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Old 11th September 2007, 07:33 AM   (permalink)
Thumbs down Electronics 101 Help, LED project

I could use some help to make sure I'm doing things right. I'm making a LED Array with a 6volt rechargable battery.

LED Stats (What the guy on Ebay supplied me with, I hope they are right)
Voltage: 1.5V~1.6V
Color: Water clear
Angle: 15~30 degree
Current: 140mA,
Peak(pulse): 700mA
Power: 200mW (think that is peak, they seam to run at 90mW)
Size: 10mm

6v rechargable battery, seems to run around 6.4v

What I'm looking at doing is running 16 LEDs. 4 in a series, so I don't have to use resistors. Then run Four series of 4 LEDs to the battery.

I tested it with one set of 4, and got 1.55 volt across the LEDs.

Is there anything I'm missing? Or anything that looks funny here?

I'm not good at figuring out the power usage. The battery says it is 4.5AH
I tested one LED, and it was pulling 90mw. Am I corrent in assuming it will last a little over 3 hours?


[Battery] (Pos term)+(LED)-+(LED)-+(LED)-+(LED)-(Neg battery terminal)
[ 6v ] (Pos term)+(LED)-+(LED)-+(LED)-+(LED)- (Neg battery terminal)
[ ] (Pos term)+(LED)-+(LED)-+(LED)-+(LED)- (Neg battery terminal)
[ ] (pos term)+(LED)-+(LED)-+(LED)-+(LED)- (Neg battery terminal)
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Old 11th September 2007, 07:48 AM   (permalink)
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Before three hours the LED's will be dim.
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Old 11th September 2007, 07:55 AM   (permalink)
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Quote:
Originally Posted by Gayan Soyza
Before three hours the LED's will be dim.
Hmmm. Any kind of formula to figure how long they will run bright? or is it a percentage of the battery life? or when I get it built I have to wait and see?
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Old 11th September 2007, 10:30 AM   (permalink)
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In theoretically for a 4.5AH battery it can supply current -

* 4500mA in one hour.
* 2250mA for two hours.
*1125mA for four hours.

In your strings the total current must be divide by these ratings.

For Ex: If your total strings get 560mA current then battery can supply 4500/560 = 8 almost 8 hours.
Note that while supplying current the battery voltage decreases so for a string the total voltage decreases & starts to dim.

Quote:
Originally Posted by ReallyConfuzed
or when I get it built I have to wait and see?
Finally this is the correct answer
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Old 11th September 2007, 10:34 AM   (permalink)
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Hi there!

This link might help you: http://led.linear1.org/led.wiz

With current-limiting resistors for each leg of the array, the calculator figures the array pulls 560mA, or 0.56 amps. I think you should be using the resistors so you know the current limit for the diode strings. Somebody who knows more about this, please jump in if I'm wrong on that.

I don't think you'll get much run time (maybe a couple of hours) off your battery. Its voltage will drop somewhat as it loses charge; how quickly it drops and when it starts to drop depends on the kind of rechargeable is it (NiCad, NiMH, LIon, etc.). Googling for "rechargeable battery discharge rate" or similar terms might help figure it out for your battery, though.

I know this isn't a lot of info, but I hope it helps.


Torben
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Old 11th September 2007, 11:15 AM   (permalink)
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Quote:
Originally Posted by Torben
Hi there!

This link might help you: http://led.linear1.org/led.wiz

With current-limiting resistors for each leg of the array, the calculator figures the array pulls 560mA, or 0.56 amps. I think you should be using the resistors so you know the current limit for the diode strings. Somebody who knows more about this, please jump in if I'm wrong on that.

I don't think you'll get much run time (maybe a couple of hours) off your battery. Its voltage will drop somewhat as it loses charge; how quickly it drops and when it starts to drop depends on the kind of rechargeable is it (NiCad, NiMH, LIon, etc.). Googling for "rechargeable battery discharge rate" or similar terms might help figure it out for your battery, though.

I know this isn't a lot of info, but I hope it helps.


Torben

Thank You, I like that link. here is what I got from it.

Solution 0: 4 x 1 array uses 4 LEDs exactly
+----|>|----|>|----|>|----|>|---/\/\/----+ R = 1 ohms

The wizard says: In solution 0:
each 1 ohm resistor dissipates 19.6 mW
the wizard thinks ¼W resistors are fine for your application
together, all resistors dissipate 19.6 mW
together, the diodes dissipate 840 mW
total power dissipated by the array is 859.6 mW
the array draws current of 140 mA from the source.



2 questions.

1. What's the reason for the 1 ohm resistor?
2. It says it uses 840mW? So 840x4 strings(total of 16 leds) = 3.6 Ah?


Blah, I'm confused. If I put my VM inline with the circuit Set it to DCA and 200m Plug into the COM, and mA. I get 90. (this is with a 1.5v battery and one LED, haven't tested the Array yet) Is this 90mW or 90mA?
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Old 11th September 2007, 05:58 PM   (permalink)
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Your voltmeter has a shunt resistor in series to measure current. The shunt resistor limits the current and makes your measurement of 90mA completely useless. The battery also has an internal resistance that limits the current and makes your measurement completely useless.

Use only 3 LEDs and a 16 ohm resistor all in series.
Then the current is up to 137.5mA when the battery is freshly charged and the LEDs are 1.50V each.
When the battery drops to 6.0V then the current is 94mA. You won't notive the slight amount of dimming.
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Old 11th September 2007, 06:58 PM   (permalink)
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It's probably a lead acid battery which will have a voltage of 7V when charged and 5V when flat.

What colour light do this LEDs emit? 1.5V to 1.6V sounds pretty low for the forward voltage.
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Old 11th September 2007, 07:38 PM   (permalink)
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Quote:
Originally Posted by Hero999
It's probably a lead acid battery which will have a voltage of 7V when charged and 5V when flat.

What colour light do this LEDs emit? 1.5V to 1.6V sounds pretty low for the forward voltage.

Yes, it's a lead acid Battery. These are IR LEDs. Looking around sites, I've seen them 1.2-1.7v are pretty common. They didn't come with a offical spec sheet. Only have the info from the ebay website. These get warm at 1.55v.
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