Single AA battery to light WHITE LED for long-long time

[Groundloop]

@All,

If you need a home made mini flash light then this is one way to make it.
The few components is soldered together inside the plastic top and bottom
from a used Duracell 9V battery. Use a side cutter to get the top plastic
with the battery connectors off, and also take out the bottom plastic. The
bottom plastic is drilled (5,5mm) for two ultra bright white LEDs. The
regulator need a little heat sink and the way to do that is by placing
the regulator at the + connector metal. I use a little heat sink silicone
compond between the regulator and the metal. The heat from the regulator
is then transferred down to the battery and dissipated to the environment.
The ultra bright LEDs has a full light output and a Duracell Ultra 9V battery
will light the LEDs for approx. 14 - 15 hours. This is not a free energy light,
just a fun project when you are bored and have nothing to do.

Groundloop.

[Pirate88179]

@ Groundloop:

Nice project man!  I remember seeing something similar (but not as nice) in a catalog a year or so ago.  It had only one led and was not bright at all.  So with your design, you can just toss the dead 9 volt when consumed, and snap on a fresh one?  Or, use a rechargeable one?  Excellent!  You should make a video on how to do this and post it on Metacafe.  If you submit it to their "Producer Rewards" program, you can earn some decent money depending upon how many view it.  This project would be perfect for that.

Fantastic!

Bill

[Groundloop]

@Bill,

Thanks for the nice comment. Yes, the black part is the connector top from the old used up Duracell.
It has the snap on connectors. The bottom of the Duracell is blue plastic and is the top of my design.
It is plenty of room for two ultrabright LEDs. I use hot glue to secure everything in place after I have
tested that the led light. You get plenty of light from two LEDs if there is a power out. It is also a handy
light to have in your car to help you if your car breaks down in the middle of the night.

I think the smart thing in my circuit is the use of a voltage regulator. The light output of the two LEDs will stay
at full power until the battery is empty. (Below 5 volt.) Then the light will dim. If one use just a resistor
then the light will be very bright when the battery is full, but after a while the led will lose strength because
the battery voltage falls. This will not happen when one use a regulator.

I'm way too lazy to make a video.  Money is not an issue. I do for the fun of this hobby.

Groundloop.

[gyulasun]

Hi Groundloop,

Very nice job!  Congratulations.

I would like to suggest an improvement for squeezing out the most power from the 9V Duracell and gain some more operating hours beyond what you indicated.
If you use an adjustable (and also a low dropout type) regulator, you can precisely set the output voltage, hence the current for the LEDs, without using the two diodes for the the voltage drop. This way the LEDs will be able to light full brightness till the 9V drops to as low as the forward voltage of the LEDs i.e to 3.6V plus the dropout voltage of  65 to 110mV.
See this link for one possible regulator type, out of many: http://www.national.com/pf/LP/LP38693-ADJ.html   (by using its ENABLE pin feature the on/off switching could be solved easily also,  though I understand the use of the snap on/off connector is good and simple).

rgds, Gyula

[Groundloop]

@gyulasun,

Yes it is possible to use an adjustable regulator. I did just use what I had in my box. LOL
With an adjustable regulator one can also more easy adjust the voltage for different
types of LEDs. If you look at the datasheet for the LM2931 then you will see:

"The LM2931 positive voltage regulator features a very low
quiescent current of 1 mA or less when supplying 10 mA
loads. This unique characteristic and the extremely low
input-output differential required for proper regulation (0.2V
for output currents of 10 mA) make the LM2931 the ideal
regulator for standby power systems. Applications include
memory standby circuits, CMOS and other low power processor
power supplies as well as systems demanding as
much as 100 mA of output current.
Designed originally for automotive applications, the LM2931
and all regulated circuitry are protected from reverse battery
installations or 2 battery jumps. During line transients, such
as a load dump (60V) when the input voltage to the regulator
can momentarily exceed the specified maximum operating
voltage, the regulator will automatically shut down to protect
both internal circuits and the load. The LM2931 cannot be
harmed by temporary mirror-image insertion. Familiar regulator
features such as short circuit and thermal overload protection
are also provided."

So the regulator uses very little power by itself. It will also provide enough output to the LEDs until the 9V battery is falling below 5 volt.
The regulator also has an built in thermal overload protection. This is nice to have because the two LEDs is using approx. 50 mA and
the regulator gets hot. I have made a "heatsink" solution for the regulator since it is glued on top of the battery connector metal.

Groundloop.