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(n00b question) I have a small (6 cell regular bike) leadacid battery and want to charge it via solar sheet of 15V - 0.3A. the voltage solar sheet changes and in cloudy day goes below 10V. does the battery keep charging that those levels (though very slowly?)

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What you asked about:

No, a nominally 12v lead acid battery will not charge at 10V unless it is essentially fully discharged.

What you didn't ask about :-) :

You MUST have a diode* between the panel and battery to prevent the battery discharging into the battery when the panel voltage is below battery voltage.
(* diode or functional equivalent - there are alternatives but a diode is simplest and cheapest and good enough.

To see if your panel is losing much energy.

The complete answer re the adequacy of your panel to charge your battery in a range of light conditions is given in the paragraph labelled ==> below BUT all the rest gives you a much better feel for your system.

Use a current meter (multimeter or other) that is able to read the maximum current the PV panel will produce.

Described below are several measurements which you can make under various conditions. Here I'll just use the terms Isc, Voc, Ichg and Vdiode. How to measure these and what they mean is detail below under "measurements".

  • Isc - warning: Note that I say below that Isc must be measured with battery disconnected. In fact, as long as you do the right thing you can measure Isc by shorting the panel with an ammeter at any point. BUT short on the wrong side of the ammeter and you will get magic smoke. Ammeter may die, Battery may die. Wiring may die. For extra points on a big system (bigger than this) you may die doing that BUT common sense should stop you shorting in the wrong place.

SO

  • Under various sun conditions measure Isc, Voc, Ichg, Vchg, Vdiode

    Record all figures including assessment of light level. If a lux meter is available so much the better.

  • ==> Note the sun conditions when Ichg is hardly more than zero.
    If Isc under those conditions is a significant portion of Isc_bright_sun then you are wasting energy and your panel would benefit from more cells and thus a higher Voc.

      -

MEASUREMENTS:

(a) Isc = panel short circuit current.

Made WITHOUT battery connected **.
Expose the panel to sunlight and connect the meter probes across the panel (Without the battery connected !!!). This essentially short circuits the panel and gives an idea of its maximum realistic output = Isc at the given light conditions.

(b) Voc = panel open circuit voltage.

(C) Ichg. Connect panel to battery via a diode and via a current meter which has low voltage drop. Measure charge current into battery.

  • Using a multimeter set to the 10 amp range will usually be OK.

    You may only be able to resolve current to 10 mA in this mode but low voltage drop is more important than accuracy.

    Measuring the voltage across the meter when it is measuring peak current in bright sunlight will be useful. You'll need a secind meter to do this.

    Vdrop across the meter will ideally be only 0.1V or les and not more than ay 0.3V max.

    You may be able to use ag a 500 mA or 200 mA range but thse will usually have too much voltage drop.

(d) Vchg = Panel voltage when charging

(e) Vdiode = diode forward voltage drop

A "magic" measurement - ie far more can be told from this one reading than may be expected.

Place a volt-meter across the diode and measure the voltage. When the battery is being charged the diode will forward conduct, & panel voltage will be above battery voltage by a diode drop = 0.6 - 0.8V for silicon and 0.3 - 0.5 V for Schottky diodes.

When the battery is not charging Vdiode will change polarity and will tell you how low panel voltage is compared to battery voltage.

Monitoring Vdiode will tell you quite a lot from a single reading about how a system is performing.

  • High Vdiode conducting = heavy charge.

  • Vdiode starting to drop off peak value but still > say 0.5V for silicon = charging at lower rates.

    • Vdiode +ve but 0 - 0.5V = trying to charg but only just - as the diode conducts the battery voltage will rise due to the tickle of current nd hold Vbattery just below Vpanel over a moderate range. This shows you that the panel is about at its starting to charge point but only just.
  • Vdiode is negative. Panel voltage is below battery voltage. How much below shows how far off charge you are.

Made without battery connected **. Measure the panel voltage when exposed unloaded to sunlight.

. Also measure the panel voltage "open circuit" in full sunlight = Voc.

Now repeat in various cloudy conditions. Note Isc at various Voc's for varying degrees of cloud.

Now connect the panel to the battery with the battery reasonably well charged from another source. Battery Voc should be well above 12V.


Blocking Diode

Using a Schottky diode rather than a silicon diode will give you a very small gain in charging capability - not liable to be worthwhile in most cases.

Using a 1N400x diode will work fine (x = 1...7 and indicates breakdown voltage. as 1N4001 = 50V any will work for you.
ie anything from here
Datasheet for 1N4001 ... 1N4007 here

If using a Schottky diode then a 1 amp rated one at 20V or better should be used. These are somewhat more sensitive to static electricity damage than silicon diodes. I'd probably use a 30V diode here rather than 20V to reduce the chances of unfortunate happenings. Anything from here rated at 20V or higher should work .

eg a 1N5817 - 1N5818 - 1N5819 rated at 20 / 30 / 40 Volt respectively will work well.

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  • \$\begingroup\$ Thanks for answering more than I asked. I needed that info. :) I had been taking readings from battery terminals but without d/c the panels. now i take two readings. Its sunshine and so panel is reading 15.4V and battery V has changed for original 8V to 10.1V. Also setup a IN4007 blocking diode. \$\endgroup\$ – thevikas Nov 28 '11 at 8:37
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The 6 cell Lead Acid battery should ideally be charged at 13.8V to 14.7V

Any lower and you wouldn't be able to reach full charge and any higher and the battery might get heated up and might get damaged .

If the battery voltage is higher than your charging voltage current will start flowing in the opposite direction and thus discharging the battery.

Think of the battery as an overhead tank of water in a building or home.

The voltage of the battery is like the pressure in the tap below.

To fill the overhead tank using that tap, you'll need to put higher pressure into the tap than the pressure of water coming out of the tap. Only then the water will start flowing upwards into the tank. The charging current is like the rate of the the flow in the tap. The slower the rate of filling it the longer it will take to fill up the tank. And similarly the volume of the tank is like the capacity of the tank, for e.g. 7Ah for a small battery and 150Ah for a larger battery. Larger the tank and longer it will take to charge it at the same current or flow rate.

Actually after giving that example, i just figured that even the battery characteristics can be understood the same way, but I'll leave that out for now so not to confuse you.

Adding a diode ensures that even when you don't have enough voltage to charge the battery, it doesn't start discharging, even though it will not charge at all at lower voltage.

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