# BeagleBone Black LED current/brightness

I'm in the process of designing a PCB that includes 10-15 LEDs. It's battery-powered, so I'd like to use as little power as possible. As such, I've been trying to figure out how little current the LEDs need to still appear relatively bright.

As an experiment, I measured the voltage across the current-limit resistor (R12) for the power LED on my BeagleBone Black. Voltage was 0.75V, so current was only 0.16mA (using Ohm's Law).

At first this seemed like great news -- the LED appears extremely bright, so I thought I'd use the same LED/resistor combination on my board. That way, I'd be able to power 10 LEDs at a total of 1.6mA.

However, upon digging around some more, I found the following graph from the LTST-C191TBKT datasheet, which shows that luminous intensity is basically zero below ~3mA.

This seems to be a contradiction ... any ideas why the datasheet claims the LED gives off no light below 3mA, but the LED appears quite bright at 0.16mA?

• Measure the 4k75. It sounds too high. I'll put a small wager that it's 47.5 ohms. May 6, 2015 at 21:24
• Just double-checked; it definitely matches the schematic (4k75). I guess I'm wondering if this seeming contradiction can be explained by how I'm interpreting the datasheet -- like maybe the LED manufacturers detail specs at much higher brightness levels than are truly needed for an indicator LED on a PCB. May 6, 2015 at 21:29
• Did you measure it and if so with what? May 6, 2015 at 21:31
• I agree with @Andyaka, 1.6ma is very unlikely to make an LED appear "extremely bright". May 6, 2015 at 22:47
• Actually, it might. Notice that this is SMD with a clear body and no diffuser. The bare LED is visible, and is very small, so it doesn't take a lot of total radiance to appear bright. Plus, given the log characteristics of human eye (1/10 th the power give half the brightness), and assuming his work area is at normal indoor levels, it seems possible. May 7, 2015 at 1:41