Forward Voltage (V) : 3.2 ~ 3.8
Reverse Current (uA) : <=30
Luminous Intensity Typ Iv (mcd) : Average in 8000
Life Rating : 100,000 Hours
Viewing Angle : 20 ~ 25 Degree
Absolute Maximum Ratings (Ta=25°C)
Max Power Dissipation : 80mw
Max Continuous Forward Current : 30mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V
say you have a 20w solar panel, specs as follows:
Peak Power (Pmpp) 20 W
Peak Power Voltage (Vmpp) 17.2 V
Peak Power Current (Impp) 1.16 A
Question 1:
how long can you run the 800 leds on the 12v battery before requiring a recharge (so say you discharge the 12v to 20% capacity)
Question 2:
how many hours of peak sunlight would it take to recharge this battery after 80% discharge
if you could show your work so I can learn, that’d be great.
also, if someone could tell me how i could educate myself so as to figure these things out without resorting to the SDMB, bonus points!
3.8 hours before requiring a recharge (assuming you’re using avoirdupois volts). 2.3 hours to recharge, assuming latitude of 45 degrees north at the vernal equinox.
Total energy in the battery = 12v * 40Ah = 480Wh. Assuming a 100% conversion efficiency, you will be extracting 80% of 480Wh out of the battery, or 384 Wh. If you run each LED at 40mW (3v x 13mA), than 800 will consume 32W, so you then have 384Wh/32W = 12 hours. Figure on 80% of that to account for losses.
Once again, with 100% efficiency, you need to put 384 Wh into the battery, so if you get 20W in full sun, it will take 384 Wh/20W = 19.2 hours. Again, you need to allow for losses.
Well, your LEDs are rated at 80mW maximum dissipation. I took 50% of that as a reasonable number, and the volts x amps figures seem right. You could probably run them at higher current, which a corresponding decrease in battery life, and still be safe.
They are just estimates - the power is the important thing. The LEDs are rated at 80mW max. You don’t want to run them at that power, so I picked 50% of that, or 40mW. Just to make sure that was reasonable, I calculated what current the LED would draw if it was running at 40mW, and 3v (which is close to the operating voltage for a white or blue LED - your spec is 3.2 - 3.8v). I got 13mA, which is a little low - you can usually run LEDs at 20mA, but the numbers worked for back-of-the-envelope calculations. So, if you wanted to run the LEDs at their maximum rating, you would get half the battery life I calculated.
Remember, nothing works out this neatly in real life - there are always losses.
Sorry if this is obvious, but you did say that you know next to nothing about EE…
Your LEDs expect a forward voltage of 3.2 to 3.8 volts, but your battery provides 12 volts. If you connect a single LED to the battery you’re gonna fry it.
Instead, you could connect sets of 3 or 4 LEDs in series across the battery. That should drop the voltage across each LED to 4 or 3 volts respectively.
There’s plenty of other things you could do to account for this problem, but that’s where I would start.
A load resistor would still be required even if you string them in series, right? My understanding is that even though there is a significant voltage drop, a LED still has very little resistance when forward-biased, and it will be like a short-circuit if there is no resistor.
Disclaimer: I’m a software engineer, not a EE, but I work with embedded designs and other EEs.
When I did my calculations, I assumed that an efficient switching-mode LED driver was used, not a simple resistor. If you use a resistor to limit the current, you are going to cut your battery life significantly.
It could very well take forever.
Lead acid batteries, even deep cycle marine batteries, do not repond well to discharging below 50% capacity or so. The lead plates begin to lose their integrity, and recharging is hampered.
IOW, while you might get 800 cycles out of a battery you only discharge to 60%, it’s likely you’ll only get 10 or so out of a battery discharged to 20%.
There’s a bit of discussion of cycle depth vs battery life here. Seems to me I read a better discussion a few years back; found it somewhere on the Internet, so you can probably find it too.
Which is why simple power calculations won’t be satisfactory for him.
Rumor_Watkins, here’s the issue with what you gave us: there’s not enough info on how you’re going to connect the pieces together. You have some LEDs which take a little over 3 V to light up, but you have a battery that supplies 12 V. How are you going to limit the voltage so as not to fry the diodes (LEDs)?
You need to decide how much current you want through the LED - how bright? Half of maximum?
That’s 40 mW, so at ~3.3 V, that’s about 12 mA. You’d need a series resistor on the battery to drop the 12 V down to 3.3 V when it’s putting out 12 mA. That comes out to about 725 ohms.
Now the problem is that the battery is sourcing 144 mW of power total, only 40 mW of which is being used to power the LED.
You need something to regulate the current. When you apply voltage to an LED, as you raise the voltage, at first almost nothing happens. Then, over a very narrow voltage range, the LED suddenly gets brighter and brighter and then POOF, it becomes a DED (dark emitting diode - a slightly humorous way to refer to a burnt out LED).
The current actually follows an exponential curve, like so:
From a practical viewpoint, what this means is that if you try to run 4 of them in series from your 12 volt source (3 volts per LED), then the LEDs will probably be so dim that they’ll appear to be off. If you try to run 3 of them in series (4 volts per LED) then they’ll probably burn out. LEDs are very difficult to drive from a voltage source because they go from off to dead over such a narrow voltage range.
There are several different ways that you can regulate the current. The simplest is a resistor in series with the LEDs. The voltage drop of the LED is going to be relatively constant over the entire working range of the LED, so you mainly have to size the resistor to limit the current to the appropriate value. A series resistor is simple and cheap, but not very efficient from a power viewpoint. As beowulff and CurtC noted, it’s going to have a fairly drastic effect on the battery life in the OP’s situation.
Another way to do it is to use a current regulator, like an LM317. This costs a bit more than a simple resistor (though LM317s are pretty cheap) but works better when you are trying to drive a string of LEDs, and also provides a constant current even when the battery voltage is changing. Your LEDs would be the same brightness regardless of whether the battery is fully charged or is on its last legs. In contrast, your LEDs would get dimmer if you were using a series resistor and the voltage dropped due to the battery dying. The LM317 is a linear regulator, so it’s still going to dump a lot of energy into heat, which means a lot of wasted battery power.
A third way is to pulse the LEDs with a switching regulator of some sort. This costs even more than the LM317, but has two main advantages. First, the pulsing means that the regulator will not be wasting energy as heat, so it’s much more efficient than an LM317 or a resistor. Second, your eyes tend to detect the peak level of the LED, not the average level, so the LEDs seem brighter for the same amount of average current (in other words, 12 mA average pulsed is going to look brighter than 12 mA that’s not pulsed). For the same apparent brightness level you can therefore use less average current, resulting in even more savings for your battery life.
ETA: By the way, it is possible to drive LEDs from a voltage source, but you have to be fairly careful about how you do it. LED Christmas lights are usually just a bunch of LEDs in series with no current regulator or limiting resistor at all.
Buy “The Art of Electronics” by Horowitz and Hill. It’s a little above the head of complete newbies, but if you are going to buy one book about electronics, that’s the book you want.
Just connecting the solar panel to the battery will not give you anywhere near 100% efficiency. To get as much efficiency as possible you will need to use something like Maximum Power Point Tracking which will allow you to get a higher efficiency. Also, I second the recommendation not to allow your battery to go down to 20%. It will reduce its life dramatically.
