you are right... sorry.. poor choice of words on my part... what I meant was "handle"... That's why I'm going into MECHANICAL engineering... :-)
-Joe-

 

GIJoe, GAH,

GI, you're almost there. As far as your calculation- don't forget to subtract the LEDs forward voltage (1.7v typical and for our examples. Always use package data for your particular diode) from the 13.8v source voltage. Then your calculation will be .121 instead of .138 watts. This subtraction is required to solve the design correctly and is due to the fact that the sum of the voltage drops (1.7v for the LED and 12.1v for the resistor) must equal the souce voltage of 13.8v for this series circuit. I admit there is not much difference but it is the correct solution. For this example use a 1/4 watt resistor and you'll be in the clear. It is always a good idea to double the required wattage to avoid overheating and subsequent failure of the resistor. Also, to be clear, there is no voltage rating on resistors so don't confuse that with the other ratings. Also good luck with your BSME!

GAH excellent explaination, right on the money regarding the wattage explaination. Don't worry about the auto mechanic, he probably has the "knowlege problem".

Feel free to write me at my email address contained in my profile anytime.

XceLenT Driver

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Y2K Oxford White XLT S.C., V6, Auto, FoMoCo Mud Flaps, Lund Bugflector, Tinted windows, Auxillary BU lights, 2x55W DRL, UNGO Keyless entry, FoMoCo/Duraliner bedliner

 

Another reason for using LEDs vs. lamps with filaments is heat. LEDs run cool. Incandescent lamps tend to heat up. There are some situations where you don't want the heat.

You also don't want water splashing on hot incandescent lamps. The thermal shock can crack the glass envelope of a bare bulb.

 

Good point Dennis.

Have you noticed the used of LED clusters in the tail lights of commercial vehicles? Not to mention the "3rd brake" lights of many "high-end" automobiles.

 

Funny you should mention that. I just saw a panel truck with round LED brake lights today. Really caught my eye. They sure do sparkle.

 

XceLenT Driver,
You are doing a wonderful job passing on the information about Ohm's law and the diode circuits, even I am understanding and able to follow along.....hehehehehehe I am sitting here wondering how you are going to explain the math of parallel resistor/diode circuit. I could not resist (pun intended) asking, I am enjoing your electronic classes.
Tubbys.

 

Those tails have another MAJOR advantage for the trucks... short of being smashed, they'll last forever. That's why the truckers love them... never have to worry about failing a DOT inspection (short of bad wiring anyhow...:-) )


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98 Explorer (Lemon Law Case in Progress)

'00 F-150, X-cab, 4x4, Lariat, Off-road, Trailer Tow, Holandia Sunroof, built 11/99 picked up 10/00!
List of planned mods: Clear corners and tails, CB wiring, Stainless brush guard, at least 1 set of KCs (probably more!), custom radar detector mount, color matched trailer hitch with stainless steel ball, torsion bar tweak, bug deflector, spray in liner, and a tonneau of some sort...

 

OK, here's the skinny... Got the catalog yesterday, the LEDs today. They are a 12 volt 2-color LED. The red has a 10mA recommended operating current, and the green has a 15 mA recommended operating current. I can list the other stuff if necessary, but it's a lot of stuf I have no clue about... Electro-optical characteristics and such....

If it's a 12 volt drop across the resistor, using V=IR, If I were to assume (however incorrectly it might be) that the input was 12 volts, the resistance I would need is 1.2 k-ohm resistor, right? (Radio shack sells them in their catalog... don't know about in the store though...) If I assume the 13.8 volts, and a voltage drop of 1.8 volts through the resistor, a 1/4 watt resistor would work, and again, using Ohm's law, I would need a ...... now I'm confused.....

There's 2 issues here... One in handling the power passing through it, another in handling the current through the LED. Which one do I address first? I need a 1.8 volt drop across the resistor to give me my 12 volt input to the LED, so using ohm's law, it means a 180 Ohm resistor, right?

Now, to find the wattage using P=IV, which value do I use for V? (the 1.8V drop that occurs over the resistor or 13.8V input voltage?)

if I assume 13.8, then I need a .138 Watt resistor, at least. if I bump up to a 1/4 watt, what effect will that have on the circuit? Now, assuming these calculations are correct for the red one, for the green half, (current = 15 mA) I need a 120 ohm resistor, and at least .207 watt resistor, right?

Please straighten me out. I think I got it this time.

-Joe-

 

GIJoeCam,
The schematic above could represent your red LED. The voltage drop across the resistor is 12.1V and the drop across the LED is 1.7V. The current through the resistor and LED is obviously the same at 10mA. The power generated by the resistor is (I^2)*R which is 0.122W. The power generated by the LED can be calculated also by solving for the resistance of the LED. R=V/I, therefore R=1.7V/0.010A which gives 168.6 ohms. So using P=(I^2)*R we get 0.017W generated by the LED.
The wattage rating of a resistor has no effect on a circuit provided that the rating is high enough to withstand the heat generated within it. If you put a ¼ W resistor in a circuit and the resistor ends up trying to generate 1W of power, then the resistor will fail and the circuit will fail.

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'00 5.4L 4X4 Lariat (Black/Silver)
'99 Ranger 4.0L 4X4 XLT



[This message has been edited by gah (edited 11-17-2000).]

 

You've got it backwards... it's a 12 volt LED, not a 1.7V. If I put 1.7 volts to it, it'll hardly light. That means I need a resistor to step down the voltage 1.7 volts at .010 A, using Ohm's law again, R=V/I, is a 170 Ohm resistor. Then, to calculate the wattage, using I^2*R, .010*.010*170 = .017 Watts, or using Power= Current*Voltage, I get Power = .010 (1.7) =.017 Watts.... which is just over 1/64 of a watt... not a chance of finding.... Now what??

So, back to the original question, what happens when I connect the 1/8 watt resistor to it? The resistor might draw more current, but it'll convert it to heat right? It should still give me the voltage drop I want, though, right? HELP!!! There's got to be a way......
-Joe-

 

Joe, GAH,

I'm proud of you guys. (Though I don't take much credit) As I said earlier it is difficult to learn anything with this "one side at a time" method we are using here. GAH your explaination and drawing is great. Joe... At the risk if sounding hypocritical, IF the LEDs are rated for 12 volts you probaly don't need any additional current limiting resistors but the 180 ohms you calculated for the 10 ma. circuit is right on! To answer your question you need to use the voltage across the resistor for the wattage calculation or: 1.8volts X .010 amps through it = .018 watts not .138 watts. Now for the 15 milliamp circuit... a similar miscalculation. The 120 ohm resistor value is exact but the wattage would be .027 watts not .207 watts (1.8 X .015). As with all the examples we've played with so far a 1/4 watt resistor will do the job. GAH covered that as well. The second question you mentioned referred to handling the power through the LED. Not a problem for us to worry about. Since we've demonstrated that Ohms, Watts, Volts and Amps are all related... If you maintain the specified forward voltage (i.e drop) and current rating of the LED you are using, the rest will take care of itself. But it is good you are learning how to calculate all these parameters for other projects. GAH's drawing illustrates Kirchoff's Law or: "The sum of the voltage drops in a series circuit must equal the souce voltage" and you could expect to measure those voltages at the points he's illustrated when referenced to ground. If you measure across the resistor you should see just about 12.1 volts (remember all components have some tolerance so the values will vary slightly).

OK, 'till we meet again... "XLT D".

 

Joe,
Be cool bud, this is simple ONCE you know it and none of us were born with this knowledge. I've "poped" a few LEDs in my earlier days but that's normal.
I posted my latest reply before seeing yours. But none the less you are OK. GAH's drawing does show how to hook up a 1.7v diode to a 13.8 volt source, but if you were to look inside the one you bought it would look very much like his drawing. Let's keep it simple here. Go hook up that diode to your battery and it will be fine. The resistors you defined are correct to do it "by the book".

 

someone pass the advil. lol.

 

I think I'm gonna take the chance and wire the first one up without the resistor, and try it and see what happens. I figured out the trouble... they actually make 2 different ones... one is a 12 volt one is a 2 volt. I got the 12 volt, so it's OK.

My brain hurts... Let me have on of those advils too.... :-)

 

GIJoeCam,
the previous figure was based on XceLenT's comments earlier in this topic.

"The most accurate way to calculate the proper series resistor value for any LED and DC voltage combo is thus... the source or DC voltage (13.8 for automobiles when running and this example) minus the forward voltage of the LED (Typically 1.7v for generic LEDs), that equals 12.1v... now to determine the resistor value, divide that voltage (12.1) by the current rating of the LED (typically 10 milliamps or .010 amps)... that equals 1210 ohms or rounded to 1.2K ohms"

My apologies if I confused the issue.

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'00 5.4L 4X4 Lariat (Black/Silver)
'99 Ranger 4.0L 4X4 XLT