piclist 2002\09\28\012834a >
www.piclist.com/techref/microchip/ios.htm?key=ir
BY : Russell McMahon email (remove spam text)

> The absolute max current for the hideously expensive IR LED I got from the
> RatShack specifies a 100ma max current. I plan to drive it at 80ma using a
> 47 ohm current limiting resistor.

V= .080 * 47 = 3.8v across the resistor leaving 1.2v for LED and transistor.
May or may not be correct depending on LED.

A safer and more flexible (and hardly any dearer) method is to use the
mickey_mouse_constant_current scheme I mentioned recently. Place LED in
collector and resistor in emitter. Iresistor = (Vin-0.6)/R which is the max
current that the LED in the collector circuit can see. This only works
"properly" if the LED has more operating voltage available than Vin. This
will be the case if you operate the PIC from a regulator and supply the LED
from the regulator input OR if you divide the Vin drive signal from the PIC.
eg Say PIC drive = 4.8v. Drive base from PIC via 2k2 to base and 1k base to
ground. Vbase-gnd max = 1.5v so Vemitter = 0.9v. If Remitter_ground = 12r
then  Ir = 75 mA (approximately of course :-) ). Adjust R or base divider to
vary current of the constant current source.

The LED and transistor then have about 4v headroom if driven from a 5v
supply. The beauty of this scheme is that it automatically compensates for
variations with LED forward voltage with type and current. A white LED with
Vf or around 3v will be driven at the same current as a red LED with a Vf of
under 2 volt. The maximum base drive to the transistor here is (4.8-1.5)/2k2
= 1.4mA. This implies a required beta of about 60 but alas tis not so as the
1k shunts all of this. As the base voltage sags more will be available for
the transistor. A beta of 200+ (eg BC337 etc) would be advised.

A better arrangement which leads to a much stiffer current source is to
replace the 1k with a zener or diode string. eg 2k2 PIC to base, 3 x 1N4148
in series from base to ground (cathodes towards ground).

All the above "playing" with a divider is avoided if the LED has a voltage >
Vdd to drive it.

Russell McMahon

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