I'm looking at the charging circuit for a lithium ion battery that was made by Adafruit. Here's an excerpt:

enter image description here

If there's a 5V rail from the USB port and a 470 Ohms resistor, is that limiting the current through the LED to 6 mA just to keep the brightness low? The datasheet for the charger has 470 Ohms resistors in the reference design too. I'm just not sure why.

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    \$\begingroup\$ Modern LEDs are quite efficient, and the application here is just a status indicator, not trying to light up a room. Why would you waste more power than you need? \$\endgroup\$
    – Dave Tweed
    Feb 3, 2019 at 22:36
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    \$\begingroup\$ If it's just an indicator, you only need it to be visibly on, and for most LEDs 5mA or so is more than enough for that. \$\endgroup\$
    – Hearth
    Feb 3, 2019 at 22:37
  • \$\begingroup\$ You may find some useful details on LED indicators here, electronics.stackexchange.com/q/378129/117785 \$\endgroup\$ Feb 4, 2019 at 3:53
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    \$\begingroup\$ 6 mA x 3V = 18 mW. Assume 150 lm/Watt from LED at low current. At 150 lm/W = 27 lm. Look at that at say 1metre and assume radiation angle of LED is such that it illuminates 1/4 say of available area. Full sphere = 4Pir^2 so 1/4 sphere at 1m = Pi^R^2 = 3.14 m^2. Illuminance = 27 lm /3.14 ~= 10 lm/m^2. That's nicely visible but not bright-distracting-bright. Sounds good. E&)E. \$\endgroup\$
    – Russell McMahon
    Feb 5, 2019 at 10:40

1 Answer 1


The MCP73831 datasheet specifies a maximum source/sink current on the STAT pin of 35 mA and 25 mA respectively - so if the designer wanted brighter LEDs she could use lower-value resistors without worrying about overloading the output drivers on the charge controller IC. Higher currents would eat into the current available for charging, but not really to a significant extent.

So yes, I think the reason is simply that 6 mA is plenty bright enough for a status LED. Some would even say it's too bright, especially with modern high-efficiency LEDs.


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