Electronic – Using capacitor as a rechargeable battery

batteriescapacitorsolar cellstorage

As part of my learning of electronics, I'm currently trying to develop my own network of sensors.

So far, I've managed to make myself a satisfying small setup :
A few nodes, built with an atmega328p, a set of sensors and a RF emitter, capture data and send it through RF to a central server, which takes care of that data.

This works well, and I'm now trying to get rid of the battery part.
My current solution works fine with either a 9v battery or a 5V USB charger, and I'd like to try to get other kind of energy inputs (like solar panels, windmills, etc…).

My circuit requirements are simple : 5v / 65mA.
However, these requirements are only necessary for 1 sec every once in a while, since i don't need the circuit to run constantly. a few times a day is largely enough, and it can sleep / power-off the rest of the time.

I got myself what should have been a 0.5W solar panel (5v/100mA). However, when I try to measure the current I get from it, fully facing sunlight, I only get 5v/5mA.
This is not enough for my circuit, so I figured I would go and try to store that energy.

So I was thinking that maybe I could use capacitors :
After reading a bit on it, I've learned how to charge and discharge it.
However, to my knowledge, this always requires some changes on the circuit in order to switch from charging to discharging : Either removing the input battery like Capacitors as a Storage for Electricity, or doing something like this https://www.arduino.cc/en/Tutorial/CapacitanceMeter . I'm not sure if the link could be suitable in my case, because this requires the Microcontroller to be powered, and at this point, it is not.

The other option was to use rechargeable batteries, but I couldn't find really small ones and I would have the same issue as above. I don't want to buy LR44/CR2032 stuff because I would have to change them quite often.

My questions are :

  1. Should I simply replace the solar panel ? Is it normal that the current I read is so low even though the voltage I get is correct ?
  2. Is there a way to make a circuit that charges a capacitor, and when it's reached a sufficient charge, automatically use the energy stored in it in order to power some other part of the circuit until it's emptied, and all over again ?
  3. Same question with a rechargeable battery.
  4. In the case where this would not be possible, which I'm starting to believe, what is the common approach for this ? If possible, I'd like to avoid using classical "AA" or "AAA" batteries, but rather use "in-circuit" batteries (I don't know their name though)

Note : Sorry if anything is unclear or seems stupid. I'm still begining in electronics. Feel free to make any suggestion relative to this project.

Thanks in advance

Best Answer

1) The first step is to determine your average current consumption
I_AV = (T_ON * I_ON + T_OFF * I_OFF) / (T_ON + T_OFF)

a) T_ON = What percentage of the time is your device active.

b) I_ON = How much current does it draw when active.

c) T_OFF = What percentage of the time is yoru device inactive.

d) I_OFF = How much current does your device draw when inactive.

2) T_RUN = Determine how long your device need to run off the capacitor.

3) V_MIN = Determine minimum allowed operating voltage for your device.

4) V_MAX = Determine minimum allowed operating voltage for your device.

5) C = Size of capacitor you need. C = I_AVG * T_RUN / (V_MAX - V_MIN).

EXAMPLE:
The device consumes 65mA when on and runs for 100ms once every minute and is asleep the rest of the time. While asleep it uses 1mA. We are running on solar power and light is available 10 hours a day and not available for 14 hours a day. The device can operate from 3.0V up to 5.5V at the extremes.

I_ON = 65mA
T_ON = 100ms
I_OFF = 1mA
T_OFF = 59.9s
I_AV = (65mA * 100ms + 1mA * 59.9s) / (59.9s + 100ms) = 1.1067mA

T_RUN = 14hr
V_MIN = 3.0V
V_MAX = 5.5V
C = 1.1067mA * 14hr / (5.5V - 3.0V) = 22F

You can get a 15F, 5.6V super capacitor from Digikey for about $8.93.
MAL219691204E3
http://www.digikey.com/product-detail/en/vishay-bc-components/MAL219691204E3/4705PHBK-ND/5015889

Using two of them you may complete your design with some spare margin.

An easy way to make your circuit automatically switch over from capacitor power to solar is to use a diode oring circuit.

Use two diodes. One attaches from the solar pannel to the load and the other attaches from the positive side of the capacitor to the load. Both diodes point to the load. The load will then be powered from whichever source has a higher voltage (either the solar pannel or the capacitor).

To trickle chare the capacitor from the solar pannel you will want to connect the positive side of the capacitor to the positive of the solar pannel through a resistor and a diode. The resistor and diode are in series. The diode points towards the capacitor. The resistor value should be low enough that the capacitor can charge at a reasonable rate, but not so low that you load down the cell. Something in the range of 100 ohms should be about right.

Finally, you should have a zener diode across the positive and negative terminals of the capacitor to make sure it doesn't over charge. The zener diode points from the ground to the positive side of the capacitor. Select a didoe with a breakdown voltage that is less than the capacitor rating. For example, 5V on a 5.5V rated capacitor.