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"
and Robert Reynolds instead replied:
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...
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.
I never through I'd say this, but in theory I think you'll do OK wiring
direct without resistors. 3 NiCads will put out around 3.6v. By the time the
voltage drops to 3v that's 1v per cell, which means the cells a are flat
No youre not ok with direct wiring. De diodes have a voltage limit. The
3 to 3.8 voltage limit dependend on the color of the LED. Below the limit
no current whatsoever, and above the limit the current goes up exponentiely.
So putting 3.6 volt on a 3.8 volt LED would lead to no current and putting
it on a 3.0 volt LED would mean you almost short circuit the source. This
leads to very high current and very short LED life. And since the voltage
limit varies even between LEDs of the same voltage limit you will see a
very large variation in brightness of the LED's.
So take the source voltage sufficient higher and ALWAYS use the resistor
to regulate the current. And each LED needs its own resistor.
For example use a 5 volt source (4.8 with 4NiMh) and a 3.0 volt led.
You want a current of 25 mA so Ohms law says you have to use a resistor
of (2 Volt/.025 A) = 80 Ohm.
If you now use 3.8 volt LED with the same resistor the current will drop
to (1.2 Volt/80 Ohm) = .015 A = 15 mA. Do you want to keep the current at
25 mA you have to change the resistor to 48 Ohm.
If you do the math for a 12 Volt source you see that you can use the same
resistor for both LED's.
Hope this helps, Peter.
On Sun, 10 Jun 2007 03:12:34 -0700, I said, "Pick a card, any card"
and IFLYJ3 instead replied:
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.
It won't work to connect any LED's in parallel. The lowest voltage
one will clamp the voltage at it's level and the others in parallel
with it will not light at all. If you have 12v. the best way is to
connect three in series and in series with one resistor across the
battery. You can parallel multiple strings this way but not multiple
LED's. Never run LED's without a current limiting resistor. Select
the resistor to give approx 20-25 ma. current to each string. Isn't
your plane an electric? If your main battery is less than 12v then
you may have to connect only 2 in series. A9v. transistor radio
battery would supply 20 ma. for quite a while and it would handle 2
Leds in series with around 270- 300 ohm 1/4watt resistor .
Hope that helps.
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.
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.
On Sun, 10 Jun 2007 14:56:52 -0700, I said, "Pick a card, any card"
and IFLYJ3 instead replied:
There's no rule that states each cell must be 2v each so any
combination is possible. Look towards backup batteries for alarm
systems which are commonly either 12v or 5v. They're out there.
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 =
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 :(
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.
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.