3
\$\begingroup\$

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.

\$\endgroup\$
  • 3
    \$\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 at 22:36
  • 2
    \$\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 at 22:37
  • \$\begingroup\$ You may find some useful details on LED indicators here, electronics.stackexchange.com/q/378129/117785 \$\endgroup\$ – Ale..chenski Feb 4 at 3:53
  • 1
    \$\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 at 10:40
6
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.