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From: John Fields
Organization: Austin Instruments,Inc.
X-Mailer: Mozilla 4.01 [en] (Win95; U)
Subject: Re: How to make/solder a circuit fast?
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NNTP-Posting-Date: Fri, 13 Sep 2002 14:39:00 CDT
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Date: Fri, 13 Sep 2002 19:39:00 GMT
The little lost angel wrote:
> On Fri, 13 Sep 2002 17:36:33 GMT, John Fields
> >> Gnd -----------+
> >> |
> >> +0 F1 0+-+------------+----------+
> >> | | | |
> >> | | +-R2--0 L2 0 0 L3 0 |
> >> | | | | | |
> >> +6~12V--+-------LM-+-R1----------+ +--+
> >> | |
> >> 0 L4 0
> >Just a little;^) After the fuse blows the circuit will start to work,
> Hmm.. I think I forgot to add in 0 F1 0 is a connector for their fan.
> >but you need a separate resistor for L3 and L4.
> No way to keep the component count down? :P
I just glossed over your circuit the first time, so there _was_ some
stuff I missed...
Basically, it's not a good idea to run LED's in parallel because the one
with the lower forward voltage will hog the current meant for the
other(s). This will result in uneven illumination or, if the current
allowed to flow through the current limiting resistor is more than the
hog is rated to carry, possible degradation or destruction of the
If you want to run LED's in parallel what you need to do is make sure
the current limiting resistor is large enough to only allow the maximum
rated current for the lowest current rated LED to pass. That means that
if you have three 20mA LED's in parallel you have to size the resistor
to allow only 20mA to pass. if you have a 20mA LED in parallel with a
2mA LED, the resistor should only allow 2mA to pass. This approach will
give you severe illumination and illumination difference problems.
The proper way to run an array of LED's is to run them in series with a
single current limiting resistor or to run each of them with its own
current limiting resistor with all the Resistor-LED circuits in
Doing it this way will allow you to compensate for the different Vf's of
the differently colored LED's without having to worry about interaction
between LED's, because there won't be any.
I don't understand how you determined the resistor values, but the right
way is to subtract the LED's Vf from the source voltage and divide that
difference by the LED current. Your LM2931 will give you a 5V output,
so for a 3V 20mA LED you'd write: R = (5V-3V)/.02A = 100 ohms. You
should also determine the wattage the resistor will be required to
dissipate. P = I*E, so in this case P = 0.02A*3V = 0.06W, so a common
quarter watt resistor would be fine. You also need to consider the
regulator's dissipation, and in this case with an input-output
differential of 7V and a 60mA load (three 20mA LED's) the regulator will
be dissipating 420mW. I don't have a data sheet in front of me, but it
wouldn't hurt if you checked to see whether that amount of power would
drive the thing's temperature into never-never land.
Professional circuit designer
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