This board provides a rechargeable power source that can be used to power a project. It works by charging up a 1F super capacitor when it is connected to a USB port. Two LEDs are used to show when the capacitor is charged (this takes about 1 minute).
Once charged the USB lead can be removed and the board used as a power source. The power connection is by two solder connections. When switched on the board provides power until the capacitor is empty. The length of time this takes will depend upon your application circuit (from a few seconds to ten minutes or more).
- Super Capacitor Charge Controller Kit.
- Use this kit as a rechargeable power source for your Electronic projects.
- Two LEDs are used to show when the capacitor is charged.
- Once charged the USB lead can be removed and the board used as a power source.
- When switched on the board provides power until the capacitor is empty.
- 1 x 1F Super Capacitor.
- 2 x 10Ω Resistor.
- 2 x 470Ω Resistors.
- 1 x 10kΩ Resistor.
- 1 x 15kΩ Resistor.
- 1 x 68kΩ Resistor.
- 1 x LM358 Op-Amp IC.
- 1 x 8 Pin IC Holder.
- 1 x Right Angle PCB Mount Slide Switch.
- 1 x BAT41 Signal Diode.
- 2 x 5mm Red LEDs.
- 1 x USB Power Cable.
- 1 x Two Way Plug In Terminal Block And Header.
- 1 x Super Capacitor Charge Controller PCB.
- PCB Length: 53.5mm.
- PCB Width: 43mm.
- This kit requires soldering.
- This board was originally developed for use on our STEM racer kit. The build instruction reflect this, but the board can be used to drive motors in other applications or to power different types of loads (not just motors).
This kit is designed and manufactured in the UK by Kitronik.
How long would the capacitor last when powering a speaker, for example the stereo amplifier circuit you stock? Also, if i swapped out the 1F capacitor for say a 10F capacitor, how long would it roughly last then when powering the speaker?
Posted byon Friday, 14 October 2016
The kit as is would only power the amplifier for a second and the 10F for around 5 seconds.
Could this board be used with a 10F or greater capacitor?
Posted byon Tuesday, 9 February 2016
There wouldn’t be any problem with swapping the capacitor to a larger one, however you would need to take into account the charging time would also increase. So the 1F capacitor takes around 1 minute to charge, so a 10F capacitor would take around 10 minutes to charge.
Would this power a high torque motor used in conjunction with a gearbox, 2508?
Posted byon Wednesday, 16 December 2015
Unfortunately the kit wouldn’t be able to power the high torque motor, if you were wanting to use a motor with this kit then we use a low solar motor with the STEM racer kit, https://www.kitronik.co.uk/4501-stem-racer.html
I've been trying to understand your circuit schematic, because I think it's quite clever, however there is one thing I don't quite get. The resistor you're using to limit the current drawn from the usb port seems to be a regular 1/4W rated resistor - how can this be? If i would want to limit the current to 500mA at 5V R=U/I gives me (5V/0.5A) 10Ohms which seems correct. However when calculting power: P=I*U=0.5A*5V=2.5W, so I would think you would at least need a resistor rated for 2.5W (usually 3W) which is much bigger. Could you please explain why you can get away with using a regular 1/4W resistor?
Posted byon Friday, 13 February 2015
The percentage of the voltage drop across the resistor would increase but the current would decrease proportionally so the power consumed by the circuit as a whole would drop and the power consumed by the resistor would remain the same, though it would be a larger percentage of the total power consumption of the circuit. In fact no value of resistor would generate more heat in that circuit. A 1 Ohm resistor would let 10 times more current through but the voltage drop would be 10 times smaller so the power dissipated would be the same. A 100 Ohm resistor would have 10 times more voltage drop across it but let 10 times less current through. Same power. The capacitor however would charge more slowly with less current allowed through so it would receive less power and the overall power of the circuit would decrease.
The key point here is that the voltage drop across the resistor is very small and the voltage drop across the capacitor is very large. P= IV
For resistor P = 500mA x a very small voltage.
For capacitor P = 500mA x almost 5V
If you add both those powers together you get 2.5W.
i just bought the Super Capacitor Charge Controller Kit and i just want to try it with super capacitor bank ,to increase the voltage any feedback thanks
Posted byon Friday, 9 January 2015
The Super Capcitor Charge Controller Kit measures the voltage across the super capacitor and cuts off charging once it reaches the peak voltage of one super capacitor. If you want to charge a bank of super capacitors you'd need a different circuit. Charging whole banks of super capacitors can be more difficult than just applying a voltage across all of them, I'd recommend researching the topic before attempting it or you may explode some capacitors.
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