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I have a smaller solar panel (say 12v, 0.5 amp and therefore 6 watts) connected to a 12v car battery (assume lead-acid) directly though with the included blocking diode, there is no charge IC attached. In full sun the panel will deliver up to 15v. What happens when a load (It must be assumed that the load isn't particularly voltage sensitive, within reason) in excess of the solar panels output is connected? For the example, lets say the load is 20 watts.

I would imagine that for a small load (below that of the solar panels output) the load would simply use the provided electricity from the solar panel (15v and whatever amperage) and the battery would charge at a lesser rate, however, once this load is exceeded, what occurs? Does the solar panel burnout? Does its voltage output drop? Or Does the battery simply provide a small quantity of current whilst the panel provides the bulk? If the latter is the case, how does this work?

Thanks for any assistance, Branyon

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When the load is off, the panel will fill the battery with the voltage above permitted for the chemistry of the battery, the electrolite in the cell will boil (gassify) and destroy the battery.
When the load is on, both battery and panel will provide current. Of course the panel will work less efficiently if the output voltage of the panel isn't held on maximum power point voltage.

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  • \$\begingroup\$ Not so, it is easy to charge a battery with a solar panel and blocking diode, provided it is a lead-acid type the battery will not boil. \$\endgroup\$ – TheSoundMan Jan 15 '16 at 9:55
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    \$\begingroup\$ @TheSoundMan, whether this is safe depends on the size of the battery and the size of the solar panel. For small panels and large batteries it may be safe. But I would recommend using an over-voltage cutoff of some sort to protect the battery anyway. \$\endgroup\$ – mkeith Jan 16 '16 at 7:30
  • \$\begingroup\$ @mkeith regardless, this configuration works. Fact. In this respect the above answer is incorrect. what I don't understand and what the above answer still fails to provide is an explanation of what happens with the load is applied. \$\endgroup\$ – TheSoundMan Jan 17 '16 at 6:56
  • \$\begingroup\$ The load will always draw current as long as either the solar panel or the battery or both are able to supply it. The solar panel will not burn out. If the load starts to draw more and more current, to the point that the panel cannot supply enough, then the battery will start to discharge into the load. If the load is tolerant of the voltage range, nothing bad will happen to the load. If the battery is charged higher than maybe 14.5 V or so, its life will surely be shortened. If it is discharged to less than 11 or 10V, its life will be shortened. \$\endgroup\$ – mkeith Jan 17 '16 at 10:11
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Battery capacity (unknown), load energy needs and PV panel wattage are needed for a full answer.

Most "12V" PV panels intended to properly charge 12V lead acid batteries have a Wmp = maximum Watts output in full sun, at Vmp = 17 to 18V. 15V Vmp would be unusual on larger panels intended for 12V LA use, but not inconceivable on smaller panels where the manufacturer "does not know and/or does not care".

Assume initially that a "constant wattage" load is available. This is not an essential assumption but removes unnecessary complexity from the example.

Assume initially that any diodes used have zero voltage drop when conducting.

If a 10 Watt load is applied to a PV (Photovoltaic) panel that is receiving enough insolation (solar input) to provide say 12 Watts at Vmp then the panel voltage will rise above its Vmp. If Vmp is say 15 Volts then, if a 12V "80% charged" lead acid battery is also attached to the panel.

  • The battery will absorb the "excess" energy. In this case 12-10 = 2 Watts.

  • The panel voltage will fall to the voltage where the battery accepts the extra 2 Watts available.

If the panel available wattage falls below 10 Watts then if the battery is also connected, the battery will make up any shortfall.

If the battery continues to supply some or all of the load current the battery voltage will fall until it reaches a safe minimum level. If discharge continues past this point damage will occur and then substantial damage.
Once discharging reduces battery voltage to the safe minimum level the battery should be recharged as soon as possible. Holding the battery in a discharged state will cause progressive damage.

A PV panel with excess energy output potential will NOT be damaged if all potentially available energy is not used.


Dangerous misunderstandings:

SM: ... it is easy to charge a battery with a solar panel and blocking diode, provided it is a lead-acid type the battery will not boil.

mkeith replied:
... whether this is safe depends on the size of the battery and the size of the solar panel. For small panels and large batteries it may be safe. But I would recommend using an over-voltage cutoff of some sort to protect the battery anyway.

SM: regardless, this configuration works.

SMs understandings are inadequate and/or wrong.
In some selected cases it is possible to simply connect a PV panel via a diode to a lead acid battery and obtain 'reasonable' results. In most real world situations this simple arrangement would be damagingly inadequate.

A suitably low energy charger MAY be able to be connected to a lead acid battery indefinitely without causing massive damage.
But, most charging systems which are designed to charge lead acid batteries in "sensible time periods" in "typical applications" will also substantially damage or destroy batteries if not managed correctly.

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