LED question for electronics wizards

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 cool.
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 resistors?
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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. -- Ray
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Thanks for the explanation. What about using ten nicads to make a 12 volt pack? Would that be better or worse than a gel cell?
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
wrote: [...] "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...
/daytripper
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Is that taking into account that the 500 ohm ones would be using a 12 volt power pack, and the 330 ohm, a 5 volt pack?
--
Jim in NC



Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
BIG SNIP

BIG SNIP

I am curious as to where you can obtain a 5 volt gell cell?
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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. -- Ray
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sun, 10 Jun 2007 14:56:52 -0700, I said, "Pick a card, any card"

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. -- Ray
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Ray Haddad wrote:

As far as I know, the chemistry of a cell dictates the voltage. Lead/acid cells are 2 volts each.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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? -- Ray
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Ray Haddad wrote:

Can you specify which audio supply company? I'm also curious as to how you get 5v from lead-acid cells, or does it use some other chemistry?
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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 properly.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Chuck wrote:

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.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Robert Reynolds wrote:

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.
--

Tim Wescott
Wescott Design Services
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

My apologies to those whom already know this, or those whose heads this goes over:
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 :(
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Ook wrote:

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.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
<ship>

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 your used.
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.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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 construction. 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 more efficiency. It may be that a combination of series and parallel leds with small value balance resistors provides the best overall solution.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.