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 that the red LEDs flicker, but
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. 

-bob mackey
 bob AT pure-flight DOT com

ps - thanks to Bob Smiley and Scott Kreuger for explaining that the
"squat" switches are really airspeed switches and don't respond to squat.