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From: will jones (no email)
Date: Mon Apr 07 2008 - 08:18:18 EDT
I love the info you give Ken. Really saves me time trying to keep up with the rapidly changing driver and led designs. It is amazing how much stuff has come out in the past 3 years alone and it is not over yet. But I would like to simplify this for readers who don't want the tech info.
The resistors get too damn hot and can cause a short and/or a fire. :)
Keep the info coming Ken. I'll be buying from you instead of building when I refit my morgan.
Ken James <> wrote:
Some of the new Lumiled Rebel LEDs can be driven at over an amp, but say
we want to use a Rebel .7 A model (that is 700 ma) to make a LED boat
cabin light.
These LEDs will suffer and soon burn out at over about 3.7 VDC, and our
light must be designed to take up to 16 VDC to be reliable, so that
means that if we use two of these LEDs connected in series the resistor
must dissipate a max of almost 9VDC at 700 ma. (We cannot soak up
more of the max voltage by using more than two of these LEDs in series
because if we do the light will be very, very dim at normal voltages.)
This works out to 6.3 watts of power that must be dumped as heat by the
resistor. The average
Radio Shack resistor is a 1/4 watt model, and you don't want to drive
them more than 75% if you want things to last, so you would need 8.4
watts worth of resistors to do the job, or that is almost forty of the
1/4 watters!
Or you could get a big power resistor from Newark Electronics online, it
is about ten inches long for this power level.
So you now have a 'toaster oven' operating behind the light, cooking the
overhead and not doing anything useful. Nice.
But we could lower the power through the LED by a bigger value resistor,
that is more ohms, right? Than we don't need as many resistors, correct?
Since the LED is a semiconductor and does not exhibit a linear (or
Ohmic) voltage/current behavior but of course the resistor does, if we
half the current through the LED by increasing the ohmic value of the
resistors enough to lower the LEDs applied voltage by about10% from say
about 3.6 to about 3.2 V and the current through the LED then goes from
700 ma to 350 ma, the LEDs will see 50% Amp X 90% Volt = 45% of the
watts or power through them as before.
If the max power of the circuit was .7A X 16V = 11.42 W, and the two
LEDs used .7 X (3.6V X two LEDs or 7.2V) or 5.1 W than the resistors
were dissipating 11.4 - 5.1 W = about 6.3W max.
Since the input voltage to the entire circuit was not changed, and the
current (amps) through the resistors and the LEDs must be the same
because they are in series, and the current through the LEDs is now at
50% of what it was, that means the total power of or through the circuit
has halved, and the LEDs now account for only dissipating or using up
only 45% of the power through the circuit in watts of what the total
value was originally or that is about 2.29W and since the total power
through the circuit must be one half the current X the original input
voltage = half the power. This means that the resistors must now be
dissipating about 11.4 W divided by one half = 5.7- 2.29W = 3.41 watts
of power compared to 6.3 watts before.
So you must use 55% of the former power dissipation value of the
resistors to get 45% of the power through the LED. Not very efficient.
Also you may well need another polarity protection diode and more LED
protection components.
Of course it should be noted that the LEDs will likely be slightly more
efficient and so brighter for its (lower) power level at the new lower
voltage.
But even so now you must use about 50% more expensive LEDs and circuit
board and resistors to get the about the same amount of light as before
while actually burning up a significant bit more power than before when
all things are considered
So in trying to use less resistors per LED we ended up actually needing
more parts and a much larger and more expensive light to get the same
light output as before. Not so good. This is why it pays to drive the
LEDs as hard as you can!
But it is very hard to do that with resistors no matter how you slice it.
What it boils down to is don't use resistors to supply power for the new
super powerful LEDs unless you also like steam driven computes! -Ken
Valhalla
Morgan 33T IOR, Hull 24
Bloomington, IN
---------------------------------
You rock. That's why Blockbuster's offering you one month of Blockbuster Total Access, No Cost.
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