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LED tail strobe



 
 
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  #4  
Old May 1st 04, 04:04 PM
BllFs6
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The criteria I had when laying out my circuit was that the lights would
operate down to 10V and still not exceed the max current limit at 15V.
I'm working from memory here, but I believe the SuperBright LEDs I used
had a forward voltage of 3.5V.

Putting 2 in series with a 220ohm, 1/4W resistor maximized the
efficiency while maintaining the range I wanted.



Okay, thats not too bad because your using a resistor to cover HALF the voltage
drop....of course your losing half your power in the resistor rather than
having it produce light in the LED...

My previous post was more concerned with folks both running the LED at absolute
max power ratings,,,,and trying to use the LED voltage drop to cover something
like 80 percent or more voltage range...thats when you have to be a little more
careful...

Your 1/4 watt resistor is NOT high powered enough....youll be running it close
to or past its maximum power rating for normal voltages....

Get something more like 1/2 watt or 1 watt resistors....or use 2,3,or or even 4
of the 1/4 watt resistors with appropriate resistance values in series to
spread the power load....resistors are cheap and there AINT no harm in using
ones that can handle significantly more power than expected....while the risk
to benefit ratio of trying to push their limits seems rather high....

Also make sure the resistors can cool effectively.....their power ratings dont
mean diddly if they are well insulated or packed into a tight space and can
warm up...

take care

Blll
  #5  
Old May 5th 04, 01:23 AM
anonymous coward
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Have you considered using a constant current regulator, instead of a
resistor? I believe there is an example circuit given in the LM337/317
datasheet showing how to build one using only the LM337 (normally used as
a voltage regulator) & one resistor.

It would need to be bolted to a heatsink, like the Luxeon Star LEDs, but
IIRC the LM337 and cousins also shut down if they overheat.

AC

  #6  
Old May 6th 04, 05:17 AM
Jay
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Jim Weir wrote in message
And as somebody else has pointed out, running a series string close to the
bottom limit of Vcc and hoping that a single resistor will provide constant
current to these devices will cook them when Vcc rises to the charging voltage.

No designer in his right mind would use a circuit in this manner.

Jim


The more the supply varies the less effective the technique is. I
picked the simple resistor design with certain assumptions. When it
comes down to it, the voltage doesn't vary all that much on a properly
operating electrical system. That big 'ole battery is like an anchor
on the supply bus. Over voltage condition blows the breaker.

What if the alternator quits? Well how far down do you want to go on
the discharge? You could get a range of zero to 14 volts if you
consider a discharged battery, thats a pretty tough range to design
for isn't it? I can think of a lot of systems that will malfunction
when supplied from a discharged battery. Even normal lights will go
out of spec on a discharged battery.
  #7  
Old April 17th 04, 01:34 PM
GeorgeB
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On 16 Apr 2004 22:15:44 -0700, (Jay) wrote:

What is typically done is the LEDs are just put in series with a
current limiting resistor. This forms a circuit akin to a kind of
voltage regulator called a "zener regulator". LEDs have a fixed
forward voltage for the recommended drive current, say for example its
2.8 volts for a green LED. Divide the power supply (e.g. 12V) by the
forward voltage of the LEDs and drop any fraction (12/2.8=4.3 make
that 4 even) Put those in series with a current limiting resistor that
will drop the fraction (.3V).


Your concept is reasonable, but there are some significant problems.
The forward voltage varies with temperature, and the power supply
LIKELY varies over time. As you have it, if the fwd voltage dropped
0.1v, and the supply did not change, you would have 0.7 vs 0.3 across
the resistor, for over 2x the current ... maybe a real problem.

Now, let's have the alternator charging the battery, and have it at
15.5V or so ... now I have 3.8 volts across that current determining
resistor ... 12 times the "design". OUCH.

Now let's have the alternator fail, the battery voltage drop to 10.5V.
Your series string will draw no current and give no light ... and you
are in an emerency situation that is exactly when someone needs to see
you.

What is the solution ...

There are "constant current" devices. I have used them, and they
work allowing operating this string with probably 3 LEDs over the
range at visually constant brightness.

You can put 2 in a string allowing the current determining resistor to
allow a power range varying over about 2:1, 10V to 15V.

You can put 3 in a string like above, power will vary over about 4:1.

You can design a pulse system turning the LED on for (maybe) 0.1ms
then off for maybe 5ms and PROBALBY not overdrive (into damage) the
LED and put "as many" as you want in parallel. The driver will likely
be a FET.

So lets say the recommended current for
the LED is 20mA. Ohms law is R=E/I, so that gives you a resistor
value of .3V/.02A=15 ohms. Check power to make sure it won't over
heat (P=IV) so thats .3V*.02A=.006W so a typicial 1/4 watt resistor is
fine. You must of course have some kind of resistor in series to limit
the current. If you math works out that you need no resistor, put one
less LED in series and then recalculate the limiting resistor. Put
the LEDs and resistor in series (in any sequence) observing the proper
polarity of the LEDS.


I was taught to allow about half the voltage for the resistor, half
for the LED string unless I had current control. In the "old days",
for current control we used an emitter resistor in a common emitter
circuit, 2 or 3 diodes to set bias (single vs darlington), and the
LEDs between collector and V+. There are other (better) ways, but
everyone understood this one.

If you hook up your entire string backwards, no
harm will be done, but if you happen to solder one LED backwards, it
will likely be toasted on power up.


I disagree that there will be damage with any in backwards. The
reverse voltage will almost certainly be higher than the forward
voltage, so there won't be any current drawn. If there is, you still
would have less than correctly wired.

Need more than 4 LEDS? Replicate this circuit in parallel as many
times as you need to get the luminous flux you need.

Of course the numbers (Vf, If) used here are for the older style
single chip LEDS. The parts that are getting everybody excited these
days are the multi-chip variety whose forward voltage and current will
vary alot from my example.

"Dean Head" wrote in message .. .
Jeff,
Would you consider sharing your design for the driver electonics?
Dean
Cozy MK4
BKV FL

"Jeff Peterson" wrote in message
om...
I am building an LED tail light for my Lancair 360. It uses 6 ea 5
watt luxeon LEDs. I can strobe 1 amp through each of these which
should give enough light to satisfy the FAA regs. Runs surprisingly
cool. I have posted photos he

http://w1.lancair.net/pix/album01\

Its not quite finished, but I thought you might like to see my
progress.

cheers,

Jeff
N273CK stilll building.

ps thanks to Eric Jones for his posts on the Lancair Mail List, his
web pages on the subject and many emails and phone calls.


  #8  
Old April 18th 04, 07:15 AM
Jay
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Read comments below...
GeorgeB wrote in message
. ..
Your concept is reasonable, but there are some significant problems.
The forward voltage varies with temperature,


Indeed it does, diodes are often used as the sense element for
temperature sensing, but it's a very small change, and you need to
amplify it when you do. Also, the self heating of the device will
swamp ambient temp effects I should guess.

and the power supply
LIKELY varies over time. As you have it, if the fwd voltage dropped
0.1v, and the supply did not change, you would have 0.7 vs 0.3 across
the resistor, for over 2x the current ... maybe a real problem.


Thats what makes it a self regulating circuit, as the current comes
up, the foward drop of the diodes go up as well, thus reducing the
drop across the resistor.

Now, let's have the alternator charging the battery, and have it at
15.5V or so ... now I have 3.8 volts across that current determining
resistor ... 12 times the "design". OUCH.


Same thing as before...

Now let's have the alternator fail, the battery voltage drop to 10.5V.
Your series string will draw no current and give no light ... and you
are in an emerency situation that is exactly when someone needs to see
you.


You can handle this case by dropping one diode off the string and
recalculating the resistor as before. Cuts your efficiency a little
but hey, some poeple drop over half the power delivered as heat into
their "current limiting device" for 28V applications.

What is the solution ...

There are "constant current" devices. I have used them, and they
work allowing operating this string with probably 3 LEDs over the
range at visually constant brightness.


You can do that, but for driving LEDs, since they form a nice self
regulating circuit with a single resistor I didn't feel it was
neccesary. Please share with the group which part you've had success
with as a "constant current" device.

You can design a pulse system turning the LED on for (maybe) 0.1ms
then off for maybe 5ms and PROBALBY not overdrive (into damage) the
LED and put "as many" as you want in parallel. The driver will likely
be a FET.


You could do this also but each LED would need its own current
limiting resistor in series because the forward drop of the LEDs vary
from part to part and with temperature as you've mentioned and the one
with the lowest drop would eat the most power without those resistors.
But again, with a pulse width modulation circuit, why so complex?

I was taught to allow about half the voltage for the resistor, half
for the LED string unless I had current control. In the "old days",
for current control we used an emitter resistor in a common emitter
circuit, 2 or 3 diodes to set bias (single vs darlington), and the
LEDs between collector and V+. There are other (better) ways, but
everyone understood this one.


So you're building a constant current supply from each group of 2 or 3
LEDS, thats pretty complex if a single resistor will work. What
you're suggesting is too complex for the average guy and I see no
practical benefit. Don't light bulbs vary in brightness with supply
voltage? Sure they do, and they vary more than the single resistor
method I've sketched out.

If you hook up your entire string backwards, no
harm will be done, but if you happen to solder one LED backwards, it
will likely be toasted on power up.


I disagree that there will be damage with any in backwards. The
reverse voltage will almost certainly be higher than the forward
voltage, so there won't be any current drawn. If there is, you still
would have less than correctly wired.


The reverse drop on the LEDS will be the supply divided by the number
of diodes. 12/4=3V. Last data sheet I looked at said the reverse
voltage limit was 5V. Thats why I also said that if you put one
backwards it will cook. It will see the full 12V.

What I've outlined is a simple method to build LEDs lights. Yes, you
could build a constant current supply, and the LEDS would see the
exact same current from 10V to 15V but your light bulbs will vary in
brightness (acnd color) over that range anyway more that my suggest
circuit due to the self limiting nature of a diode(s) in series with a
resistor.
  #9  
Old April 19th 04, 03:24 AM
TaxSrv
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"GeorgeB" wrote in message

Now, let's have the alternator charging the battery, and have it at
15.5V or so ... now I have 3.8 volts across that current determining
resistor ... 12 times the "design". OUCH.

15.5V is not a normal charging voltage, and if sustained will ruin the
battery. It would only be a fault condition with a regulator gone
bad. I think you can safely calculate dropping resistors based on
14.5V max.

Fred F.


  #10  
Old April 20th 04, 10:08 PM
Blueskies
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That is a good point...bad regulator putting out high voltage. How high can the alternator go with a runaway regulator?

--
Dan D.
http://www.ameritech.net/users/ddevillers/start.html


..
"TaxSrv" wrote in message ...
"GeorgeB" wrote in message

Now, let's have the alternator charging the battery, and have it at
15.5V or so ... now I have 3.8 volts across that current determining
resistor ... 12 times the "design". OUCH.

15.5V is not a normal charging voltage, and if sustained will ruin the
battery. It would only be a fault condition with a regulator gone
bad. I think you can safely calculate dropping resistors based on
14.5V max.

Fred F.





 




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