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113752 Views 25 Replies Latest reply: Oct 21, 2012 9:50 PM by JoeNikk RSS
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Jun 2, 2011 3:13 PM

How to use a voltage regulator with a 12v solar panel

I have a smaller solar panel used to "trickle charge" 12v auto batteries that I want to use to power car cell phone and gps chargers.  I simply added a female cigarette lighet recepticle on the end of the panel, however the cvhargers never work.  Turns out the output ranges from 0 to 20 plus volts depending on the available sunlight.  How can I use a simple regulator keep the output at 12v or less?  Which part(s) should I use?

 

Thanks!

Dan

  • Novice 1 posts since
    Jun 12, 2011
    Currently Being Moderated
    Jun 12, 2011 1:35 PM (in response to dsmall)
    Re: How to use a voltage regulator with a 12v solar panel

    What is the current ratting that the PV can supply?

     

    This circuit may help

     

    http://www.den-uijl.nl/solar.html

  • Novice 3 posts since
    May 23, 2011
    Currently Being Moderated
    Jun 17, 2011 12:30 AM (in response to dsmall)
    Re: How to use a voltage regulator with a 12v solar panel

    Whatever regulator you may want to choose, the important factor is efficiency. The solar PV will already have a pretty low conversion rate, therefore go for a regulator which has a high efficiency.

    #TWiPad

    • Novice 3 posts since
      Jun 13, 2011
      Currently Being Moderated
      Jun 19, 2011 12:17 PM (in response to dsmall)
      Re: How to use a voltage regulator with a 12v solar panel

      Dear dsmall,


      You've got it!


      I didn't realize you wanted 5V output, since your original post stated that you wanted a 12 V output.

      The engineering solution finds the optimal solution by evaluating trade-offs.  You have discovered that.

      If you are up to the challenge, you could design a similar one using a switching regulator, which would,

      in a twisted ironic way, waste less solar energy... =D ... but generate LF-RF noise pollution instead... albeit

      at a low level.


      Happy absorbing!


      Dan

      • Novice 3 posts since
        Jun 23, 2011

        I'm looking at building that solar phone charger, but I have one question/concern. Why does it call for a DC-DC converter rated to 2.5 watts max @ 5V when the solar panel they use is 10 watts? Maybe I'm unclear on how DC-DC converters work. Do they not maintain the same power level (minus inefficiencies) as AC transformers do?

        • designspecialist-power 62 posts since
          Mar 22, 2011

          Hi eeyorebill04,

           

          You are correct that the solar panel is capable of providing much more power than the DC/DC converter can output downstream, and that the input power of the DC/DC converter is the output power plus any inefficiencies (and I doubt there is 7.5W of loss).  I think that particular solar panel was used to promote it and show it off a little.  The DC/DC converter was only limited to 500mA because that is the standard USB power max current (there are variations that allow more, but not all devices can necessarily handle that when charging).

           

          That being said, if you wanted to, you could either:

               1) Use a smaller solar panel with less output.

               2) Use a larger DC/DC converter with more output (just make sure the charging device can handle the load).

               3) Parallel more DC/DC converters in parallel with the one already there for multiple standard 500mA USB charging connections all off of the same solar panel.

          • Novice 3 posts since
            Jun 23, 2011
            Currently Being Moderated
            Jun 23, 2011 8:21 PM (in response to Alec)
            How to use a voltage regulator with a 12v solar panel

            ohhh, I see. I was thinking that if you tried to put much more than 2.5W into the DC-DC converter could damage/fry it. I guess that's not the case? So is the extra power dissipated by the DC-DC converter as heat?

            • designspecialist-power 62 posts since
              Mar 22, 2011

              eeyorebill04,

               

              No, you were correct the first time.  You won't be able to put much more than 2.5W into the DC/DC converter listed for the project (just 2.5W plus the inefficiencies will go in like you mentioned).  If you put too much in it could damage it because the excess is dissipated by the DC/DC converter and that will heat it up (the excess power is due to those inefficiencies).  I think maybe the part you are confused on is that the solar panel will only produce the power that it needs in order to supply the load attached to it.  The 10W rating of the solar panel is more of a "can output up to 10W" number.

               

              Let me throw some fake numbers out to illustrate an example.  Let's say you are charging a 5V USB device at the limit of 500mA, so (5V * 0.500A) = 2.5W.  The solar panel voltage output will change with light level, and in turn the efficiency of the DC/DC switcher will change depending on the voltage input, but to simplify the calculations let's just assume that the efficiency for the  converter is 80% across the whole voltage range of the solar panel.  Now with the efficiency as such, the total input power to the converter would be (2.5W / .80) = 3.125W.  So this means the solar panel outputs 3.125W, the charging device is getting 2.5W of that, and the DC/DC converter is forced to internally dissipate (3.125W - 2.5W) = 0.625W.  That extra 0.625W causes the DC/DC converter to heat up above the ambient temperature according to its degrees Celsius per Watt thermal characteristics.  If the ambient temperature plus the temperature rise associated with the dissipated power exceeds the maximum operating temperature of the device, then it can be damaged.

               

              To illustrate it another way, let's say that you are blasting a solar panel with direct sunlight and it is therefore at it's maximum possible voltage.  But let's add that the output is an open circuit (no load).  Then the current out of the solar panel will be essentially zero since there is no load to conduct through.  Even if you have a solar panel that can reach a thousand volts and is capable of producing thousands of Watts, the output power would still be zero in this case.  (1000V * 0A) = 0W.

              • Novice 3 posts since
                Jun 23, 2011
                Currently Being Moderated
                Jun 24, 2011 10:44 AM (in response to Alec)
                Re: How to use a voltage regulator with a 12v solar panel

                So what keeps the DC-DC converter from being damaged if you're blasting the solar panel with direct sunlight and you have a full load (500mA at 5V) connected?

                 

                Sorry for all the questions. I think I must be overlooking something here.

                • designspecialist-power 62 posts since
                  Mar 22, 2011

                  eeyorebill04,

                   

                  The solar panel only outputs the amount of power that the DC/DC converter demands, whether it is getting just barely enough sunlight or getting many times more than it needs doesn't matter.  Like in the example I gave above, if the DC/DC converter only demands 3.125W, then the solar panel will only give it 3.125W.  The DC/DC converter would only be damaged if it became too hot from the sum of the ambient temperature and the heat produced due to internal inefficiencies.

                   

                  Unless you have a very inefficient DC/DC converter, and you are trying to operate it in a very hot environment, this shouldn't be a problem.

                  • Novice 4 posts since
                    Jun 22, 2011
                    Currently Being Moderated
                    Jun 28, 2011 9:32 PM (in response to Alec)
                    Re: How to use a voltage regulator with a 12v solar panel

                    Now I have a question.  Obviously solar panels cannot make more power than is available in sunshine.  So, disregarding battery backup, would a larger panel produce more power for a low light condition than a smaller panel? 

                    • designspecialist-power 62 posts since
                      Mar 22, 2011

                      jlewis,

                       

                      Yes, from what I understand a larger solar panel (of the same type) could produce more power.  A solar panel with more area can absorb more ambient light, think in terms of (power / unit_area).  If we make the assumption that the light from the sun has a uniform distribution over the area of the solar panel, then more area ultimately equates to more potential power output.

                       

                      The other thing to think about is that solar panels are still fairly inefficient in their ability to convert ambient light to electrical power.  The ambient light from the sun could be measured in (power / unit_area) on the surface of the panel, but the majority of this won't make it to the output because the panel won't be perfect at absorbing all the available energy, and there are losses involved in the conversion.

    • designspecialist-power 62 posts since
      Mar 22, 2011
      Currently Being Moderated
      Jun 29, 2011 10:13 AM (in response to dsmall)
      Re: How to use a voltage regulator with a 12v solar panel

      dsmall,

       

      My best guess there is that during the lower sunlight  condition, even though the panel has enough voltage to be in range of  the DC/DC converter input spec, the DC/DC converter was probably asking  for more power from the solar panel than it had available to supply.   Subsequently, by demanding too much current at 5V from the DC/DC  converter, the charging device was likely overloading it and therefore  causing the regulated voltage output to droop.

       

      The question of what panel voltage output level  equates to enough available power to properly charge your device at 5V  is a tricky one since I can't find any documentation for the solar panel  that might shed some light (no pun intended) on the matter.  I was  hoping the datasheet would have a graph that gave available output power  vs. output voltage, but I might as well hope to win the lottery I guess  because I can't even find an actual datasheet.  I also was curious  about absorption/conversion efficiencies of the panel after my response  to jlewis above, but again with no datasheet that has the information...

       

      I'm  also thinking the solar panel voltage was likely drooping as well due  to being overloaded.  If you measured 10-15V when the load was connected and  attempting to be charged, I wonder if that voltage wouldn't go up by a  good margin if you disconnected the panel from any load and measured it  again.  The only spec I really have is that the panel is 18V @ 10W.  Without any other information I would have to assume that is a direct sunlight spec and already takes into account absorption/conversion losses.   If you measured 25V during your successful charge in full direct  sunlight, it's probably because you were only drawing a fraction of the  available power (which was probably close to 10W) during those  environmental conditions.  If you had been demanding power closer to the  full possible output of the panel at the time, I'd bet you would  measure closer to the spec of 18V.

       

      You should be able to figure this out experimentally if you have the patience to monitor things for a good portion of the day.  Just monitor the regulator output as the sun rises and sets, and measure the voltage output of the solar panel at both ends of the window of time in the day where you get a solid 5V and proper charge.  I'm not sure how stable the numbers would be though over different conditions.  It could potentially vary with weather conditions like temperature and humidity, time of year, solar behavior such as sunspots and flares, obviously how overcast the sky is in the area, and also with the current charge state of your battery.  The last one matters because usually a battery charging system will demand more current (and therefore more power) the more depleted the battery being charged is, and then taper off as the battery gets closer to full capacity.

        • designspecialist-power 62 posts since
          Mar 22, 2011
          Currently Being Moderated
          Jun 29, 2011 3:57 PM (in response to dsmall)
          Re: How to use a voltage regulator with a 12v solar panel

          dsmall,

           

          Yes, you are correct, the current changes depending on the load, but because a raw solar panel voltage is unregulated the voltage is likely to drop as well as the panel is loaded more.  I know that may seem a little confusing, but that's actually part (the other larger part being voltage varies with sunlight level) of the reason you feed the solar panel into a DC/DC regulator.  As long as your input voltage is in range and you have a large enough power source available so as not to starve it, the regulator will basically maintain its set  voltage independent of loading, unless of course you overload it by trying to have it supply more current then it is designed for (which  can also potentially damage/destroy it depending on the existence of  its own internal safety features).

           

          There may have been  a regulator inside of the cigarette lighter assembly on the end of the  solar panel you have, which regulated 12V (extremely poorly at the  +-20% spec listed) from the bare solar panel, but since you are saying  you measured 25V at the output I'm assuming you cut this off and are  just using the bare panel?  If that's the case then the specs listed  are probably not applicable anymore because they likely describe the  regulator, i.e. the regulator  was approximately 12V +- 20% output with  max power output of 1.5W at 125mA max current.  I'm confused by the nominal voltage spec of 17.5V, that one might actually describe the  bare panel maximum efficiency point or something like that, I don't  know for sure.

           

          Regardless of whether you have a 12V regulator in the cigarette lighter or not, you might be power starving your DC/DC regulator under certain conditions, and that might be why you were only seeing 2.5V at the output when you first tried it in  indirect sunlight.  The engineer that choose the regulator for the  Digi-Key USB solar charger did so because it met the exact spec for a  standard 5V 500mA max (2.5W) USB charging system.  The solar panel for  that project is capable of providing much more than the 2.5W, but this  helps prevent the DC/DC switcher from being power starved at it's input  even in decreased ambient sunlight.  If your charging device is  demanding more current than the DC/DC regulator you have is capable of supplying from your solar panel, things won't work correctly.  Since the current the device demands will likely change with the current  charge state of the battery (more current when it is at low charge, less current as it gets closer to full charge), the solar panel you choose  may be adequate for the low current situations but not the high.  Even  in full direct sunlight this may be true, it depends on what the device  you are charging is asking for at the given point in time, and that's why it may be worthwhile to  choose a solar panel and DC/DC regulator pair that are both capable of  meeting the standard USB spec limits.

           

          P.S.

           

          I kept trying to fix the formatting issues in this post as I found them, but after clicking update for the 6th time or so I decided it would be easier to just apologize for them.  The forum software doesn't seem to like it when you copy and paste a lot, even from within the forum itself, and it especially likes to randomly add double spaces in between words when you do.  There were too many of these for me to try and find and fix.

           

          Message was edited by: Alec

  • JoeNikk Novice 3 posts since
    Oct 21, 2012
    Currently Being Moderated
    Oct 21, 2012 7:42 AM (in response to dsmall)
    Re: How to use a voltage regulator with a 12v solar panel

    Figured I'd leave a few pointers for the next guy how finds this page by Google search.......regulator efficiency is a fascinating and usually misunderstood subject.

     

    The Switching regulator is a pretty fancy and pretty "new"  technology...assume it's not for beginners...

     

    Usually the recommended regulator for beginners is the "Shunt regulator."   You can use the same device for either current regulation or voltage regulation.  When it is small batteries we are charging, it is current we are first concerned about.

     

     

    This post speaks of charging an auto battery with a small 1.5 watt solar panel.  You really don't  need to "regulate that, and neither did the maker of the panel.  A lead/acid battery of that size will never boil over from a panel that tiny.  That size battery, even if fully charged, will just shunt regulate itself against that little potential, it is smaller than the recommended trickle current(which is based on wattage).  In fact that panel is so small that you actually need the battery to act as the charging regulator, otherwise you'd have to wait till tomorrow to answer your phone.  Don't leave a auto battery on trickle indefinably, the plates will "Sulfate".(That means it will need to be shorted out and then jumpstarted, just to blow the scum off the plates, otherwise, it wont take a charge.)

     

    Efficiency?----well here is a thought for you teckies.    For shunt regulation, efficiency is proportional to drop voltage, regardless if your drop voltage is due to diodes, transistors, or diodes. For switching regulation, you are trying to build a class C or class AB linear amplifier.  Method of oscillation aside, a "Switching circuit" is actually a very well thought through effort to use a transformer on DC.  Although the currently popular method of "Driving" this doodad is called a PM oscillator(That thing is really cool.)  You don't really need to buy the chip, you can even use an astable multivibrator instead of that fancy digital switching. 

     

    BTW, if you also had an undersized solar panel, remember that it does "see" load impedance, and open circuits ALWAYS do severe damage, but it's not obvious, but just like any "Simple" Zener shunt, if you don't "regulate" the current,(usually with a resistor) something pops.....Look up the concept of measuring voltage in series, current in parallel, it would probably make more sense to you......my best advice, if you really don't have all day to answer the phone, it should be because you bought TWO, of the eight pin switching "regulators", and realized your circuit would be more efficient if you left the Darlingtons out of the circuit.  That is that stuff that "dropout voltage stuff" that the factory Tecs hurl on is al about. 

     

     

     

    Charging batteries really isnt a process ment to be too efficient, the battery itself usually looses about the same amount of power that it stores.  Have you ever calculated the performance of a solar panel that isnt being used?

     

    "If you really want that old vacuum tube box to get out, boy, you'd study heck out of the tanks."

  • JoeNikk Novice 3 posts since
    Oct 21, 2012

    Figured I'd leave a few pointers for the next guy how finds this page by Google search.......regulator efficiency is a fascinating and usually misunderstood subject.

     

    The Switching regulator is a pretty fancy and pretty "new"  technology...assume it's not for beginners...

     

    Usually the recommended regulator for beginners is the "Shunt regulator."   You can use the same device for either current regulation or voltage regulation.  When it is small batteries we are charging, it is current we are first concerned about.

     

     

    This post speaks of charging an auto battery with a small 1.5 watt solar panel.  You really don't  need to "regulate that, and neither did the maker of the panel.  A lead/acid battery of that size will never boil over from a panel that tiny.  That size battery, even if fully charged, will just shunt regulate itself against that little potential, it is smaller than the recommended trickle current(which is based on wattage).  In fact that panel is so small that you actually need the battery to act as the charging regulator, otherwise you'd have to wait till tomorrow to answer your phone.  Don't leave a auto battery on trickle indefinably, the plates will "Sulfate".(That means it will need to be shorted out and then jumpstarted, just to blow the scum off the plates, otherwise, it wont take a charge.)

     

    Efficiency?----well here is a thought for you teckies.    For shunt regulation, efficiency is proportional to drop voltage, regardless if your drop voltage is due to diodes, transistors, or diodes. For switching regulation, you are trying to build a class C or class AB linear amplifier.  Method of oscillation aside, a "Switching circuit" is actually a very well thought through effort to use a transformer on DC.  Although the currently popular method of "Driving" this doodad is called a PM oscillator(That thing is really cool.)  You don't really need to buy the chip, you can even use an astable multivibrator instead of that fancy digital switching. 

     

    BTW, if you also had an undersized solar panel, remember that it does "see" load impedance, and open circuits ALWAYS do severe damage, but it's not obvious, but just like any "Simple" Zener shunt, if you don't "regulate" the current,(usually with a resistor) something pops.....Look up the concept of measuring voltage in series, current in parallel, it would probably make more sense to you......my best advice, if you really don't have all day to answer the phone, it should be because you bought TWO, of the eight pin switching "regulators", and realized your circuit would be more efficient if you left the Darlingtons out of the circuit.  That is that stuff that "dropout voltage stuff" that the factory Tecs hurl is all about. 

     

     

     

    Charging batteries really isnt a process ment to be too efficient, the battery itself usually looses about the same amount of power that it stores.  Have you ever calculated the performance of a solar panel that isnt being used?

     

    "If you really want that old vacuum tube box to get out, boy, you'd study heck out of the tanks."

    • edprice Novice 1 posts since
      Oct 21, 2012

      Thanks for your technical comments. It would be easier on readers if you used common English as well as common sense.

      • JoeNikk Novice 3 posts since
        Oct 21, 2012

        Thanks ED.  I'll admit, I defiantly bought this computer as a necessary, "work on my writing skills," tool.  I was taking a break from filtering the color of a fresh batch of Chinese made indicators when I posted that.  I'll never recover from the spitball that stuck to the back of my head in Phonics.  I'll think up a few more excuses and post a link, would you prefer JPEG to RAW files?

         

        If you have any technical concerns, I'll refer it to my technical writing department, they already have orders to print a lower case "th" "'behind' every reference 'that engineers may have had'" about any voltage comparator inputs' involved polarity.  My CW radio is down right now, if you would like to know what "th" positive current is, just fax it in derivative form, but please leave the polarity indicator out of your algebraic shorthand, she'll understand. 

         

         

         

         

        Any one want to try it?*        Any current that has polarity, also has a parallel x2 -+ implied voltage.  My "Algebra" teacher forgot about the "skip over that 'finding the terms thing'" and jumped right to the summation of a "derivative thing".

         

        Note:

        *1)  For plain English teckies:  All proper paragraphs(abbreviation P), have a differential subject (abbreviation S) >> +-// V / Iout.  All others must use the simple algebaic sum.

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