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!
HERE IS THE FINAL PRODUCT:
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.
THE WALL TRANSFORMER CAN BE SUBSTITUTED WITH A SOLAR PANEL! SEE NEXT STEP!
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!
HERE IS THE FINAL PRODUCT:
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.
THE WALL TRANSFORMER CAN BE SUBSTITUTED WITH A SOLAR PANEL! SEE NEXT STEP!
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Signing UpStep 1: Circuit Background
HOW LONG WILL IT LAST?
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!
HOW LONG WILL IT TAKE TO CHARGE?
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)
DO YOU WANT TO CHANGE YOUR RESISTOR VALUES AND INPUT VOLTAGE SOURCE TO CHARGE FASTER?
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).
THINGS TO REMEMBER:
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.
CAN I USE SOLAR PANELS TO POWER THIS?
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!
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!
HOW LONG WILL IT TAKE TO CHARGE?
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)
DO YOU WANT TO CHANGE YOUR RESISTOR VALUES AND INPUT VOLTAGE SOURCE TO CHARGE FASTER?
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).
THINGS TO REMEMBER:
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.
CAN I USE SOLAR PANELS TO POWER THIS?
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!
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R=E/I (12.5v/2.5A) = 5 Ohms
Power = E x I = 12.5v x 2.5A = 31.25 Watts
I'd use a 5 Ohm 40W resistor
I'd use a 16v 4A wall transformer, and a well heat sinked LM338 based variable power supply to tune to my charge voltage.
It would be very helpful if we could get the part numbers for the LED bank and DC converter.
In the video you show the boost PC board, that puts out 8 volts. I have been looking through the steps and I didn't see any information about this boost board. Am I looking at the wrong instructable? Or did I miss something? Thanks.
i thought I mentioned it? Sorry if not. The booster board takes any voltage between 3.4Min-34vMax at the input, can can boost up to 34VDC.
:)
Anyway, thanks for sharing it.
ITS CHALLENGING TO ME................