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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: Sat, 14 Sep 2002 05:34:06 CDT
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Date: Sat, 14 Sep 2002 10:34:06 GMT
The little lost angel wrote:
> On Fri, 13 Sep 2002 19:39:00 GMT, John Fields
> >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.
> Since these are all super bright LED, I think they are all rated for
> 20mA typical and max 30mA? At least the only specs I can find says
> that :(
> >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
> The problem why I didn't do it this way is I have no idea what LED
> will be plugged in or not. I'm only making the widget, the boys get to
> plug in either 1,2 or 3 LEDs of any colour depending on their fancy.
> And I'm trying to keep this thing under US$3 if possible.
> >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
> Yup, I did this part.
> >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.
> I forgot about this though :(
> Accordingly to the LM2931 To92 datasheet, at 0.4W, ambient temp max is
> around 50~60C. Since I expect the ambient temp to be up to 60~70C, I
> guess that means I can't use the To92 package but must up to the
> To-220 which can do like 1W at 100C without a heatsink. *sighz*
> The small T-92 cost me like US$1.5, the To-220 will probably be at
> least $2.50 :(
Maybe there's an easier way out. Think about this:
| | | |
| [R1] [R2] [R3]
[ZENER] | | |
| [L1] [L2] [L3]
| | | |
For a first cut, let's make the Zener 3.9V with an Iz of 20mA. It must
then dissipate 78mW. Piece of cake. If we make each of the loads draw
20mA max, then the current through R1 with a 6V supply will be 80mA and
R1 must be = (6V-3.9V)/80mA = 26.25 ohms. Since this thing is only
driving LED's, let's starve the Zener a little and make R1 = 27ohms
since that's a standard value. The power R1 must dissipate with a 6V
supply is (6V-3.9V)*80mA = 168mW. Not bad! A standard 5% carbon film
1/4W resistor will do it. Now let's look at the grim side: With a 12V
supply the current through the resistor will be (12V-3.9V)/27 ohms =
300mA, so the resistor must dissipate (12V-3.9V)*300mA = 2.43W. Not
great, but not awful either. Now let's look at the Zener: With a 12V
supply and 60mA flowing through the loads the Zener current will be
240mA and it will be dissipating 0.936W. Also not awful. Now we've got
to make some decisions about the loads. First, if _any_ LED can be
plugged into _any_ port, then each port must be able to reliably drive
the lowest-current, lowest Vf LED. The higher current, higher Vf LED
outputs will suffer in terms of illumination, but that's how it goes...
Probably the lowest common denominator is the standard red LED with a Vf
of (say) 1.2V at 20mA. Such being the case, the series resistors will
be = (3.9V-1.2V)/20mA = 135 ohms. That's pretty close to a standard 5%
130 ohm resistor, so if we use them the LED current will rise to
2.7V/130ohms = 20.77mA and the resistor dissipation will be 2.7V*20.77mA
= 56mW. A quarter watter will work fine.
Just for grins, what will happen with a 12V supply if all the LED's are
unplugged? That extra 60 mA will have to flow through the Zener so it
will dissipate 3.9V*300mA = 1.17W. Seems like a 2 watt Zener is what's
A couple of caveats: If you want to go this way you'll need to know
where on the Zener's knee you'll be operating at the low and high ends.
20mA was chosen arbitrarily as the Zener current for a 6V input, so the
Zener's output will be somewhat lower than 3.9V with its spec'd test
current. Similarly, with a 12V supply the Zener voltage will be higher
than with its spec'd current.
Maybe there's even an easier way out...
Put a regulator in front of each LED?^)
Run the numbers, it might be interesting...
Professional circuit designer
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