## Step 3: Assembly Notes

You can really go about this in your own way, but here is how my girlfriend chose to put this device together.  She wrote this section.

1) I soldered the circuit together.  I then made sure to add long wires in between each component so that placing each component into the flashlight would be easy.  I also made sure that the wires from the jack to the resistor, and from the resistor to the capacitors was thick and well insulated.  This is because the wires will be hot termporarily during charge.

2) I utilized the on board switch on the flash light to apply/disconnect power from the capacitors to the LED flash light head.

3) I glued the DC back in the battery bank section, so that I could remove the battery cover and plug in my transformer.  It just makes the device look more professional.

4) I glued the capacitors together length wise, and placed them in the shaft of the flashlight.  I then added some glue in the shaft to secure them in place.

5) I placed my resistor at the top of the shaft.

7) I put the flash light back together.

Then, we both had a beer to celebrate this new fast-charging super capacitor flash light!

<p>It's relatively easy to get an idea of the run time of these types <br>of circuits. The energy stored in a capacitor is 0.5xCxVsquared <br>(CxVxV/2). The capacitance is in farads, the energy in joules.<br></p><p>So the energy in this case is a maximum of 0.5x100(total capacitance)x2.7x2.7=364.5joules.</p><p>So<br> how do we decide the runtime? We divide the power by the power of the <br>LEDs in watts and that gives us the runtime in seconds. So, if we are <br>using a 1W array, the run time will be 364.5seconds, or roughly 6 <br>minutes. If we used a 3W array it would be only 2minutes.</p><p>However, although this is the calculated time, it will vary in reality based on two conflicting issues:</p><p> Firstly,<br> unless we use a sophisticated dc-dc converter, not all of the energy <br>will be used. If the LEDs fail to light below e.g.80 of the voltage, we <br>could only get 20% of the runtime. Capacitors, unlike batteries, don't <br>store all the power and then have a fairly stable voltage until the <br>battery is almost empty, but decay away proportionately to the current <br>taken, so the voltage is continuously dropping.</p><p>Secondly, working <br>against the first is that an LED doesn't take the same current as the <br>voltage drops. It continues to take less and less so therefore the <br>runtime gets extended. The cost though is that the LEDs will fade <br>continuously before finally disappearing. This is why some old design <br>LED torches without a DC-DC converter seemed to have long quoted battery<br> lives, but were actually unusable by the end in terms of any useful <br>light output.</p><p>So, I said it was relatively easy to calculate the <br>runtime? It is, but the actual runtime will vary quite dramatically <br>depending on how you define &quot;runtime&quot;. The calculated runtime is correct<br> for full brightness but it may be much longer for a &quot;useable&quot; <br>brightness.</p><p>In comparison, if you want to work out how many joules<br> a typical battery holds, simply multiply the Ah rating by the voltage <br>by 3600 (to convert to seconds). Thus, an AA rechargeable 1.2v battery <br>with 1000mAH holds 1.2x1x3600=4320 joules, which is roughly 12x the <br>energy stored in these caps.</p><p>That's one of the reasons <br>batteries are used in appliances. Although supercaps have many <br>advantages in terms of recharge speed and potential current delivery and<br> may be the preferred storage mechanism of the future, they <br>unfortunately currently fall short in terms of economic power storage.<br></p><p>ref:</p><p>Capacitors Energy<br></p><p><a href="http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html" rel="nofollow">http://hyperphysics.phy-astr.gsu.edu/hbase/electri...</a></p><p><a href="http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html" rel="nofollow">Batteries Energy</a></p><p><a href="http://www.allaboutbatteries.com/Energy-tables.html" rel="nofollow">http://www.allaboutbatteries.com/Energy-tables.htm...</a></p>
Pretty simple circuit it is, but I still have some questions:<br><br>1-does this circuit FULLY CHARGE the capacitors ( like they'll make a big spark when shorted )?<br><br>2-what does make a bigger spark with fast emptying: higher voltage cap or higher capacitance cap? <br><br>3-is it possible to fully charge these super capacitors using disposable camera circuit?<br><br>4-what it I used a combination of capacitors that'll make 0.00086 farad with 400v instead of the super capacitors used here?<br><br>Thank you.
OK, this is nice and 'simple', but what am I missing?<br> <br> Your resistance calculation seems to say that 5 divided by 2.5 is 2.5 - not 2.<br> <br> Your power calculation appears to claim that 5 multiplied by 2.5 equals 10, instead of 12.5.
What kind of runtime do you get out of this?
Yes, this is simpler than former. Thanks for do it.