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You should be able to obtain adequate visibility using a ultra bright LED and shunt between 1% and 10% of the 100ma available from the 100 ohm ballast.
10v - 3v = 7v, 7v/0.001 amps = 7k. = 7 mW (0.010A * 7v = 70mW) 1/4watt parts OK.
EDIT P=IE should have said .001 Amps * 3v = 3mw 0.010Amps * 3v = 30mW.
Try a 6.8k resistor in series with the Ultra Bright with these two components connected across (in shunt with) the 100 ohm ballast.
680 ohms should be quite bright if the 6.8 k is too dim. You probably will want something in between.
10v - 3v = 7v, 7v/0.001 amps = 7k. = 7 mW (0.010A * 7v = 70mW) 1/4watt parts OK.
EDIT P=IE should have said .001 Amps * 3v = 3mw 0.010Amps * 3v = 30mW.
Try a 6.8k resistor in series with the Ultra Bright with these two components connected across (in shunt with) the 100 ohm ballast.
680 ohms should be quite bright if the 6.8 k is too dim. You probably will want something in between.
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