LED tail strobe

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.

I think someone may have already pointed this out, and maybe I didn't
make it as clear as I should have... I stacked the forward drop of
MULTIPLE LEDs up until I got somewhere near the bottom end of the
supply voltage. So for the example I gave, I got to 4 LEDS in series.
Why waste all that power as long IR (heat) off a big resistor when we
want red and green light right?

Regarding 2.8V- The forward drop of these devices now-a-days is all
over the place. The new chemistries seem to be making higher forward
drops, plus the trend is to package multiple die into one larger
device and this can effect the forward drop of the composite device.

By the way, anyone building my circuit should try one instance of it
(4 LEDS and resistor) on your bench supply before you go fly at night
cross country.



Jim Weir wrote in message . ..
Before everybody in the Western Hemisphere blows a bucket full of light emitting
diodes, would you care to calculate the resistor one more time? And perhaps
post a retraction?


(Jay)
shared these priceless pearls of wisdom:

-You can find examples on how to power the LEDs on the manufacturer web
-site.
-
-Having said that...


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.


Um, no. Suppose the diode has a forward voltage drop of 2.8 volts (that's not a
common value, but I'll give it to you for argument.

Now the power supply (battery) is a 12 volt supply, but 14.2 volts at full
charge with the alternator going, so the drop across the series resistor is
going to be

14.2 minus 2.8 equals 11.4 volts, which is the voltage across the resistor.
This current limiting resistor is going to have 20 mA flowing through it, so Ohm
tells us that resistance equals voltage divided by current. In this case, 11.4
volts divided by 20 mA gives us a resistor of 570 ohms (560 is the nearest
standard value).

You put your calculated 15 ohm resistor in series with this diode and I
guarantee you that the SNAP you hear is the gallium aluminum arsenide
semiconductor of the diode being sacrificed on Ohm's altar.

I'm serious. You owe the newsgroup a correction before somebody takes your
error and blows up a whole bunch of LEDs.

Jim


Jim Weir (A&P/IA, CFI, & other good alphabet soup)
VP Eng RST Pres. Cyberchapter EAA Tech. Counselor
http://www.rst-engr.com

"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.

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.

Something I've been pondering...

4 Luxeon star 1W emmitters in parallel will draw 1A or so. Your 555 based
timer circuit will draw large(ish) currents for short periods of time,
because of the small mark:space ratio. Might this interfere with other
systems powered by the battery?

The switching regulators for standard Xenon flash tubes draw a lower
but much more constant current (though goodness knows they can generate
radio noise if they're not shielded right).

AC




On Sat, 17 Apr 2004 23:25:24 -0700, Jay wrote:

I think someone may have already pointed this out, and maybe I didn't
make it as clear as I should have... I stacked the forward drop of
MULTIPLE LEDs up until I got somewhere near the bottom end of the
supply voltage. So for the example I gave, I got to 4 LEDS in series.
Why waste all that power as long IR (heat) off a big resistor when we
want red and green light right?

Regarding 2.8V- The forward drop of these devices now-a-days is all
over the place. The new chemistries seem to be making higher forward
drops, plus the trend is to package multiple die into one larger
device and this can effect the forward drop of the composite device.

By the way, anyone building my circuit should try one instance of it
(4 LEDS and resistor) on your bench supply before you go fly at night
cross country.



Jim Weir wrote in message . ..
Before everybody in the Western Hemisphere blows a bucket full of light emitting
diodes, would you care to calculate the resistor one more time? And perhaps
post a retraction?


(Jay)
shared these priceless pearls of wisdom:

-You can find examples on how to power the LEDs on the manufacturer web
-site.
-
-Having said that...


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.


Um, no. Suppose the diode has a forward voltage drop of 2.8 volts (that's not a
common value, but I'll give it to you for argument.

Now the power supply (battery) is a 12 volt supply, but 14.2 volts at full
charge with the alternator going, so the drop across the series resistor is
going to be

14.2 minus 2.8 equals 11.4 volts, which is the voltage across the resistor.
This current limiting resistor is going to have 20 mA flowing through it, so Ohm
tells us that resistance equals voltage divided by current. In this case, 11.4
volts divided by 20 mA gives us a resistor of 570 ohms (560 is the nearest
standard value).

You put your calculated 15 ohm resistor in series with this diode and I
guarantee you that the SNAP you hear is the gallium aluminum arsenide
semiconductor of the diode being sacrificed on Ohm's altar.

I'm serious. You owe the newsgroup a correction before somebody takes your
error and blows up a whole bunch of LEDs.

Jim


Jim Weir (A&P/IA, CFI, & other good alphabet soup)
VP Eng RST Pres. Cyberchapter EAA Tech. Counselor
http://www.rst-engr.com

anonymous coward wrote:
Something I've been pondering...

4 Luxeon star 1W emmitters in parallel will draw 1A or so. Your 555 based
timer circuit will draw large(ish) currents for short periods of time,
because of the small mark:space ratio. Might this interfere with other
systems powered by the battery?

The switching regulators for standard Xenon flash tubes draw a lower
but much more constant current (though goodness knows they can generate
radio noise if they're not shielded right).

AC


So? Stick a large capacitor and a low-value, one watt resister in series
with the circuit. Still cost less that $5(US) at the RatShack.

--
http://www.ernest.isa-geek.org/
"Ignorance is mankinds normal state,
alleviated by information and experience."
Veeduber

At full field and normal cruise RPM with low load the alternator will
put out about 90 volts DC. This is enough to fry all of your
electronic goodies!
This is how you convert your 12 volt auto alternator to power your 110
volt electric drill with the flip of a switch.

On Tue, 20 Apr 2004 21:08:07 GMT, "Blueskies" wrote:

That is a good point...bad regulator putting out high voltage. How high can the alternator go with a runaway regulator?

On Thu, 22 Apr 2004 01:14:13 +0000, Ernest Christley wrote:

anonymous coward wrote:
Something I've been pondering...

4 Luxeon star 1W emmitters in parallel will draw 1A or so. Your 555 based
timer circuit will draw large(ish) currents for short periods of time,
because of the small mark:space ratio. Might this interfere with other
systems powered by the battery?

The switching regulators for standard Xenon flash tubes draw a lower
but much more constant current (though goodness knows they can generate
radio noise if they're not shielded right).

AC


So? Stick a large capacitor and a low-value, one watt resister in series
with the circuit. Still cost less that $5(US) at the RatShack.

No reason why not, but that would have to be one whopping capacitor.

e.g. 4V voltage drop over 1/5 second @ 1A = 50,000 microFarads (OK the
battery will still be contributing some juice).

The aircraft on your website is truly neat.

AC

Hopefully breakers and/or fuses pop before your electronics do.

(John) wrote in message ...
At full field and normal cruise RPM with low load the alternator will
put out about 90 volts DC. This is enough to fry all of your
electronic goodies!
This is how you convert your 12 volt auto alternator to power your 110
volt electric drill with the flip of a switch.

On Tue, 20 Apr 2004 21:08:07 GMT, "Blueskies" wrote:

That is a good point...bad regulator putting out high voltage. How high can the alternator go with a runaway regulator?

anonymous coward wrote:


The aircraft on your website is truly neat.

AC

That it is, nameless one.

Looks like georgous work, doesn't it.


Richard