If you wanna see something illuminated by one or more LEDs at night, I strongly advise choosing white over red.
White ones have roughly double the luminous efficacy of red ones, and that is according to "photopic vision" / "day vision". The disadvantage of red for illuminating things from low power worsens when the application is a nighttime/"dark" one. Human "night vision" ("scotopic vision") peaks at a shorter wavelength than human photopic vision does.
-- - Don Klipstein (don@misty.com)
If the LED voltage drop is less than the supply voltage by about or more than one "diode drop" (roughly .6 volt, .3 for less-lossy Schottky diode), then the circuit topology seriously changes.
If the load voltage is below the supply voltage by more than a "diode drop", then the favored circuit topology is a "buck regulator" or "buck / bucking switching regulator". Preferably load current as opposed to load voltage is what is regulated.
-- - Don Klipstein (don@misty.com)
At 10 mA, red LEDs tend to have voltage drop of 1.6 to 1.95 volts. Even longer-wavelength lower-efficiency ones with GaAs substrate tend to have voltage drop around 1.6 volts, slightly exceeding 1.5 volts, at 10 or even 5 mA.
-- - Don Klipstein (don@misty.com)
Thanks for the correction - it's been a while since I've actually used an LED, so my numbers were kind of pulled out of a hat, primarily for the sake of example.
In either case, the formula still applies. :-)
Thanks, Rich
I have seen this phenomenon also. I am still wondering whether the mechanism is stimulated emission (something lasers depend on) or the usually-less-desired "quenching".
-- - Don Klipstein (don@misty.com)
I think that a 1 watt "potentiometer" ("variable resistor") is sufficient. If even 11V after LED drop appears across total resistance, at 20 mA this is .22 watt.
Heck, since I know some super-efficient LEDs, I would suggest 10 or 5-6 milliamps max, which means ~1K to 1.8K ohms fixed series resistor (go for less here), and the variable one then only needs to pass at most an amount of current that would produce less than 1/8 watt of heat in full resistance of its resistive part.
For that matter, I would choose a "potentiometer" (a "pot") of 5K to 10K ohms, in case a milliamp or 2 is sufficient with a super-efficient white LED (such as the Digi-Key stocked Cree C535A-WJN-CU0V0231).
6 degrees sounds to me so narrow that the LED needs to be distant from the flag by 6-8 times the width of the flag.Bright white "low-power" white LEDs tend to have beam angle (2-theta-1/2) of at least 15 degrees - good for placing no farther than 4 times the width of the flag, maybe only 3 times. Ones with better optical efficiency appear to me to be wider beam ones
40-plus degrees that want to have distance from the flag not greatly more than the width of the flag.If the size of the flag is small, then the LED that I named can illuminate it at nighttime at 1 or 2 milliamps.
-- - Don Klipstein (don@misty.com)
Although this is true, I would suggest *not* worrying about mere milliwatts of power consumption waste, especially "in light of" LEDs that are both "widebeam" and capable of putting spots in people's eyes at a few mA. Put one or 2-3 series-wired puppies somewhat-in-front-of-the-flag by a distance roughly the width of the flag (with plenty of "give-or-take"), and you probably have all the light you need from around 2 mA, fair chance
1 mA.Such as with the "in-stock" "Digi-Key-available in-stock" Cree C535A-WJN-CU0V0231. That one is a nominally 110 degree model, known to me to make itself visibly glowing in "direct high-noon sunlight" at 2.5 milliamps.
Nichia makes even better "low-power" LEDs, but it appears to me that you need to buy those 100 at a time from one of their sales offices, such as the one close to Detroit.
-- - Don Klipstein (don@misty.com)
In , petrus bitbyter wrote (slightly edited for space by me):
Though I could agree with this suggestion to use market-available products, I also suggest usage of 1-3 LEDs "in series string" (such as Digi-Key-available C535A-WJN-CU0V0231) to work from ~1 to ~2.5 milliamps) along with a sauitable "dropping resistor" as high as around
1.0 kiilo-ohm, maybe 2.2 kilo-ohm (a common rersistor resistance).In the likely event the current drawn by LED(s) gets down to ~2.5 or ~1 milliamp "ballpark", then I have doubt about practical improvement over the "obviously wasteful" "dropping resistors" that "more obviosly" "waste power". At ~2.5 mA or less, I would re-consider "energy efficiciency requirement" due to small figures for enegy and power requirements.
-- - Don Klipstein (don@misty.com)
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