What kind of resistor?

I'm thinking about hooking up a 3w lumiled star to a battery designed to jump start your car. Just when the power goes off. What kind of resistor(s) should be used? On both +/-? If its two much power what kind of resistor for a single LED and 6v lantern battery? Specs of jump starter

12V
7AH
200A

more information
http://www.batteryweb.com/schumacher-charger-detail.cfm?Model=PS-70A

BTW I know its powerfull because I killed one in less than 1s by playing around with it Also I have 3 living LuxIII's so hooking them all up might be something I want to do as well.

Give a 12 ohm or 15 ohm 10W resistor a go with the 12V option. For 6V try a 4.7 ohm or 3.3 ohm 5W resistor.

Standard values quoted. Very inefficient, but it's what you're after I think.

All options will drive the LIII close to max rating with some +/- for voltage drops and battery resistance etc. It's a wishy-washy solution to a LED's real need for constant current, but it'll do the trick for what you are asking.

Chris

One on + and another on - or another config? What would be a more efficient method? The 12v battery is rechargeable.

Eric

ericjwi said:

One on + and another on - or another config? What would be a more efficient method? The 12v battery is rechargeable.

Eric

Click to expand...

The resistor would be in SERIES with the Lux3.


More efficient would be a "driver", i.e. an electronic circuit that steps the voltage down AND regulates the current.

Some numbers for you...

Lux3 at 1A, has ~3.7V across it. To hook to a 12V battery let's assume it is at 12.8V, that means you have to drop 12.8 - 3.7V = 9.1V. At 1A, you have 9.1W in the resistor and about 3.7W in the Luxeon. Efficiency of 3.7W/12.8W = 29%, yuk! You have a lot of heat to dump and you're pulling 1A from the battery.

Now, the numbers with a Driver, usually around 85% or better efficiency. So, now we have 3.7V @ 1A at the Lux3 or 3.7W and 3.7/0.85 = 4.4W from the battery. 4.4W/12.8V = 0.34A.

With the driver you'll get about 3x the runtime of a 'resistor' scheme and you have a lot less heat to dump.

Either way, you MUST heatsink the Lux3.

george.

George is dead on - for this kind of application, you really need a constant current driver. A 10 watt resistor will just barely be enough, and they are usually quite large. (and will get hot)

You mentioned that you have 3 Lux IIIs - that might help the situation. Put the three Lux IIIs in series with a 10 ohm, 1 watt resistor and you will have lots of light and a much more efficient setup. Not exactly cheap, and not really regulated, but workable.

George did not mention it in his post, but he (taskled.com) sells some driver boards all setup to do exactly what you need. I can personally vouch for them, as will many CPFrs. Just as a comparison - your 1 x Lux III + big resistor will have about 1/3rd the run time of doing the same thing with one of his driver setups, and will be a much more consistent brightness level.

georges80 said:

The resistor would be in SERIES with the Lux3.


More efficient would be a "driver", i.e. an electronic circuit that steps the voltage down AND regulates the current.

Some numbers for you...

Lux3 at 1A, has ~3.7V across it. To hook to a 12V battery let's assume it is at 12.8V, that means you have to drop 12.8 - 3.7V = 9.1V. At 1A, you have 9.1W in the resistor and about 3.7W in the Luxeon. Efficiency of 3.7W/12.8W = 29%, yuk! You have a lot of heat to dump and you're pulling 1A from the battery.

Click to expand...

What about a setup using 3 Lux3's? Would that be the same math if setup correctly? 12.8v - 11.1v = 1.7V. Parallel or series? Could that possibly be a lot more efficient? A small ascii diagram might help

Eric

NM that got answered in the next post

How exactly is that all hooked up for series? Is such a setup workable with 200A input?

For the Lux IIIs

(battery + terminal) - resistor - (lux III anode in, Lux III cathode out) - (repeat lux III setup) - (repeat lux III) - ( battery - terminal)

For the driver, there are instructions on the web site.

Assuming you are using the Lux IIIs in series
- approx 12 V in (can vary from 10 - 16 v, which is the real challenge)
- approx 3.5 V Vf x 3 = approx 10.5 V

So at a nominal 13 V battery - 10.5 Vf = 2.5 V to "throw away"

2.5 V / 10 ohms = 250 ma.

But, if V bat = 16 V, then you have 16 - 10.5 = 5.5 V extra

5.5 V / 10 ohms = 550 ma

This is a lot of light, and driving a Lux III beyond this point requires a great deal of heat sinking care or you actually do not gain more light.

I can throw in an 80mm PC fan to cover some of the difference and cool the lights at the same time if it helps I ran tests on the battery and it showed about 12v + or - 1v. Maybe run the fan off the 3.3v line and put it in series as well?

13 - 3.3v = 9.7v / 3 = 3.23v per led

What kind of resistor in that instance if any?

Eric

ericjwi said:

I can throw in an 80mm PC fan to cover some of the difference and cool the lights at the same time if it helps I ran tests on the battery and it showed about 12v + or - 1v. Maybe run the fan off the 3.3v line and put it in series as well?

13 - 3.3v = 9.7v / 3 = 3.23v per led

What kind of resistor in that instance if any?

Eric

Click to expand...

Umm... you can't just put different things in series - the voltage across them depends on the current through them...

V = I x R
P = V * I

with those 2 equations you should be able to determine you resistor value and power rating. Note also, the voltage across a Luxeon will vary with temperature and it will also vary over the first few hundred hours (it'll drop a little) and it will vary between bin codes. i.e. you will need to get the Luxeons, run some current through them and tune the resistor to suit your desired LED current.

george.

I used george's regulator in my car and could not be happier.

Here's your cost breakdown of the different options for you.

=== Option #1 - Direct Drive (series) ===
3x LuxIII ... $27.00
1x Resistor ... $0.50
+ Cheapest in cost
- Inefficient (~80%)
- Light output decrease as battery depleats

=== Connection Diagram ===
Battery (+) -> LuxIII-#1 (+) -> LuxIII-#1 (-) -> LuxIII-#2 (+) -> LuxIII-#2 (-) -> LuxIII-#3 (+) -> LuxIII-#3 (-) -> Resistor-> Battery (-)

=== Option #2 - MultiLux Regulated ===
Xx LuxIII ... $9.00/each
1x Current Regulator ... $15.00 & up
+ Most efficient
+ Constant light output
- More $$$

=== Option #3 - Direct Drive (series) ===
4x LuxIII ... $36
* you need a fairly low (H/J/K) Vf for this option

driver is always best.. since the vbat will change a lot during draining.

-awr

I'll try to make everything temporary. Going driver is not something I'd be able to get done by the weekend. I'm wanting something for a little light if they happen to go off for a few days this weekend. That sucker probably has enough juice for a week straight

Eric

Eric,

I have a power outage set up in my kitchen similar to what you're talking about. It uses 4 LuxIII's mounted under a cabinet over a counter top controlled by a nFlex driver. At the highest setting (light hog) it puts out plenty of light to get the chores done in the kitchen area or just read a newspaper. On low (conservation mode) it makes a nice navigation light or to just set a mood





At this point the project is uncompleted because I have been waiting on a boost driver from somebody that will drive 8 LuxIII's in series from a 12V source if I ever get that I have 4 more LuxIII's to put up on the other side of the room for more even lighting.

I understand your in a hurry right now but in the future I would highly recommend looking at the nFlex it should work very nicely with 3 T**J LuxIII's in series from a 12V SLA, if you need to go over that number of LED's guess you will have to wait with the rest of us for the Hulking Boost Driver (HBD) :wave:


Later
Kelly

Quick and dirty for this weekend with reasonable run time is the 3 x Lux IIIs with a 10 - 20 ohm resistor. At 20 ohms, you can (almost) ignore the heat sink with a star. It will be bright in a dark room.

No more trying to use that 12v battery unless I use the car electrical jack. I've killed too many things One more star gone. I also killed a multimeter with it. Question. Is running two LUXIII's in series off of a 6v lantern ok without a resistor? It seems to work really well and the stars heatsinks are not even noticeably warmer. I've got the resistor but it does not seem that much warmer and there is a lot more light without it. How long should two run off in that setup? A 6v lantern has a 26000mah estimated life.

Eric

You are fast approaching the weekend, so there is less time to be an experimentalist. I have run 6 V battery setups into a Lux III, and I also have plenty of experience killing Lux IIIs as well. (mostly due to deciding the heat sink was not important or I could push it a little harder.

Any 1 watt resistor in the 2 - 10 ohm range will work, and you can be sure that it will run longer and happier. Can you push it - sure - we have all done it, but then again, I usually do the "experimenting" when it is not important.

6V lantern with two series i doubt would put more than 1/2 power into a lux3... you have to measure the current but you said you killed your multimeter.. i'm guessing you blew the fuse trying to measure current in parallel vs series.. get a new fuse, if that's the case and put your MM on HIGH amps mode (not mA) and measure the current through the 2 luxes and the MM in series.. i would bet you are at like 300-500mA.. and well 26000/500 = 52 hours continuous runtime.. if the heatsink isn't even getting warm you aren't even over 300mA that puts you closer to 90 hours.

-awr

Didn't even use it. I left like everyone else. Or at least 3 million others.

I'm playing with renewable energy. With a bank of 6 70AH deep cycle batteries I can bump up to a 30W T8 flourescent on an inverter and last about 75 hours before the batteries are at 50% charge. With about 6 hours per day usage I could have light for 12.5 days That would also be about 3000 lumens versus about 100 off of the LED.

Eric