I was wondering, how can I calculate the average life of the battery?
Side question:
When I have all 50 wired, the 9V keeps them at a decent brightness but as soon as I put a 1k ohm resistor in the mix they go super dim (duh). How can I calculate the required (if needed at all) resistance for this amount of LEDs? From my initial tests there doesn't seam to be a need but I don't want to burn the LEDs out or anything.
The LED's want 25 mA each. 50 of them 1250 mA (1.25 A).
Find the capacity of your battery in amp hours and divide by 1.25 to get maximum life. Make sure your battery is capable of producing 1.25 A.
A typical 9V alkaline transistor battery as a capacity of 0.565 Ah. That would give an active life of 0.452 Hours (27 minutes) assuming such a battery can survive a draw of 1.25 A.
johnwasser:
The LED's want 25 mA each. 50 of them 1250 mA (1.25 A).
Find the capacity of your battery in amp hours and divide by 1.25 to get maximum life. Make sure your battery is capable of producing 1.25 A.
A typical 9V alkaline transistor battery as a capacity of 0.565 Ah. That would give an active life of 0.452 Hours (27 minutes) assuming such a battery can survive a draw of 1.25 A.
Ouch. Ok, I am understanding so if I put 2 9V in serial, I could (theoretically) push almost an hour.
What would be the best battery combo / setup to get maximum life for this?
Datasheet says forward current 20mA (max rating of 25mA).
So 50 in parallel you're multiplying the current usage. 20mA x 50 = 1A
A 9V battery has somewhere between 500-800mA/h so IF your LEDs were getting the full 20mA, on a 9V, assuming a theoretical 800mA/h on that battery, you're looking at around 48 minutes. 800mA / 1000mA = 0.8h = 48 minutes.
But I highly doubt you're giving each LED its full 20mA needed. It will try, but the battery's just going to say 'oh hell no' ...
You're better off using four AA batteries in series (for 6V). AA batteries can push almost a full 2A.
Ouch. Ok, I am understanding so if I put 2 9V in serial, I could (theoretically) push almost an hour.
Do you mean parallel? That is the only way to get more current. What resistor value are you using on each LED?
If the answere is none the you are doing it wrong.lease explain your circuit.
ok so if I understand correctly, the best bet would be using the 4 AA batteries in series which should give 3.2 hours of battery with 50 LEDs assuming each battery has 800mAh. If I make it 100 LEDs, then I could get 1.6 hours of life? Also, at 2A I could get max (or near to) brightness with the 4 AA batteries and 100 LEDs - yes?
If I add a few resisters, would this help with battery life at the cost of brightness?
Grumpy_mike - the circuit is just parallel leds in a row with a 9V attached to the end.
shiznatix:
ok so if I understand correctly, the best bet would be using the 4 AA batteries in series which should give 3.2 hours of battery with 50 LEDs assuming each battery has 800mAh.
NO NO NO >_<
If you put batteries in SERIES you will not increase the capacity, you will increase the voltage. So 4 AA's in Series will give you 6V, 800mAh.
If you put them in PARALLEL, you will increase capacity, but not voltage. So 4AA's in Parallel will give you 1.5V, 3.2Ah (Though there will be losses if a battery has a lower voltage than the others as they will start to try and charge it).
the circuit is just parallel leds in a row with a 9V attached to the end
Then if you do that with a battery of any real current capacity you will blow them all. The reason why you have not blown them already is that the battery you are using is so poor that its internal resistance is limiting the current.
Read this:- http://www.thebox.myzen.co.uk/Tutorial/LEDs.html
ok, sorry for my very noobish questions / statements. I am learning (obviously).
So, best bet is to put 4 AA batteries in PARALLEL which should give me multiple hours (roughly 3.2 hours with 100 LEDs) of run time. How can I determine the amount of resistance that I would need and would adding enough help increase battery life?
Also, for resistors, should I have 1 resistor per LED or would 1 resistor for the whole circuit be enough? I am sure it is 1 per LED but why would be great to know.
Ok, I think I don't understand all the math going on here.
So, if we have 4 AA batteries wired in parallel this would equal:
Amp Hours: 8 (~2 * 4)
Volts: 2
Milliamps: 4000 (~1000 * 4 with no resistors)
Each LED wants 25mA so 100 LEDs want 2500mA.
So, with 4 batteries wired in parallel and 100 LEDs wired in parallel, this would supply more than enough Amps to each LED and would require some resistance to keep them from burning out.
For the resistance, we have V / I = R so 2 / 4000 = 0.0005 which is 0.5ohms? So, I would need 0.5ohm resistor for each LED or 0.5ohm resistor for the whole thing?
I am almost certain I am doing something wrong here and I would love some help with my math if someone could explain since I don't believe that I could wire this up without the LEDs being super amazingly dim (but I could be mistaken since I haven't tried this configuration yet)
I have attached the initial version of my circuit (sorry for the amazing crudeness of it, I am at work and have limit resources and skill with drawing).
@fungus - I am not sure how this would help? If they are all in parallel what good does splitting them up do?
NO 25mA is the absolute maximum, the point you must not exceed, you need to give them 20mA
At that they have a typical forward voltage drop of 2V, that means the resistor must drop the reaming voltage.
That is supply voltage - 2 = voltage across the resistor.
Suppose that supply voltage is 9V then you will have 9 - 2= 7V
A resistor with 7V across it and 20mA flowing through it has to have a resistance of 7 / 0.02 = 350 Ohms
So EVERY LED needs to have a 350R resistor in series with it.
So, if we have 4 AA batteries wired in parallel this would equal:
Amp Hours: 8 (~2 * 4)
Volts: 2
Milliamps: 4000 (~1000 * 4 with no resistors)
Just so much wrong here. An AA battery has a voltage of 1.5V. If you have four in parallel you will have a voltage of 1.5V so there will not be enough to light up the LED.
If you wire some LEDs in series you can wire 3 in series with a 9V battery and so have 3V left to drop across your resistor, giving you a resistor value of 150 ohms. So you then wire your LEDs in parallel but in groups of 3 and a resistor in series.
4 of your LEDs in series needs 8.8V - pretty close to what you have available.
Therefore you do not have enough voltage drop to allow a series resistor to approximate to a constant current supply. That is why I suggested three.
Even with a resistor you should be able to run them at a decent brightness while reducing the current draw on the battery by a third (compared to groups of three).
OTOH ... the voltage drop on the battery might still be enough that you're much closer to three LEDs than four. You'd have to experiment.
So, if I use any amount of AA batteries in parallel, ill still only be able to get 1.5V (with the parallel config) thus I can't even light up a single LED since I need a minimum of 2V for an LED to light up - correct?
In this case, I must use either 2 AA batteries in serial (3V) with a resistor to bring the mA to an acceptable level or use a 9V with groups of LEDs with a resistor in serial and 3 LEDs in parallel.
Since from what I can find online AA batteries have a better mAh than 9V batteries do, it seams to make more sense to use the AAs. If I do use the AAs though, would there be any benefit in splitting everything into groups (as with the 9V) or can I just put a bunch in parallel and be done with it?
