Right resistor for high power LED for sound reader?

Hi, You have a device (LED) that will drop 3.5 volts at 300 mA. Your supply is 7.5 volts. You need a resistor that will drop 4.0 volts at

300 mA. R=E/I P=EI, Closest standard values 10-15 Ohms 2 Watt. Tom

Thanks Tom! So your saying that:

4V / 300mA = 13.33 Ohms, and 4V * 300mA = 1.2W

and the only resistor I can use is a standard 10-15 Ohms 2 Watt. Can I use any kind of variable resistor and set it at 13.33 Ohms? If I can, what kind would be best? Would this be a waste of time, and would the fixed value resistor be fine?

I know this is really basic stuff for you people, but this is the first time I've ever done this.

Thanks, Mitch

--
spamsuckaWell-Worn-Hat at yeahhoo dott caum
Take off the Well-Worn-Hat to respond.

Mitch,

Best to keep your life simple. Just use 15 ohms, and don't worry about messing around with potentiometer settings. You'll be well below the max current (around 270 mA), so burning the thing out shouldn't be a concern, and you'll still get a reasonable amount of light output.

Regards & good luck,

Mark

Oh man, thanks guys! That's what I was hoping to hear! I'm going to have at least 300 people counting on me to put on a good show, so it's a big deal. I've been trying to figure this out for myself and am making some small progress. I'm just really stuck for time and appreciate your help a lot. This is my very first foray into the world of electronics.

I live in a small town and went to the local Radio Shack. They don't have any 2 watt resistors. They have

15 ohm 1/2 watt, 10 ohm 1/2 watt, and 10 ohm 1 watt. What combinations of common resistors can I put together to get a 10 ohm 2 watt?

Someone told me it has to do with "Kirchoff's Law." I'm trying to learn about that from some badly written web pages. It's not making a dent yet.

-Mitch

--
spamsuckaWell-Worn-Hat at yeahhoo dott caum
Take off the Well-Worn-Hat to respond.

I agree with Mark (AKA redbelly). Keep it simple. There is no need to set the resistance to 13.33 ohms. Any resistor between 10 and 15 ohms will be fine for this application. Two watt resistors are easy to come by so I would use one. Tom

Excellent info! I can dig some heatsinks out of my old computers. The power supply is a Kelmar 8604-B 5-10vdc

5a. It is made for sound "exciter" lamps, so I'm fairly certain it is regulated.

I also got 4 LXHL-LD3C Luxeon Star III LEDs to experiment with. They are 140 max lumens, 3.51v 1400ma. If I can figure out the proper resistors tomorrow morning, I may make the lamp with those instead. I was just planning on using the older LXHL-MD1Ds because others had done it and it seemed safer to follow in their footsteps.

Assuming your power supply will be or can easily be set to your favored

7.5 volts DC when drawing only 300-350 mA, then:

The LED typical voltage drop is 2.95 volts according to the "DS23" datasheet available from the Lumileds website.

Subtract this from 7.5 volts and this leaves 4.55 volts across the dropping resistor. Plus .64 volt, minus .56 volt tolerance worked out from subtracting the minimum and maximum voltage drops mentioned in this datasheet.

To get 350 mA with 4.55 volts across the dropping resistor: Divide 4.55 by .35, and you get 13 ohms. The next common value up is 15 ohms, although 13 ohms is a semi-standard value. Power dissipation will typically be 4.55 volts times .35 amp, or about 1.59 watts. I would use a

5 watt "sandstone" style 15 ohm resistor.

To be extra conservative, let's use 300 mA (.3 amp) and 2.25 volts LED voltage (minimum at .35 amp is 2.31 volts), leaving a worst case of 5.25 volts across the droping resistor assuming a 7.5V supply.

5.25 volts divided by .3 amps is 17.5 ohms. The next higher "standard" resistor value is 18 ohms. Again, I recommend the 5 watt "sandstone" type.

For "bad worst case" designing, use 22 ohms 5 watt rectangular sandstone style. Expect typical LED current then of about 230 mA with a 7.5V supply. And the resistor will typically be dissipating 1.15 watts, which is enough to get a 5 watt rectangular sandstone style resistor very warm.

Another note: Performance of Lumileds LEDs is with the "junction temperature" at 25 degrees C, and the red one has quite a high sensitivity of light output to temperature. The DS23 datasheet gives a thermal resistance of 23 degrees C per watt for this model, and at 2.95 volts .35 amp (1.03 watt) expect to require a heatsink temperature of 1-2 degrees C to achieve this! And this red model has output slightly less than 70% of "full" when the "junction temperature" is 25 degrees C warmer than this!

Not that I don't think it will work, but I advise to drive these things conservatively, heatsink them adequately to excessively, and make your expectations of light output realistic - I would say 30 lumens at most, plan on 25 lumens even with a heatsink, and less if you have a worse-than-average LED - worst case is about 30.4% below "typical" according to the datasheet. This may well be plenty good for your application however!

- Don Klipstein ( snipped-for-privacy@misty.com)

I would use a fixed resistor with value at the high end of the range for possible answers - at least 15 ohms.

Two reasons:

1. The chips in these LEDs have a nonlinearity, with efficiency being maximized at currents in the general ballpark of 50-60% of "full current"

1. These LEDs (red Luxeons) have light output very sensitive to temperature, with output doubling by having the junction 45 degrees C cooler than the 25 C "characterizing temperature", and halved by having the junction 45 C warmer than this. This means a 1 degree C temperature change causes light output to change about 1.5%, with higher temperature being unfavorable. So I consider it good to operate these LEDs conservatively, and expect little to gain in light output from pushing them with current past about 300 mA. In addition, I recommend heatsinking them to an extent many would call excessive.

- Don Klipstein ( snipped-for-privacy@misty.com)

[snip]

...and let's hope that that's a fairly well regulated supply, at that: I would imagine that any variations in the intensity of the LED (due to line regulation) would show up as audible noise!

I've worked with the Luxeon Star LEDs and I heartily recommend a heatsink. The LED life goes down very quickly as Tj rises, so keeping it cool will improve the reliability of your projector retrofits.

I have a red Luxeon Star on an old i486 (passive) heatsink and that seems to work pretty well. I also didn't want to worry about a regulated supply, so I built a "constant current regulator" onto the same heatsink using an LM317 adjustable regulator IC. The circuit is right out of the LM317 datasheet (see figure 26 of the On Semi datasheet located at

) and I used a 1W 3.6 ohm resistor to set the current at approx 300mA. The benefit of this circuit is that the LED will be fed with a constant 300mA current without depending on the input voltage. Not bad for a three component circuit (counting the LED)!

TJL

--
No, you don\'t.  You get a 15 ohm 2 watt resistor.

--- Two 10 ohm half watters in series, in parallel with two 10 ohm half watters in series, like this:

+---[10R]-----[10R]---+ | | +---[10R]-----[10R]---+ ||

Or, two 10 ohm half watters in parallel, in series with two 10 ohm half watters in parallel, like this:

+---[10R]--+--[10R]---+ | | | +---[10R]--+--[10R]---+ ||

-- John Fields Professional Circuit Designer

If the're out of 10 ohm half watters when you get there you can do this:

+---[15R]--+--[15R]---+ | | +---[15R]--+--[15R]---+ | | +---[15R]--+--[15R]---+ ||

Or this:

+---[15R]--+--[15R]---+ | | | +---[15R]--+--[15R]---+ | | | +---[15R]--+--[15R]---+ ||

--
John Fields
Professional Circuit Designer

--
Oops...

You get a 15 ohm 1.5W resistor.


       +--[10R]--+--[10R]---
       |         |
       +--[10R]--+

equals

       ---[5R]---[10R]---

so, since:


       P = I²R

rearranging and solving for the current in the 10 ohm resistor when
it\'s dissipating 1 watt:

                 P              1W
       I = sqrt ---(PR) = sqrt ----- = 0.316 amperes 
                 R              10R


Now, with 316mA through 5 ohms,


       P = I²R = 0.316A² * 5R = 0.5W

So, since the 10 ohm resistor is dissipating one watt and the set of
paralleled resistors is dissipating half a watt, the array is
dissipating 1.5 watts.

^

--- | Oops again... (PR)----+ shouldn't be there.

--

>Now, with 316mA through 5 ohms,
>
>
>       P = I²R = 0.316A² * 5R = 0.5W
>
>So, since the 10 ohm resistor is dissipating one watt and the set of
>paralleled resistors is dissipating half a watt, the array is
>dissipating 1.5 watts.

Four 15 ohm ones in a 2-by-2 series-parallel arrangement is a 15 ohm 2 watt resistor.

Put two 10 ohm 1 watt ones in parallel with each other and put that combo in series with a third 10 ohm one, and you get a 10 ohm 2 watt resistor.

Put four 10 ohm 1 watt ones in series-parallel 2-by-2 arrangement and you get a 10 ohm 4 watt resistor.

With a 2-by-2 series-parallel arrangement of four identical resistors, it does not matter whether or not you put two parallel pairs in series or put two series pairs in parallel.

- Don Klipstein ( snipped-for-privacy@misty.com)

Oops, better keep my eye on the screen better. It's obviously 15 ohms, and that's what I meant to type.

- Don ( snipped-for-privacy@misty.com)

Thanks... I did screw that one up.

- Don ( snipped-for-privacy@misty.com)

...

This is the kind of question we really like around here - you gave us what information you have, and said, "This is the goal".

Yes, a power rheostat (variable resistor) would work here, but a new one would be astronomical:

for an example.

But you should be able to find a 12 ohm, 2 watt resistor! I'm sure that will be fine.

Break a Leg! Rich

But Ten Ohms will fry his LED! Take SIX pcs. 10 ohm, two watt, and put them 3x2 for 15 ohms at SIX watts! :-)

Cheers! Rich