Shop OBEX P1 Docs P2 Docs Learn Events
BS2 flash high power Luxeon LED's — Parallax Forums

BS2 flash high power Luxeon LED's

Jeff RJeff R Posts: 9
edited 2009-10-23 02:46 in BASIC Stamp
Hello, I have a project with the Basic Stamp 2 that flashes 2 or 3 Luxeon 3 LED's.
A single 3.7v Li Ion battery will power the Luxeons and also the Basic Stamp (voltage boosted to 5v or this).

There are LED drivers available that buck or boost current to the LED, but I wanted to use less hardware and still have as much allowable current available.
Or, a 1 ohm, 2.4 watt resistor could be used.
  • Is there an alternative to using a buck or boost converter for the LED's?

I will use small MosFETs as the transistors connected to the output of the micro.

3.7v Li Ion battery (approx 2200mAh)
5v microcontroller (5v boosted from 3.7v)
Two or three flashing Luxeon 3's (2.95v Vf, 1540mA max.) approximately 25ms each (powered by the 3.7v Li Ion battery)

Thank you for any input, Jeff
This is modified from the example "PULSOUT.bs2" (Thank you)
Scale CON 500         ' to ms for 2 us per unit
Flash CON 25 * Scale  ' 25 milliseconds

Setup:
  LOW 0               ' make P0 low (LED off)
  LOW 1               ' make P1 low (LED off)

Main:

  GOSUB LED_1         ' go to the LED 1 routine, then return to next task below
  GOSUB LED_2         ' go to the LED 2 routine, then return to next task below
  GOSUB LED_1         ' go to the LED 1 routine, then return to next task below
  GOSUB LED_2         ' go to the LED 2 routine, then return to next task below
  GOSUB LED_2         ' go to the LED 2 routine, then return to next task below
  GOSUB LED_2         ' go to the LED 2 routine, then return to next task below

 LED_1:
    PULSOUT 0, Flash  ' flash LED 1
    PAUSE 30          ' delay
    PULSOUT 0, Flash  ' flash LED 1
    PAUSE 100         ' delay
  RETURN

 LED_2:
    PULSOUT 1, Flash  ' flash LED 2
    PAUSE 30          ' delay
    PULSOUT 1, Flash  ' flash LED 2
    PAUSE 100         ' delay
  RETURN

Post Edited (Jeff R) : 10/23/2009 2:43:12 AM GMT

Comments

  • Mike GreenMike Green Posts: 23,101
    edited 2009-10-23 02:40
    3.7V - 2.95V = 0.75V. An MOSFET might have as little as a 0.3V drop across it, certainly no more than 0.6V when turned on. You'll need some kind of current limiting resistor whose value depends on actual voltages and voltage drops. The problem is that this is all under ideal conditions. The LiIon battery voltage will begin to drop lower and the MOSFET voltage drop may increase as it heats up and you've got very little "headroom". The details depend on the specifics of the battery, MOSFET, and Luxeons. Do your homework and try it if the numbers turn up ok.
  • Jeff RJeff R Posts: 9
    edited 2009-10-23 02:46
    Wow, thanks for the fast reply! smile.gif
    Yes, the voltage range of the battery will be from about 4.2 max down to about 2.5 volts.
    Jeff
Sign In or Register to comment.