dkshema
rrMaster
Location: Cedar Rapids, IA
My Posts This: Topic Forum | OK kids, here goes. I looked up the specs on a high intensity, white LED, rated at 3.8 watts. I've got one of those flashlights, runs on 3 AAA cells, advertised as having a 3 watt LED.
The 3.8 watt, high-intensity white LED was rated to have a forward voltage of 3.2 typical with 350 ma running through it.
The forward voltage is the voltage you'll see dropped across the diode junction when it is "on".
A germanium diode has a forward voltage of about 0.3 volts. A silicon diode has a forward voltage of about 0.7 volts. LEDs get their color from their doping and construction, and the forward voltage varies with the color of the LED and a few other design goodies.
So, lets assume that you're running a white LED with a forward voltage of 3.2 volts, typical (could be as high as 3.8 volts, could be as low as 2.6 volts, depending upon the particular batch its die was processed in).
You're running it from a battery that has a fully charged voltage of 4.2 volts, a nominal discharge voltage of 3.7 volts, and a fully discharged voltage of 3.0 volts.
With a fully charged battery of 4.2 volts, the difference between the typical forward voltage of 3.2 volts and the fully charged pack is 1.0 volt (4.2 - 3.2).
As the pack runs down, it averages about 3.7 volts until it's flat. 3.7 - 3.2 gives you a half a volt of headroom between the voltage in and the forward ("ON" ) voltage of the LED.
As the pack continues to run down its voltage actually drops BELOW the required voltage to turn the LED ON (3.0 volts - 3.2 = -0.2 volts.
Luminosity is a function of the forward voltage and the current through the LED. As current goes down, so does light output.
If you don't limit the current THROUGH the LED, then there is a good chance you will burn it out.
Back to the LED. It has a forward voltage of 3.2 volts. Hooking it directly up to a power source that turns it on without a current limiting device is bad news. The diode turns on, conducts current, and without something to limit the amount of juice going through it, the diode rapidly heats up and the internal junction is destroyed.
Hook the LED up to your 4.2 volt fully charged pack. The difference in voltage between the pack and the forward voltage of the LED is 1.0 volt. The data sheet includes a chart showing the maximum permissible current through the LED vs ambient temperature. The LED has a maximum current rating of one amp, a minimum rating of 100 ma.
Assume you want to limit the current through the LED to 500 ma (1/2 amp) max.
Ohm's law, E = I x R (volts = current x resistance) is handy here. It can be rearranged so you can figure voltage, current, or resistance as the unknown value.
E = I X R
E/I = R
E/R = I
We know the battery is fully charged at 4.2 volts. We also know that out of that 4.2 volts, the LED will have typically 3.2 volts dropped across its leads simply due to the physics involved.
E, in the equation above is figured as battery voltage minus the LED forward voltage. 4.2 - 3.2 = 1.0 volt. To limit the current through the LED to 500 ma (0.5 amp), with a fully charged pack, solve:
E / 0.5 = R, where E is the battery voltage minus the LED forward voltage (1V).
1 volt / 0.5 amp = 2 ohms
You would need a 2 ohm resistor wired in series with the battery and the LED. What kind of power would the resistor need to handle?
P = V x I
2 ohms, 0.5 amps, 1 volt across the resistor.
V x I --> 1 volt x 0.5 amp, you'll want at least a 1/2 watt resistor. You would actually want at least a one watt resistor, to add some margin to add some life to the circuit.
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Great, you now have a healthy bright white LED with a 1/2 amp coursing through its veins. But the battery begins to discharge.
The resistance stays fixed at 2 ohms. The forward voltage of the LED remains relatively constant at 3.5 volts.
That 1 volt dropped across the resistor from a fully charged pack drops to zero as the battery reaches the diode's forward voltage of 3.2 volts. As the battery voltage drops toward the forward voltage of the LED, the current through the LED drops. At some point, it drops to zero. Before that happens, the LED has all but gone out.
In between, the LED gets dimmer and dimmer and...
And if you have no current limit resistor in the mix, the LEDs simply burn out.
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What you REALLY need is a constant current source to run the LED. You need a source whose voltage is at least 3.2 volts (to turn on the LED), but it needs to be a source the can continuously provide a CONSTANT current through the LED regardless of the overall source voltage.
And you need to insure that the LED devices are properly heat-sinked to keep them from simply burning up due to high internal die temperatures.
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* Making the World a Better Place -- One Helicopter at a time! *
Dave |
BTW I said this before, NO I CANNOT PUT MORE LIGHTS ON IT! 
Worst case it burns out again...
