## Step 3: Schematic Diagram and Explanation

Before I begin talking about the schematic, let it be known that there are four hardware blocks to this circuit:
1) The wall transformer
2) The charger board (Connected to the capacitors)
3) The voltage booster board
4) The LED bank

You can follow along with the picture, as I will be EXTREMELY THOROUGH with my hardware Explanation.  The wall transformer, booster, and LED bank are all stand-alone products.  The charger board is the circuit that has to be created in order to get everything up and running, so let's talk in detail about the charger circuit.

THE POWER SUPPLY:
As you can see, the power supply circuit is relatively simple.  We have our DC power jack that connects to the wall transformer.  This voltage source provides power to the capacitor bank through a current limiting resistor bank and a diode.  As well, this 9v source is connected to a 7805 5v regulator.  The output of this regulator is a smooth 5v, which acts to power our micro controller (PIC10F222), as well as our 5v relay.   This regulated 5v source is labelled VCC in the schematic.   There is a 10uf capacitor on the input of the regulator, as well as on the output.  These caps are just used to stabilize the output.  There is also a 0.1uf capacitor on the output that is used as a decoupling capacitors.  It acts to filter out any unwanted noise at the output; high frequency spikes, and what not.

THE CAPACITOR CHARGING RELAY AND DRIVER:
When power is applied, pin#4 of the PIC10F222 goes from 0v to 5v, which activates the relay through a simple driver circuit.  I've used an NPN transistor (2N2222) and a 10k resistor.  When 5v is applied to the base of the transistor through the 10k protective resistor, power from the collector sinks through to the emitter of the transistor, which is connected to ground.  The collector is connected to the relay coil, and to the anode of a 1N4001 diode.  I'll get to the diode in a second.  The secondary end of the relay coil is connected to the regulated 5v supply (VCC).  When the base of the transistor is activated, power sourced by our VCC line sinks through the coil and through the transistor to ground, and therefore completes the circuit.  The coil is then magnetized, which acts to switch the common pin of the relay (CO) to the normally open pin of the relay (NO).  When the microprocessor turns the relay off, the magnetic field along the relay collapses, and a large voltage spike occurs.   The cathode (negative end) of the diode is connected also to our VCC 5v source.  The diode acts to protect the circuit from this voltage spike.  It is very necessary.  Make sure that you don't place the diode in the wrong way, or else you're going to have a short circuit when the relay turns on, and that will reset your device.

The common pin of the relay is connected to the cathode (negative end) of the 1N4001 diode on the power line, just after the resistor bank.    This diode is to ensure that there is no back powering from the capacitor bank.  This diode will ensure that current will only flow into the capacitor bank from the power supply, and not backwards from the cap bank into the power supply or regulator.  When the relay is activated, the common pin connects to the normally open pin (NO), which is connected to our capacitor bank, which enables the charging of the capacitors.  When the relay turns off, the common pin is re-connected to the normally connected (NC) pin of the relay, which cuts off the charge to the caps.

THE MICRO-CONTROLLER AND THE ADC:
I've chosen the PIC10F222 for this project.  This is a programmable microchip that I've programmed in my lab at home.  It requires a regulated 5v (VCC) on pin#2, and our DC ground to pin#7.  I employ only three of the four onboard I/O ports.  The first is  GPIO1, which is configured as an output (Pin#4).  This acts to activate/deactivate our relay.  The second I/O port is GPIO0, which is pin#5.  This pin is configured as an input, and is programmed to work as an analog to digital converter (ADC).  When the power turns on, the relay turns on and charging commences.  From there, the ADC is constantly sampling the charge on the caps through a resistor divider network made up of 2x 10k resistors along the charge line after the diode.  Why is this needed, you ask?  Since we'll be requiring that the capacitor bank be charged to more than 5v, we need to divide the voltage value on the caps in half, as if you place a value higher than 5v on the ADC line, you're going to damage your chip.  With the resistor divider in place, the ADC will see half of that voltage only.  In programming, the ADC is looking for a voltage higher than 5.2v, so as soon as the voltage  on the caps reaches 5.2v, there will be 2.6v on the ADC line.  The second the charge on the caps reaches this charge level, the MCU instructs GPIO1 to go LOW, which turns the relay off and disables the charge to the super caps.  Lastly, GPIO2, which is pin#3 acts as an output, which acts to flicker the green LED indicator on and off when charging is about to commence, and when charging has completed.

THE INTERFACES:
There has to be a good way of connecting our capacitors to the charger board, as well as our booster board and LED bank.  There are two (2-pin) terminal blocks that can be used for the two capacitors, and a 3-pin terminal block for the LED bank and the booster.  The 3-pin terminal block has two grounds, and a lead connected to the positive lead of the capacitor bank.  The positive lead of the capacitor bank will be connected to the VIN (Voltage input) of the voltage booster.  One of the ground pins can be connected to the GND (ground) pin of the booster board.  The output pin of the voltage booster can be connected to the RED (Positive) wire of the LED bank, and the BLACK (negative) wire of the LED bank can be connected to the secondary ground pin of the 3-pin terminal block on the charge board.  So now ALL ground are connected.  This is very important.  The power of the capacitor bank is connected to the input of the voltage booster, which we have calibrated to 8v.  The output of the booster, which is boosted to 8v, is connected to the LED bank, which provides it power.  The booster will keep boosting to 8v until the voltage on the capacitors runs below 3.4v.

SUPERLYK N AWESOME AND<br> ITS CHALLENGING TO ME................
<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

41,359views

60favorites

License:

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 »
More by EngineeringShock:
Add instructable to: