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From: bob mackey <n103md@yahoo.com>
Subject: LED Position Lights
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As many have mentioned, there is some non-linearity
in the human eye's response that favors pulsed bright 
lights over continuous dim lights. But that effect is 
rather small. 

The much greater effect is the power efficiency of the
circuit when used in pulse mode. There are two effects
here. The first is that a smaller ballast resistor is
used along with a short duty cycle. More power goes
to the LED and less to the resistor. 

A second effect should also be considered. The light
output 
of a RED LED increases faster than linearly with
current. 
Some portion of the current always goes to
non-emissive
leakage. That is, the first milliamp is wasted. The
second
milliamp makes light. The third milliamp makes light. 
So 3mA is twice as bright as 2mA, not 3/2 as bright 

With blue, green, (and white) LEDs, there is less
leakage
due to different materials and higher bandgap voltage.

So there is less motivation to pulse the current. If
you look 
at common handheld consumer electronics you'll usually
see 
the red LEDs flicker, and the blue ones don't.  There
are 
very good reasons to work that way when you have only
a 
small battery. 

For aircraft uses, it really makes very little
difference
whether pulse width modulation or a large ballast
resistor
is used. Either way will be much more efficient than
an 
incandescent lamp with a color filter. 

There was also the suggestion to use LEDs in series. 
That's a good idea, but don't take it too far. If
there
is no ballast resistor, the current will vary greatly
with small variations in system voltage. Let's assume
that we want the nav lights to stay lit when the
alternator
is off. The system voltage might be 12V with the
engine off
and 14.6V with the engine running. 

   I = (Vsys - Vled) / R

I usually use:  
4 red LEDs (Vled = 4 * 1.8V = 7.2V),
3 green LEDs (Vled = 3 * 2.2V = 6.6V), or
2 white LEDs (Vled = 2 * 3.8V = 7.6V). 

In the red case, that leaves
   I = (Vsys - 7.2V) / R

If we want 30mA per strand at the higher voltage
(14.6V), 
then choose 
   R = (14.6V - 7.2V) / 0.030A = 247 Ohm 
[might actually choose 300 or 330 ohm because it is 
readily available]

When the voltage drops to 12V, the current will be: 
   I = (12V - 7.2V) / 247 ohm = 19.5 mA
So it will be about 2/3 as bright with the alternator
offline. 

If we isntead used 8 red LEDs (8 * 1.8V = 14.4V)
and a small ballast resistor to limit it to 30mA at
14.6V
   R = (14.6V - 14.4V) / 0.030A = 0.67 ohm 
   [about equal to the wires and contacts]
Then, when the voltage dropped to below 14.4V, the
lights
would go completely off. If the voltage surged to 15V,

the current would increase to almost an Amp, burning
out 
the LEDs.  


So.... 
stack a few LEDs, but don't get rid of the ballast
resistors!

That still applies if you use PWM, unless the PWM
circuit
is peak current limiting. 


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