I'm building a Whimpy for dusk and night flying. The plan is to install
a bunch of LEDs on the inside so the whole plane will light up like a
light bulb. I saw a Telemaster set up this way, and the effect was very
The LEDs are rated for 3 to 3.8 volts. My plan is to set up a
three-cell nicad pack to be turned on and off with a micro switch on a
spare radio channel, with strings of LEDs wired in parallel.
The LEDs were shipped with a big packet of 500 ohm resistors, one for
each LED. I was assuming that these were probably included for use with
12 volt power supplies. Am I correct in thinking that I should simply
wire the LEDs straight to the correct voltage and forget about the
On Sat, 09 Jun 2007 16:33:43 -0600, I said, "Pick a card, any card"
You can certainly do that but the resistors and higher voltage will
allow the LEDs to work longer. The resistors limit current only so
that the LED doesn't draw too much and burn out. It's an active
device and can do that if you exceed the rated voltage. It throttles
the current, as it were, much like a fixed carburetor would.
Imagine a 3 volt battery pack powering a string of LEDs. If it takes
3 volts to get them to light, as soon as you drop below that, they
go dark. The 12 volt package you bought was probably made for use in
cars where, for all intents and purposes, you have unlimited power
for the LEDs available. In your model, you have only what you can
carry as payload plus the weight of the LEDs, resistors and wire.
Placing a 12 battery pack or 12 volt gel cell onboard would be a
good option but not as efficient as changing the resistors to 330
ohm and using a 5 volt gel cell.
Resistors of 1/8 watt 330 ohms can be purchase for around 1c each. A
5 volt gel cell is very light and comes in a variety of shapes. The
discharge curve for lead/acid gel cells will pretty much keep the
voltage above 3 volts until the battery is nearly empty. Very
efficient and more cost effective than any other voltage. That cost
effectiveness comes from the factors of availability and overall
power consumed. Small 5 volt batteries are easier to find in the
light weight you need and the humble 330 ohm resistor has the honor
of being the most manufactured value because of all the LED devices
we use in our daily lives and the humble 5 volt power supplies that
drive them in stereos, computers and everything electronic. The 330
ohm resistor consumes a mere trickle of power while the 500 ohm
resistor will consume a bit of a larger trickle.
In short, don't use 3 volts unless you want a very short time for
the lights to be on. Use at least 5 volts, which I consider optimum,
and 12 volts if you can carry the payload.
I must add a disclaimer for the engineers out there. I deliberately
avoided getting too technical and used a lot of references that were
completely non-technical as this forum cannot guarantee that all
readers are up to understanding the complexity of electronics for
the purposes of our hobby. Please don't nitpick this post or you'll
confuse those who really don't want to get into valence levels,
quantum mechanics and electron versus hole flow theory.
"The 330 ohm resistor consumes a mere trickle of power while the 500 ohm
resistor will consume a bit of a larger trickle."
Not trying to nitpick, but Ohm's Law says this is backward at best...
On Sun, 10 Jun 2007 03:12:34 -0700, I said, "Pick a card, any card"
Been a long time since I bought one in the USA but the last time I
did it was from Jameco by mail order. Digikey probably has them. Any
electronics wholesaler would have them. Over here in Oz, I get mine
from an audio supply company. I use a pair of 6 volt beauties for a
long charge on another project unrelated to R/C flying.
I am always willing to learn, but IIRC gell cells are lead-acid in
make up and they are 2 volts per cell. That means either a 4 or 6 volt
battery. I can't locate a 5 volt one as you have indicated. Do you
have a web site reference? A 5 volt cell would certainly elimate me
have to use a 5 volt regulator on projects.
Yah - nicads have lower voltage then NiMH, which are lower (I think) then
alkaline. For lead-acid cells, the open circuit voltage of a full charged
cell is ~2.1. A lead-acid car battery with 12.0V across the terminals is
probably too discharged to start your car. ~12.6 is normal if it hasn't been
charged recently but is in a good state of charge.
On Sun, 10 Jun 2007 19:11:16 -0600, I said, "Pick a card, any card"
Ok. I dug out my Panasonic 5v Gel Cell (took me two days of digging
around in the garage to find a no-longer-used item) and in
parenthesis is this: (5.8v). So, even though the gel cell is marked
as a 5v cell it is in reality closer to 6 volts. Used in an alarm
system as backup for the microcontroller and relay boards. Charged
with a 6.2 volt charger. After two years or more in storage, it
still has a charge of 5.5 volts. Gotta love lead acid, eh?
I would connect one of the LEDs in series with the 500 ohm resistor to your
proposed power source, and see if the results are what you need.
Then, I might try two 500 ohm resistors in series, and if really brave two
resistors in parallel. (This may or may not cause the LED to fail.)
LEDs have a brightness range and a current range. The idea is to have them
as bright as you need but no brighter.
Conventional LEDs normally need between 20-30 milliamps of current to work
These are rated at 3 to 3.8 volts and 25 mA. I appreciate Ray's advice
about using a larger source that's limited by resistors, for greater
endurance. That makes sense. I'll do the math and see what the optimum
voltage source is for 25 mA and 500 ohms.
You don't _have_ to use 500 ohms -- you just need to limit the current
appropriately. 500 ohms may waste a lot of voltage, which would make
your batteries necessarily heavy.
The idea with an LED is that you need to give it the right _current_
(hang the voltage), and when you do that it'll impede that current with
_some_ voltage in it's range. Just what that voltage will be depends on
the LED and the temperature. So you calculate the right resistance for
your voltage source over it's range, and the LED's over theirs.
My apologies to those whom already know this, or those whose heads this goes
As others have said - NEVER use an LED without a current limiting resistor
in series. Period.
When I design an LED in a circut, I need to know a couple of things: how
much current does the LED require (I ususally use the 1/4" and 1/8" round
LEDs, which require 15mA), and what is the forward voltage at that current
(2V in my case). When I design the circuit, I know the voltage and current
through the LED, and I know my source voltage. A bit of math will tell me
everything else I need to know.
This is an example you can use in the above case:
Say my LED is 3.5V at 25mA. And my battery is 12V. This tells me that there
will be 25mA going through the resistor (because the resistor and LED are in
series, therefore they have the same amount of current going through them).
12V (my battery) - 3.5V (the voltage across the LED) = 8.5V (because this is
all that is left), which is the voltage that will be across the resistor.
Voltage across resistor = batter voltage - LED voltage = 12 - 3.5 = 8.5V
Now I know my resistor has 25mA going through it, and 8.5V across it. I need
to know what size of a resistor this is:
Resistance = Voltage/Current = 8.5/.025 = 340 Ohms.
I've never seen a 340 ohm resistor, but 330 is very common, so either use
that or use a 330 ohm resistor in series with a 10 ohm resistor if you
really have to have 340 ohms. I'd just use the 330 ohm resistor.
Now we move from the theoritical world into the real world. Now we take a
330 ohm resistor and plug it into our real world circuit. Since we are using
a the 330 ohm, not the 340 ohm resistor, the current will be a tad more then
25mA, so let's refigure the actual current. The voltage across the resistor
stays the same (for all pracitcal purposes), so:
Current = Voltage / Resistance = 8.5V / 330 ohm = 25.75mA.
Now, we need to know how hot it is going to get:
Resistor power dissapation in watts = Voltage times current = 8.5 * .02575 .219 Watts.
A tadd less then a quarter of a watt. In this case you can use a quarter
watt resistor, but I would recommend against it. It will get quite warm, and
I don't like to push the tolerance of components. If your battery voltage is
a bit higher then expected, or if the specific LED you use has a lower
voltage, you could exceed the power rating of the resistor. I would use a
half watt resistor. It will still get a bit warm, but not hot, and there is
no danger of it burning out. I've burned out cheap resistors before, though
in the real world a quarter watt resistor will take a lot more then a
quarter watt before burning out - I just don't like them to get that hot.
If you want to be more efficient, hook up 2 or 3 LEDS in series. 3 of them
at 3.5 V would be 10.5V, leaving 1.5V across the resistor. The total current
(power) used is the same, but now, instead of heating up a resistor and
lighting 1 LED, we are lighting up 3 LEDs instead of one. I leave finding
the size of the resistor in this case as an exercise for anyone one that
really wants to find out because I gotta go and don't have any more time :(
Thanks. That's about the most practical suggestion so far, considering
that the plane in question has front and rear spars with shear webs.
That would require 3 LEDs per rib to light up each bay, which would be
very convenient to do with three in series with each resistor. The next
thing to do is a monokote test, to see which colors will light up the
right amount with LEDs turned on inside. I don't want to use
transparent colors. That would be too much. Maybe yellow and red.
Here is another thought. Using the math I did, if you had a 12V
battery and 3 LEDS, each 3.5V, then your resistor would be 60 ohm, and
you could use a 1/4 or 1/8 watt resistor. However! If your LEDs turned
out to be 3.0 V and not 3.5, then the current through them would be
50mA, not 25mA. You may be pushing their design limitations and risk
burning them out, and the resistor may get hot depending on the size
If you want to get fancy, you could build a current regulator and set
it at 25mA. Use this in place of your current limiting resistor. Then
you could use any combination of 1, 2, or 3 LEDs, and the current
would still be properly regulated. An LM317 will do this quite nicely.
The downside of this is that you need to understand a bit about
electronics, or find a friend that does to design it for you. I could
whip one out pretty quickly, but if you don't know a bit about
electronics, you won't know what to do if it's not working, or how to
test that it is indeed working correctly.
The ONLY reason I limited the discussed resistance to 500 ohm resistors is
that the poster said that the LEDs came with them.
LEDS also vary in the voltage drop across the LED. So, you generally need to
read the complete spec sheet, or do a little testing.
I.E. lets say that the LED drops 3/10v at rated current. So, the resistor
only drops 3.3v instead of the 3.6.
Actually, the LED may have a much greater voltage drop due to internal
Another issue is brightness matching, esp. between different color LEDs.
Maximum applied voltage may be 3,6v, or it may have been the working voltage
spec. I have no way to know. Some LEDs have internal regulation
and or series led chips.
Yes, the dissipation of the resistor must be taken into account.
Certainly series operation will result in less resistor dissapation, thus
It may be that a combination of series and parallel leds with small value
balance resistors provides the best overall solution.
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