Hello, I want to make a project using an attiny 85 that gets powered with solar panels and supercapacitors. The goal of this first step is to understand how do i charge my supercapacitor to then power a basic led when there is no light. I tried using a 100μF capacitor the following schema and everything works fine, when there is light the led is on and if i cover up the solar panel it slowly fades out discharging the capacitor :
But when using my 4F super capacitor nothing happens, the light doesn't even lights up, i feel like the supercapacitor might be taking all of the power to charge up ? Here is the schema:
Hi,
Do you have a DMM? (Digital MultiMeter)
The supercapacitor will take much longer then a 100uF capacitor to charge up to a voltage high enough to get the LED to conduct.
Yes i measured 2.6 v out of the pins of the 100μF one and 0.4 V out of the 4F one. When putting the solar panel very close to a source of light this 0.4 value slowly rises up. I think you are right, i have a second solar pannel i might try to use both to charge it, I saw some people talking about a diode to not let the current flow back to the solar panel is this right ?
A discharged capacitor is, essentially, a short circuit.
So connecting a discharged capacitor will short-out your solar panel, until the capacitor voltage rises as it charges. With a supercapacitor, it will take a very long time to charge - so the voltage will remain low for a long time.
Until the capacitor has charged to at least the forward voltage of the LED, the LED is not going to light
A 4F capacitor is has 40 thousand times the capacitance of a 100uF capacitor - so expect it to be taking ~ 40000 times as long to charge!
You would see much the same thing with a rechargeable battery...
You will find that, unlike a battery, the energy stored in a capacitor is proportional to the voltage across it, or more accurately proportional to the square of the voltage across it. This means that as you start to charge it the voltage across it is low and increases slowly. For the LED to light you have to charge the capacitor to at least the LED forward voltage. One effect of this is that all the charge up to that point is effectively wasted, you can't use it to light your LED. A battery on the other hand has a near constant voltage from hardly charged at all to fully charged, meaning most if not all the charge that went in can be used to power the load.
Please do carry out your experiments rather than relying on what I and other tell you, you'll learn more that way.
Another thing with capacitors is that the voltage drops off linearly as they discharge - unlike a battery, which (more or less) maintains a constant voltage until it dies.
So powering electronics from a supercapacitor is a bit more involved than using a battery.
There are chips available for managing supercapacitors ...
Another term to look at is MPPT = Maximum Power Point Tracking.
This is about keeping the solar cell operating at its peak efficiency (the maximum-power point); eg, not shorting it out with a discharged supercapacitor.
Maybe I should charge the supercapacitor first with a regular source of power then making the whole circuit charge back for itself using the solar panel ?
charging the supercapacitor with another source of power makes it work like said, with that method the voltage gets to 2.6 and powers the led, using the multimeter the voltage keeps rising up, how could i calculate the maximum voltage for the supercapa and will it automatically cap ?
The maximum voltage is printed on the capacitor. If you exceed the maximum voltage then the capacitor will cap it for you by exploding (I'm not joking!)
The maximum voltage that the panel will charge the capacitor to is whatever voltage you get from the panel when it is open circuit.
The maximum is can ever possibly reach is the open-circuit voltage of the solar panel.
But it will never (quite) reach that if you keep drawing power from it.
The capacitor equation is: Q = C x V
Where:
Q is the amount of charge (ie, energy) stored in the capacitor (in joules);
C is the capacitance (in farads);
V is the voltage across the capacitor (in volts)
You can rearrange that to V = Q / C
ie, the voltage across the capacitor is proportional to the charge in it.
That's another difference between capacitors & batteries:
a battery has a defined "charged" voltage;
the voltage across a capacitor (in theory) will just increase indefinitely as it charges.
(in practice, of course, an actual capacitor will have a maximum voltage rating)
thanks again for all your help, i tried to power a bigger circuit and everything is working fine, i will have to adjust my program to enable periodic sleep modes and lower refreshing rate to limit the power consumption !
