For the most part, it is eye-safe. I wear sunglasses for the final test, and welding goggles while building becasue I am right next to the lights. More than a foot or so away, and it's not very dangerous for a short glance. Only when you are RIGHT on the LEDs does it really start to cause damage (I used a light-sensitive resistor and the threshold of a 3mW laser to determine what would cause eye damage).
Basically, if you don't have something right in front of the LEDs, and no one is looking directly at it, it isn't dangerous.
First off let me give John a thumbs up for the good work.
I have been reading this article over and over and cannot understand how Design News publishes a project that falls into a category of a flash light that can cause eye damage if looked at accidentally. This is not a novice project and Design News should considered not publishing projects of this type.
Nice project;it's great to see this lad working hands-on on this project. Perhaps a good follow-on project would be to design a switching regulator and avoid the power loss in the resistor bank. Keep up the the good work you've started.
My humble apologies. I just cranked through the numbers (Using E^2/R, 90W across 1-ohm means voltage of ~9.48V, and current of ~9.48V; break that into thirds, carry the zero, hold yer tongue right, etc...) and John (along with the rest of you) is correct! Somehow, 30 years as a BSEE and all, I let my mind take over before doing the (full) analysis.
"it is definately not much more than 3A, or else the LEDs would burn out"
I wonder. They can probably take dobule the nominal current for a while, and BTW, put out considerably over 1000 lumens each while doing so.
This could be explained by looking at a few assumptions. Your battery voltage is probably actually more like 12.6 V or even a bit higher, at full charge, which will drive up both the current the LEDs draw and the resulting power. The I/V curve for these devices increases pretty sharply! And - the LEDs are in series, correct?
As Phil commented, power = heat, and is not dependent upon either voltage or current more than the other. In the case of the 100 kV at 20 mA load, what's the source? It could be that what you are driving with has limited current capbility, and the terminal voltage drops drastically under a 20 mA load. Otherwise, you would be dissipating 5 Megawatts!!
Naw, the power being dissipated IS the heat being given off. Remember that law of thermodynamics? Energy shall neither be created nor destroyed. Here elecrtrical energy (the power x the time it is applied to the resistor) is entirely converted to heat.
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