TripleStar, Luxeon Star Bike Light!

I decided to put 3 Luxeon Stars on my large bike light heat sink, and wire them in series, and use 8 AA (12 volts) to power them. Whoah! Very bright! Seems like the perfect combination. No resistance, and not really overdriven that much.

Now I have to get a good bike to mount it on!



Wayne www.elektrolumens.com

Put in fresh batteries last night, and I checked the current draw. It was pulling 1 amp. So each LS is pulling 333mA. Quite amazing how bright it is. Went out for a walk with it, it really lights up the night! (Akward to carry though, as it is a bike light.) Heat sink gets mildly warm after about 20 - 30 minutes. Would work fine for slow street bike riding, or for slow off road riding at night. Probably not suitable for racing at night, though.

Wayne

Series right? Wouldn't each Luxeon be pulling 1 amp rather than 1/3 of the current?

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Guy Kuo:
Series right? Wouldn't each Luxeon be pulling 1 amp rather than 1/3 of the current?<HR></BLOCKQUOTE>

No, three in series, so each one is pulling 1/3 of the load. If each one were drawing 1 amp, the total would be 3 amps. Twelve volts is a good choice for running 3 in series. No need to regulate the voltage. Looks crude, but it works fine. I don't really have a good bike right now, or I'd mount it and take photos of it.

Wayne www.elektrolumens.com

Hi Wayne,

How much the 8 AA 12 Volt battery voltage dropped when powering those LS ?

From the pic looks like each LS was running at 1 Amp , in series right ?

Vic

I don't think I agree. In series they should all pull the same as the total current (1 amp) since current is the electron flow rate. The same number of electrons pass through each Luxeon in series connection. The voltage drop across them, however, would be 1/3 of the 12 volt total so the power (volts x amps) for each would be 1/3 of the total. Only in parallel would the current pull be divided.

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by vicbin:
Hi Wayne,

How much the 8 AA 12 Volt battery voltage dropped when powering those LS ?

From the pic looks like each LS was running at 1 Amp , in series right ?

Vic<HR></BLOCKQUOTE>

Hi Vic,

I only checked the current draw, and not the voltage drop. I usually think in terms of current usage with these. In series, each LS is connected to the others. The negative from the first goes into the positive lead on the second, so on and so forth. This is in series. Each individual LS, if I check the voltage, will be drawing 1/3 of the total voltage. So, if the voltage under load were, say, 10 volts, each LS would be pulling 3.3 volts. If the total current draw is 1 amp, each individual LS would be pulling 1/3 of the total, or, 333mA. I did not check the current draw on each individual LS, as I would have to undo the wiring to do this test. As the current is 333mA or so, probably the voltage drop is 2 volts, to 10 volts, which makes a lot of sense, based upon the current being drawn. I suspect there is some amount of internal resistence from the 8 AA's. Probably if I used 8 'D' cells, or if I used 10 NimH's, the combined current would be about 1.5 amps. Just an asumption here.

Wayne www.elektrolumens.com

Heya Wayne,

When not loaded (open) the battery were 12 Volt (1.5 x 8 pcs), just curious what was those battery's terminal voltage when powering that 3 LS ? Just switch it on and measure the battery terminal, that's it ! From that we can roughly calculate those alkaline's internal resistance.

About the current , you don't need to undo the wiring since they're in series, each LS was driven at 1 amp. If they're parallel then it is 333ma.

Cheers,
Vic

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by ElektroLumens:
Hi Vic,

I only checked the current draw, and not the voltage drop. I usually think in terms of current usage with these. In series, each LS is connected to the others. The negative from the first goes into the positive lead on the second, so on and so forth. This is in series. Each individual LS, if I check the voltage, will be drawing 1/3 of the total voltage. So, if the voltage under load were, say, 10 volts, each LS would be pulling 3.3 volts. If the total current draw is 1 amp, each individual LS would be pulling 1/3 of the total, or, 333mA. I did not check the current draw on each individual LS, as I would have to undo the wiring to do this test.....<snip>
Wayne <A HREF="http://www.elektrolumens.com" TARGET=_blank>www.elektrolumens.com
</A><HR></BLOCKQUOTE>

In a series connection, each LS has the same current flowing through it.

From the picture your LS's do look like they are in series. Therefore each one will have roughly 1/3 of 12 volts across it, or 4 volts. You can picture this in your head - 8 batteries x 1.5v = 12V, and 3 LS's x 4V = 12V. The numbers balance like a check book. Now the batteries might be more like 10V under load -> 3-1/3V per LS, but you get the idea.

Now, if you measured one Amp leaving the batteries, and the LS's are in series - then each LS has one amp passing through it. Think of it this way, you need to connect the meter in series with the LS's to measure the current. Pick a spot, cut the wire and put the meter in. You'll measure one amp any place you chose to cut that wire to insert the meter. Even between two LS's. Say you have new batteries that can maintain 12V with the LS's on. That's 12Vx1A=12Watts of power leaving the batteries. Let's balance the check book again. Each LS will have 1Ax4V=4W. Add it up 4W+4W+4W=12W.

If you had the LS's in parallel, then each LS would roughly take 1/3 of the total current (assuming the LS's were identical), and have the same voltage across them. In series, though, each LS will have the same current flowing through it as the others, and roughly 1/3 the voltage.

Having said that - you're seriously overdriving those LS's !!

Yup the amperage is the same for each if they are in series and the voltage is divided among them. In parallel it is the opposite, the voltage is the same and the amperage is divided among them.

In any case I want to see this in the head of a 3 C or 3 D Mag light

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Brock:
Yup the amperage is the same for each if they are in series and the voltage is divided among them. In parallel it is the opposite, the voltage is the same and the amperage is divided among them.

In any case I want to see this in the head of a 3 C or 3 D Mag light

<HR></BLOCKQUOTE>

The settup would be different, though, because I would be using 3 cells at 4.5 volts, and the LS's are in parallel. This particular bike light is 3 in series, and I'm using 8 AA's to power it.

Anyway, I'll have a triple done this weekend, and I'll test the current draw, and voltage. The doubles I've been making have been drawing from 1.5 to 1.7 amps. I haven't tested the drop in voltage.

Wayne www.elektrolumens.com

Yes, the 3 in a 3 C or D Mag will have to be in parallel. I bet it will be right about 2.0 to 2.25 amps on D cells. But it should be brighter because the LS's will driven right where they are rated, assuming the batteries don't drop to much under that load.