From: "Frank Bemelman"
References: <email@example.com> <firstname.lastname@example.org> <email@example.com>
Subject: Re: current source for array of infrared leds
Date: Tue, 3 Dec 2002 17:58:34 +0100
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Organization: EuroNet Internet
NNTP-Posting-Date: 03 Dec 2002 17:00:01 GMT
"John Woodgate" schreef in bericht
> I read in sci.electronics.design that Frank Bemelman
> wrote (in <firstname.lastname@example.org
> uronet.nl>) about 'current source for array of infrared leds', on Mon, 2
> Dec 2002:
> >That's why I wanted to start with a currentsource, and compensate
> >the current, when the voltagedrop sets in. Doesn't need to be
> >perfect, just a bit better than a simple currentsource.
> If you use an 'over-simple' current source, i.e. too low a voltage and
> series resistor for a 'real' current source, you get the compensation
> effect mentioned by Harry. As the diode drop voltage goes down with
> temperature, the current goes up a bit. But having no external
> resistance at all seems to me to be a very dodgy solution.
> For example, take a 4 V supply, a diode drop of 2 V and a current of 20
> mA. The resistor is 100 ohms. Now suppose the diode drop falls to 1.8 V.
> the current is now 22 mA. 10% drop in voltage, 10 % increase in current.
Since I use 7 leds, dropping 2V each, I need a resistor 10V/20mA -> 500 ohm,
with a supply of 24V. After a while, when they drop 7x1.8V, or 12.6V, the
resistor gets 11.4V, so the current rises to 22.8 mA. That could be just
fine. My input voltage actually varies between 21-24V, but I could make
a regulated 19.4V, and use a 270 ohm resistor or so. That gives 20mA.
After the warming-up, the diodes take 12.6V and the current rises to
25mA. First impression: not bad at all.
> But nothing needs to be held to +/-0.25%.
Right, a constant voltage without a resistor or using a very small
resistor requires too much attention.
I'll run some experiments (later) to see what component values
give me the best compromise. Thanks!
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