Measuring solar panel output power

Discussion in 'General Electronics Chat' started by alexx_88, Nov 14, 2010.

  1. alexx_88

    Thread Starter New Member

    May 24, 2010
    17
    0
    Hello!

    In short, the aim is to create a device that is completely autonomous. It would sleep ~ 99% of the time, wake up, do some processing, and transmit the data. It would be powered at 3.3 V.

    Average consumption would be somewhere around 1mW, with 100 mW peaks when the device is processing.

    Going beyond this introduction, I'm interested at this stage to do some tests to determine the degree to which the solar cell + battery can power the device in an autonomous manner during a whole day. In other words, if the energy captured by the cell in 24 hours is sufficient to operate the device.

    As the power source will go through a 3.3V voltage regulator, I guess I must use a battery with at least 4V. At this stage I didn't do any research regarding the type of battery which I should use, but it should be one able to handle multiple charge/discharge cycles as the battery will function as a buffer between the solar cell and the device.

    The first thing I want to do now is find a resistor which when connected to the solar cell simulates the load which charging a battery would imply. After that, I could measure the voltage on the resistor and determine the output power of the solar cell in different lighting conditions.

    So I have some questions:
    1) How can I determine the resistor value to simulate a charging battery?
    2) I saw that each solar cell has a maximum power output at a specific voltage and current. How can I choose the solar cell so that when connecting it to the battery, the power generated is maximal?
    3) Considering that the battery is permanently connected to the solar cell, does it lose much of its life cycle?

    Thank you!
    Alex
     
  2. spinnaker

    AAC Fanatic!

    Oct 29, 2009
    4,866
    990
    I am working on a very similar project. In my project my battery has powers a garden light. The PIC monitors the panel voltage, When the voltage drops to zero, a user settable timer starts. The light goes on when the timer times out. When the light goes on a new timer is started. The light goes off when that timer expires or the the voltage of the battery drops below a user settable level.

    You do not want to charge the battery direct off the panel. You will want to use a charge controller. Battery charging is fairly sophisticated. You can build your own or buy an off the shelf controller (this is what I did). The controller will help to extend the live of the battery.

    You will also save the battery by shutting it down before it becomes 100% discharged.

    Your other problem is going to be powering your PIC. You will need a high efficiency regulator such as a buck regulator. No sense having a high efficiency PIC when your regulator is gobbling up power via heat lose.

    Once again these regulators are not exactly easy to build for the newbie like me. I solved the problem by choosing an off the shelf buck regulator for my LED lamp. It has a +5 supply for the purpose of powering micro controllers.
     
    alexx_88 likes this.
  3. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    Differrent levels of average power consumption generally lead to different solutions.

    Very low levels of average load consumption, say 50-100mWh / day tend to suggest a very simple solution is in order. I would suggest a string of supercapacitors, and a high-efficiency step down buck switchmode, perhaps using a hysteretic type control mode suited to low standby power levels. That avoids the use of a battery, and the buck adjusts for the fall in capacitor voltage. You may be able to connect the PV panel just through a diode to the cap, depending on max voltage levels under all conditions.
     
    alexx_88 likes this.
  4. alexx_88

    Thread Starter New Member

    May 24, 2010
    17
    0
    Basically, when your solar panel voltage drops to zero means that there isn't any light available and the light must be turned on? I saw similar devices which used a photodiode or a photoresistor.

    I was thinking the same thing. Space can be a problem in my device so I am planning to do the same thing: buy a high-efficiency regulator and, after I decide on a battery, purchase an IC to handle the charging for me.

    Thanks!

    I may have said it wrong, the power absorbed by the device it's 0.2mW when sleeping and 100mW when active.

    Given the fact that on a 5 minute cycle, the device works only for 2 seconds and sleeps for 298, then the energy consumed in these 5 minutes would be: 2s*100mW + 298s*0.2mW =~ 260 mJ / 300s. In an hour: 3120 mWh.

    I would prefer using supercapacitors, but can they store enough energy to operate the device for 10-12hrs / day without too much power coming from the solar panel (night, poor weather etc) ? I will make some calculations and see if they are a viable option.

    For now, I will buy the solar panel, some resistors and start measuring the power output in real life situations.

    Thank you for your help!
    Alex
     
  5. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    0.2mW for 24 hours is 4.8mWh. The additional energy for 2 seconds is only a small additional increment.

    Ballpark design would be to use a nominal 6V small PV panel then you could directly (through a diode) connect to nominal say 8V supercap, with a shunt zener to limit voltage to supercap max. You can then swing the supercap voltage from say 4-8V through a low drop out regulator to 3V3. Your energy store is 0.5C(8x8 - 4x4)= 24C. Work through the maths and see what you need for PV panel power - but you will need to make a call on light per day min that you can accept.

    Ciao, Tim
     
  6. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    Oops - should have done my maths first. 0.2mW load requires about 4.8mWh, 100mW load requires about 16mWh, so total about 21mWh.
     
  7. alexx_88

    Thread Starter New Member

    May 24, 2010
    17
    0
    Ok, but if I put in a 6V solar panel then how could it charge the capacitor up to 8V?

    If I have a 6V solar panel will the shunt zener diode be necessary given the fact that the supercap will be rated with a higher voltage?

    From what I understand, the energy stored in the capacitor between Vmax (of the capacitor) and V min (of the voltage regulator - 4V I presume ) must be enough to power my system through the hours when there's no light available. So, I either go for a higher voltage system (with bigger costs in terms of solar panel purchase) or with a bigger capacity supercap. Is that right?

    I am thinking of getting this solar panel and match it up either with 2 * 2.5V supercaps in series or with a bigger 5V one.
     
  8. alexx_88

    Thread Starter New Member

    May 24, 2010
    17
    0
    I have done some more research and I will go for a boost converter to deliver the 3V3 to the system. This will allow me to discharge the supercaps lower than the 4V needed in low drop out regulator.

    I will design the power circuit in the next few days and then come back for feedback.

    Thank you very much for your help!

    Alex
     
  9. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    I suggest you would benefit from a little more research and pre-design. Look at the voltage characteristics of a PV panel at different temperatures and different currents - a "6V" panel may be able to output much higher voltage. Look at a suitable LDO reg for its voltage drop at the current levels you are working at (3.3V 100mW is ~30mA). Look at a suitable commercial supercap specification for max voltage, self discharge, and effective capacitance, and determine if it will supply your energy requirements. You have a very low power requirement system.
     
  10. alexx_88

    Thread Starter New Member

    May 24, 2010
    17
    0
    Thank you very much for your help! I will certainly continue to study on my own. For now, I will start testing the solar panel and then I will do the math too see what supercaps and what regulator I need.

    One last question though: wouldn't it be better for me to use a boost converter instead of a low dropout regulator? I am thinking that I can make better use of my supercaps by discharging them to a lower voltage and thus providing more energy.

    Alex
     
  11. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    At such low power transfer levels, you are likely to consume more energy in the control of the dc/dc; have much more complexity (if it isn't an integrated hybrid); be a lot more costly; and possibly be no more efficient.
     
  12. alexx_88

    Thread Starter New Member

    May 24, 2010
    17
    0
    I have just received the solar cell I ordered, but didn't receive any schematics on how to connect wires to it. This is the cell: http://www.adelaida.ro/product_info.php?products_id=11852&osCsid=c305bb0e5b53797ca3626341 and I have attached a photo (sorry for the bad quality) of the back. I have done some google searching about how to connect this type of solar cell but so far I didn't find anything.

    PS: I have measured the voltage between the two oval shaped areas in the picture and is 0V. I think that I must scratch those areas to get to the conductive part, but I am afraid I will irreversibly damage the cell if that's not correct.

    Regards,
    Alex
     
  13. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,779
    932
    They expect you to solder them together yourself.

    Look at this and then google for more info.
    http://www.greenjoyment.com/step-by-step-diy-solar-cell-soldering-green-power-science

    It isn't that hard, but it is confusing at first. Use a voltmeter and identify the polarity of a single cell. (should be .5 volt) All the other cells should be arranged identically to that one, but use the meter to be sure. If there are two rows they could be mirror imaged in polarity
     
  14. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Your Romanian link shows a different solar panel.
    Your photo shows a solar panel with only 5 cells. Each cell is only 0.5V so the solar panel has an output rated at only 2.5V. It might be 3.8V at noon in summer with no load.

    My cheap Chinese solar garden lights use a solar panel with 4 cells (2.0V) and a series Schottky diode (0.3V) to charge a single Ni-Cad cell to 1.4V when fully charged. It uses a simple voltage stepup circuit to power white or colour-changing LEDs.
    In summer the garden lights glow all night (8 hours) following a sunny day. In winter they glow dimly for only 1 hour following a sunny day. They do not glow in winter following a cloudy day.
     
  15. alexx_88

    Thread Starter New Member

    May 24, 2010
    17
    0
    Last edited: Nov 23, 2010
  16. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    I'm liking this thread as it mirrors some of my efforts.

    What made you decide on a nicad?

    My project has a much higher power requirement (camera). I've been working with batteries but a cold environment complicates the matter. Maybe the cap approach would help that.

    Anyone have experience with this device;
    http://cds.linear.com/docs/Datasheet/3625f.pdf
     
    Last edited: Nov 23, 2010
  17. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Your "docs.google" didn't show me an ordinary PNG file type shcematic that is attached here to your reply.
     
  18. alexx_88

    Thread Starter New Member

    May 24, 2010
    17
    0
    Why should it show a schematic file? It contains the results of some little experiments that I've done with that solar cell. I have now bought a solar garden lamp and I will do some testing on that too.

    Thank you!
    Alex
     
Loading...