Hi Everyone!


I'm a bit of a super capacitor fanatic, and I've made dozens of circuits that employ them.  This circuit is a prototype that I'm turning into a DIY kit.  Itis relatively simple, and is pretty darn efficient.  There is also a lot of room for customization!  When I get my custom PCBs made, I'll be throwing this device into an old flash light housing.  For the time being, I'll be talking about the circuit as it is.

This is my entry for the MAD SCIENCE FAIR contest, as well as the MAKE IT GLOW contest, so if you liked this instructable, I'd sincerely appreciate your vote or a rating =)  I've done my best to be AS THOROUGH AS POSSIBLE!


What The Circuit Does:
Unfortunately, super capacitors can only be charged to lower voltages; typically around 2.5v or 2.7v as a standard.  If you place some super capacitors in series, you can charge to higher voltages, but you lose a tremendous amount of capacitance.  When you plug this device into a wall transformer ((I designed this device around a 9v@1A transformer),  the on-board microprocessor turns on a relay that connects power to the capacitor bank.  The series super capacitors then charge to 5.2v through the relay contacts.   The capacitors an be interchanged to use higher or lower values, depending on how much you want to spend.   The voltage on the capacitor bank is constantly being sampled by an ADC (Analog to Digital Converter) that is embedded in the microprocessor.  When the voltage exceeds a value of roughly 5.2v, a flag trips in software, and the MCU turns off the charging relay, at which point the green LED indicator will start and continue to blink as an indicator to show the user that the caps are charged.  You can leave this device plugged in for as long as you want, and the caps will be very much safe and sound.

When the caps are charged, the user can flip a switch that connects power from the capacitors into a DC-DC voltage booster.  The voltage booster takes the 5.2v from the capacitor bank and boosts it to a calibrated 8v.  The output voltage from the booster can be boosted anywhere from 3.4v to 34v, and is easily calibrated by an on-board 10-turn variable resistor.

Since I've calibrated the booster to output 8v, as soon as you flip the switch, the output of the booster will provide a constant 8v to the LED bank that acts to emit light.  The LED bank is meant for 12v, but works great at 8v, and consumes MUCH less current. However, the LED bank is much brighter at 12v.  The booster will continue to source power to the LED bank until the capacitors drain down to 3.4v, at which point the circuit shuts down.  At this point, if you can plug it in again, to charge back up to 5.2v.  

When the booster is tuned to output 12v, the circuit consumes quite a bit more current, but the light output is much greater.   If you're going to consider maximum brightness, you're going to want to use 2x 400f 2.7v caps in parallel with one another.  I also took the liberty of hooking up a $1 LED flash light head that I purchased from the dollar store directly to the capacitors as opposed to the booster, and it lasts MUCH longer.  See the video.


Step 1: Circuit Background

That is a trick question =)  In regards to how long the caps can last, or how many times they can be charged..... In all likelihood, they're going to outlive you . Super capacitors, if treated properly, can be charged 100,000s of times.  Batteries can not, which makes this such an interesting project.   How long will the light stay on after charging the cap bank to 5.2v?  That depends on three things:

1) What value of capacitors are you using?  If you're using two 400f 2.7v caps in series (200f 5.4v), then your flash light will last well over an hour in the current configuration.  If you use 2x 50f 2.7v caps (25f 5.4v), then the flash light will last for about 15 minutes.  Keep in mind that since the booster can only operated with 3.4v at the input, you only have about 1.8v of dissipation from the 5.2v on the capacitor bank.  Charging back up to 5.2v from 3.4v will only take a short time.  This project can easily be modified to work with more than 2x series capacitors, but this is merely a prototype!

2) What kind of light are you using?  In my current configuration, the booster is actually consuming more current than the LED bank.  If you choose to use a single high bright LED with no booster, and you use the proper current limiting resistor, you can really make this circuit last.  I've offered an example of this in the video.  You can customize this circuit so that it requires extremely little power, which will make your flash light work for hours on end.

3) If you are using the booster, the power consumption will vary depending on how much voltage you're applying to your bank.  Keep in mind that you don't need to use a bank like mine.  My bank is very bright, but it also requires more current.  DO NOT use normal flash light bulbs.  They require a bunch of current, and they're no good for this application.  USE HIGH BRIGHT LEDs!
You can also connect some parallel LEDs and a current limiting resistor directly to the capacitors when charged (AS SEEN IN THE VIDEO).  This won't give you as much of an opportunity to customize your circuit, but it certainly is cheaper, and it will consume less current, and also last much longer!

Another interesting question.  It depends on three things as well:

1) What is your bank size?  If you're using two 50f 2.7v caps in series, it will take about 5 minutes in the current configuration.  If you're using two 400f 2.7v caps in series, it will take about an hour to charge, but it will last much longer than an hour when emitting constant light.

2) I designed this prototype around a 9v 1A transformer.  Since I need to save some power for the digital circuitry, I am only supplying around 500mA to the capacitor bank, which is not a lot.  If you have a bigger wall transformer that can output 9v or higher, and can support more than 1A of current, then you can change the current limiting resistor bank so that is allows for more current to charge the capacitor bank, and therefore will charge it faster.

3) The current limiting resistor has a HUGE impact.  I only had a bunch of 1.8 ohm power resistors in house, but what I needed was an 18 ohm 5 watt resistor.  I worked with what I had.  With an 18 ohm 5 watt resistor, I would be charging the bank to 500mA.  Here is the logic and the calculation:

The power supply is 9v
The transformer can support 1A of current. 
I want to save a few hundred mA for the digital circuitry and the relay (Even through it requires less than 100mA)
To determine which resistor to use, I need to know what current I want to charge the caps at.  I chose 500mA (0.5A).
Resistance = Voltage / Current
R = 9v / 0.5A
R = 18 Ohms
Now, we need to determine the power requirement for the resistor. 
Power = Voltage x Current
P = 9v x 0.5A
P = 4.5W  (So I'll use an 18R 5W resistor)


Here is another example that you can follow if you want to change how fast your bank charges.
Power supply = 12v @ 3A
I need to save 500mA for my digital circuitry and my relay, so I can charge my caps at 2.5A.
R = 12v / 2.5A
R = 4.8 Ohms
P = 12v x 2.5A
P = 57.6 Watts
So I need a 4.8 Ohm resistor rated for 57.6 Watts.  (A more likely value would be 5 Ohm 60W resistor). 

Since there is a resistor bank limiting current to the capacitor bank, you don't need to worry about the cap bank being charged at a higher voltage, because as the caps charge, there will be a bunch of voltage along the resistor bank.  If you place a multimeter on the cap bank, you will see the voltage slowly charge up from 0v.  Make sure to place your resistors at least 0.5cm above the board, as they will get HOT.  You don't want to damage your board!  Since this circuit employs a 5v regulator for the microprocessor and the 5v relay, you don't need to worry about your input voltage source.  It must be more than 7v, and lower than 28v.

YES!  If you have a solar panel, or a series/parallel bank of solar cells as a power supply, as long as you consider the calculations above, you can use them to charge this circuit as opposed to a wall transformer!

<p>Well, yes. I have yet to meet a cat that could hold a soldering iron, let alone phathom ohms law.</p>
I'm really intrigued by the possibility of powering things from capacitors instead of batteries.
<p>Pros:</p><p>* Super Caps have a longer life: can be charged/discharged far more times than any battery -- on the order of millions, compared to 500 to 1000 times for a secondary battery.<br>* No &quot;memory effect&quot;.<br>* Super Caps can be discharged all the way to zero volts with no damage.<br>* Because of their incredibly low internal resistance, Super Caps can deliver tremendously high current at very little loss. Pound per pound, they pretty much kick the butt of any battery chemistry in terms of current delivery.<br>* Super Caps have a far better Power Density than any battery chemistry.<br>* Super Caps have a slightly better working temperature range than any battery chemistry.<br>* Super Caps tend to be safer to handle and far more &quot;forgiving&quot; to abuse [e.g. Lithium battery fires, Lead Acid hydrogen explosions, the tendency of batteries to leak corrosive chemicals]. Of course, a short across a fully charged Super Cap can be pretty hazardous -- shower of sparks, hot molten metal, etc. Also, a Super Cap cares not about it's orientation -- no concern for chemical leakage like with some batteries.</p><p>Cons:</p><p>* Batteries, in general, have a much greater Energy Density than Super Caps, mainly because the energy that a chemical battery can store per unit of weight, is far greater than that of a Super Cap.<br>* A Super Cap's linear discharge voltage requires, either a shortened discharge time, or some sort of active compensation. Most battery chemistries are capable of keeping the discharge voltage relatively stable across the discharge curve (when drained at a nominal current rate for that battery).<br>* A Super Cap has a higher self discharge rate then most battery chemistries.<br>* Batteries/Cells can easily be ganged in series for higher voltages. Super Caps can also be ganged in series, but require some sort of charge balancing.<br>* Super Caps have a much higher cost/watt.</p>
<p>Super Capacitors are difficult to charge efficiently. They present such a low impedance, they behave, essentially, like a hefty short, especially when they are, or nearly are, fully discharged. So, to charge them efficiently, the charger needs to have an extremely low output impedance [highest energy transfer occurs when the input and output impedances match]. </p><p>The best way to do that is with a switch mode buck converter. Program the output voltage of the converter to the maximum charge voltage [which is really, the &quot;over charge protection&quot; feature]. The converter will repeatedly charge an inductor, and then dump the inductors energy, through a switching diode, into the SuperCap, until the capacitor's voltage reaches the converter's output voltage set point. If you use a converter that is capable of &quot;burst mode&quot;, it will, then, only transfer energy to the super cap, if the cap's voltage sags -- which is likely, since SuperCaps, inherently, have relatively high leakage current [the leakage becomes greater, the higher the voltage across the cap].</p><p>Also, there is a reverse proportionality between voltage across the capacitor [i.e. the charge voltage] and the life of the capacitor. Holding a SuperCap that is rated at 2.7V, AT 2.7V, will significantly reduce the life of the capacitor over charging it to only 2.5V. This from the Maxwell Technologies BOOSTCAP Ultracapacitors Product Guide &ndash; Doc. No. 1014627.1:</p><p>----------------------------------------------------------<br>30% reduction in rated capacitance may occur for an ultracapacitor held at 2.7 V after<br>5,500 hrs @ 65 oC<br>11,000 hrs @ 55 oC<br>22,000 hrs @ 45 oC<br>44,000 hrs @ 35 oC<br>88,000 hrs @ 25 oC<br>15% reduction in rated capacitance may occur for an ultracapacitor held at 2.5 V after<br>5,500 hrs @ 65 oC<br>11,000 hrs @ 55 oC<br>22,000 hrs @ 45 oC<br>44,000 hrs @ 35 oC<br>88,000 hrs @ 25 oC<br>-----------------------------------------------------------</p><p>Something to consider, when making the claim that the thing will last &quot;forever&quot;. 88,000 hrs IS 10 years, but that's still less than 'forever'.</p><p>Solar panels, BTW, are excellent at charging SuperCaps. Since there are, essentially, current sources, they can deliver peak current, even into a short. The trick is to use a solar array with an open circuit [OC] voltage equal to or less than the highest voltage you want the capacitor to charge to. So, say you want your SuperCap to charge no higher than 2.5V, and lets say the OC voltage of a set of solar cells is 0.56V peak [at a 90 degree angle to full summer sunlight] and a blocking diode with a forward voltage of 0.4V at maximum current. (2.5 + 0.4)/0.56 = 5.17, thus, the panel will have no more than 5 cells in series and output a max voltage of 5 * 0.56 = 2.8V and with the blocking diode loss, the final highest voltage applied to the SuperCap will be: 2.8 - 0.4 = 2.4V.</p><p>Thus, the solar panel will efficiently charge the SuperCap to 2.4V and stop [and, actually, it will go a little higher, because as it approaches 2.4V, the charging current will drop off, thus lowering the forward voltage of the diode. The system will reach equilibrium when the charging current is equal to the SuperCap leakage current. So, a wise engineer will determine, via experimentation, what the forward voltage on that diode is, at the equilibrium current, across the projected range of temperatures the Cap+charger will be subjected to in the field [or at least the range specified in the instruction manual ;) ]</p>
<p>Hi great project thanks for sharing,just wondering can i use PIC12F683 chip instead</p>
<p>Pretty cool but I just want to point out that while you lose capacitance by putting capacitors in series, you don't lose energy. J=C/2(V^2). So with two 2.7 V, 50 F caps in series that's 25F/2(5.4V^2) = 364.5 J = 100F/2(2.7V^2). Whether you put them in series or parallel, their energy storage potential remains additive. You are however losing energy through the power converter.</p><p>Also, why so many LEDs? You could improve the optics greatly just by arranging the LED bank on a round board, and positioning it centered at the very very bottom of the reflector. Try it with a light meter, you'd be surprised. I'll bet it'll be brighter with half the LEDs running at full power</p>
wooo, jonny 5, 2nd movie is the best. <br>:)
Awesome project, but too difficult for me...<br><br>Anyway, thanks for sharing it.
I just made a very simple version of this flashlight, and an instructable if you're interested. No software, and no complicated circuitry =) Check out my channel if you're interested!
Oh, thanks. I will see that.
Agreed. It is a bit complicated. I'll make an easier one next week, and cut down on the theory and such =) Thanks for having a look!

About This Instructable




Bio: Hi there! My name is Patrick, and I am an electronics engineering technician who works full time as a lab tech, and part time as ... More »
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