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n38139

Well Known Member
Can I put in a simple resistor with an on off switch to control about five led indicator lights? And if so, how do I figure out what size/type to install in line? If there is a yes answer, please make it as easy as possible so a simpleton like me can understand it. :D

thanks
 
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LEDs

Yes, but you need to share the circuit diagram to determine the proper resistance.
The best thing i can recommend is to Join the AeroElectric matronics list, and buy a copy of the Nuckolls 'bible'.
LEDs are current limited. Find the max current for the LEDs being used. V=IR,
So R = V/I. V= 12 volts, I is the max current for the LED.
 
I'll take a shot at it: you can put a 500 ohm resistor in series with your voltage source and your switch (assuming the normal 14 volts), and then put your LEDs in parallel. That way if one burns out, the remaining LEDs will continue to work. Let me know if you need a sketch and I'll be glad to send it to you. What you're trying for is about 20 to 30 mA of current for the LED's, which most are rated at. Some come with the resistors built in so be aware of that. If you go to Radio Shack, for example, you can buy the LEDs rated for 12 volts, with or without the resistor. If you have the ones with the resistor, just put them in parallel without the 500 ohm resistor. Send me a PM if you need a schematic.
 
It's a little more complicated than your question proposes. EACH LED sould be limited to its max current rating with a current limiting rersistor. If you connect multiple LED's in parallel with a single current limiting resistor and one LED fails, or is turned off, then the remaining LED's will be subject to higher voltage (thus higher current).

You will probably face the issue of some LED indicators requiring a ground to activate and some requiring 12 vdc. A single switch (or dimmer) in this case is more complicated.

If an LED requires 12 vdc (instead of ground) to activate, then consider using a fusible link (see Aerolectic Connection, mentioned earlier) or a fuse in line to protect the wiring in case of a short.

I highly recommend installig the ability to dim any indicator LED's if they may be lit during normal operations, especially at night. Do a search on LED DIMMER to find more info.

If you are a "simpleton" as you claim then I would highly advise getting help directly from someone with knowledge of electronics. This could save you from a lot of headache, or possibly worse.
 
OK its not too complicated. I'll try to explain. LED indicators typically come in two types.

The first type includes an internal resistor and they are typically rated for the voltage of the application so 5V, 12V, 24V, etc. The manufacturer has figured out the resistor for you.

The second type requires an external resistor for each LED. I think this is the type you were originally referring to. If you don't include the external current limiting resistor the the LED will be very bright for a very short lifetime (probably about 2 seconds). The resistance is calculated using Ohms Law but you need a couple of pieces of information first. The first piece of information you need is the forward voltage of the LED. Typical values are 2.0 - 2.5 volts. Then you need to know the rated current for the LED. More current means its brighter, less current more dim. A typical current for a bright LED would be 20mA or 0.020 amps.

Now for the first calculation. The voltage across the resistor is the supply voltage 14 volts minus the forward voltage of the LED. So for example lets say the forward voltage (Vf) is 2.5 volts the the voltage across the resistor is 14 - 2.5 = 11.5 volts. Then by using Ohms Law we have V = I * R.
Solving for R we get R =V / I. In our example R = 11.5 / 0.02 = 575 ohms. Then you need to select the next larger available value. In our example this would be 600 ohms. 5% tolerance resistors are good enough you don't need 1% values.

The second calculation is the required power rating for the resistor. This can be calculated by the equation. P = I*I*R in our example P = 0.02 *0.02*600 =0.24 Watts. So a 1/4 resistor would be a little close, 1/2 would be a more conservative selection. Remember this is just an example. You will need to get the specs for your LED.

The resistor calculation is independent on how the indicator circuit is connected with regard to the active voltage to turn on the LED. If the status output is +14V or 0V it doesn't matter.

Now the fuse. LEDs typically burn out and open up. They do not normally short when they fail. I would connect all of the common supply signal pins together and then connect a fuse between that point and the supply. Again that supply could be ground or +14V. If you have a mix of active low (ground) or active high (+14V) indicator signals then you will have two common supply points and two fuses will be needed.

I hope this helps. Just my two cents follow at your own risk.
 
A strip of window tinting with velcro coins takes care of dimming my 6 LED's. How often do I fly at night, not much, so I like it simple, cheap and lightweight. If I flew alot then I would want something different.
 
Thanks

OK, I might not be a complete simpleton. I appreciate all of the input and will try and figure it out.

The LED's I have I bought at a Street Rod show and they are 12V. They are pretty bright, I used some of them in my 37 Ford :cool: for various things and they look good and still work well.

thanks
 
If the LEDs are 12V and they are too bright you can always add a small resistor in series. This will reduce the current and dim the LED. The value of the resistor is a little difficult to determine. You might buy a variable resistor (pot) from radio shack and put it in series with the LED. Then adjust it the brightness you want. Then remove the pot and measure the resistance. Then buy an equivalent resistor and add it in series.
 
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