4VDC 0.100 Amp Polycrystalline Solar Panel

Thread Starter

mrel

Joined Jan 20, 2009
185
Hello
I purchase five 4VDC 0.100 Amp Polycrystalline Solar Panel.
Question when i connect five of these solar cell in series I should get 20 volts but i don't understand how to read current 0.100 amp how many( MA)) would five 0.100 amp total be in (MA).
If I connect the five solar cell in parallel the voltage will only be 4 volts and what would current be?
Is this solar cell (five solar cell) big enough to charge a D size 9000 MAH rechargeable battery in 8 hours time
Thank for the help mrel
 
Last edited:

Markd77

Joined Sep 7, 2009
2,806
5 in series the current would be the same: 0.1A which is 0.0000001 MA or to use a more convenient unit 100mA
In parallel the current is 500mA
Assuming you mean 9000 mAh you would need over 1A to charge it from empty in 8 hours.
 

wayneh

Joined Sep 9, 2010
17,493
Assuming you mean 9000 mAh you would need over 1A to charge it from empty in 8 hours.
+1
In fact it's probably worse than that, since 1) You won't have noon sun for all 8 hrs and 2) Power conversion will not be ideal. Assuming the panels are wired in parallel to provide maximum current (500mA), they would supposedly deliver 2W at 4 volts. But the battery will pull the voltage down to ~1.5, and the current won't go up proportionately, so the power will drop to maybe only 1W. You need about 9AH * 1.45v = 13.05 watt-hours to charge the D cell, so possibly more than 12 hours of full sun. Ain't gonna happen. Putting 2 D cells in series would actually capture energy more efficiently, but then you're very near having to worry about the blocking diode's voltage drop.
 

Thread Starter

mrel

Joined Jan 20, 2009
185
+1
In fact it's probably worse than that, since 1) You won't have noon sun for all 8 hrs and 2) Power conversion will not be ideal. Assuming the panels are wired in parallel to provide maximum current (500mA), they would supposedly deliver 2W at 4 volts. But the battery will pull the voltage down to ~1.5, and the current won't go up proportionately, so the power will drop to maybe only 1W. You need about 9AH * 1.45v = 13.05 watt-hours to charge the D cell, so possibly more than 12 hours of full sun. Ain't gonna happen. Putting 2 D cells in series would actually capture energy more efficiently, but then you're very near having to worry about the blocking diode's voltage drop.
Could you please explain what you mean Putting 2 D cells in series would actually capture energy more efficiently, but then you're very near having to worry about the blocking diode's voltage drop .
Are you saying i should have two 9000 mha battery connect in series,would it not take long to charge both batterys plus i would have 2.4 volts that will be more than 1.2 volts ?
What is this about blocking diode voltage drop are you saying i should put diode into the circuit to stop battery dischargeing into solar panal when there no sunshine?
mrel
 

iONic

Joined Nov 16, 2007
1,662
I wouldn't bother with trying to charge two D Cells in series with those solar cells. you will need a minimum of ~3.0V to charge them, 4V will be better. And given the sun, you may not get 3V for too long during the day. I'd stick with the Theoretical 4V @ 500mA to charge the single battery, but you have to be very careful not to overcharge it (more than 1.45V or so).
 

wayneh

Joined Sep 9, 2010
17,493
Could you please explain what you mean Putting 2 D cells in series would actually capture energy more efficiently, but then you're very near having to worry about the blocking diode's voltage drop .
Are you saying i should have two 9000 mha battery connect in series,would it not take long to charge both batterys plus i would have 2.4 volts that will be more than 1.2 volts ?
What is this about blocking diode voltage drop are you saying i should put diode into the circuit to stop battery dischargeing into solar panal when there no sunshine?
mrel
To charge a battery, you must first overcome its voltage, say 1.4v for a single NiMH cell or 2.8v for two in series. If you want to avoid reverse current in low light, you also need a blocking diode, which will require another ~0.7v to start conducting. So if you have enough light to get more than 3.5 volts from your panel, you could charge two cells in series. In fact a solar cell is most efficiently used when it's providing current at about 90% of its unloaded voltage. So if you're getting 4.0 volts unloaded, and ~3.5 volts when loaded, you're running about as efficiently as you can.

But instead of efficiency, you're goal may be to just get the battery charged ASAP while the sun is shining. In that case, I agree that wasting some efficiency by charging just one cell will ensure that you have more hours of charging as the lighting varies, and a higher current during the peak hours. You'd just need more panels to charge a given number of batteries, but maybe that's not your problem.
 
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