From: John Popelish
Organization: This space not available for advertising.
X-Mailer: Mozilla 4.7 [en] (Win98; U)
Subject: Re: Driving an LED from 9v battery
Date: Wed, 20 Nov 2002 23:33:41 GMT
NNTP-Posting-Date: Wed, 20 Nov 2002 18:33:41 EST
Walter Harley wrote:
> "John Popelish" wrote in message
> > As long as the LED string has lower forward drop than the battery, it
> > is hard ot beat the efficiency of a buck regulator, with a low value
> > of series resistor, to provide current regulation feedback. There are
> > single chips that will do the whole job
> If I understand properly (which is unlikely), you're basically saying, use a
> buck regulator to create a low voltage that is a bit higher than the cutoff
> voltage of the LED, with the additional drop being taken up by a series
> resistor. In other words,
> 9v ---> regulator -----> resistor -----> LED -----> ground
> But it seems like if I go that route, it seems like I need the voltage drop
> across the R to be comparable to the voltage across the LED, in order to be
> reasonably stable across changes in temp, manufacturing variation of LED,
> and so forth; which means I'm down to 50% efficiency, not even counting the
> efficiency of the regulator itself.
> Or do these chips have a way to sense currents without involving a
> (comparatively) large voltage drop? Or do I just not understand what you
You are almost right.
9v ---> regulator ----->LED -----> +resistor -----> ground
The resistor can drop as little as .1 volt and produce very good
So at 20 ma, a 10 ohm resistor provides a .2 volt current feedback,
while wasting only 4% of the power. 5 ohms should work with a 2%
If you need to add more LEDs just parallel this resistor LED string
with similar sets, and if you drop about .1 volt across the resistor,
they will share current pretty well, if they are from the same batch.
The regulator can be as simple as a comparator with a little
hysterisis and a .1 volt reference.