Info required about working principle of solar panel.

1. Jun 16, 2013

signode

Hi;

Normally, the solar panel charges a 12V battery and battery provides backup to load via an inverter. The solar panel produces voltage which is directly proportional to sun light. So early in the morning and in evening, solar panel produces less voltage which is insufficient to charge the battery but with more bright of sunlight, solar panel produces more voltage which then start recharge of battery. So what is produced voltage range of solar panel for low wattage applications? Any link/e-paper on it? I want to know the detail/working principle of SP.

2. Jun 16, 2013

Staff: Mentor

Are you sure that voltage depends on the amount of sunlight (for reasonable illuminations), and not current?

I don't understand that question.

3. Jun 16, 2013

OmCheeto

Interesting question. I can't answer that off the top of my head.

(runs off and does an experiment)

Cloudy day at ≈ 9am

Panel name plate ratings:
Kyocera LA361J48
48 watts
16.7 vdc
2.88 amps

hooking various sized resistors across my solar panel yields the following data:

Ω______VDC____Amps____Watts
∞______15.87__0.0_____0.0
981____14.59__0.015___0.22
469____10.20__0.022___0.22
266_____6.55__0.025___0.16
101_____3.21__0.032___0.10
_11_____0.32__0.029___0.0093

Ok. That tells me a little bit. My panel is producing about 1/240 of its rated capacity. But I'm going to guess that my panel will charge my battery, based on the fact that the voltage under the 981 Ω load exceeds maximum charging voltage: 14.4 vdc

(runs off and does another experiment)

Measured battery voltage, no load, no supply: 12.51 vdc
Lead Acid, Marine, Deep Cycle, purchased in 2008

Hooking the panel up to my battery in series with a 1 ohm resistor and measuring voltage across the resistor yields 0.023 vdc
0.023 vdc / 1 ohm = 0.023 amps
0.023 vdc * 0.023 amps = 0.0053 watts
Wow. That's ≈1/9000 of the panels rated capacity.

So there's your answer. The panel will very slowly charge your battery, even on a cloudy day, morning, and evening.

I believe my battery is rated at ≈ 100 amp hours, which at 12.75 nominal fully charged voltage yields 1275 watt hours.

1275 watt hours / 0.0053 watts = 240,000 hours

Ha! Good thing the sun comes out, once in awhile.
1275 watt hours / 48 watts = 27 hours

But I never run my battery to zero charge. It shortens their lifespan, significantly.

The measured 12.51 vdc indicates that the battery was charged to 85% this morning.
Meaning that it only requires 190 watt hours to fully charge it.
190 watt hours / 48 watts = 4 hours.

4. Jun 16, 2013

Staff: Mentor

But that is just the power in your resistor, not the energy going to the battery.

A diode could be interesting to control current flow, if a reverse flow is possible.

5. Jun 16, 2013

OmCheeto

You are absolutely correct. Thank you.
The power yielded to the battery would be 12.51 vdc * 0.023 amps = 0.29 watts.

That makes more sense. Roughly what the panel was supplying the resistors.

I'm not sure. Locked in a closet, my panel measures ≈3600 ohms in both directions.
They'll put out a few millivolts with even the tiniest amount of light present.

12.52 v / 3600 Ω = 0.043 watts.

I suppose I'll have to wait for the sun to go down before I can do that experiment.

The neighbors should get a kick out of that. Why would anyone install solar panels at night and remove them during the day?

6. Jun 16, 2013

OmCheeto

The results of this experiment are:

The battery discharges through the 1 Ω resistor and zero voltage panel with 0.011 volts being dropped across the resistor.

Yielding 0.011 amps
and 12.5 volts / 0.011 amps = 1100 Ω
I'm blaming the disparity in ohmmeter measured resistance and battery powered calculated resistance on the fact that solar panels are semiconductors, and don't like following the rules.

12.5 volts * 0.011 amps = 0.14 watts

1275 watt hours / 0.14 watts = 9100 hours = 380 days > 1 year to completely discharge the battery.

I'll just disconnect the panel in the winter, and forgo the diode.

Thank you signode for the question.
And thank you mfb for your assistance.

Last edited: Jun 17, 2013