm not mistaken, with this arrangement it
would be possible to store enough energy to power that LED for around an
hour. Although it still lags behind what can be had with batteries, it is
already a decent amount of time. If it was possible to recharge the
capacitors through some means, such as a mechanical generator, then it
wouldn't be a bad thing to have.
Rui Maciel
Kaz Kylheku
Guest
Wed Sep 07, 2011 8:52 pm
On 2011-09-06, Wanderer <wanderer_at_dialup4less.com> wrote:
Quote:
On Sep 6, 6:33Â pm, Rui Maciel <rui.mac...@gmail.com> wrote:
John Larkin wrote:
The energy storage ratio is extreme, four or six orders of magnitude.
Yes, it appears that those values are coherent with the information which
has been presented in this thread. Â Yet, in some applications energy density
may not be the decisive property of a energy storage device. Â For example,
in applications such as small LED flashlights energy density may be less
important than recharging speed and even the ability to recharge them
through simpler means, such as mechanical generators. Â Is there any reason
that makes this sort of applications impractical?
Rui Maciel
Lets just do it.
Here's a supercap from digikey 1 Farad 6.3V around 4 dollars.
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=604-1018-ND
Lets say we can charge it to 6 volts and discharge it to 2 volts and
we build a constant current circuit to drive 10mA into a 2 volt Led.
Energy in a cap is (1/2)*C*V^2 so at 6 volts the cap has 18 joules and
at 2 volts it has 2 joules so we can supply 16 joules.
The LED at 10mA and 2 volts uses 0.02W.
A joule is a Watt*second so 2/0.02 = 100 seconds = 1.6 minutes
An easier way to look at this is to consider that an Ampere is a Coulomb
per second (C/s) (how much of a charge of electrons flow past a point
per unit time) whereas a Farad is a Coulomb per Volt (how much charge is
crammed into a capacitor for a given potential difference).
At 6V, our 1F cap stores 6 coulombs, and discharges to 2 coulombs, so we
lose 4 coulombs. These 4 coulombs are being doled out using an active
current limiter at a steady 10 mA trickle (in other words 0.01 C/s), so
we get a linear discharge, rather than the typical RC discharge.
We thus easily calculate the discharge time:
Thus 4C / 0.01 C/s) = 400 s.
We get 6 minutes and forty seconds. (Of course, a real current limiter
will have a voltage drop, such as the VCE(sat) of its transistor or
whatever: it will stop working somewhere above 2V).
From this we can also see how much energy is wasted. We have 16J, right?
But we got to run our 0.02W LED for 400 seconds, which gave us 8J of
energy. Watts is Joules per second, so 0.02 (J/s) * 400s = 8J.
8J of light and heat out of the LED, 8J of heat out of the limiter.
Bret Cahill
Guest
Fri Sep 09, 2011 4:01 pm
Quote:
At first glance it appears that, at least for some applications, employing
capacitors as a means to store energy has considerable advantages over the
need to rely on batteries. Yet, batteries tend to be used almost
universally. Is there a reason for this? Can anyone more knowledgeable on
the subject comment on the practical implications of relying on capacitors
as a power source instead of other standard means such as batteries?
Thanks in advance,
Rui Maciel
For some applications capacitors are eminently suitable as a power
source today. As for replacing high power storage batteries - they
still have a way to go.
They are already experimenting with using super caps in conjunction
with car batteries to save on battery costs or to allow the battery be
located further from the engine (where the heat won't kill it as
quickly).
A cap is perfect for getting you started going through the
intersection. That way the $15,000 battery isn't damaged by 100 kW
loads.
An EV battery only needs to put out 10 kW on the freeway.
Bret Cahill
Bob F
Guest
Tue Sep 13, 2011 4:58 am
Rui Maciel wrote:
Quote:
Wanderer wrote:
Lets just do it.
Here's a supercap from digikey 1 Farad 6.3V around 4 dollars.
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=604-1018-
ND
Lets say we can charge it to 6 volts and discharge it to 2 volts and
we build a constant current circuit to drive 10mA into a 2 volt Led.
Energy in a cap is (1/2)*C*V^2 so at 6 volts the cap has 18 joules
and at 2 volts it has 2 joules so we can supply 16 joules.
The LED at 10mA and 2 volts uses 0.02W.
A joule is a Watt*second so 2/0.02 = 100 seconds = 1.6 minutes
Now a AAA alkaline has about 1000mAh at 1.5v. or 5400 joules. Let say
we boost that to 2v and we're 50% efficient. So, we supply 2700
joules to the LED. 2700/0.02 = 135000 seconds = 37.5 hours.
So you will need a supercap with lots of Farads of capacitance and a
high voltage rating to compete with the battery.
I see what you mean. Yet, the capacitor you've shown is a somewhat
smaller than a AA battery and it appears to be possible to stack
around 5 of those to reach the same height. If I'm not mistaken,
with this arrangement it would be possible to store enough energy to
power that LED for around an hour. Although it still lags behind
what can be had with batteries, it is already a decent amount of
time. If it was possible to recharge the capacitors through some
means, such as a mechanical generator, then it wouldn't be a bad
thing to have.
But batteries would regulate the voltage. A capacitor running a lightbulb would
have a huge varience in light output as the voltage drops. A batterys voltage
would be much more constant. Plus it would last way longer per charge. But then,
the capacitor could be charged/drained thousands of times more.
We just need better capacitors.
Rui Maciel
Guest
Tue Sep 13, 2011 12:47 pm
Bob F wrote:
Quote:
But batteries would regulate the voltage. A capacitor running a lightbulb
would have a huge varience in light output as the voltage drops. A
batterys voltage would be much more constant. Plus it would last way
longer per charge. But then, the capacitor could be charged/drained
thousands of times more.
We just need better capacitors.
Isn't it possible to at least limit the voltage output of a capacitor?
Rui Maciel
Winston
Guest
Tue Sep 13, 2011 1:07 pm
Bob F wrote:
(...)
Quote:
We just need better capacitors.
The capacitors are all right.
We need better batteries more.
:)
--Winston
Frnak McKenney
Guest
Tue Sep 13, 2011 1:37 pm
On Mon, 12 Sep 2011 21:58:28 -0700, Bob F <bobnospam_at_gmail.com> wrote:
Quote:
Hey, Bob.
How about a 2600F 2.5V cap for $10?
Maxwell 2600 Farad 2.5VDC Boostcap
http://www.goldmine-elec-products.com/prodinfo.asp?number=G17960
A couple of these hould be enough to power a white LED for some
time, yes?
Enjoy...
Frank McKenney
--
Reading achievement will nor advance significantly until schools
recognize and act on the fact that it depends on the possession of
a broad but definable range of diverse knowledge. The effective
teaching of reading will require schools to teach the diverse,
enabling knowledge that reading requires.
-- E.D. Hirsch, Jr./The Knowledge Deficit
--
Frank McKenney, McKenney Associates
Richmond, Virginia / (804) 320-4887
Munged E-mail: frank uscore mckenney aatt mindspring ddoott com
Michael A. Terrell
Guest
Tue Sep 13, 2011 2:59 pm
Rui Maciel wrote:
Quote:
Bob F wrote:
But batteries would regulate the voltage. A capacitor running a lightbulb
would have a huge varience in light output as the voltage drops. A
batterys voltage would be much more constant. Plus it would last way
longer per charge. But then, the capacitor could be charged/drained
thousands of times more.
We just need better capacitors.
Isn't it possible to at least limit the voltage output of a capacitor?
How much of the stored energy do you want to waste?
--
You can't have a sense of humor, if you have no sense.
Michael A. Terrell
Guest
Tue Sep 13, 2011 3:01 pm
Bob F wrote:
Quote:
We just need better capacitors.
First, you'll need to get rid of those pesky laws of physics that
tell you how they work.
--
You can't have a sense of humor, if you have no sense.
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